http://openelectrical.org/api.php?action=feedcontributions&user=Jules&feedformat=atomOpen Electrical - User contributions [en]2024-03-28T14:46:48ZUser contributionsMediaWiki 1.35.0http://openelectrical.org/index.php?title=Motor_Starting_Calculation&diff=206Motor Starting Calculation2021-05-04T05:27:17Z<p>Jules: /* Computer Software */</p>
<hr />
<div>== Introduction ==<br />
<br />
[[Image:HV_motor.jpg|right|thumb|300px|Figure 1. High voltage motor (courtesy of ABB)]]<br />
<br />
This article considers the transient effects of motor starting on the system voltage. Usually only the largest motor on a bus or system is modelled, but the calculation can in principle be used for any motor. It's important to note that motor starting is a transient power flow problem and is normally done iteratively by computer software. However a static method is shown here for first-pass estimates only.<br />
<br />
=== Why do the calculation? ===<br />
<br />
When a motor is started, it typically draws a current 6-7 times its full load current for a short duration (commonly called the locked rotor current). During this transient period, the source impedance is generally assumed to be fixed and therefore, a large increase in current will result in a larger voltage drop across the source impedance. This means that there can be large momentary voltage drops system-wide, from the power source (e.g. transformer or generator) through the intermediary buses, all the way to the motor terminals. <br />
<br />
A system-wide voltage drop can have a number of adverse effects, for example:<br />
<br />
:* Equipment with minimum voltage tolerances (e.g. electronics) may malfunction or behave aberrantly<br />
:* Undervoltage protection may be tripped<br />
:* The motor itself may not start as torque is proportional to the square of the stator voltage, so a reduced voltage equals lower torque. Induction motors are typically designed to start with a terminal voltage >80%<br />
<br />
=== When to do the calculation? ===<br />
<br />
This calculation is more or less done to verify that the largest motor does not cause system wide problems upon starting. Therefore it should be done after preliminary system design is complete. The following prerequisite information is required: <br />
<br />
:* Key single line diagrams<br />
:* Preliminary load schedule<br />
:* Tolerable voltage drop limits during motor starting, which are typically prescribed by the client<br />
<br />
== Calculation Methodology ==<br />
<br />
This calculation is based on standard impedance formulae and Ohm's law. To the author's knowledge, there are no international standards that govern voltage drop calculations during motor start. <br />
<br />
It should be noted that the proposed method is not 100% accurate because it is a static calculation. In reality, the voltage levels are fluctuating during a transient condition, and therefore so are the load currents drawn by the standing loads. This makes it essentially a load flow problem and a more precise solution would solve the load flow problem iteratively, for example using the Newton-Rhapson or Gauss-Siedel algorithms. Notwithstanding, the proposed method is suitably accurate for a first pass solution.<br />
<br />
The calculation has the following six general steps:<br />
<br />
:* Step 1: Construct the system model and assemble the relevant equipment parameters<br />
:* Step 2: Calculate the relevant impedances for each equipment item in the model<br />
:* Step 3: Refer all impedances to a reference voltage<br />
:* Step 4: Construct the equivalent circuit for the voltage levels of interest<br />
:* Step 5: Calculate the initial steady-state source emf before motor starting<br />
:* Step 6: Calculate the system voltages during motor start<br />
<br />
=== Step 1: Construct System Model and Collect Equipment Parameters ===<br />
<br />
The first step is to construct a simplified model of the system single line diagram, and then collect the relevant equipment parameters. The model of the single line diagram need only show the buses of interest in the motor starting calculation, e.g. the upstream source bus, the motor bus and possibly any intermediate or downstream buses that may be affected. All running loads are shown as lumped loads except for the motor to be started as it is assumed that the system is in a steady-state before motor start. <br />
<br />
The relevant equipment parameters to be collected are as follows:<br />
<br />
:* Network feeders: fault capacity of the network (VA), X/R ratio of the network<br />
:* Generators: per-unit transient reactance, rated generator capacity (VA)<br />
:* Transformers: transformer impedance voltage (%), rated transformer capacity (VA), rated current (A), total copper loss (W)<br />
:* Cables: length of cable (m), resistance and reactance of cable (<math> \Omega \ km </math>)<br />
:* Standing loads: rated load capacity (VA), average load power factor (pu)<br />
:* Motor: full load current (A), locked rotor current (A), rated power (W), full load power factor (pu), starting power factor (pu)<br />
<br />
=== Step 2: Calculate Equipment Impedances ===<br />
<br />
Using the collected parameters, each of the equipment item impedances can be calculated for later use in the motor starting calculations.<br />
<br />
==== Network Feeders ====<br />
<br />
Given the approximate fault level of the network feeder at the connection point (or point of common coupling), the impedance, resistance and reactance of the network feeder is calculated as follows:<br />
<br />
: <math> Z_{f} = \frac{c V_{n}^{2}}{S_{f}} \, </math><br />
: <math> R_{f} = \frac{Z_{f}}{\sqrt{1 + \left( \frac{X}{R} \right)^{2}}} \, </math><br />
: <math> X_{f} = \frac{X}{R} \times R_{f} \, </math><br />
<br />
Where <math> Z_{f} \, </math> is impedance of the network feeder (<math> \Omega </math>)<br />
:: <math> R_{f} \, </math> is resistance of the network feeder (<math> \Omega </math>)<br />
:: <math> X_{f} \, </math> is reactance of the network feeder (<math> \Omega </math>)<br />
:: <math> V_{n} \, </math> is the nominal voltage at the connection point (Vac)<br />
:: <math> S_{f} \, </math> is the fault level of the network feeder (VA)<br />
:: <math> c \, </math> is a voltage factor which accounts for the maximum system voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV)<br />
:: <math> \frac{X}{R} \, </math> is X/R ratio of the network feeder (pu)<br />
<br />
==== Synchronous Generators ====<br />
<br />
The transient resistance and reactance of a synchronous generator can be estimated by the following:<br />
<br />
: <math> X_{d}^{'} = \chi_{d}{'} \times K_{g} \times \frac{V_{g}^{2}}{S_{g}} \, </math><br />
: <math> R_{g} = \frac{X_{d}^{'}}{\frac{X}{R}} \, </math><br />
: <math> K_{g} = \frac{V_{n}}{V_{g}} \frac{c}{1 + \chi_{d}{'} \sin \phi_{g}} \, </math><br />
<br />
Where <math> X_{d}^{'} \, </math> is the transient reactance of the generator (<math> \Omega </math>)<br />
:: <math> R_{g} \, </math> is the resistance of the generator (<math> \Omega </math>)<br />
:: <math> K_{g} \, </math> is a voltage correction factor (pu)<br />
:: <math> \chi_{d}{'} \, </math> is the per-unit transient reactance of the generator (pu)<br />
:: <math> V_{g} \, </math> is the nominal generator voltage (Vac)<br />
:: <math> V_{n} \, </math> is the nominal system voltage (Vac)<br />
:: <math> S_{g} \, </math> is the rated generator capacity (VA)<br />
:: <math> \frac{X}{R} \, </math> is the X/R ratio, typically 20 for <math> S_{g} \geq \, </math> 100MVA, 14.29 for <math> S_{g} < \, </math> 100MVA, and 6.67 for all generators with nominal voltage <math> V_{g} \leq \, </math> 1kV<br />
:: <math> c \, </math> is a voltage factor which accounts for the maximum system voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV)<br />
:: <math> \cos \phi_{g} \, </math> is the power factor of the generator (pu)<br />
<br />
==== Transformers ====<br />
<br />
The impedance, resistance and reactance of two-winding transformers can be calculated as follows:<br />
<br />
: <math> Z_{t} = u_{k} \frac{V_{t}^{2}}{S_{t}} \, </math><br />
: <math> R_{t} = \frac{P_{kt}}{3 I_{t}^{2}} \, </math><br />
: <math> X_{t} = \sqrt{Z_{t}^{2} - R_{t}^{2}} \, </math><br />
<br />
Where <math> Z_{t} \, </math> is the impedance of the transformer (<math> \Omega </math>)<br />
:: <math> R_{t} \, </math> is the resistance of the transformer (<math> \Omega </math>)<br />
:: <math> X_{t} \, </math> is the reactance of the transformer (<math> \Omega </math>)<br />
:: <math> u_{k} \, </math> is the impedance voltage of the transformer (pu)<br />
:: <math> S_{t} \, </math> is the rated capacity of the transformer (VA)<br />
:: <math> V_{t} \, </math> is the nominal voltage of the transformer at the high or low voltage side (Vac)<br />
:: <math> I_{t} \, </math> is the rated current of the transformer at the high or low voltage side (I)<br />
:: <math> P_{kt} \, </math> is the total copper loss in the transformer windings (W)<br />
<br />
==== Cables ====<br />
<br />
Cable impedances are usually quoted by manufacturers in terms of Ohms per km. These need to be converted to Ohms based on the length of the cables:<br />
<br />
: <math> R_{c} = R \times \frac{L_{c}}{1000} \, </math><br />
: <math> X_{c} = X \times \frac{L_{c}}{1000} \, </math><br />
<br />
Where <math> R_{c} \, </math> is the resistance of the cable {<math> \Omega </math>)<br />
:: <math> X_{c} \, </math> is the reactance of the cable {<math> \Omega </math>)<br />
:: <math> R \, </math> is the quoted resistance of the cable {<math> \Omega /km </math>)<br />
:: <math> X \, </math> is the quoted reactance of the cable {<math> \Omega /km </math>)<br />
:: <math> L_{c} \, </math> is the length of the cable {m)<br />
<br />
==== Standing Loads ====<br />
<br />
Standing loads are lumped loads comprising all loads that are operating on a particular bus, excluding the motor to be started. Standing loads for each bus need to be calculated.<br />
<br />
The impedance, resistance and reactance of the standing load is calculated by:<br />
<br />
: <math> Z_{l} = \frac{V_{l}^{2}}{S_{l}} \, </math><br />
: <math> R_{l} = Z_{l} \cos \phi \, </math><br />
: <math> X_{l} = \sqrt{Z_{l}^{2} - R_{l}^{2}} \, </math><br />
<br />
Where <math> Z_{l} \, </math> is the impedance of the standing load {<math> \Omega </math>)<br />
:: <math> R_{l} \, </math> is the resistance of the standing load {<math> \Omega </math>)<br />
:: <math> X_{l} \, </math> is the reactance of the standing load {<math> \Omega </math>)<br />
:: <math> V_{l} \, </math> is the standing load nominal voltage (Vac)<br />
:: <math> S_{l} \, </math> is the standing load apparent power (VA)<br />
:: <math> \cos \phi \, </math> is the average load power factor (pu)<br />
<br />
==== Motors ====<br />
<br />
The motor's transient impedance, resistance and reactance is calculated as follows:<br />
<br />
: <math> Z_{m} = \frac{1}{I_{LRC}/I_{FLC}} \times \frac{V_{m}^{2} \cos \phi_{m}}{P_{m}} \, </math><br />
: <math> R_{m} = \frac{P_{m} \times I_{LRC}/I_{FLC} \times \cos \phi_{s}}{3 I_{LRC}^{2} \times \cos \phi_{m}} \, </math><br />
: <math> X_{m} = \sqrt{Z_{m}^{2} - R_{m}^{2}} \, </math><br />
<br />
Where <math> Z_{m} \, </math> is transient impedance of the motor (<math> \Omega </math>)<br />
:: <math> R_{m} \, </math> is transient resistance of the motor (<math> \Omega </math>)<br />
:: <math> X_{m} \, </math> is transient reactance of the motor (<math> \Omega </math>)<br />
:: <math> I_{LRC}/I_{FLC} \, </math> is ratio of the locked rotor to full load current<br />
:: <math> I_{LRC} \, </math> is the motor locked rotor current (A)<br />
:: <math> V_{m} \, </math> is the motor nominal voltage (Vac)<br />
:: <math> P_{m} \, </math> is the motor rated power (W)<br />
:: <math> \cos \phi_{m} \, </math> is the motor full load power factor (pu)<br />
:: <math> \cos \phi_{s} \, </math> is the motor starting power factor (pu)<br />
<br />
=== Step 3: Referring Impedances ===<br />
<br />
Where there are multiple voltage levels, the equipment impedances calculated earlier need to be converted to a reference voltage (typically the HV side) in order for them to be used in a single equivalent circuit.<br />
<br />
The winding ratio of a transformer can be calculated as follows:<br />
<br />
: <math> n = \frac{V_{t2} \left( 1 + t_{p} \right)}{V_{t1}} \, </math><br />
<br />
Where <math> n \, </math> is the transformer winding ratio<br />
:: <math> V_{t2} \, </math> is the transformer nominal secondary voltage at the principal tap (Vac)<br />
:: <math> V_{t1} \, </math> is the transformer nominal primary voltage (Vac)<br />
:: <math> t_{p} \, </math> is the specified tap setting (%)<br />
<br />
Using the winding ratio, impedances (as well as resistances and reactances) can be referred to the primary (HV) side of the transformer by the following relation:<br />
<br />
: <math> Z_{HV} = \frac{Z_{LV}}{n^{2}} \, </math><br />
<br />
Where <math> Z_{HV} \, </math> is the impedance referred to the primary (HV) side (<math> \Omega </math>)<br />
:: <math> Z_{LV} \, </math> is the impedance at the secondary (LV) side (<math> \Omega </math>)<br />
:: <math> n \, </math> is the transformer winding ratio (pu)<br />
<br />
Conversely, by re-arranging the equation above, impedances can be referred to the LV side:<br />
<br />
: <math> Z_{LV} = Z_{HV} \times n^{2} \, </math><br />
<br />
=== Step 4: Construct the Equivalent Circuit ===<br />
<br />
[[Image:MS_Near_Thevenin1.jpg|right|thumb|300px|Figure 2. "Near" Thévenin equivalent circuit]]<br />
<br />
The equivalent circuit essentially consists of a voltage source (from a network feeder or generator) plus a set of complex impedances representing the power system equipment and load impedances.<br />
<br />
The next step is to simplify the circuit into a form that is nearly the [http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem Thévenin equivalent circuit], with a circuit containing only a voltage source (<math> V_{s} \, </math>), source impedance (<math> Z_{s} \, </math>) and equivalent load impedance (<math> Z_{eq} \, </math>). <br />
<br />
This can be done using the standard formulae for [[Complex_Impedance|series and parallel impedances]], keeping in mind that the rules of [[Complex_Impedance|complex arithmetic]] must be used throughout. This simplification to a "Near" Thévenin equivalent circuit should be done both with the motor off (open circuit) and the motor in a starting condition.<br />
<br />
=== Step 5: Calculate the Initial Source EMF ===<br />
<br />
Assuming that the system is initially in a steady-state condition, we need to first calculate the initial emf produced by the power source (i.e. feeder connection point or generator terminals). This voltage will be used in the transient calculations (Step 6) as the initial source voltage.<br />
<br />
Assumptions regarding the steady-state condition:<br />
<br />
:* The source point of common coupling (PCC) is at its nominal voltage<br />
:* The motor is switched off<br />
:* All standing loads are operating at the capacity calculated in Step 2<br />
:* All transformer taps are set at those specified in Step 2<br />
:* The system is at a steady-state, i.e. there is no switching taking place throughout the system<br />
<br />
Since we assume that there is nominal voltage at the PCC, the initial source emf can be calculated by voltage divider:<br />
<br />
: <math> E_{0} = V_{n} \left( 1 + \frac{Z_{s}}{Z_{eq}} \right) \, </math><br />
<br />
Where <math> E_{0} \, </math> is the initial emf of the power source (Vac)<br />
:: <math> V_{n} \, </math> is the nominal voltage (Vac)<br />
:: <math> Z_{s} \, </math> is the source impedance (<math> \Omega </math>)<br />
:: <math> Z_{eq} \, </math> is the equivalent load impedance with the motor switched off (<math> \Omega </math>)<br />
<br />
=== Step 6: Calculate System Voltages During Motor Start ===<br />
<br />
It is assumed in this calculation that during motor starting, the initial source emf calculated in Step 5 remains constant; that is, the power source does not react during the transient period. This is a simplifying assumption in order to avoid having to model the transient behaviour of the power source.<br />
<br />
Next, we need to calculate the overall system current that is supplied by the power source during the motor starting period. To do this, we use the "Near" Thevenin equivalent circuit derived earlier, but now include the motor starting impedance. A new equivalent load impedance during motor starting <math> Z_{eq,s} \, </math> will be calculated. <br />
<br />
The current supplied by the power source is therefore:<br />
<br />
: <math> I_{s} = \frac{E_{0}}{Z_{eq,s} + Z_{s}} \, </math><br />
<br />
Where <math> I_{s} \, </math> is the system current supplied by the source (A)<br />
:: <math> E_{0} \, </math> is the initial source emf (Vac)<br />
:: <math> Z_{eq,s} \, </math> is the equivalent load impedance during motor start (<math> \Omega </math>)<br />
:: <math> Z_{s} \, </math> is the source impedance (<math> \Omega </math>)<br />
<br />
The voltage at the source point of common coupling (PCC) is:<br />
<br />
: <math> V_{PCC} = E_{0} - I_{s} Z_{s} \, </math><br />
<br />
Where <math> V_{PCC} \, </math> is the voltage at the point of common coupling (Vac)<br />
:: <math> E_{0} \, </math> is the initial source emf (Vac)<br />
:: <math> I_{s} \, </math> is the system current supplied by the source (A)<br />
:: <math> Z_{s} \, </math> is the source impedance (<math> \Omega </math>)<br />
<br />
The downstream voltages can now be calculated by voltage division and simple application of Ohm's law. Specifically, we'd like to know the voltage at the motor terminals and any buses of interest that could be affected. Ensure that the voltages are acceptably within the prescribed limits, otherwise further action needs to be taken (refer to the [[Motor_Starting#What_Next.3F|What's Next?]] section).<br />
<br />
== Worked Example ==<br />
<br />
The worked example here is a very simple power system with two voltage levels and supplied by a single generator. While unrealistic, it does manage to demonstrate the key concepts pertaining to motor starting calculations.<br />
<br />
=== Step 1: Construct System Model and Collect Equipment Parameters ===<br />
<br />
[[Image:MS_Model.jpg|right|thumb|400px|Figure 3. Simplified system model for motor starting example]]<br />
<br />
The power system has two voltage levels, 11kV and 415V, and is fed via a single 4MVA generator (G1). The 11kV bus has a standing load of 950kVA (S1) and we want to model the effects of starting a 250kW motor (M1). There is a standing load of 600kVA at 415V (S2), supplied by a 1.6MVA transformer (TX1). The equipment and cable parameters are as follows:<br />
<br />
{| border = "1" cellpadding="5" cellspacing="0" style = "text-align:center;"<br />
|-<br />
!style="background:#efefef;" width="200px" | Equipment<br />
!style="background:#efefef;" width="300px" | Parameters<br />
|-<br />
| Generator G1<br />
| <br />
* <math> S_{g1} \, </math> = 4,000 kVA<br />
* <math> V_{g1} \, </math> = 11,000 V<br />
* <math> \chi_{d}^{'} \, </math> = 0.33 pu<br />
* <math> \cos \phi \, </math> = 0.85 pu<br />
|-<br />
| Generator Cable C1<br />
| <br />
* Length = 50m<br />
* Size = 500 <math> mm^{2} </math> <br />
(R = 0.0522 <math> \Omega </math>\km, X = 0.0826 <math> \Omega </math>\km)<br />
|-<br />
| 11kV Standing Load S1<br />
| <br />
* <math> S_{s1} \, </math> = 950 kVA<br />
* <math> V_{s1} \, </math> = 11,000 V<br />
* <math> \cos \phi \, </math> = 0.84 pu<br />
|-<br />
| Motor M1<br />
| <br />
* <math> P_{m1} \, </math> = 250 kW<br />
* <math> V_{m1} \, </math> = 11,000 V<br />
* <math> I_{LRC} \, </math> = 106.7 A<br />
* <math> I_{LRC}/I_{FLC} \, </math> = 6.5 pu<br />
* <math> \cos \phi_{m} \, </math> = 0.85 pu<br />
* <math> \cos \phi_{s} \, </math> = 0.30 pu<br />
|-<br />
| Motor Cable C2<br />
| <br />
* Length = 150m<br />
* Size = 35 <math> mm^{2} </math> <br />
(R = 0.668 <math> \Omega </math>\km, X = 0.115 <math> \Omega </math>\km)<br />
|-<br />
| Transformer TX1 <br />
| <br />
* <math> S_{tx1} \, </math> = 1,600 kVA<br />
* <math> V_{t1} \, </math> = 11,000 V<br />
* <math> V_{t2} \, </math> = 415 V<br />
* <math> u_{k} \, </math> = 0.06 pu<br />
* <math> P_{kt} \, </math> = 12,700 W<br />
* <math> t_{p} \, </math> = 0%<br />
|-<br />
| Transformer Cable C3 <br />
| <br />
* Length = 60m<br />
* Size = 120 <math> mm^{2} </math> <br />
(R = 0.196 <math> \Omega </math>\km, X = 0.096 <math> \Omega </math>\km)<br />
|-<br />
| 415V Standing Load S2<br />
| <br />
* <math> S_{s2} \, </math> = 600 kVA<br />
* <math> V_{s2} \, </math> = 415 V<br />
* <math> \cos \phi \, </math> = 0.80 pu<br />
|}<br />
<br />
=== Step 2: Calculate Equipment Impedances ===<br />
<br />
Using the patameters above and the equations outlined earlier in the methodology, the following impedances were calculated:<br />
<br />
{| border = "1" cellpadding="5" cellspacing="0" style = "text-align:center;"<br />
|-<br />
!style="background:#efefef;" width="250px" | Equipment<br />
!style="background:#efefef;" width="100px" | Resistance (<math> \Omega </math>)<br />
!style="background:#efefef;" width="100px" | Reactance (<math> \Omega </math>)<br />
|-<br />
| Generator G1 || 0.65462 || 9.35457<br />
|-<br />
| Generator Cable C1 || 0.00261 || 0.00413<br />
|-<br />
| 11kV Standing Load S1 || 106.98947 || 69.10837<br />
|-<br />
| Motor M1 || 16.77752 || 61.02812<br />
|-<br />
| Motor Cable C2 || 0.1002 || 0.01725<br />
|-<br />
| Transformer TX1 (Primary Side) || 0.60027 || 4.49762<br />
|-<br />
| Transformer Cable C3 || 0.01176 || 0.00576<br />
|-<br />
| 415V Standing Load S2 || 0.22963 || 0.17223<br />
|}<br />
<br />
=== Step 3: Referring Impedances ===<br />
<br />
11kV will be used as the reference voltage. The only impedance that needs to be referred to this reference voltage is the 415V Standing Load (S2). Knowing that the transformer is set at principal tap, we can calculate the winding ratio and apply it to refer the 415V Standing Load impedance to the 11kV side:<br />
<br />
: <math> n = \frac{415 \left( 1 + 0% \right)}{11,000} = 0.03773 \, </math><br />
<br />
The resistance and reactance of the standing load referred to the 11kV side is now, R = 161.33333 <math> \Omega </math> and X = 121.00 <math> \Omega </math>.<br />
<br />
=== Step 4: Construct the Equivalent Circuit ===<br />
<br />
[[Image:MS_Equiv_Circuit.jpg|right|thumb|400px|Figure 4. Equivalent circuit for motor starting example]]<br />
<br />
The equivalent circuit for the system is shown in the figure to the right. The "Near" Thevenin equivalent circuit is also shown, and we now calculate the equivalent load impedance <math> Z_{eq} \, </math> in the steady-state condition (i.e. without the motor and motor cable impedances included):<br />
<br />
: <math> Z_{eq} = Z_{C1} + \left[ Z_{S1} || \left( Z_{C3} + Z_{TX1} + Z_{S2} \right) \right] \, </math><br />
::: <math> = 64.59747 + j 44.80458 \, </math><br />
<br />
Similarly the equivalent load impedance during motor starting (with the motor impedances included) can be calculated as as follows:<br />
<br />
: <math> Z_{eq,s} = Z_{C1} + \left[ Z_{S1} || \left( Z_{C3} + Z_{TX1} + Z_{S2} \right) || Z_{C2} + Z_{M1} \right] \, </math><br />
::: <math> = 20.371997 + j 31.22116 \, </math><br />
<br />
=== Step 5: Calculate the Initial Source EMF ===<br />
<br />
[[Image:MS_Near_Thevenin.jpg|right|thumb|300px|Figure 5. "Near" Thevenin equivalent circuit for motor starting example]]<br />
<br />
Assuming that there is nominal voltage at the 11kV bus in the steady-state condition, the initial generator emf can be calculated by voltage divider:<br />
<br />
: <math> E_{0} = V_{n} \left( 1 + \frac{Z_{G1}}{Z_{eq}} \right) \, </math><br />
::: <math> = 11,821.25 + j 1,023.33 = 11,865 \, </math> Vac<br />
<br />
=== Step 6: Calculate System Voltages During Motor Start ===<br />
<br />
Now we can calculate the transient effects of motor starting on the system voltages. Firstly, the current supplied by the generator during motor start is calculated:<br />
<br />
: <math> I_{G1} = \frac{E_{0}}{Z_{eq,s} + Z_{G1}} \, </math><br />
::: <math> = 138.8949 - j 219.36166 = 259.64A \, </math><br />
<br />
Next, the voltage at the 11kV bus can be found:<br />
<br />
: <math> V_{11kV} = E_{0} - I_{G1} ( Z_{G1} + Z_{C1} ) \, </math><br />
::: <math> = 9,677.024 - j 132.375 = 9,677.9 \, </math> Vac (or 87.98% of nominal voltage)<br />
<br />
The voltage at the motor terminals can then be found by voltage divider:<br />
<br />
: <math> V_{M1} = V_{11kV} \frac{Z_{M1}}{Z_{C2} + Z_{M1}} \, </math><br />
::: <math> = 9,670.597 - j 118.231 = 9,671.3 \, </math> Vac (or 87.92% of nominal voltage)<br />
<br />
The voltage at the low voltage bus is:<br />
<br />
: <math> V_{415V} = V_{11kV} \frac{Z_{S2}}{Z_{C3} + + Z_{TX1} + Z_{S2}} \, </math><br />
::: <math> = 9,521.825 - j 280.698 = 9,525.6 \, </math> Vac, then referred to the LV side = 359.39Vac (or 86.60% of nominal voltage)<br />
<br />
Any other voltages of interest on the system can be determined using the same methods as above.<br />
<br />
Suppose that our maximum voltage drop at the motor terminals is 15%. From above, we have found that the voltage drop is 12.08% at the motor terminals. This is a slightly marginal result and it may be prudent to simulate the system in a software package to confirm the results.<br />
<br />
== Computer Software ==<br />
<br />
Motor starting is a standard component of most Power_Systems_Analysis_Software|power systems analysis software (e.g. ETAP, PTW, PowerFactory, etc) and this calculation is really intended to be done using this software. The numerical calculation performed by the software should also solve the power flow problem through an iterative algorithm (e.g. such as Newton-Rhapson).<br />
<br />
== What Next? ==<br />
<br />
If the results of the calculation confirm that starting the largest motor does not cause any unacceptable voltage levels within the system, then that's the end of it (or perhaps it could be simulated in a power systems analysis software package to be doubly sure!). Otherwise, the issue needs to be addressed, for example by:<br />
<br />
:* Reduce the motor starting current, e.g. via soft-starters, star-delta starters, etc<br />
:* Reduce the source impedances, e.g. increase the size of the generator, transformer, supply cables, etc<br />
<br />
The calculation should be performed iteratively until the results are acceptable.<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=File:MS_Near_Thevenin1.jpg&diff=205File:MS Near Thevenin1.jpg2021-05-04T05:26:29Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=File:MS_Near_Thevenin.jpg&diff=204File:MS Near Thevenin.jpg2021-05-04T05:26:18Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=File:MS_Model.jpg&diff=203File:MS Model.jpg2021-05-04T05:26:04Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=File:MS_Equiv_Circuit.jpg&diff=202File:MS Equiv Circuit.jpg2021-05-04T05:25:31Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=File:HV_motor.jpg&diff=201File:HV motor.jpg2021-05-04T05:24:56Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=Motor_Starting_Calculation&diff=200Motor Starting Calculation2021-05-04T05:18:24Z<p>Jules: Created page with "== Introduction == Figure 1. High voltage motor (courtesy of ABB) This article considers the transient effects of motor starting on..."</p>
<hr />
<div>== Introduction ==<br />
<br />
[[Image:HV_motor.jpg|right|thumb|300px|Figure 1. High voltage motor (courtesy of ABB)]]<br />
<br />
This article considers the transient effects of motor starting on the system voltage. Usually only the largest motor on a bus or system is modelled, but the calculation can in principle be used for any motor. It's important to note that motor starting is a transient power flow problem and is normally done iteratively by computer software. However a static method is shown here for first-pass estimates only.<br />
<br />
=== Why do the calculation? ===<br />
<br />
When a motor is started, it typically draws a current 6-7 times its full load current for a short duration (commonly called the locked rotor current). During this transient period, the source impedance is generally assumed to be fixed and therefore, a large increase in current will result in a larger voltage drop across the source impedance. This means that there can be large momentary voltage drops system-wide, from the power source (e.g. transformer or generator) through the intermediary buses, all the way to the motor terminals. <br />
<br />
A system-wide voltage drop can have a number of adverse effects, for example:<br />
<br />
:* Equipment with minimum voltage tolerances (e.g. electronics) may malfunction or behave aberrantly<br />
:* Undervoltage protection may be tripped<br />
:* The motor itself may not start as torque is proportional to the square of the stator voltage, so a reduced voltage equals lower torque. Induction motors are typically designed to start with a terminal voltage >80%<br />
<br />
=== When to do the calculation? ===<br />
<br />
This calculation is more or less done to verify that the largest motor does not cause system wide problems upon starting. Therefore it should be done after preliminary system design is complete. The following prerequisite information is required: <br />
<br />
:* Key single line diagrams<br />
:* Preliminary load schedule<br />
:* Tolerable voltage drop limits during motor starting, which are typically prescribed by the client<br />
<br />
== Calculation Methodology ==<br />
<br />
This calculation is based on standard impedance formulae and Ohm's law. To the author's knowledge, there are no international standards that govern voltage drop calculations during motor start. <br />
<br />
It should be noted that the proposed method is not 100% accurate because it is a static calculation. In reality, the voltage levels are fluctuating during a transient condition, and therefore so are the load currents drawn by the standing loads. This makes it essentially a load flow problem and a more precise solution would solve the load flow problem iteratively, for example using the Newton-Rhapson or Gauss-Siedel algorithms. Notwithstanding, the proposed method is suitably accurate for a first pass solution.<br />
<br />
The calculation has the following six general steps:<br />
<br />
:* Step 1: Construct the system model and assemble the relevant equipment parameters<br />
:* Step 2: Calculate the relevant impedances for each equipment item in the model<br />
:* Step 3: Refer all impedances to a reference voltage<br />
:* Step 4: Construct the equivalent circuit for the voltage levels of interest<br />
:* Step 5: Calculate the initial steady-state source emf before motor starting<br />
:* Step 6: Calculate the system voltages during motor start<br />
<br />
=== Step 1: Construct System Model and Collect Equipment Parameters ===<br />
<br />
The first step is to construct a simplified model of the system single line diagram, and then collect the relevant equipment parameters. The model of the single line diagram need only show the buses of interest in the motor starting calculation, e.g. the upstream source bus, the motor bus and possibly any intermediate or downstream buses that may be affected. All running loads are shown as lumped loads except for the motor to be started as it is assumed that the system is in a steady-state before motor start. <br />
<br />
The relevant equipment parameters to be collected are as follows:<br />
<br />
:* Network feeders: fault capacity of the network (VA), X/R ratio of the network<br />
:* Generators: per-unit transient reactance, rated generator capacity (VA)<br />
:* Transformers: transformer impedance voltage (%), rated transformer capacity (VA), rated current (A), total copper loss (W)<br />
:* Cables: length of cable (m), resistance and reactance of cable (<math> \Omega \ km </math>)<br />
:* Standing loads: rated load capacity (VA), average load power factor (pu)<br />
:* Motor: full load current (A), locked rotor current (A), rated power (W), full load power factor (pu), starting power factor (pu)<br />
<br />
=== Step 2: Calculate Equipment Impedances ===<br />
<br />
Using the collected parameters, each of the equipment item impedances can be calculated for later use in the motor starting calculations.<br />
<br />
==== Network Feeders ====<br />
<br />
Given the approximate fault level of the network feeder at the connection point (or point of common coupling), the impedance, resistance and reactance of the network feeder is calculated as follows:<br />
<br />
: <math> Z_{f} = \frac{c V_{n}^{2}}{S_{f}} \, </math><br />
: <math> R_{f} = \frac{Z_{f}}{\sqrt{1 + \left( \frac{X}{R} \right)^{2}}} \, </math><br />
: <math> X_{f} = \frac{X}{R} \times R_{f} \, </math><br />
<br />
Where <math> Z_{f} \, </math> is impedance of the network feeder (<math> \Omega </math>)<br />
:: <math> R_{f} \, </math> is resistance of the network feeder (<math> \Omega </math>)<br />
:: <math> X_{f} \, </math> is reactance of the network feeder (<math> \Omega </math>)<br />
:: <math> V_{n} \, </math> is the nominal voltage at the connection point (Vac)<br />
:: <math> S_{f} \, </math> is the fault level of the network feeder (VA)<br />
:: <math> c \, </math> is a voltage factor which accounts for the maximum system voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV)<br />
:: <math> \frac{X}{R} \, </math> is X/R ratio of the network feeder (pu)<br />
<br />
==== Synchronous Generators ====<br />
<br />
The transient resistance and reactance of a synchronous generator can be estimated by the following:<br />
<br />
: <math> X_{d}^{'} = \chi_{d}{'} \times K_{g} \times \frac{V_{g}^{2}}{S_{g}} \, </math><br />
: <math> R_{g} = \frac{X_{d}^{'}}{\frac{X}{R}} \, </math><br />
: <math> K_{g} = \frac{V_{n}}{V_{g}} \frac{c}{1 + \chi_{d}{'} \sin \phi_{g}} \, </math><br />
<br />
Where <math> X_{d}^{'} \, </math> is the transient reactance of the generator (<math> \Omega </math>)<br />
:: <math> R_{g} \, </math> is the resistance of the generator (<math> \Omega </math>)<br />
:: <math> K_{g} \, </math> is a voltage correction factor (pu)<br />
:: <math> \chi_{d}{'} \, </math> is the per-unit transient reactance of the generator (pu)<br />
:: <math> V_{g} \, </math> is the nominal generator voltage (Vac)<br />
:: <math> V_{n} \, </math> is the nominal system voltage (Vac)<br />
:: <math> S_{g} \, </math> is the rated generator capacity (VA)<br />
:: <math> \frac{X}{R} \, </math> is the X/R ratio, typically 20 for <math> S_{g} \geq \, </math> 100MVA, 14.29 for <math> S_{g} < \, </math> 100MVA, and 6.67 for all generators with nominal voltage <math> V_{g} \leq \, </math> 1kV<br />
:: <math> c \, </math> is a voltage factor which accounts for the maximum system voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV)<br />
:: <math> \cos \phi_{g} \, </math> is the power factor of the generator (pu)<br />
<br />
==== Transformers ====<br />
<br />
The impedance, resistance and reactance of two-winding transformers can be calculated as follows:<br />
<br />
: <math> Z_{t} = u_{k} \frac{V_{t}^{2}}{S_{t}} \, </math><br />
: <math> R_{t} = \frac{P_{kt}}{3 I_{t}^{2}} \, </math><br />
: <math> X_{t} = \sqrt{Z_{t}^{2} - R_{t}^{2}} \, </math><br />
<br />
Where <math> Z_{t} \, </math> is the impedance of the transformer (<math> \Omega </math>)<br />
:: <math> R_{t} \, </math> is the resistance of the transformer (<math> \Omega </math>)<br />
:: <math> X_{t} \, </math> is the reactance of the transformer (<math> \Omega </math>)<br />
:: <math> u_{k} \, </math> is the impedance voltage of the transformer (pu)<br />
:: <math> S_{t} \, </math> is the rated capacity of the transformer (VA)<br />
:: <math> V_{t} \, </math> is the nominal voltage of the transformer at the high or low voltage side (Vac)<br />
:: <math> I_{t} \, </math> is the rated current of the transformer at the high or low voltage side (I)<br />
:: <math> P_{kt} \, </math> is the total copper loss in the transformer windings (W)<br />
<br />
==== Cables ====<br />
<br />
Cable impedances are usually quoted by manufacturers in terms of Ohms per km. These need to be converted to Ohms based on the length of the cables:<br />
<br />
: <math> R_{c} = R \times \frac{L_{c}}{1000} \, </math><br />
: <math> X_{c} = X \times \frac{L_{c}}{1000} \, </math><br />
<br />
Where <math> R_{c} \, </math> is the resistance of the cable {<math> \Omega </math>)<br />
:: <math> X_{c} \, </math> is the reactance of the cable {<math> \Omega </math>)<br />
:: <math> R \, </math> is the quoted resistance of the cable {<math> \Omega /km </math>)<br />
:: <math> X \, </math> is the quoted reactance of the cable {<math> \Omega /km </math>)<br />
:: <math> L_{c} \, </math> is the length of the cable {m)<br />
<br />
==== Standing Loads ====<br />
<br />
Standing loads are lumped loads comprising all loads that are operating on a particular bus, excluding the motor to be started. Standing loads for each bus need to be calculated.<br />
<br />
The impedance, resistance and reactance of the standing load is calculated by:<br />
<br />
: <math> Z_{l} = \frac{V_{l}^{2}}{S_{l}} \, </math><br />
: <math> R_{l} = Z_{l} \cos \phi \, </math><br />
: <math> X_{l} = \sqrt{Z_{l}^{2} - R_{l}^{2}} \, </math><br />
<br />
Where <math> Z_{l} \, </math> is the impedance of the standing load {<math> \Omega </math>)<br />
:: <math> R_{l} \, </math> is the resistance of the standing load {<math> \Omega </math>)<br />
:: <math> X_{l} \, </math> is the reactance of the standing load {<math> \Omega </math>)<br />
:: <math> V_{l} \, </math> is the standing load nominal voltage (Vac)<br />
:: <math> S_{l} \, </math> is the standing load apparent power (VA)<br />
:: <math> \cos \phi \, </math> is the average load power factor (pu)<br />
<br />
==== Motors ====<br />
<br />
The motor's transient impedance, resistance and reactance is calculated as follows:<br />
<br />
: <math> Z_{m} = \frac{1}{I_{LRC}/I_{FLC}} \times \frac{V_{m}^{2} \cos \phi_{m}}{P_{m}} \, </math><br />
: <math> R_{m} = \frac{P_{m} \times I_{LRC}/I_{FLC} \times \cos \phi_{s}}{3 I_{LRC}^{2} \times \cos \phi_{m}} \, </math><br />
: <math> X_{m} = \sqrt{Z_{m}^{2} - R_{m}^{2}} \, </math><br />
<br />
Where <math> Z_{m} \, </math> is transient impedance of the motor (<math> \Omega </math>)<br />
:: <math> R_{m} \, </math> is transient resistance of the motor (<math> \Omega </math>)<br />
:: <math> X_{m} \, </math> is transient reactance of the motor (<math> \Omega </math>)<br />
:: <math> I_{LRC}/I_{FLC} \, </math> is ratio of the locked rotor to full load current<br />
:: <math> I_{LRC} \, </math> is the motor locked rotor current (A)<br />
:: <math> V_{m} \, </math> is the motor nominal voltage (Vac)<br />
:: <math> P_{m} \, </math> is the motor rated power (W)<br />
:: <math> \cos \phi_{m} \, </math> is the motor full load power factor (pu)<br />
:: <math> \cos \phi_{s} \, </math> is the motor starting power factor (pu)<br />
<br />
=== Step 3: Referring Impedances ===<br />
<br />
Where there are multiple voltage levels, the equipment impedances calculated earlier need to be converted to a reference voltage (typically the HV side) in order for them to be used in a single equivalent circuit.<br />
<br />
The winding ratio of a transformer can be calculated as follows:<br />
<br />
: <math> n = \frac{V_{t2} \left( 1 + t_{p} \right)}{V_{t1}} \, </math><br />
<br />
Where <math> n \, </math> is the transformer winding ratio<br />
:: <math> V_{t2} \, </math> is the transformer nominal secondary voltage at the principal tap (Vac)<br />
:: <math> V_{t1} \, </math> is the transformer nominal primary voltage (Vac)<br />
:: <math> t_{p} \, </math> is the specified tap setting (%)<br />
<br />
Using the winding ratio, impedances (as well as resistances and reactances) can be referred to the primary (HV) side of the transformer by the following relation:<br />
<br />
: <math> Z_{HV} = \frac{Z_{LV}}{n^{2}} \, </math><br />
<br />
Where <math> Z_{HV} \, </math> is the impedance referred to the primary (HV) side (<math> \Omega </math>)<br />
:: <math> Z_{LV} \, </math> is the impedance at the secondary (LV) side (<math> \Omega </math>)<br />
:: <math> n \, </math> is the transformer winding ratio (pu)<br />
<br />
Conversely, by re-arranging the equation above, impedances can be referred to the LV side:<br />
<br />
: <math> Z_{LV} = Z_{HV} \times n^{2} \, </math><br />
<br />
=== Step 4: Construct the Equivalent Circuit ===<br />
<br />
[[Image:MS_Near_Thevenin1.jpg|right|thumb|300px|Figure 2. "Near" Thévenin equivalent circuit]]<br />
<br />
The equivalent circuit essentially consists of a voltage source (from a network feeder or generator) plus a set of complex impedances representing the power system equipment and load impedances.<br />
<br />
The next step is to simplify the circuit into a form that is nearly the [http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem Thévenin equivalent circuit], with a circuit containing only a voltage source (<math> V_{s} \, </math>), source impedance (<math> Z_{s} \, </math>) and equivalent load impedance (<math> Z_{eq} \, </math>). <br />
<br />
This can be done using the standard formulae for [[Complex_Impedance|series and parallel impedances]], keeping in mind that the rules of [[Complex_Impedance|complex arithmetic]] must be used throughout. This simplification to a "Near" Thévenin equivalent circuit should be done both with the motor off (open circuit) and the motor in a starting condition.<br />
<br />
=== Step 5: Calculate the Initial Source EMF ===<br />
<br />
Assuming that the system is initially in a steady-state condition, we need to first calculate the initial emf produced by the power source (i.e. feeder connection point or generator terminals). This voltage will be used in the transient calculations (Step 6) as the initial source voltage.<br />
<br />
Assumptions regarding the steady-state condition:<br />
<br />
:* The source point of common coupling (PCC) is at its nominal voltage<br />
:* The motor is switched off<br />
:* All standing loads are operating at the capacity calculated in Step 2<br />
:* All transformer taps are set at those specified in Step 2<br />
:* The system is at a steady-state, i.e. there is no switching taking place throughout the system<br />
<br />
Since we assume that there is nominal voltage at the PCC, the initial source emf can be calculated by voltage divider:<br />
<br />
: <math> E_{0} = V_{n} \left( 1 + \frac{Z_{s}}{Z_{eq}} \right) \, </math><br />
<br />
Where <math> E_{0} \, </math> is the initial emf of the power source (Vac)<br />
:: <math> V_{n} \, </math> is the nominal voltage (Vac)<br />
:: <math> Z_{s} \, </math> is the source impedance (<math> \Omega </math>)<br />
:: <math> Z_{eq} \, </math> is the equivalent load impedance with the motor switched off (<math> \Omega </math>)<br />
<br />
=== Step 6: Calculate System Voltages During Motor Start ===<br />
<br />
It is assumed in this calculation that during motor starting, the initial source emf calculated in Step 5 remains constant; that is, the power source does not react during the transient period. This is a simplifying assumption in order to avoid having to model the transient behaviour of the power source.<br />
<br />
Next, we need to calculate the overall system current that is supplied by the power source during the motor starting period. To do this, we use the "Near" Thevenin equivalent circuit derived earlier, but now include the motor starting impedance. A new equivalent load impedance during motor starting <math> Z_{eq,s} \, </math> will be calculated. <br />
<br />
The current supplied by the power source is therefore:<br />
<br />
: <math> I_{s} = \frac{E_{0}}{Z_{eq,s} + Z_{s}} \, </math><br />
<br />
Where <math> I_{s} \, </math> is the system current supplied by the source (A)<br />
:: <math> E_{0} \, </math> is the initial source emf (Vac)<br />
:: <math> Z_{eq,s} \, </math> is the equivalent load impedance during motor start (<math> \Omega </math>)<br />
:: <math> Z_{s} \, </math> is the source impedance (<math> \Omega </math>)<br />
<br />
The voltage at the source point of common coupling (PCC) is:<br />
<br />
: <math> V_{PCC} = E_{0} - I_{s} Z_{s} \, </math><br />
<br />
Where <math> V_{PCC} \, </math> is the voltage at the point of common coupling (Vac)<br />
:: <math> E_{0} \, </math> is the initial source emf (Vac)<br />
:: <math> I_{s} \, </math> is the system current supplied by the source (A)<br />
:: <math> Z_{s} \, </math> is the source impedance (<math> \Omega </math>)<br />
<br />
The downstream voltages can now be calculated by voltage division and simple application of Ohm's law. Specifically, we'd like to know the voltage at the motor terminals and any buses of interest that could be affected. Ensure that the voltages are acceptably within the prescribed limits, otherwise further action needs to be taken (refer to the [[Motor_Starting#What_Next.3F|What's Next?]] section).<br />
<br />
== Worked Example ==<br />
<br />
The worked example here is a very simple power system with two voltage levels and supplied by a single generator. While unrealistic, it does manage to demonstrate the key concepts pertaining to motor starting calculations.<br />
<br />
=== Step 1: Construct System Model and Collect Equipment Parameters ===<br />
<br />
[[Image:MS_Model.jpg|right|thumb|400px|Figure 3. Simplified system model for motor starting example]]<br />
<br />
The power system has two voltage levels, 11kV and 415V, and is fed via a single 4MVA generator (G1). The 11kV bus has a standing load of 950kVA (S1) and we want to model the effects of starting a 250kW motor (M1). There is a standing load of 600kVA at 415V (S2), supplied by a 1.6MVA transformer (TX1). The equipment and cable parameters are as follows:<br />
<br />
{| border = "1" cellpadding="5" cellspacing="0" style = "text-align:center;"<br />
|-<br />
!style="background:#efefef;" width="200px" | Equipment<br />
!style="background:#efefef;" width="300px" | Parameters<br />
|-<br />
| Generator G1<br />
| <br />
* <math> S_{g1} \, </math> = 4,000 kVA<br />
* <math> V_{g1} \, </math> = 11,000 V<br />
* <math> \chi_{d}^{'} \, </math> = 0.33 pu<br />
* <math> \cos \phi \, </math> = 0.85 pu<br />
|-<br />
| Generator Cable C1<br />
| <br />
* Length = 50m<br />
* Size = 500 <math> mm^{2} </math> <br />
(R = 0.0522 <math> \Omega </math>\km, X = 0.0826 <math> \Omega </math>\km)<br />
|-<br />
| 11kV Standing Load S1<br />
| <br />
* <math> S_{s1} \, </math> = 950 kVA<br />
* <math> V_{s1} \, </math> = 11,000 V<br />
* <math> \cos \phi \, </math> = 0.84 pu<br />
|-<br />
| Motor M1<br />
| <br />
* <math> P_{m1} \, </math> = 250 kW<br />
* <math> V_{m1} \, </math> = 11,000 V<br />
* <math> I_{LRC} \, </math> = 106.7 A<br />
* <math> I_{LRC}/I_{FLC} \, </math> = 6.5 pu<br />
* <math> \cos \phi_{m} \, </math> = 0.85 pu<br />
* <math> \cos \phi_{s} \, </math> = 0.30 pu<br />
|-<br />
| Motor Cable C2<br />
| <br />
* Length = 150m<br />
* Size = 35 <math> mm^{2} </math> <br />
(R = 0.668 <math> \Omega </math>\km, X = 0.115 <math> \Omega </math>\km)<br />
|-<br />
| Transformer TX1 <br />
| <br />
* <math> S_{tx1} \, </math> = 1,600 kVA<br />
* <math> V_{t1} \, </math> = 11,000 V<br />
* <math> V_{t2} \, </math> = 415 V<br />
* <math> u_{k} \, </math> = 0.06 pu<br />
* <math> P_{kt} \, </math> = 12,700 W<br />
* <math> t_{p} \, </math> = 0%<br />
|-<br />
| Transformer Cable C3 <br />
| <br />
* Length = 60m<br />
* Size = 120 <math> mm^{2} </math> <br />
(R = 0.196 <math> \Omega </math>\km, X = 0.096 <math> \Omega </math>\km)<br />
|-<br />
| 415V Standing Load S2<br />
| <br />
* <math> S_{s2} \, </math> = 600 kVA<br />
* <math> V_{s2} \, </math> = 415 V<br />
* <math> \cos \phi \, </math> = 0.80 pu<br />
|}<br />
<br />
=== Step 2: Calculate Equipment Impedances ===<br />
<br />
Using the patameters above and the equations outlined earlier in the methodology, the following impedances were calculated:<br />
<br />
{| border = "1" cellpadding="5" cellspacing="0" style = "text-align:center;"<br />
|-<br />
!style="background:#efefef;" width="250px" | Equipment<br />
!style="background:#efefef;" width="100px" | Resistance (<math> \Omega </math>)<br />
!style="background:#efefef;" width="100px" | Reactance (<math> \Omega </math>)<br />
|-<br />
| Generator G1 || 0.65462 || 9.35457<br />
|-<br />
| Generator Cable C1 || 0.00261 || 0.00413<br />
|-<br />
| 11kV Standing Load S1 || 106.98947 || 69.10837<br />
|-<br />
| Motor M1 || 16.77752 || 61.02812<br />
|-<br />
| Motor Cable C2 || 0.1002 || 0.01725<br />
|-<br />
| Transformer TX1 (Primary Side) || 0.60027 || 4.49762<br />
|-<br />
| Transformer Cable C3 || 0.01176 || 0.00576<br />
|-<br />
| 415V Standing Load S2 || 0.22963 || 0.17223<br />
|}<br />
<br />
=== Step 3: Referring Impedances ===<br />
<br />
11kV will be used as the reference voltage. The only impedance that needs to be referred to this reference voltage is the 415V Standing Load (S2). Knowing that the transformer is set at principal tap, we can calculate the winding ratio and apply it to refer the 415V Standing Load impedance to the 11kV side:<br />
<br />
: <math> n = \frac{415 \left( 1 + 0% \right)}{11,000} = 0.03773 \, </math><br />
<br />
The resistance and reactance of the standing load referred to the 11kV side is now, R = 161.33333 <math> \Omega </math> and X = 121.00 <math> \Omega </math>.<br />
<br />
=== Step 4: Construct the Equivalent Circuit ===<br />
<br />
[[Image:MS_Equiv_Circuit.jpg|right|thumb|400px|Figure 4. Equivalent circuit for motor starting example]]<br />
<br />
The equivalent circuit for the system is shown in the figure to the right. The "Near" Thevenin equivalent circuit is also shown, and we now calculate the equivalent load impedance <math> Z_{eq} \, </math> in the steady-state condition (i.e. without the motor and motor cable impedances included):<br />
<br />
: <math> Z_{eq} = Z_{C1} + \left[ Z_{S1} || \left( Z_{C3} + Z_{TX1} + Z_{S2} \right) \right] \, </math><br />
::: <math> = 64.59747 + j 44.80458 \, </math><br />
<br />
Similarly the equivalent load impedance during motor starting (with the motor impedances included) can be calculated as as follows:<br />
<br />
: <math> Z_{eq,s} = Z_{C1} + \left[ Z_{S1} || \left( Z_{C3} + Z_{TX1} + Z_{S2} \right) || Z_{C2} + Z_{M1} \right] \, </math><br />
::: <math> = 20.371997 + j 31.22116 \, </math><br />
<br />
=== Step 5: Calculate the Initial Source EMF ===<br />
<br />
[[Image:MS_Near_Thevenin.jpg|right|thumb|300px|Figure 5. "Near" Thevenin equivalent circuit for motor starting example]]<br />
<br />
Assuming that there is nominal voltage at the 11kV bus in the steady-state condition, the initial generator emf can be calculated by voltage divider:<br />
<br />
: <math> E_{0} = V_{n} \left( 1 + \frac{Z_{G1}}{Z_{eq}} \right) \, </math><br />
::: <math> = 11,821.25 + j 1,023.33 = 11,865 \, </math> Vac<br />
<br />
=== Step 6: Calculate System Voltages During Motor Start ===<br />
<br />
Now we can calculate the transient effects of motor starting on the system voltages. Firstly, the current supplied by the generator during motor start is calculated:<br />
<br />
: <math> I_{G1} = \frac{E_{0}}{Z_{eq,s} + Z_{G1}} \, </math><br />
::: <math> = 138.8949 - j 219.36166 = 259.64A \, </math><br />
<br />
Next, the voltage at the 11kV bus can be found:<br />
<br />
: <math> V_{11kV} = E_{0} - I_{G1} ( Z_{G1} + Z_{C1} ) \, </math><br />
::: <math> = 9,677.024 - j 132.375 = 9,677.9 \, </math> Vac (or 87.98% of nominal voltage)<br />
<br />
The voltage at the motor terminals can then be found by voltage divider:<br />
<br />
: <math> V_{M1} = V_{11kV} \frac{Z_{M1}}{Z_{C2} + Z_{M1}} \, </math><br />
::: <math> = 9,670.597 - j 118.231 = 9,671.3 \, </math> Vac (or 87.92% of nominal voltage)<br />
<br />
The voltage at the low voltage bus is:<br />
<br />
: <math> V_{415V} = V_{11kV} \frac{Z_{S2}}{Z_{C3} + + Z_{TX1} + Z_{S2}} \, </math><br />
::: <math> = 9,521.825 - j 280.698 = 9,525.6 \, </math> Vac, then referred to the LV side = 359.39Vac (or 86.60% of nominal voltage)<br />
<br />
Any other voltages of interest on the system can be determined using the same methods as above.<br />
<br />
Suppose that our maximum voltage drop at the motor terminals is 15%. From above, we have found that the voltage drop is 12.08% at the motor terminals. This is a slightly marginal result and it may be prudent to simulate the system in a software package to confirm the results.<br />
<br />
== Computer Software ==<br />
<br />
Motor starting is a standard component of most [[Power_Systems_Analysis_Software|power systems analysis software]] (e.g. ETAP, PTW, ERAC, etc) and this calculation is really intended to be done using this software. The numerical calculation performed by the software should also solve the power flow problem through an iterative algorithm (e.g. such as Newton-Rhapson).<br />
<br />
== What Next? ==<br />
<br />
If the results of the calculation confirm that starting the largest motor does not cause any unacceptable voltage levels within the system, then that's the end of it (or perhaps it could be simulated in a power systems analysis software package to be doubly sure!). Otherwise, the issue needs to be addressed, for example by:<br />
<br />
:* Reduce the motor starting current, e.g. via soft-starters, star-delta starters, etc<br />
:* Reduce the source impedances, e.g. increase the size of the generator, transformer, supply cables, etc<br />
<br />
The calculation should be performed iteratively until the results are acceptable.<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=Short_Circuit_Calculation&diff=199Short Circuit Calculation2021-01-25T22:54:26Z<p>Jules: </p>
<hr />
<div>=== Introduction ===<br />
<br />
Some of the related terms defined by the IEC's ''Electropedia'':<br />
<br />
[[Image:Power-line-short-circuit.jpg|right|thumb|350px|Figure 1. 110 kV power line short-circuit]]<br />
<br />
[[Image:Fotolia 1481286 M-798471.jpg|right|thumb|350px|Figure 2. Short-Circuit Arc]]<br />
<br />
*''Short-Circuit'': accidental or intentional conductive path between two or more conductive parts forcing the electric potential differences between these conductive parts to be equal to or close to zero.<br />
*''Short-Circuit Current'': an over-current resulting from a short circuit due to a fault or an incorrect connection in an electric circuit.<br />
*''Short-Circuit Operation'': no-load operation with zero output voltage (Note – Zero output voltage can be obtained when the output terminals are short-circuited). <br />
<br />
*''Line-to-Earth Short-Circuit'': short-circuit between a line conductor and the Earth, in a solidly earthed neutral system or in an impedance earthed neutral system (Note – The short-circuit can be established, for example, through an earthing conductor and an earth electrode). <br />
<br />
*''Line-to-Line Short-Circuit'': short-circuit between two or more line conductors, combined or not with a line-to-earth short-circuit at the same place <br />
<br />
*''Short-Circuit Current Capability'': the permissible value of the short-circuit current in a given network component for a specified duration.<br />
<br />
=== AC Short-Circuit Analysis ===<br />
<br />
* [[IEC 60909 AC Short Circuit Analysis]]<br />
<br />
=== DC Short-Circuit Analysis ===<br />
<br />
* [[IEC 61660 DC Short Circuit Analysis]]<br />
<br />
* [[ANSI/IEEE 946 DC Short Circuit Analysis]]<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=Short_Circuit_Calculation&diff=198Short Circuit Calculation2021-01-25T22:54:10Z<p>Jules: </p>
<hr />
<div>=== Introduction ===<br />
<br />
Some of the related terms defined by the IEC's ''Electropedia'':<br />
<br />
[[Image:Power-line-short-circuit.jpg|right|thumb|350px|Figure 1. 110 kV power line short-circuit]]<br />
<br />
[[Image:Fotolia 1481286 M-798471.jpg|right|thumb|350px|Figure 2. Short-Circuit Arc]]<br />
<br />
*''Short-Circuit'': accidental or intentional conductive path between two or more conductive parts forcing the electric potential differences between these conductive parts to be equal to or close to zero.<br />
*''Short-Circuit Current'': an over-current resulting from a short circuit due to a fault or an incorrect connection in an electric circuit.<br />
*''Short-Circuit Operation'': no-load operation with zero output voltage (Note – Zero output voltage can be obtained when the output terminals are short-circuited). <br />
<br />
*''Line-to-Earth Short-Circuit'': short-circuit between a line conductor and the Earth, in a solidly earthed neutral system or in an impedance earthed neutral system (Note – The short-circuit can be established, for example, through an earthing conductor and an earth electrode). <br />
<br />
*''Line-to-Line Short-Circuit'': short-circuit between two or more line conductors, combined or not with a line-to-earth short-circuit at the same place <br />
<br />
*''Short-Circuit Current Capability'': the permissible value of the short-circuit current in a given network component for a specified duration.<br />
<br />
=== AC Short-Circuit Analysis ===<br />
<br />
[[IEC 60909 AC Short Circuit Analysis]]<br />
<br />
=== DC Short-Circuit Analysis ===<br />
<br />
[[IEC 61660 DC Short Circuit Analysis]]<br />
<br />
[[ANSI/IEEE 946 DC Short Circuit Analysis]]<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=File:Power-line-short-circuit.jpg&diff=197File:Power-line-short-circuit.jpg2021-01-25T22:52:57Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=File:Fotolia_1481286_M-798471.jpg&diff=196File:Fotolia 1481286 M-798471.jpg2021-01-25T22:52:45Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=Short_Circuit_Calculation&diff=195Short Circuit Calculation2021-01-25T22:48:59Z<p>Jules: Created page with "=== Introduction === Some of the related terms defined by the IEC's ''Electropedia'': Image:Power-line-short-circuit.jpg|right|thumb|350px|Figure 1. 110 kV power line shor..."</p>
<hr />
<div>=== Introduction ===<br />
<br />
Some of the related terms defined by the IEC's ''Electropedia'':<br />
<br />
[[Image:Power-line-short-circuit.jpg|right|thumb|350px|Figure 1. 110 kV power line short-circuit]]<br />
<br />
[[Image:Fotolia 1481286 M-798471.jpg|right|thumb|350px|Figure 2. Short-Circuit Arc]]<br />
<br />
*''Short-Circuit'': accidental or intentional conductive path between two or more conductive parts forcing the electric potential differences between these conductive parts to be equal to or close to zero.<br />
*''Short-Circuit Current'': an over-current resulting from a short circuit due to a fault or an incorrect connection in an electric circuit.<br />
*''Short-Circuit Operation'': no-load operation with zero output voltage (Note – Zero output voltage can be obtained when the output terminals are short-circuited). <br />
<br />
*''Line-to-Earth Short-Circuit'': short-circuit between a line conductor and the Earth, in a solidly earthed neutral system or in an impedance earthed neutral system (Note – The short-circuit can be established, for example, through an earthing conductor and an earth electrode). <br />
<br />
*''Line-to-Line Short-Circuit'': short-circuit between two or more line conductors, combined or not with a line-to-earth short-circuit at the same place <br />
<br />
*''Short-Circuit Current Capability'': the permissible value of the short-circuit current in a given network component for a specified duration.<br />
<br />
=== AC Short-Circuit Analysis ===<br />
<br />
[[according to the IEC 60909]]<br />
<br />
=== DC Short-Circuit Analysis ===<br />
<br />
[[according to the IEC 61660]]<br />
<br />
[[according to the ANSI/IEEE 946]]<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=File:Load_Schedule.jpg&diff=194File:Load Schedule.jpg2020-12-25T03:13:52Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=Electrical_Load_Schedule&diff=193Electrical Load Schedule2020-12-25T03:13:27Z<p>Jules: Created page with "== Introduction == Figure 1. Example of an electrical load schedule The electrical load schedule is an estimate of the instant..."</p>
<hr />
<div>== Introduction ==<br />
<br />
[[Image:Load_Schedule.jpg|400px|thumb|right|Figure 1. Example of an electrical load schedule]] <br />
<br />
The electrical load schedule is an estimate of the instantaneous electrical loads operating in a facility, in terms of active, reactive and apparent power (measured in kW, kVAR and kVA respectively). The load schedule is usually categorised by switchboard or occasionally by sub-facility / area.<br />
<br />
=== Why do the calculation? ===<br />
<br />
Preparing the load schedule is one of the earliest tasks that needs to be done as it is essentially a pre-requisite for some of the key electrical design activities (such as equipment sizing and power system studies). <br />
<br />
=== When to do the calculation? ===<br />
<br />
The electrical load schedule can typically be started with a preliminary key single line diagram (or at least an idea of the main voltage levels in the system) and any preliminary details of process / building / facility loads. It is recommended that the load schedule is started as soon as practically possible.<br />
<br />
== Calculation Methodology ==<br />
<br />
There are no standards governing load schedules and therefore this calculation is based on generally accepted industry practice. The following methodology assumes that the load schedule is being created for the first time and is also biased towards industrial plants. The basic steps for creating a load schedule are:<br />
<br />
:* Step 1: Collect a list of the expected electrical loads in the facility<br />
:* Step 2: For each load, collect the electrical parameters, e.g. nominal / absorbed ratings, power factor, efficiency, etc<br />
:* Step 3: Classify each of the loads in terms of switchboard location, load duty and load criticality<br />
:* Step 4: For each load, calculate the expected consumed load<br />
:* Step 5: For each switchboard and the overall system, calculate operating, peak and design load<br />
<br />
=== Step 1: Collect list of loads ===<br />
<br />
The first step is to gather a list of all the electrical loads that will be supplied by the power system affected by the load schedule. There are generally two types of loads that need to be collected:<br />
<br />
:* Process loads - are the loads that are directly relevant to the facility. In factories and industrial plants, process loads are the motors, heaters, compressors, conveyors, etc that form the main business of the plant. Process loads can normally be found on either Mechanical Equipment Lists or Process and Instrumentation Diagrams (P&ID's). <br />
<br />
:* Non-process loads - are the auxiliary loads that are necessary to run the facility, e.g. lighting, HVAC, utility systems (power and water), DCS/PLC control systems, fire safety systems, etc. These loads are usually taken from a number of sources, for example HVAC engineers, instruments, telecoms and control systems engineers, safety engineers, etc. Some loads such as lighting, UPS, power generation auxiliaries, etc need to be estimated by the electrical engineer. <br />
<br />
=== Step 2: Collect electrical load parameters ===<br />
<br />
A number of electrical load parameters are necessary to construct the load schedule:<br />
<br />
:* '''Rated power''' is the full load or nameplate rating of the load and represents the maximum continuous power output of the load. For motor loads, the rated power corresponds to the [[Standard_IEC_Ratings|standard motor size]] (e.g. 11kW, 37kW, 75kW, etc). For load items that contain sub-loads (e.g. distribution boards, package equipment, etc), the rated power is typically the maximum power output of the item (i.e. with all its sub-loads in service). <br />
<br />
:* '''Absorbed power''' is the expected power that will be drawn by the load. Most loads will not operate at its rated capacity, but at a lower point. For example, absorbed motor loads are based on the mechanical power input to the shaft of the driven equipment at its duty point. The motor is typically sized so that the rated capacity of the motor exceeds the expected absorbed load by some conservative design margin. Where information regarding the absorbed loads is not available, then a [[load_factor|load factor]] of between 0.8 and 0.9 is normally applied. <br />
<br />
:* '''Power factor''' of the load is necessary to determine the reactive components of the load schedule. Normally the load power factor at full load is used, but the power factor at the duty point can also be used for increased accuracy. Where power factors are not readily available, then estimates can be used (typically 0.85 for motor loads >7.5kW, 1.0 for heater loads and 0.8 for all other loads).<br />
<br />
:* '''Efficiency''' accounts for the losses incurred when converting electrical energy to mechanical energy (or whatever type of energy the load outputs). Some of the electrical power drawn by the load is lost, usually in the form of heat to the ambient environment. Where information regarding efficiencies is not available, then estimates of between 0.8 and 1 can be used (typically 0.85 or 0.9 is used when efficiencies are unknown).<br />
<br />
=== Step 3: Classify the loads ===<br />
<br />
Once the loads have been identified, they need to be classified accordingly:<br />
<br />
==== Voltage Level ====<br />
What voltage level and which switchboard should the load be located? Large loads may need to be on MV or HV switchboards depending on the size of the load and how many voltage levels are available. Typically, loads <150kW tend to be on the LV system (400V - 690V), loads between 150kW and 10MW tend to be on an intermediate MV system (3.3kV - 6.6kV) where available and loads >10MW are usually on the HV distribution system (11kV - 33kV). Some consideration should also be made for grouping the loads on a switchboard in terms of sub-facilities, areas or sub-systems (e.g. a switchboard for the compression train sub-system or the drying area). <br />
<br />
==== Load duty ==== <br />
Loads are classified according to their duty as either continuous, intermittent and standby loads:<br />
<br />
: 1) '''Continuous''' loads are those that normally operate continuously over a 24 hour period, e.g. process loads, control systems, lighting and small power distribution boards, UPS systems, etc<br />
: 2) '''Intermittent''' loads that only operate a fraction of a 24 hour period, e.g. intermittent pumps and process loads, automatic doors and gates, etc<br />
: 3) '''Standby''' loads are those that are on standby or rarely operate under normal conditions, e.g. standby loads, emergency systems, etc<br />
<br />
Note that for redundant loads (e.g. 2 x 100% duty / standby motors), one is usually classified as continuous and the other classified as standby. This if purely for the purposes of the load schedule and does not reflect the actual operating conditions of the loads, i.e. both redundant loads will be equally used even though one is classified as a standby load.<br />
<br />
==== Load criticality ====<br />
Loads are typically classified as either normal, essential and critical:<br />
<br />
: 1) '''Normal''' loads are those that run under normal operating conditions, e.g. main process loads, normal lighting and small power, ordinary office and workshop loads, etc<br />
: 2) '''Essential''' loads are those necessary under emergency conditions, when the main power supply is disconnected and the system is being supported by an emergency generator, e.g. emergency lighting, key process loads that operate during emergency conditions, fire and safety systems, etc<br />
: 3) '''Critical''' are those critical for the operation of safety systems and for facilitating or assisting evacuation from the plant, and would normally be supplied from a UPS or battery system, e.g. safety-critical shutdown systems, escape lighting, etc<br />
<br />
=== Step 4: Calculate consumed load ===<br />
<br />
The consumed load is the quantity of electrical power that the load is expected to consume. For each load, calculate the consumed active and reactive loading, derived as follows:<br />
<br />
: <math> P_{l} = \frac{P_{abs}}{\eta} \,</math><br />
: <math> Q_{l} = P_{l} \sqrt{\frac{1}{\cos^{2} \phi} - 1} \,</math><br />
<br />
Where <math> P_{l} \, </math> is the consumed active load (kW)<br />
:: <math> Q_{l} \, </math> is the consumed reactive load (kVAr)<br />
:: <math> P_{abs} \, </math> is the absorbed load (kW)<br />
:: <math> \eta \, </math> is the load efficiency (pu)<br />
:: <math> \cos \phi \, </math> is the load power factor (pu)<br />
<br />
Notice that the loads have been categorised into three columns depending on their load duty (continuous, intermittent or standby). This is done in order to make it visually easier to see the load duty and more importantly, to make it easier to sum the loads according to their duty (e.g. sum of all continuous loads), which is necessary to calculate the operating, peak and design loads.<br />
<br />
=== Step 5: Calculate operating, peak and design loads ===<br />
<br />
Many organisations / clients have their own distinct method for calculating operating, peak and design loads, but a generic method is presented as follows:<br />
<br />
==== Operating load ====<br />
<br />
The operating load is the expected load during normal operation. The operating load is calculated as follows:<br />
<br />
: <math> OL = \sum L_{c} + 0.5 \times \sum L_{i} \, </math><br />
<br />
Where <math> OL \, </math> is the operating load (kW or kVAr)<br />
:: <math> \sum L_{c} \, </math> is the sum of all continuous loads (kW or kVAr)<br />
:: <math> \sum L_{i} \, </math> is the sum of all intermittent loads (kW or kVAr)<br />
<br />
==== Peak load ====<br />
<br />
The peak load is the expected maximum load during normal operation. Peak loading is typically infrequent and of short duration, occurring when standby loads are operated (e.g. for changeover of redundant machines, testing of safety equipment, etc). The peak load is calculated as the larger of either:<br />
<br />
: <math> PL = \sum L_{c} + 0.5 \times \sum L_{i} + 0.1 \times \sum L_{s} \, </math><br />
<br />
or<br />
<br />
: <math> PL = \sum L_{c} + 0.5 \times \sum L_{i} + L_{s,max} \, </math><br />
<br />
Where <math> PL \, </math> is the peak load (kW or kVAr)<br />
:: <math> \sum L_{c} \, </math> is the sum of all continuous loads (kW or kVAr)<br />
:: <math> \sum L_{i} \, </math> is the sum of all intermittent loads (kW or kVAr)<br />
:: <math> \sum L_{s} \, </math> is the sum of all standby loads (kW or kVAr)<br />
:: <math> L_{s,max} \, </math> is the largest standby load (kW or kVAr)<br />
<br />
==== Design load ====<br />
<br />
The design load is the load to be used for the design for equipment sizing, electrical studies, etc. The design load is generically calculated as the larger of either:<br />
<br />
: <math> DL = 1.1 \times OL + 0.1 \times \sum L_{s} \, </math><br />
<br />
or <br />
<br />
: <math> DL = 1.1 \times OL + L_{s,max} \, </math><br />
<br />
Where <math> DL \, </math> is the design load (kW or kVAr)<br />
:: <math> OL \, </math> is the operating load (kW or kVAr)<br />
:: <math> \sum L_{s} \, </math> is the sum of all standby loads (kW or kVAr)<br />
:: <math> L_{s,max} \, </math> is the largest standby load (kW or kVAr)<br />
<br />
The design load includes a margin for any errors in load estimation, load growth or the addition of unforeseen loads that may appear after the design phase. The load schedule is thus more conservative and robust to errors. On the other hand however, equipment is often over-sized as a result. Sometimes the design load is not calculated and the peak load is used for design purposes.<br />
<br />
== Worked Example ==<br />
<br />
=== Step 1: Collect list of loads ===<br />
<br />
Consider a small facility with the following loads identified:<br />
<br />
:* [[Load_Redundancy|2 x 100%]] vapour recovery compressors (process)<br />
:* [[Load_Redundancy|2 x 100%]] recirculation pumps (process)<br />
:* [[Load_Redundancy|1 x 100%]] sump pump (process)<br />
:* [[Load_Redundancy|2 x 50%]] firewater pumps (safety)<br />
:* [[Load_Redundancy|1 x 100%]] HVAC unit (HVAC)<br />
:* [[Load_Redundancy|1 x 100%]] AC UPS system (electrical)<br />
:* 1 x Normal lighting distribution board (electrical)<br />
:* 1 x Essential lighting distribution board (electrical)<br />
<br />
=== Step 2: Collect electrical load parameters ===<br />
<br />
The following electrical load parameters were collected for the loads identified in Step 1:<br />
<br />
{| class="wikitable"<br />
|-<br />
!style="background:#efefef;" width="250px" | Load Description<br />
!style="background:#efefef;" width="80px" | Abs. Load<br />
!style="background:#efefef;" width="80px" | Rated Load<br />
!style="background:#efefef;" width="80px" | PF<br />
!style="background:#efefef;" width="80px" | Eff.<br />
|-<br />
|Vapour recovery compressor A || 750kW ||800kW || 0.87 || 0.95<br />
|-<br />
|Vapour recovery compressor B || 750kW ||800kW || 0.87 || 0.95<br />
|-<br />
|Recirculation pump A || 31kW || 37kW || 0.83 || 0.86<br />
|-<br />
|Recirculation pump B || 31kW || 37kW || 0.83 || 0.86<br />
|-<br />
|Sump pump || 9kW || 11kW || 0.81 || 0.83<br />
|-<br />
|Firewater pump A || 65kW || 75kW || 0.88 || 0.88<br />
|-<br />
|Firewater pump B || 65kW || 75kW || 0.88 || 0.88<br />
|-<br />
|HVAC unit || 80kW || 90kW || 0.85 || 0.9<br />
|-<br />
|AC UPS System || 9kW || 12kW || 0.85 || 0.9<br />
|-<br />
|Normal lighting distribution board || 7kW || 10kW || 0.8 || 0.9<br />
|-<br />
|Essential lighting distribution board || 4kW || 5kW || 0.8 || 0.9<br />
|}<br />
<br />
=== Step 3: Classify the loads ===<br />
<br />
Suppose we have two voltage levels, 6.6kV and 415V. The loads can be classified as follows:<br />
<br />
{| class="wikitable"<br />
|-<br />
!style="background:#efefef;" width="250px" | Load Description<br />
!style="background:#efefef;" width="80px" | Rated Load<br />
!style="background:#efefef;" width="80px" | Voltage<br />
!style="background:#efefef;" width="90px" | Duty<br />
!style="background:#efefef;" width="90px" | Criticality<br />
|-<br />
|Vapour recovery compressor A || 800kW || 6.6kV || Continuous || Normal<br />
|-<br />
|Vapour recovery compressor B || 800kW || 6.6kV || Standby || Normal<br />
|-<br />
|Recirculation pump A || 37kW || 415V || Continuous || Normal<br />
|-<br />
|Recirculation pump B || 37kW || 415V || Standby || Normal<br />
|-<br />
|Sump pump || 11kW || 415V || Intermittent|| Normal<br />
|-<br />
|Firewater pump A || 75kW || 415V || Standby || Essential<br />
|-<br />
|Firewater pump B || 75kW || 415V || Standby || Essential<br />
|-<br />
|HVAC unit || 90kW || 415V || Continuous || Normal<br />
|-<br />
|AC UPS System || 12kW || 415V || Continuous || Critical<br />
|-<br />
|Normal lighting distribution board || 10kW || 415V || Continuous || Normal<br />
|-<br />
|Essential lighting distribution board || 5kW || 415V || Continuous || Essential<br />
|}<br />
<br />
=== Step 4: Calculate consumed load ===<br />
<br />
Calculating the consumed loads for each of the loads in this example gives:<br />
<br />
{| class="wikitable"<br />
|-<br />
!style="background:#efefef;" width="250px" rowspan="2" | Load Description<br />
!style="background:#efefef;" width="80px" rowspan="2" | Abs Load<br />
!style="background:#efefef;" width="70px" rowspan="2" | PF<br />
!style="background:#efefef;" width="70px" rowspan="2" | Eff.<br />
!style="background:#efefef;" width="160px" colspan="2" | Continuous<br />
!style="background:#efefef;" width="160px" colspan="2" | Intermittent<br />
!style="background:#efefef;" width="160px" colspan="2" | Standby<br />
|-<br />
!style="background:#efefef;" width="80px" | P (kW)<br />
!style="background:#efefef;" width="80px" | Q (kVAr)<br />
!style="background:#efefef;" width="80px" | P (kW)<br />
!style="background:#efefef;" width="80px" | Q (kVAr)<br />
!style="background:#efefef;" width="80px" | P (kW)<br />
!style="background:#efefef;" width="80px" | Q (kVAr)<br />
|-<br />
|Vapour recovery compressor A || 750kW || 0.87 || 0.95 || 789.5 || 447.4 || - || - || - || -<br />
|-<br />
|Vapour recovery compressor B || 750kW || 0.87 || 0.95 || - || - || - || - || 789.5 || 447.4<br />
|-<br />
|Recirculation pump A || 31kW || 0.83 || 0.86 || 36.0 || 24.2 || - || - || - || -<br />
|-<br />
|Recirculation pump B || 31kW || 0.83 || 0.86 || - || - || - || - || 36.0 || 24.2<br />
|-<br />
|Sump pump || 9kW || 0.81 || 0.83 || - || - || 10.8 || 7.9 || - || -<br />
|-<br />
|Firewater pump A || 65kW || 0.88 || 0.88 || - || - || - || - || 73.9 || 39.9<br />
|-<br />
|Firewater pump B || 65kW || 0.88 || 0.88 || - || - || - || - || 73.9 || 39.9<br />
|-<br />
|HVAC unit || 80kW || 0.85 || 0.9 || 88.9 || 55.1 || - || - || - || -<br />
|-<br />
|AC UPS System || 9kW || 0.85 || 0.9 || 10.0 || 6.2 || - || - || - || -<br />
|-<br />
|Normal lighting distribution board || 7kW || 0.8 || 0.9 || 7.8 || 5.8 || - || - || - || -<br />
|-<br />
|Essential lighting distribution board || 4kW || 0.8 || 0.9 || 4.4 || 3.3 || - || - || - || -<br />
|- <br />
|colspan="4" | '''SUM TOTAL''' || '''936.6''' || '''542.0''' || '''10.8''' || '''7.9''' || '''973.3''' || '''551.4'''<br />
|}<br />
<br />
=== Step 5: Calculate operating, peak and design loads ===<br />
<br />
The operating, peak and design loads are calculated as follows:<br />
<br />
{| border = "1" cellpadding="5" cellspacing="0" style = "text-align:center;"<br />
|-<br />
!style="background:#efefef;" width="250px" | <br />
!style="background:#efefef;" width="80px" | P (kW)<br />
!style="background:#efefef;" width="80px" | Q (kW)<br />
|-<br />
|Sum of continuous loads || 936.6 || 542.0<br />
|-<br />
|50% x Sum of intermittent loads || 5.4 || 4.0<br />
|-<br />
|10% x Sum of standby loads || 97.3 || 55.1<br />
|-<br />
|Largest standby load || 789.5 || 447.4<br />
|-<br />
|Operating load || 942 || 546.0<br />
|-<br />
|Peak load || 1,731.5 || 993.4<br />
|-<br />
|Design load || 1,825.7 || 1,047.9<br />
|}<br />
<br />
Normally you would separate the loads by switchboard and calculate operating, peak and design loads for each switchboard and one for the overall system. However for the sake of simplicity, the loads in this example are all lumped together and only one set of operating, peak and design loads are calculated.<br />
<br />
== Operating Scenarios ==<br />
<br />
It may be necessary to construct load schedules for different operating scenarios. For example, in order to size an emergency diesel generator, it would be necessary to construct a load schedule for emergency scenarios. The classification of the loads by criticality will help in constructing alternative scenarios, especially those that use alternative power sources.<br />
<br />
== Computer Software ==<br />
<br />
In the past, the load schedule has typically been done manually by hand or with the help of an Excel spreadsheet. However, this type of calculation is extremely well-suited for database driven software packages (such as [http://www.intergraph.com/products/ppm/smartplant/electrical/default.aspx Smartplant Electrical]), especially for very large projects. For smaller projects, it may be far easier to simply perform this calculation manually.<br />
<br />
== What Next? ==<br />
<br />
The electrical load schedule is the basis for the sizing of most major electrical equipment, from generators to switchgear to transformers. Using the load schedule, major equipment sizing can be started, as well as the power system studies. A preliminary load schedule will also indicate if there will be problems with available power supply / generation, and whether alternative power sources or even process designs will need to be investigated. <br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=Low_Frequency_Induction_(LFI)_Calculation&diff=192Low Frequency Induction (LFI) Calculation2020-12-25T03:12:10Z<p>Jules: /* Step 3: Define Pipeline Sections */</p>
<hr />
<div>== Introduction ==<br />
<br />
Low frequency induction (LFI) in metal pipelines is a form of electromagnetic interference that occurs when high voltage transmission lines are run in parallel with metallic pipelines. <br />
<br />
The loaded phases on a transmission line and the pipeline act like single-turn windings on a large air-core transformer. Current flowing through the transmission line induces a voltage at the pipeline.<br />
<br />
During normal operation, each loaded phase of the transmission line induces a voltage on the pipeline. If the three-phases are balanced, then most of the induced voltage will cancel each other out, but the spatial assymetry of the phases on the transmission line will prevent the induced voltages from fully cancelling out. Thus a non-zero induced voltage results on the pipeline. This condition is usually called "Load LFI".<br />
<br />
During an earth fault condition, the phases of the transmission line are no longer balanced and as a result, a more significant voltage is induced on the pipeline. This condition is usually called "Fault LFI".<br />
<br />
=== Why do the calculation? ===<br />
<br />
LFI calculations are typically done for personnel safety reasons, in order to ensure that induced voltages are not hazardous to someone in contact with the pipeline. <br />
<br />
In cases where such a situation is not possible, then an approach minimising the risk of electric shock could be followed. This could involve restricting access to the pipeline or using a probability-based methodology to estimate the risk of exposure, etc.<br />
<br />
=== When to do the calculation? ===<br />
<br />
The calculation should be done whenever HV transmission lines are installed in the vicinity of pipelines, and vice versa. Of particular concern is when the pipeline is run nearby and parallel to the transmission line for long distances. On the other hand, if a pipeline crosses the transmission line perpendicularly, then the magnitude of LFI would be low. <br />
<br />
== Calculation Methodology ==<br />
<br />
The LFI calculation has nine general steps:<br />
<br />
:* Step 1: Data Gathering<br />
:* Step 2: Define the Zone of Influence <br />
:* Step 3: Define Pipeline Sections<br />
:* Step 4: Calculate Effective Distances<br />
:* Step 5: Calculate Pipeline Impedances<br />
:* Step 6: Calculate Mutual Coupling Impedances<br />
:* Step 7: Compute Load LFI<br />
:* Step 8: Compute Fault LFI<br />
:* Step 9: Calculate Pipeline-to-Earth Touch Voltages (if necessary)<br />
:* Step 10: Apply Mitigation Works (if necessary)<br />
<br />
=== Step 1: Data Gathering ===<br />
<br />
Before beginning the calculation, the following data needs to be gathered:<br />
<br />
:* Plan layouts of pipeline(s) and transmission line(s), to scale with enough resolution to measure horizontal distances between pipeline and transmission line<br />
:* General arrangement of transmission line tower or underground spatial arrangement<br />
:* Pipeline data, e.g. diameter, metal resistivity, coating details, etc<br />
:* Soil resistivity data around pipeline<br />
:* Forecast (or actual) load currents in transmission line<br />
:* Worst case prospective fault currents on transmission line<br />
<br />
=== Step 2: Define the Zone of Influence ===<br />
<br />
Based on the plan layouts of the pipeline and transmission line routes, the first step is to define the zone(s) of influence in which you will perform the LFI study. This zone is typically a corridor where the pipeline and transmission line are run close together in parallel. <br />
<br />
As a general rule of thumb, the zone extends along this corridor until the horizontal separation between the pipeline and transmission line exceeds 1km. It is deemed that there is minimal LFI effect on the pipeline beyond a 1km pipeline-transmission line separation.<br />
<br />
=== Step 3: Define Pipeline Sections ===<br />
<br />
Once the zone of influence has been defined, the transmission line or pipeline should be broken up into small sections and the horizontal separations calculated for each section. The length of the indivudal sections do not have to be constant and can vary along the the route, depending on the rate of change of the horizontal separation. <br />
<br />
For example, long section lengths can be applied when the pipeline and transmission line are roughly parallel. However, where the two lines converge, diverge or cross, smaller sectional lengths should be used (typically, a maximum ratio of 3:1 between the start of section and end of section separations is used to determine sectional length). <br />
<br />
A table showing the sectional lengths and horizontal distances can then be developed, for example:<br />
<br />
<table><br />
{| class="wikitable"<br />
|-<br />
!style="background:#efefef;" width="80px" rowspan="2" | Section<br />
!style="background:#efefef;" width="100px" rowspan="2" | Distance along pipeline (km)<br />
!style="background:#efefef;" width="100px" rowspan="2" | Section Length (m)<br />
!style="background:#efefef;" width="200px" colspan="2" | Horizontal distance from phase A to pipeline (m)<br />
|-<br />
!style="background:#efefef;" width="80px" | Start<br />
!style="background:#efefef;" width="80px" | End<br />
|-<br />
| 1 || 0.0 || 300 || 140 || 190<br />
|-<br />
| 2 || 0.3 || 200 || 190 || 200<br />
|}<br />
<br />
=== Step 4: Calculate Effective Distances ===<br />
<br />
The table we developed in Step 2 shows the horizontal separations between phase A and the pipeline at the start and end of each pipeline section. However, the transmission line will have three phases (and possibly two sets of conductors per phase) and the horizontal distances also do not take into account the geometry of the transmission tower.<br />
<br />
Therefore, we'd like to further refine these distances. Ideally, we want a single effective distance between the pipeline and each phase (and if applicable, the earth wire). By using an effective distance, this makes it more convenient for us to calculate the induced voltages on a pipeline section. <br />
<br />
This effective distance should incorporate the following:<br />
<br />
:* The average horizontal distance along the pipeline section<br />
:* The geometry of the transmission line towers (and the spatial geometry of the individual conductors on the tower)<br />
:* An equivalent distance for parallel conductors of the same phase<br />
<br />
==== Transmission Line Geometry ====<br />
<br />
[[Image:Towerline.jpg|thumb|300px|Figure 1. Double conductor tower example]]<br />
<br />
In order to construct the effective distances, we first need to define the spatial geometry of the transmission line with respect to the pipeline. This could be the geometry of transmission line towers or the spatial arrangement of underground conductors.<br />
<br />
The figure to the right, showing a double conductor tower with two earth wires, gives an example of the kind of data that is required:<br />
<br />
:* L1 = Distance from pipeline to phase a<br />
:* L<sub>b</sub> = Distance from phase a to phase b<br />
:* L<sub>c</sub> = Distance from phase a to phase c<br />
:* L<sub>w</sub> = Distance from phase a to the earth wire<br />
:* L<sub>aa</sub> = Distance between phase a double conductors<br />
:* L<sub>bb</sub> = Distance between phase b double conductors<br />
:* L<sub>cc</sub> = Distance between phase c double conductors<br />
:* H<sub>a</sub> = Height from pipeline to phase a<br />
:* H<sub>b</sub> = Height from pipeline to phase b<br />
:* H<sub>c</sub> = Height from pipeline to phase c<br />
:* H<sub>w</sub> = Height from pipeline to earth wire<br />
<br />
Of course your data requirements may vary depending on the transmission line type, but you need enough data to fully describe the spatial relationships between the phase conductors and the pipeline geometrically.<br />
<br />
==== Effective Distances ====<br />
<br />
As the horizontal distance between the transmission line and the pipeline is not necessarly constant along each section, we want to calculate an average horizontal distance for each line section. Typically, the geometric mean distance is used:<br />
<br />
: <math> L_{m} = \sqrt{L1 \times L2} \, </math><br />
<br />
Where <math> L_{m} \, </math> is the effective horizontal distance between phase a and the pipeline<br />
:: <math> L1 \, </math> is the horizontal distance (between phase a and the pipeline) at the start of the section (m)<br />
:: <math> L2 \, </math> is the horizontal distance (between phase a and the pipeline) at the end of the section (m)<br />
<br />
Similarly for double conductors arranged horizontally, we use the geomtric mean distance to obtain an effective distance between double conductors and the pipeline:<br />
<br />
: <math> L_{a} = \sqrt{L_{m} \times (L_{m} + L_{aa}} \, </math><br />
<br />
Where <math> L_{a} \, </math> is the effective horizontal distance between double conductors (of phase a) and the pipeline (m)<br />
:: <math> L_{m} \, </math> is the effective horizontal distance calculated above (m)<br />
:: <math> L_{aa} \, </math> is the horizontal distance between the two conductors of phase a (m)<br />
<br />
Note that the above show only horizontal distances. We can use simple Pythagoras to calculate the overall effective distance between phase a and the pipeline, taking into account the vertical arrangement of the conductors (relative to the pipeline). Below, we do this for a tower line:<br />
<br />
: <math> D_{ap} = \sqrt{L_{a}^{2} + H_{a}^{2}} \, </math><br />
<br />
Where <math> D_{ap} \, </math> is the overall effective distance between the phase a and the pipeline<br />
:: <math> L_{a} \, </math> is the effective horizontal distance between double conductors (of phase a) and the pipeline (m)<br />
:: <math> H_{a} \, </math> is the height from the pipeline to phase a (m)<br />
<br />
The same process above is repeated to calculate the effective distances for the other phases and the earth wire (if applicable). <br />
<br />
The effective distances can be calculated also for other spatial conductor arrangements, using the geometric mean as a basis. Ultimately for each line section, we'd like to end up with only a single effective distance between the pipeline and a phase of the tower line. <br />
<br />
=== Step 5: Calculate Pipeline Impedances ===<br />
<br />
To assess the safety of a joint right-of-way installation, we want to calculate pipeline-to-earth touch voltages. However a buried pipeline cannot be treated like another overhead conductor in air because a pipeline has a finite impedance to earth that is distributed along its entire length. <br />
<br />
Therefore the open-circuit voltages induced in the pipeline that we calculated earlier are not equivalent to the pipeline-to-earth touch voltage. There are continuous leakages to earth along the pipeline and the open-circuit induced voltages can be up to 10 times higher than the pipeline-to-earth touch voltages.<br />
<br />
In order to calculate a more accurate pipeline-to-earth touch voltage, we need to consider the electrical characteristics of the pipeline. <br />
<br />
The pipeline impedances described in this section are simplified approximations of Sunde's method [5] and are based on Appendix G of CIGRE Guide 95 "Guide on the Influence of High Voltage AC Power Systems on Metallic Pipelines" [2]. Refer to Sunde's original work [5] for more details on accurately modelling pipeline and other buried conductor impedances.<br />
<br />
''' a) Pipeline longitudinal impedance '''<br />
<br />
For a buried and coated metallic pipeline, the longitudinal impedance per metre can be approximated as follows:<br />
<br />
: <math> Z_{p} = \frac{1}{\pi D} \sqrt{\pi f \rho_{p} \mu_{0} \mu_{p}} + \frac{\pi f \mu_{0}}{4} + j \left[ \frac{1}{\pi D} \sqrt{\pi f \rho_{p} \mu_{0} \mu_{p}} + \mu_{0} f \ln \left( \frac{3.7}{D} \sqrt{\frac{\rho}{\mu_{0} 2 \pi f}} \right) \right] \, </math><br />
<br />
Where <math> Z_{p} \, </math> is the pipeline longitudinal (or series) impedance (<math>\Omega /m </math>)<br />
:: <math> D \, </math> is the diamter of the pipeline (m)<br />
:: <math> \rho_{p} \, </math> is the resistivity of the pipeline metal (<math>\Omega .m </math>)<br />
:: <math> \rho \, </math> is the resistivity of the soil (<math>\Omega .m </math>)<br />
:: <math> \mu_{0} = 4 \pi \times 10^{-7} \, </math> is the permeability of free space (H/m)<br />
:: <math> \mu_{p} \, </math> is the relative permeability of the pipeline metal (H/m)<br />
:: <math> f \, </math> is the nominal frequency of the transmission line (Hz)<br />
<br />
''' b) Pipeline shunt admittance '''<br />
<br />
For a buried and coated metallic pipeline, the shut admittance per metre can be approximated as follows:<br />
<br />
: <math> Y_{p} = \frac{\pi D}{r_{c} \delta_{c}} + j \left[ 2 \pi f \frac {\epsilon_{0} \epsilon_{c} \pi D}{\delta_{c}} \right] \, </math><br />
<br />
Where <math> Y_{p} \, </math> is the pipeline shunt admittance (<math>\Omega^{-1} /m </math>)<br />
:: <math> D \, </math> is the diameter of the pipeline (m)<br />
:: <math> r_{c} \, </math> is the resistivity of the pipeline coating (<math>\Omega .m </math>)<br />
:: <math> \delta_{c} \, </math> is the thickness of the pipeline coating (m)<br />
:: <math> \epsilon_{c} \, </math> is the relative permeability of the coating (H/m)<br />
:: <math> \epsilon_{0} = 8.85 \times 10^{-12} \, </math> is the permittivity of free space (F/m)<br />
<br />
The pipeline shunt impedance is simply the reciprocal of the admittance (remember that it is complex quantity):<br />
<br />
: <math> Z_{p,sh} = \frac{1}{Y_{p}} \, </math> <br />
<br />
''' c) Pipeline characteristic impedance '''<br />
<br />
: <math> Z_{c} = \sqrt{\frac{Z_{i}}{Y_{i}}} \, </math><br />
<br />
Where <math> Z_{c} \, </math> is the pipeline characteristic impedance (<math>\Omega </math>)<br />
:: <math> Z_{i} \, </math> is the pipeline longitudinal impedance (<math>\Omega /m </math>)<br />
:: <math> Y_{c} \, </math> is the pipeline longitudinal conductance (<math>\Omega^{-1} /m </math>)<br />
<br />
''' d) Pipeline effective length '''<br />
<br />
: <math> l_{e} = \frac{1}{ Re \left[ \sqrt{Y_{i} Z_{i}} \right] } \, </math><br />
<br />
Where <math> l_{e} \, </math> is the pipeline effective length (m)<br />
:: <math> Z_{i} \, </math> is the pipeline longitudinal impedance (<math>\Omega /m </math>)<br />
:: <math> Y_{c} \, </math> is the pipeline longitudinal conductance (<math>\Omega^{-1} /m </math>)<br />
<br />
=== Step 6: Calculate Mutual Coupling Impedances ===<br />
<br />
The mutual coupling impedance between a line conductor and the pipeline (with earth return) can be given by the general formula below (based on Carson's equations [4]):<br />
<br />
: <math> Z_{lp} = \pi^{2} \times 10^{-4} f + j 4 \pi \times 10^{-4} \ln \left( \frac{D_{e}}{D_{lp}} \right) \, </math><br />
<br />
Where <math> Z_{lp} \, </math> is the mutual impedance between the line conductor <math>l \,</math> and pipeline <math>p \,</math> (<math>\Omega /km </math>)<br />
:: <math> D_{lp} \, </math> is the equivalent distance between the line conductor <math>l \,</math> and pipeline <math>p \,</math> (m)<br />
:: <math> f \, </math> is the nominal system frequency (Hz)<br />
:: <math> D_{e} \, </math> is the depth of equivalent earth return given by:<br />
<br />
::: <math> D_{e} = 658.37 \sqrt{\frac{\rho}{f}} \, </math><br />
<br />
:: Where <math> \rho \, </math> is the resistivity of the soil (<math>\Omega .m </math>)<br />
<br />
Using the formula above, you can calculate the mutual impedances <math> Z_{ap} \, </math>, <math> Z_{bp} \, </math> and <math> Z_{cp} \, </math>, between the pipeline and phase a, b and c of the transmission line respectively. To calculate the impedances in Ohms, multiply the mutual impedances by the sectional length of the pipeline.<br />
<br />
=== Step 7: Compute Load LFI ===<br />
<br />
Load LFI results from either unbalanced load currents or spatial differences between the phase conductors relative to the pipeline. Given the load current phasors for the transmission line, the total open-circuit induced LFI voltage on the pipeline is simply the vector sum:<br />
<br />
: <math> V_{p} = I_{a} Z_{ap} + I_{b} Z_{bp} + I_{c} Z_{cp} \, </math><br />
<br />
Where <math> V_{p} \, </math> is the overall induced load LFI voltage (V)<br />
:: <math> I_{a} \, </math>, <math> I_{b} \, </math> and <math> I_{c} \, </math> are the load current for phase a, b and c respectively (A). Note that these quantities are complex phasors<br />
:: <math> Z_{ap} \, </math>, <math> Z_{bp} \, </math> and <math> Z_{cp} \, </math> are the mutual coupling impedances between the pipeline and phase a, b and c respectively for a section of the pipeline (<math>\Omega </math>)<br />
<br />
==== Allowable Load LFI Limits ====<br />
<br />
Induced voltages on pipelines represent are electric shock hazards. Because load LFI can be regarded as a continuous hazard (i.e. the hazard is always present), then the allowable pipeline-to-earth touch voltage limits are lower than in the fault LFI case. Limits are typically in the range of 32V to 64V (see the section on international standards below for more guidance on allowable limits). <br />
<br />
Note that the open-circuit LFI voltage calculated earlier is '''higher''' than the pipeline-to-earth touch voltage as it doesn't take into account any voltage leakages to earth along the pipeline. Therefore, if the load LFI voltage you have calculated is lower than the allowable limits, then no further analysis is necessary.<br />
<br />
==== Effect of Overhead Earth Wires and Counterpoise Earths ====<br />
<br />
Overhead earth wires and counterpoise earth conductors provide a shielding effect (since a voltage is induced in them and they in turn induce an opposing voltage in the pipeline). The interaction of the overhead earth wires or counterpoise earth conductors with the pipeline can be modelled and a shielding factor computed (more on this later).<br />
<br />
=== Step 8: Compute Fault LFI ===<br />
<br />
During a balanced three-phase fault, the fault current will still be balanced across all three phases (i.e. vector sum of the faulted phase currents is close to zero). Therefore any induced voltage in the pipeline will be due primarily to the spatial assymetry of the transmission line relative to the pipeline, and this will generally be small. <br />
<br />
However, an unbalanced fault such as a line-to-earth or line-to-line fault will produce a fault current that is unbalanced and can induce much higher LFI voltages on the pipeline (until the fault is cleared). The case of a line-to-earth fault will induce the highest LFI voltage and is the type of fault that will be examined hereafter.<br />
<br />
The general formula for fault LFI is as follows:<br />
<br />
: <math> V_{f} = 3 I_{0} K_{sf} Z_{lp} \, </math><br />
<br />
Where <math> V_{f} \, </math> is the overall induced fault LFI voltage (V)<br />
:: <math> 3 I_{0} \, </math> is the earth fault current, expressed as a complex phasor (A)<br />
:: <math> K_{sf} \, </math> is the shielding factor from earth wires, counterpoise earth conductors, etc<br />
:: <math> Z_{lp} \, </math> is the mutual coupling impedances between the pipeline and the faulted line for a section of the pipeline (<math>\Omega </math>)<br />
<br />
==== Shielding Factor ====<br />
<br />
The shielding factor is the shielding effect caused by earth wires, counterpoise earths and perhaps other metallic structures, which serves to lower the LFI voltage induced on the pipeline. There a number of mechanisms that bring about this shielding effect, for example:<br />
<br />
:* If there are overhead earth wires, then not all of the earth fault current will return to the source via the earth. Some of the fault current will return via the earth wires. The current flowing through the earth wire will induce a voltage on the pipeline opposed to the voltage induced by the faulted line.<br />
:* There is the interaction between the faulted line and overhead earth wires, where the faulted line induces a voltage on the overhead earth wires, which in turn induce an opposing voltage on the pipeline.<br />
:* There are similar inductive coupling interactions between the faulted line, the pipeline and counterpoise earths or other metallic structures located in the vicinity<br />
<br />
For an overhead earth wire, the shielding factor can be calculated as follows:<br />
<br />
: <math> K_{sf} = 1 - \frac{Z_{lw}}{Z_{ww}} \times \frac{Z_{wp}}{Z_{lp}} \, </math><br />
<br />
Where <math> Z_{lw} \, </math> is the mutual coupling impedances between the faulted line and the earth wire (<math>\Omega </math>)<br />
:: <math> Z_{ww} \, </math> is the self impedance of the earth wire (<math>\Omega </math>)<br />
:: <math> Z_{wp} \, </math> is the mutual coupling impedances between the pipeline and the earth wire (<math>\Omega </math>)<br />
:: <math> Z_{lp} \, </math> is the mutual coupling impedances between the pipeline and the faulted line (<math>\Omega </math>)<br />
<br />
==== Considerations for Calculating Mutual Coupling Impedances ====<br />
<br />
Calculating the mutual coupling impedances between the pipeline and the faulted line, <math> Z_{lp} \, </math>, depends on the phase of the faulted line and it's spatial orientation relative to the pipeline. You could calculate the mutual coupling impedance between the closest phase and the pipeline, which would represent the worst case scenario. <br />
<br />
Alternatively, you could calculate the mutual coupling impedance for a group of conductors (e.g. all three phases) by considering the geometric mean separation distance between the pipeline and the group of conductors. This will result in an induced LFI voltage that is an average of the conductor group.<br />
<br />
Similarly for groups of earth wires (e.g. two overhead earth wires), a geometric mean can be selected to represent the group rather than selecting a single wire, or modelling interactions between the wires, faulted line and the pipeline (which becomes increasingly complicated). <br />
<br />
==== Allowable Fault LFI Limits ====<br />
<br />
Like the case with load LFI, allowable limits for fault LFI are typically stipulated to prevent electric shock hazards from injuring or killing personnel. Because fault LFI is temporary (lasting only until the fault is cleared), the allowable limits are normally higher than in the load LFI case. The limits are generally based on some kind risk analysis and are typically in the order of 500V to 1000V (see the section on international standards below for more guidance on allowable limits). <br />
<br />
Note that the open-circuit LFI voltage calculated earlier is '''higher''' than the pipeline-to-earth touch voltage as it doesn't take into account any voltage leakages to earth along the pipeline. Therefore, if the fault LFI voltage you have calculated is lower than the allowable limits, then no further analysis is necessary.<br />
<br />
=== Step 9: Analysis of Pipeline-to-Earth Touch Voltages (if necessary) ===<br />
<br />
If the open-circuit load or fault LFI voltage is above the permissible touch voltage, then the pipeline-to-earth touch voltages along each section of the pipeline need to be calculated. In order to calculate the pipeline-to-earth touch voltages, we need an equivalent circuit for the pipeline.<br />
<br />
A common approach is to model the pipeline as a lossy transmission line, with the following equivalent circuit for each pipeline section:<br />
<br />
[[Image:Pipeline model 1.png]]<br />
<br />
Where <math> V_{p} \, </math> is the induced LFI voltage on the pipeline section (Vac)<br />
:: <math> Y_{s} \, </math> is the longitudinal series admittance of the pipeline section (<math>\Omega^{-1} </math>)<br />
:: <math> Y_{sh} \, </math> is the shunt admittance of the pipeline section (<math>\Omega^{-1} </math>)<br />
<br />
For a pipeline with "n" linear sections, the overall equivalent circuit model is therefore:<br />
<br />
[[Image:Pipeline model 2.png]]<br />
<br />
Where <math> Y_{g} \, </math> is the admittance of a pipeline shunt earthing conductor (<math>\Omega^{-1} </math>). The inverse of the pipeline characteristic impedance can be used if no shunt earthing conductors are installed at the ends of the pipeline. In the model above, shunt earthing conductors can be connected to any of the linear pipeline sections, modelled in series with the pipeline section shunt admittance. <br />
<br />
The equivalent circuit can now be readily analysed using [http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchhoff's laws] and converted into a system of linear equations. The unknown pipeline-to-earth node voltages can then be solved using matrix operations. Depending on the number of line sections, you will probably need to use a computer program to solve this linear algebra problem.<br />
<br />
=== Step 10: Apply Mitigation Works (if necessary) ===<br />
<br />
If the pipeline-to-earth touch voltages are still above the allowable limits, then the design of the right-of-way needs to be modified or mitigation works installed. Design modifications and mitigation works can include:<br />
<br />
:* Installation of shunt earthing conductors on the pipeline to underground earthing systems, to allow LFI voltage to drain to earth along sections of the pipeline<br />
:* Increasing the distance between the pipeline and the transmission line<br />
:* Installation of overhead earth wires or counterpoise earthing conductors<br />
:* Modifying the design of the pipeline, e.g. coating specification, diameter, etc<br />
:* Installation of gradient control wires alongside the pipeline, typically of zinc<br />
<br />
== International Standards ==<br />
<br />
Most countries have their own standards for electromagnetic interference on pipelines.<br />
<br />
In Europe, [http://www.cenelec.eu/dyn/www/f?p=104:110:788050941426229 EN 50443:2011] is the standard for low frequency induction, conductive coupling and capacitive coupling. This has also been adopted by the British Standard [http://shop.bsigroup.com/ProductDetail/?pid=000000000030125297 BS EN 50443:2011].<br />
<br />
In Australia, [http://infostore.saiglobal.com/store/details.aspx?ProductID=386659 AS/NZS 4853:2000] stipulates the limits on pipeline-to-earth touch voltages. There are two main categories: Category A for pipelines with access to the public or unskilled staff, and Category B for pipelines with restricted access to skill personnel. For Category A, the load LFI limit is 32 Vac and fault LFI limit is between 32 and 350 Vac depending on the fault clearing time. For Category B, the load LFI limit is also 32 Vac, but the fault LFI limit is 1000 Vac (for faults cleared in less than 1s).<br />
<br />
== Computer Software ==<br />
<br />
The LFI calculation is greatly simplified by using a computer software package, especially for more complicated joint rights-of-way. The following commercial software packages are among the most popular: <br />
<br />
:* [http://www.sestech.com/Products/SoftPackages/CDEGS.htm SES Technologies CDEGS] and in particular, the [http://www.sestech.com/Products/SoftPackages/ROW.htm Right of Way module]<br />
:* [http://www.ttoolboxes.com/products/PRCI-AC-Mitigation/ PRCI Pipeline A/C Interference & Mitigation Toolbox]<br />
:* [http://www.elsyca.com/Home/CathodicProtectionACMitigation/Home/Software/PipelineACMitigation.aspx Elsyca IRIS]<br />
<br />
A free alternative for single pipeline and tower line rights-of-way:<br />
<br />
:* [https://www.sigmapower.com.au/software/ Sigma Power Low-Fi]<br />
<br />
== What Next? ==<br />
<br />
The LFI calculation is typically done to confirm the pipeline and / or transmission line design and to determine if mitigation is required. If mitigation measures are required, then the next step would be to design and specify these mitigation works.<br />
<br />
== Selected References ==<br />
<br />
:* [1] AS/NZS 4853:2000, "Electrical hazards on metallic pipelines"<br />
:* [2] CIGRE Guide 95, WG 36.02, "Guide on the Influence of High Voltage AC Power Systems on Metallic Pipelines", 1995<br />
:* [3] Schlabbach, J., "Short-circuit Currents (IET Power and Energy Series 51)", 2005, IEE<br />
:* [4] Tleis, N. D., "Power System Modelling and Fault Analysis", 2008, Elsevier Ltd<br />
:* [5] Sunde, E. D., "Earth Conduction Effects in Transmission Systems", 1968, Dover Publications<br />
:* [6] Carson, J., "Wave Propagation in Overhead Wires with Ground Return", 1926, Bell System Technology Journal, Vol 5, pp. 539-554<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=File:Towerline.jpg&diff=191File:Towerline.jpg2020-12-25T03:11:00Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:Pipeline_model_2.png&diff=190File:Pipeline model 2.png2020-12-25T03:10:47Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:Pipeline_model_1.png&diff=189File:Pipeline model 1.png2020-12-25T03:10:32Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=Low_Frequency_Induction_(LFI)_Calculation&diff=188Low Frequency Induction (LFI) Calculation2020-12-25T03:10:00Z<p>Jules: Created page with "== Introduction == Low frequency induction (LFI) in metal pipelines is a form of electromagnetic interference that occurs when high voltage transmission lines are run in para..."</p>
<hr />
<div>== Introduction ==<br />
<br />
Low frequency induction (LFI) in metal pipelines is a form of electromagnetic interference that occurs when high voltage transmission lines are run in parallel with metallic pipelines. <br />
<br />
The loaded phases on a transmission line and the pipeline act like single-turn windings on a large air-core transformer. Current flowing through the transmission line induces a voltage at the pipeline.<br />
<br />
During normal operation, each loaded phase of the transmission line induces a voltage on the pipeline. If the three-phases are balanced, then most of the induced voltage will cancel each other out, but the spatial assymetry of the phases on the transmission line will prevent the induced voltages from fully cancelling out. Thus a non-zero induced voltage results on the pipeline. This condition is usually called "Load LFI".<br />
<br />
During an earth fault condition, the phases of the transmission line are no longer balanced and as a result, a more significant voltage is induced on the pipeline. This condition is usually called "Fault LFI".<br />
<br />
=== Why do the calculation? ===<br />
<br />
LFI calculations are typically done for personnel safety reasons, in order to ensure that induced voltages are not hazardous to someone in contact with the pipeline. <br />
<br />
In cases where such a situation is not possible, then an approach minimising the risk of electric shock could be followed. This could involve restricting access to the pipeline or using a probability-based methodology to estimate the risk of exposure, etc.<br />
<br />
=== When to do the calculation? ===<br />
<br />
The calculation should be done whenever HV transmission lines are installed in the vicinity of pipelines, and vice versa. Of particular concern is when the pipeline is run nearby and parallel to the transmission line for long distances. On the other hand, if a pipeline crosses the transmission line perpendicularly, then the magnitude of LFI would be low. <br />
<br />
== Calculation Methodology ==<br />
<br />
The LFI calculation has nine general steps:<br />
<br />
:* Step 1: Data Gathering<br />
:* Step 2: Define the Zone of Influence <br />
:* Step 3: Define Pipeline Sections<br />
:* Step 4: Calculate Effective Distances<br />
:* Step 5: Calculate Pipeline Impedances<br />
:* Step 6: Calculate Mutual Coupling Impedances<br />
:* Step 7: Compute Load LFI<br />
:* Step 8: Compute Fault LFI<br />
:* Step 9: Calculate Pipeline-to-Earth Touch Voltages (if necessary)<br />
:* Step 10: Apply Mitigation Works (if necessary)<br />
<br />
=== Step 1: Data Gathering ===<br />
<br />
Before beginning the calculation, the following data needs to be gathered:<br />
<br />
:* Plan layouts of pipeline(s) and transmission line(s), to scale with enough resolution to measure horizontal distances between pipeline and transmission line<br />
:* General arrangement of transmission line tower or underground spatial arrangement<br />
:* Pipeline data, e.g. diameter, metal resistivity, coating details, etc<br />
:* Soil resistivity data around pipeline<br />
:* Forecast (or actual) load currents in transmission line<br />
:* Worst case prospective fault currents on transmission line<br />
<br />
=== Step 2: Define the Zone of Influence ===<br />
<br />
Based on the plan layouts of the pipeline and transmission line routes, the first step is to define the zone(s) of influence in which you will perform the LFI study. This zone is typically a corridor where the pipeline and transmission line are run close together in parallel. <br />
<br />
As a general rule of thumb, the zone extends along this corridor until the horizontal separation between the pipeline and transmission line exceeds 1km. It is deemed that there is minimal LFI effect on the pipeline beyond a 1km pipeline-transmission line separation.<br />
<br />
=== Step 3: Define Pipeline Sections ===<br />
<br />
Once the zone of influence has been defined, the transmission line or pipeline should be broken up into small sections and the horizontal separations calculated for each section. The length of the indivudal sections do not have to be constant and can vary along the the route, depending on the rate of change of the horizontal separation. <br />
<br />
For example, long section lengths can be applied when the pipeline and transmission line are roughly parallel. However, where the two lines converge, diverge or cross, smaller sectional lengths should be used (typically, a maximum ratio of 3:1 between the start of section and end of section separations is used to determine sectional length). <br />
<br />
A table showing the sectional lengths and horizontal distances can then be developed, for example:<br />
<br />
<table><br />
{| border = "1" cellpadding="5" cellspacing="0" style = "text-align:center;"<br />
|-<br />
!style="background:#efefef;" width="80px" rowspan="2" | Section<br />
!style="background:#efefef;" width="100px" rowspan="2" | Distance along pipeline (km)<br />
!style="background:#efefef;" width="100px" rowspan="2" | Section Length (m)<br />
!style="background:#efefef;" width="200px" colspan="2" | Horizontal distance from phase A to pipeline (m)<br />
|-<br />
!style="background:#efefef;" width="80px" | Start<br />
!style="background:#efefef;" width="80px" | End<br />
|-<br />
| 1 || 0.0 || 300 || 140 || 190<br />
|-<br />
| 2 || 0.3 || 200 || 190 || 200<br />
|}<br />
<br />
=== Step 4: Calculate Effective Distances ===<br />
<br />
The table we developed in Step 2 shows the horizontal separations between phase A and the pipeline at the start and end of each pipeline section. However, the transmission line will have three phases (and possibly two sets of conductors per phase) and the horizontal distances also do not take into account the geometry of the transmission tower.<br />
<br />
Therefore, we'd like to further refine these distances. Ideally, we want a single effective distance between the pipeline and each phase (and if applicable, the earth wire). By using an effective distance, this makes it more convenient for us to calculate the induced voltages on a pipeline section. <br />
<br />
This effective distance should incorporate the following:<br />
<br />
:* The average horizontal distance along the pipeline section<br />
:* The geometry of the transmission line towers (and the spatial geometry of the individual conductors on the tower)<br />
:* An equivalent distance for parallel conductors of the same phase<br />
<br />
==== Transmission Line Geometry ====<br />
<br />
[[Image:Towerline.jpg|thumb|300px|Figure 1. Double conductor tower example]]<br />
<br />
In order to construct the effective distances, we first need to define the spatial geometry of the transmission line with respect to the pipeline. This could be the geometry of transmission line towers or the spatial arrangement of underground conductors.<br />
<br />
The figure to the right, showing a double conductor tower with two earth wires, gives an example of the kind of data that is required:<br />
<br />
:* L1 = Distance from pipeline to phase a<br />
:* L<sub>b</sub> = Distance from phase a to phase b<br />
:* L<sub>c</sub> = Distance from phase a to phase c<br />
:* L<sub>w</sub> = Distance from phase a to the earth wire<br />
:* L<sub>aa</sub> = Distance between phase a double conductors<br />
:* L<sub>bb</sub> = Distance between phase b double conductors<br />
:* L<sub>cc</sub> = Distance between phase c double conductors<br />
:* H<sub>a</sub> = Height from pipeline to phase a<br />
:* H<sub>b</sub> = Height from pipeline to phase b<br />
:* H<sub>c</sub> = Height from pipeline to phase c<br />
:* H<sub>w</sub> = Height from pipeline to earth wire<br />
<br />
Of course your data requirements may vary depending on the transmission line type, but you need enough data to fully describe the spatial relationships between the phase conductors and the pipeline geometrically.<br />
<br />
==== Effective Distances ====<br />
<br />
As the horizontal distance between the transmission line and the pipeline is not necessarly constant along each section, we want to calculate an average horizontal distance for each line section. Typically, the geometric mean distance is used:<br />
<br />
: <math> L_{m} = \sqrt{L1 \times L2} \, </math><br />
<br />
Where <math> L_{m} \, </math> is the effective horizontal distance between phase a and the pipeline<br />
:: <math> L1 \, </math> is the horizontal distance (between phase a and the pipeline) at the start of the section (m)<br />
:: <math> L2 \, </math> is the horizontal distance (between phase a and the pipeline) at the end of the section (m)<br />
<br />
Similarly for double conductors arranged horizontally, we use the geomtric mean distance to obtain an effective distance between double conductors and the pipeline:<br />
<br />
: <math> L_{a} = \sqrt{L_{m} \times (L_{m} + L_{aa}} \, </math><br />
<br />
Where <math> L_{a} \, </math> is the effective horizontal distance between double conductors (of phase a) and the pipeline (m)<br />
:: <math> L_{m} \, </math> is the effective horizontal distance calculated above (m)<br />
:: <math> L_{aa} \, </math> is the horizontal distance between the two conductors of phase a (m)<br />
<br />
Note that the above show only horizontal distances. We can use simple Pythagoras to calculate the overall effective distance between phase a and the pipeline, taking into account the vertical arrangement of the conductors (relative to the pipeline). Below, we do this for a tower line:<br />
<br />
: <math> D_{ap} = \sqrt{L_{a}^{2} + H_{a}^{2}} \, </math><br />
<br />
Where <math> D_{ap} \, </math> is the overall effective distance between the phase a and the pipeline<br />
:: <math> L_{a} \, </math> is the effective horizontal distance between double conductors (of phase a) and the pipeline (m)<br />
:: <math> H_{a} \, </math> is the height from the pipeline to phase a (m)<br />
<br />
The same process above is repeated to calculate the effective distances for the other phases and the earth wire (if applicable). <br />
<br />
The effective distances can be calculated also for other spatial conductor arrangements, using the geometric mean as a basis. Ultimately for each line section, we'd like to end up with only a single effective distance between the pipeline and a phase of the tower line. <br />
<br />
=== Step 5: Calculate Pipeline Impedances ===<br />
<br />
To assess the safety of a joint right-of-way installation, we want to calculate pipeline-to-earth touch voltages. However a buried pipeline cannot be treated like another overhead conductor in air because a pipeline has a finite impedance to earth that is distributed along its entire length. <br />
<br />
Therefore the open-circuit voltages induced in the pipeline that we calculated earlier are not equivalent to the pipeline-to-earth touch voltage. There are continuous leakages to earth along the pipeline and the open-circuit induced voltages can be up to 10 times higher than the pipeline-to-earth touch voltages.<br />
<br />
In order to calculate a more accurate pipeline-to-earth touch voltage, we need to consider the electrical characteristics of the pipeline. <br />
<br />
The pipeline impedances described in this section are simplified approximations of Sunde's method [5] and are based on Appendix G of CIGRE Guide 95 "Guide on the Influence of High Voltage AC Power Systems on Metallic Pipelines" [2]. Refer to Sunde's original work [5] for more details on accurately modelling pipeline and other buried conductor impedances.<br />
<br />
''' a) Pipeline longitudinal impedance '''<br />
<br />
For a buried and coated metallic pipeline, the longitudinal impedance per metre can be approximated as follows:<br />
<br />
: <math> Z_{p} = \frac{1}{\pi D} \sqrt{\pi f \rho_{p} \mu_{0} \mu_{p}} + \frac{\pi f \mu_{0}}{4} + j \left[ \frac{1}{\pi D} \sqrt{\pi f \rho_{p} \mu_{0} \mu_{p}} + \mu_{0} f \ln \left( \frac{3.7}{D} \sqrt{\frac{\rho}{\mu_{0} 2 \pi f}} \right) \right] \, </math><br />
<br />
Where <math> Z_{p} \, </math> is the pipeline longitudinal (or series) impedance (<math>\Omega /m </math>)<br />
:: <math> D \, </math> is the diamter of the pipeline (m)<br />
:: <math> \rho_{p} \, </math> is the resistivity of the pipeline metal (<math>\Omega .m </math>)<br />
:: <math> \rho \, </math> is the resistivity of the soil (<math>\Omega .m </math>)<br />
:: <math> \mu_{0} = 4 \pi \times 10^{-7} \, </math> is the permeability of free space (H/m)<br />
:: <math> \mu_{p} \, </math> is the relative permeability of the pipeline metal (H/m)<br />
:: <math> f \, </math> is the nominal frequency of the transmission line (Hz)<br />
<br />
''' b) Pipeline shunt admittance '''<br />
<br />
For a buried and coated metallic pipeline, the shut admittance per metre can be approximated as follows:<br />
<br />
: <math> Y_{p} = \frac{\pi D}{r_{c} \delta_{c}} + j \left[ 2 \pi f \frac {\epsilon_{0} \epsilon_{c} \pi D}{\delta_{c}} \right] \, </math><br />
<br />
Where <math> Y_{p} \, </math> is the pipeline shunt admittance (<math>\Omega^{-1} /m </math>)<br />
:: <math> D \, </math> is the diameter of the pipeline (m)<br />
:: <math> r_{c} \, </math> is the resistivity of the pipeline coating (<math>\Omega .m </math>)<br />
:: <math> \delta_{c} \, </math> is the thickness of the pipeline coating (m)<br />
:: <math> \epsilon_{c} \, </math> is the relative permeability of the coating (H/m)<br />
:: <math> \epsilon_{0} = 8.85 \times 10^{-12} \, </math> is the permittivity of free space (F/m)<br />
<br />
The pipeline shunt impedance is simply the reciprocal of the admittance (remember that it is complex quantity):<br />
<br />
: <math> Z_{p,sh} = \frac{1}{Y_{p}} \, </math> <br />
<br />
''' c) Pipeline characteristic impedance '''<br />
<br />
: <math> Z_{c} = \sqrt{\frac{Z_{i}}{Y_{i}}} \, </math><br />
<br />
Where <math> Z_{c} \, </math> is the pipeline characteristic impedance (<math>\Omega </math>)<br />
:: <math> Z_{i} \, </math> is the pipeline longitudinal impedance (<math>\Omega /m </math>)<br />
:: <math> Y_{c} \, </math> is the pipeline longitudinal conductance (<math>\Omega^{-1} /m </math>)<br />
<br />
''' d) Pipeline effective length '''<br />
<br />
: <math> l_{e} = \frac{1}{ Re \left[ \sqrt{Y_{i} Z_{i}} \right] } \, </math><br />
<br />
Where <math> l_{e} \, </math> is the pipeline effective length (m)<br />
:: <math> Z_{i} \, </math> is the pipeline longitudinal impedance (<math>\Omega /m </math>)<br />
:: <math> Y_{c} \, </math> is the pipeline longitudinal conductance (<math>\Omega^{-1} /m </math>)<br />
<br />
=== Step 6: Calculate Mutual Coupling Impedances ===<br />
<br />
The mutual coupling impedance between a line conductor and the pipeline (with earth return) can be given by the general formula below (based on Carson's equations [4]):<br />
<br />
: <math> Z_{lp} = \pi^{2} \times 10^{-4} f + j 4 \pi \times 10^{-4} \ln \left( \frac{D_{e}}{D_{lp}} \right) \, </math><br />
<br />
Where <math> Z_{lp} \, </math> is the mutual impedance between the line conductor <math>l \,</math> and pipeline <math>p \,</math> (<math>\Omega /km </math>)<br />
:: <math> D_{lp} \, </math> is the equivalent distance between the line conductor <math>l \,</math> and pipeline <math>p \,</math> (m)<br />
:: <math> f \, </math> is the nominal system frequency (Hz)<br />
:: <math> D_{e} \, </math> is the depth of equivalent earth return given by:<br />
<br />
::: <math> D_{e} = 658.37 \sqrt{\frac{\rho}{f}} \, </math><br />
<br />
:: Where <math> \rho \, </math> is the resistivity of the soil (<math>\Omega .m </math>)<br />
<br />
Using the formula above, you can calculate the mutual impedances <math> Z_{ap} \, </math>, <math> Z_{bp} \, </math> and <math> Z_{cp} \, </math>, between the pipeline and phase a, b and c of the transmission line respectively. To calculate the impedances in Ohms, multiply the mutual impedances by the sectional length of the pipeline.<br />
<br />
=== Step 7: Compute Load LFI ===<br />
<br />
Load LFI results from either unbalanced load currents or spatial differences between the phase conductors relative to the pipeline. Given the load current phasors for the transmission line, the total open-circuit induced LFI voltage on the pipeline is simply the vector sum:<br />
<br />
: <math> V_{p} = I_{a} Z_{ap} + I_{b} Z_{bp} + I_{c} Z_{cp} \, </math><br />
<br />
Where <math> V_{p} \, </math> is the overall induced load LFI voltage (V)<br />
:: <math> I_{a} \, </math>, <math> I_{b} \, </math> and <math> I_{c} \, </math> are the load current for phase a, b and c respectively (A). Note that these quantities are complex phasors<br />
:: <math> Z_{ap} \, </math>, <math> Z_{bp} \, </math> and <math> Z_{cp} \, </math> are the mutual coupling impedances between the pipeline and phase a, b and c respectively for a section of the pipeline (<math>\Omega </math>)<br />
<br />
==== Allowable Load LFI Limits ====<br />
<br />
Induced voltages on pipelines represent are electric shock hazards. Because load LFI can be regarded as a continuous hazard (i.e. the hazard is always present), then the allowable pipeline-to-earth touch voltage limits are lower than in the fault LFI case. Limits are typically in the range of 32V to 64V (see the section on international standards below for more guidance on allowable limits). <br />
<br />
Note that the open-circuit LFI voltage calculated earlier is '''higher''' than the pipeline-to-earth touch voltage as it doesn't take into account any voltage leakages to earth along the pipeline. Therefore, if the load LFI voltage you have calculated is lower than the allowable limits, then no further analysis is necessary.<br />
<br />
==== Effect of Overhead Earth Wires and Counterpoise Earths ====<br />
<br />
Overhead earth wires and counterpoise earth conductors provide a shielding effect (since a voltage is induced in them and they in turn induce an opposing voltage in the pipeline). The interaction of the overhead earth wires or counterpoise earth conductors with the pipeline can be modelled and a shielding factor computed (more on this later).<br />
<br />
=== Step 8: Compute Fault LFI ===<br />
<br />
During a balanced three-phase fault, the fault current will still be balanced across all three phases (i.e. vector sum of the faulted phase currents is close to zero). Therefore any induced voltage in the pipeline will be due primarily to the spatial assymetry of the transmission line relative to the pipeline, and this will generally be small. <br />
<br />
However, an unbalanced fault such as a line-to-earth or line-to-line fault will produce a fault current that is unbalanced and can induce much higher LFI voltages on the pipeline (until the fault is cleared). The case of a line-to-earth fault will induce the highest LFI voltage and is the type of fault that will be examined hereafter.<br />
<br />
The general formula for fault LFI is as follows:<br />
<br />
: <math> V_{f} = 3 I_{0} K_{sf} Z_{lp} \, </math><br />
<br />
Where <math> V_{f} \, </math> is the overall induced fault LFI voltage (V)<br />
:: <math> 3 I_{0} \, </math> is the earth fault current, expressed as a complex phasor (A)<br />
:: <math> K_{sf} \, </math> is the shielding factor from earth wires, counterpoise earth conductors, etc<br />
:: <math> Z_{lp} \, </math> is the mutual coupling impedances between the pipeline and the faulted line for a section of the pipeline (<math>\Omega </math>)<br />
<br />
==== Shielding Factor ====<br />
<br />
The shielding factor is the shielding effect caused by earth wires, counterpoise earths and perhaps other metallic structures, which serves to lower the LFI voltage induced on the pipeline. There a number of mechanisms that bring about this shielding effect, for example:<br />
<br />
:* If there are overhead earth wires, then not all of the earth fault current will return to the source via the earth. Some of the fault current will return via the earth wires. The current flowing through the earth wire will induce a voltage on the pipeline opposed to the voltage induced by the faulted line.<br />
:* There is the interaction between the faulted line and overhead earth wires, where the faulted line induces a voltage on the overhead earth wires, which in turn induce an opposing voltage on the pipeline.<br />
:* There are similar inductive coupling interactions between the faulted line, the pipeline and counterpoise earths or other metallic structures located in the vicinity<br />
<br />
For an overhead earth wire, the shielding factor can be calculated as follows:<br />
<br />
: <math> K_{sf} = 1 - \frac{Z_{lw}}{Z_{ww}} \times \frac{Z_{wp}}{Z_{lp}} \, </math><br />
<br />
Where <math> Z_{lw} \, </math> is the mutual coupling impedances between the faulted line and the earth wire (<math>\Omega </math>)<br />
:: <math> Z_{ww} \, </math> is the self impedance of the earth wire (<math>\Omega </math>)<br />
:: <math> Z_{wp} \, </math> is the mutual coupling impedances between the pipeline and the earth wire (<math>\Omega </math>)<br />
:: <math> Z_{lp} \, </math> is the mutual coupling impedances between the pipeline and the faulted line (<math>\Omega </math>)<br />
<br />
==== Considerations for Calculating Mutual Coupling Impedances ====<br />
<br />
Calculating the mutual coupling impedances between the pipeline and the faulted line, <math> Z_{lp} \, </math>, depends on the phase of the faulted line and it's spatial orientation relative to the pipeline. You could calculate the mutual coupling impedance between the closest phase and the pipeline, which would represent the worst case scenario. <br />
<br />
Alternatively, you could calculate the mutual coupling impedance for a group of conductors (e.g. all three phases) by considering the geometric mean separation distance between the pipeline and the group of conductors. This will result in an induced LFI voltage that is an average of the conductor group.<br />
<br />
Similarly for groups of earth wires (e.g. two overhead earth wires), a geometric mean can be selected to represent the group rather than selecting a single wire, or modelling interactions between the wires, faulted line and the pipeline (which becomes increasingly complicated). <br />
<br />
==== Allowable Fault LFI Limits ====<br />
<br />
Like the case with load LFI, allowable limits for fault LFI are typically stipulated to prevent electric shock hazards from injuring or killing personnel. Because fault LFI is temporary (lasting only until the fault is cleared), the allowable limits are normally higher than in the load LFI case. The limits are generally based on some kind risk analysis and are typically in the order of 500V to 1000V (see the section on international standards below for more guidance on allowable limits). <br />
<br />
Note that the open-circuit LFI voltage calculated earlier is '''higher''' than the pipeline-to-earth touch voltage as it doesn't take into account any voltage leakages to earth along the pipeline. Therefore, if the fault LFI voltage you have calculated is lower than the allowable limits, then no further analysis is necessary.<br />
<br />
=== Step 9: Analysis of Pipeline-to-Earth Touch Voltages (if necessary) ===<br />
<br />
If the open-circuit load or fault LFI voltage is above the permissible touch voltage, then the pipeline-to-earth touch voltages along each section of the pipeline need to be calculated. In order to calculate the pipeline-to-earth touch voltages, we need an equivalent circuit for the pipeline.<br />
<br />
A common approach is to model the pipeline as a lossy transmission line, with the following equivalent circuit for each pipeline section:<br />
<br />
[[Image:Pipeline model 1.png]]<br />
<br />
Where <math> V_{p} \, </math> is the induced LFI voltage on the pipeline section (Vac)<br />
:: <math> Y_{s} \, </math> is the longitudinal series admittance of the pipeline section (<math>\Omega^{-1} </math>)<br />
:: <math> Y_{sh} \, </math> is the shunt admittance of the pipeline section (<math>\Omega^{-1} </math>)<br />
<br />
For a pipeline with "n" linear sections, the overall equivalent circuit model is therefore:<br />
<br />
[[Image:Pipeline model 2.png]]<br />
<br />
Where <math> Y_{g} \, </math> is the admittance of a pipeline shunt earthing conductor (<math>\Omega^{-1} </math>). The inverse of the pipeline characteristic impedance can be used if no shunt earthing conductors are installed at the ends of the pipeline. In the model above, shunt earthing conductors can be connected to any of the linear pipeline sections, modelled in series with the pipeline section shunt admittance. <br />
<br />
The equivalent circuit can now be readily analysed using [http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchhoff's laws] and converted into a system of linear equations. The unknown pipeline-to-earth node voltages can then be solved using matrix operations. Depending on the number of line sections, you will probably need to use a computer program to solve this linear algebra problem.<br />
<br />
=== Step 10: Apply Mitigation Works (if necessary) ===<br />
<br />
If the pipeline-to-earth touch voltages are still above the allowable limits, then the design of the right-of-way needs to be modified or mitigation works installed. Design modifications and mitigation works can include:<br />
<br />
:* Installation of shunt earthing conductors on the pipeline to underground earthing systems, to allow LFI voltage to drain to earth along sections of the pipeline<br />
:* Increasing the distance between the pipeline and the transmission line<br />
:* Installation of overhead earth wires or counterpoise earthing conductors<br />
:* Modifying the design of the pipeline, e.g. coating specification, diameter, etc<br />
:* Installation of gradient control wires alongside the pipeline, typically of zinc<br />
<br />
== International Standards ==<br />
<br />
Most countries have their own standards for electromagnetic interference on pipelines.<br />
<br />
In Europe, [http://www.cenelec.eu/dyn/www/f?p=104:110:788050941426229 EN 50443:2011] is the standard for low frequency induction, conductive coupling and capacitive coupling. This has also been adopted by the British Standard [http://shop.bsigroup.com/ProductDetail/?pid=000000000030125297 BS EN 50443:2011].<br />
<br />
In Australia, [http://infostore.saiglobal.com/store/details.aspx?ProductID=386659 AS/NZS 4853:2000] stipulates the limits on pipeline-to-earth touch voltages. There are two main categories: Category A for pipelines with access to the public or unskilled staff, and Category B for pipelines with restricted access to skill personnel. For Category A, the load LFI limit is 32 Vac and fault LFI limit is between 32 and 350 Vac depending on the fault clearing time. For Category B, the load LFI limit is also 32 Vac, but the fault LFI limit is 1000 Vac (for faults cleared in less than 1s).<br />
<br />
== Computer Software ==<br />
<br />
The LFI calculation is greatly simplified by using a computer software package, especially for more complicated joint rights-of-way. The following commercial software packages are among the most popular: <br />
<br />
:* [http://www.sestech.com/Products/SoftPackages/CDEGS.htm SES Technologies CDEGS] and in particular, the [http://www.sestech.com/Products/SoftPackages/ROW.htm Right of Way module]<br />
:* [http://www.ttoolboxes.com/products/PRCI-AC-Mitigation/ PRCI Pipeline A/C Interference & Mitigation Toolbox]<br />
:* [http://www.elsyca.com/Home/CathodicProtectionACMitigation/Home/Software/PipelineACMitigation.aspx Elsyca IRIS]<br />
<br />
A free alternative for single pipeline and tower line rights-of-way:<br />
<br />
:* [https://www.sigmapower.com.au/software/ Sigma Power Low-Fi]<br />
<br />
== What Next? ==<br />
<br />
The LFI calculation is typically done to confirm the pipeline and / or transmission line design and to determine if mitigation is required. If mitigation measures are required, then the next step would be to design and specify these mitigation works.<br />
<br />
== Selected References ==<br />
<br />
:* [1] AS/NZS 4853:2000, "Electrical hazards on metallic pipelines"<br />
:* [2] CIGRE Guide 95, WG 36.02, "Guide on the Influence of High Voltage AC Power Systems on Metallic Pipelines", 1995<br />
:* [3] Schlabbach, J., "Short-circuit Currents (IET Power and Energy Series 51)", 2005, IEE<br />
:* [4] Tleis, N. D., "Power System Modelling and Fault Analysis", 2008, Elsevier Ltd<br />
:* [5] Sunde, E. D., "Earth Conduction Effects in Transmission Systems", 1968, Dover Publications<br />
:* [6] Carson, J., "Wave Propagation in Overhead Wires with Ground Return", 1926, Bell System Technology Journal, Vol 5, pp. 539-554<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=File:SKM_Groundmat.jpg&diff=187File:SKM Groundmat.jpg2020-12-25T03:06:48Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:Earthing_grid_design.jpg&diff=186File:Earthing grid design.jpg2020-12-25T03:06:35Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=Earthing_Calculation&diff=185Earthing Calculation2020-12-25T03:05:47Z<p>Jules: Created page with "== Introduction == The earthing system in a plant / facility is very important for a few reasons, all of which are related to either the protection of people and equipment an..."</p>
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<div>== Introduction ==<br />
<br />
The earthing system in a plant / facility is very important for a few reasons, all of which are related to either the protection of people and equipment and/or the optimal operation of the electrical system. These include:<br />
<br />
:* Equipotential bonding of conductive objects (e.g. metallic equipment, buildings, piping etc) to the earthing system prevent the presence of dangerous voltages between objects (and earth). <br />
:* The earthing system provides a low resistance return path for earth faults within the plant, which protects both personnel and equipment<br />
:* For earth faults with return paths to offsite generation sources, a low resistance earthing grid relative to remote earth prevents dangerous ground potential rises (touch and step potentials)<br />
:* The earthing system provides a low resistance path (relative to [[Remote_Earth|remote earth]]) for voltage transients such as lightning and surges / overvoltages<br />
:* Equipotential bonding helps prevent electrostatic buildup and discharge, which can cause sparks with enough energy to ignite flammable atmospheres<br />
:* The earthing system provides a reference potential for electronic circuits and helps reduce electrical noise for electronic, instrumentation and communication systems<br />
<br />
This calculation is based primarily on the guidelines provided by [http://standards.ieee.org/findstds/interps/80-2000.html IEEE Std 80 (2000)], "Guide for safety in AC substation grounding". Lightning protection is excluded from the scope of this calculation (refer to the specific [[lightning_protection|lightning protection calculation]] for more details). <br />
<br />
=== Why do the calculation? ===<br />
<br />
The earthing calculation aids in the proper design of the earthing system. Using the results of this calculation, you can:<br />
<br />
:* Determine the minimum size of the earthing conductors required for the main earth grid<br />
:* Ensure that the earthing design is appropriate to prevent dangerous step and touch potentials (if this is necessary)<br />
<br />
=== When to do the calculation? ===<br />
<br />
This calculation should be performed when the earthing system is being designed. It could also be done after the preliminary design has been completed to confirm that the earthing system is adequate, or highlight the need for improvement / redesign. Ideally, soil resistivity test results from the site will be available for use in touch and step potential calculations (if necessary).<br />
<br />
=== When is the calculation unnecessary? ===<br />
<br />
The sizing of earthing conductors should always be performed, but touch and step potential calculations (per IEEE Std 80 for earth faults with a return path through remote earth) are not always necessary.<br />
<br />
For example, when all electricity is generated on-site and the HV/MV/LV earthing systems are interconnected, then there is no need to do a touch and step potential calculation. In such a case, all earth faults would return to the source via the earthing system (notwithstanding some small leakage through earth). <br />
<br />
However, where there are decoupled networks (e.g. long transmission lines to remote areas of the plant), then touch and step potential calculations should be performed for the remote area only.<br />
<br />
== Calculation Methodology ==<br />
<br />
This calculation is based on IEEE Std 80 (2000), "Guide for safety in AC substation grounding". There are two main parts to this calculation:<br />
<br />
:* Earthing grid conductor sizing<br />
:* Touch and step potential calculations<br />
<br />
IEEE Std 80 is quite descriptive, detailed and easy to follow, so only an overview will be presented here and IEEE Std 80 should be consulted for further details (although references will be given herein).<br />
<br />
=== Prerequisites ===<br />
<br />
The following information is required / desirable before starting the calculation:<br />
<br />
:* A layout of the site<br />
:* Maximum earth fault current into the earthing grid<br />
:* Maximum fault clearing time<br />
:* Ambient (or soil) temperature at the site<br />
:* Soil resistivity measurements at the site (for touch and step only)<br />
:* Resistivity of any surface layers intended to be laid (for touch and step only)<br />
<br />
=== Earthing Grid Conductor Sizing ===<br />
<br />
Determining the minimum size of the earthing grid conductors is necessary to ensure that the earthing grid will be able to withstand the maximum earth fault current. Like a normal power cable under fault, the earthing grid conductors experience an [[Adiabatic_Short_Circuit_Temperature_Rise|adiabatic short circuit temperature rise]]. However unlike a fault on a normal cable, where the limiting temperature is that which would cause permanent damage to the cable's insulation, the temperature limit for earthing grid conductors is the melting point of the conductor. In other words, during the worst case earth fault, we don't want the earthing grid conductors to start melting!<br />
<br />
The minimum conductor size capable of withstanding the adiabatic temperature rise associated with an earth fault is given by re-arranging IEEE Std 80 Equation 37:<br />
<br />
: <math> A = \sqrt{ i^{2}t \left( \frac{\frac{\alpha_{r} \rho_{r} 10^{4}}{TCAP}}{\ln \left[ 1+ \left( \frac{T_{m}-T_{a}}{K_{0}+T_{a}} \right) \right]} \right)} \, </math><br />
<br />
Where <math>A \,</math> is the minimum cross-sectional area of the earthing grid conductor (<math>mm^{2}</math>)<br />
: <math>i^{2}t \,</math> is the energy of the maximum earth fault (<math>A^{2}s</math>)<br />
: <math>T_{m} \,</math> is the maximum allowable (fusing) temperature (ºC)<br />
: <math>T_{a} \,</math> is the ambient temperature (ºC)<br />
: <math>\alpha_{r} \,</math> is the thermal coefficient of resistivity (º<math>C^{-1}</math>)<br />
: <math>\rho_{r} \,</math> is the resistivity of the earthing conductor (<math>\mu \Omega . cm </math>)<br />
: <math>K_{0} \,</math> is <math> \left( \frac{1}{\alpha_{r}} - 20 \deg C \right)</math><br />
: <math>TCAP \,</math> is the thermal capacity of the conductor per unit volume(<math>J cm^{-3}</math>º<math>C^{-1}</math>)<br />
<br />
The material constants <math>T_{m}</math>, <math>\alpha_{r}</math>, <math>\rho_{r}</math> and <math>TCAP</math> for common conductor materials can be found in IEEE Std 80 Table 1. For example. commercial hard-drawn copper has material constants:<br />
:* <math>T_{m}=</math>1084 ºC<br />
:* <math>\alpha_{r}=</math>0.00381 º<math>C^{-1}</math><br />
:* <math>\rho_{r}=</math>1.78 <math> \mu \Omega . cm </math><br />
:* <math>TCAP=</math>3.42 <math> J cm^{-3}</math>º<math>C^{-1}</math>.<br />
<br />
As described in IEEE Std 80 Section 11.3.1.1, there are alternative methods to formulate this equation, all of which can also be derived from [[Adiabatic_Short_Circuit_Temperature_Rise#Derivation|first principles]]).<br />
<br />
There are also additional factors that should be considered (e.g. taking into account future growth in fault levels), as discussed in IEEE Std 80 Section 11.3.3.<br />
<br />
=== Touch and Step Potential Calculations ===<br />
<br />
When electricity is generated remotely and there are no return paths for earth faults other than the earth itself, then there is a risk that earth faults can cause dangerous voltage gradients in the earth around the site of the fault (called ground potential rises). This means that someone standing near the fault can receive a dangerous electrical shock due to:<br />
* Touch voltages - there is a dangerous potential difference between the earth and a metallic object that a person is touching<br />
* Step voltages - there is a dangerous voltage gradient between the feet of a person standing on earth<br />
<br />
The earthing grid can be used to dissipate fault currents to remote earth and reduce the voltage gradients in the earth. The touch and step potential calculations are performed in order to assess whether the earthing grid can dissipate the fault currents so that dangerous touch and step voltages cannot exist.<br />
<br />
==== Step 1: Soil Resistivity ====<br />
<br />
The resistivity properties of the soil where the earthing grid will be laid is an important factor in determining the earthing grid's resistance with respect to [[remote_Earth|remote earth]]. Soils with lower resistivity lead to lower overall grid resistances and potentially smaller earthing grid configurations can be designed (i.e. that comply with safe step and touch potentials).<br />
<br />
It is good practice to perform soil resistivity tests on the site. There are a few standard methods for measuring soil resistivity (e.g. Wenner four-pin method). A good discussion on the interpretation of soil resistivity test measurements is found in IEEE Std 80 Section 13.4. <br />
<br />
Sometimes it isn't possible to conduct soil resistivity tests and an estimate must suffice. When estimating soil resistivity, it goes without saying that one should err on the side of caution and select a higher resistivity. IEEE Std 80 Table 8 gives some guidance on range of soil resistivities based on the general characteristics of the soil (i.e. wet organic soil = 10 <math>\Omega</math>.m, moist soil = 100 <math>\Omega</math>.m, dry soil = 1,000 <math>\Omega</math>.m and bedrock = 10,000 <math>\Omega</math>.m).<br />
<br />
==== Step 2: Surface Layer Materials ====<br />
<br />
Applying a thin layer (0.08m - 0.15m) of high resistivity material (such as gravel, blue metal, crushed rock, etc) over the surface of the ground is commonly used to help protect against dangerous touch and step voltages. This is because the surface layer material increases the contact resistance between the soil (i.e. earth) and the feet of a person standing on it, thereby lowering the current flowing through the person in the event of a fault. <br />
<br />
IEEE Std 80 Table 7 gives typical values for surface layer material resistivity in dry and wet conditions (e.g. 40mm crushed granite = 4,000 <math>\Omega</math>.m (dry) and 1,200 <math>\Omega</math>.m (wet)). <br />
<br />
The effective resistance of a person's feet (with respect to earth) when standing on a surface layer is not the same as the surface layer resistance because the layer is not thick enough to have uniform resistivity in all directions. A surface layer derating factor needs to be applied in order to compute the effective foot resistance (with respect to earth) in the presence of a finite thickness of surface layer material. This derating factor can be approximated by an empirical formula as per IEEE Std 80 Equation 27:<br />
<br />
: <math> C_{s} = 1 - \frac{0.09 \left(1 - \frac{\rho}{\rho_{s}} \right)}{2h_{s} + 0.09} \,</math><br />
<br />
Where <math>C_{s} \,</math> is the surface layer derating factor<br />
:: <math>\rho \,</math> is the soil resistivity (<math>\Omega</math>.m)<br />
:: <math>\rho_{s} \,</math> is the resistivity of the surface layer material (<math>\Omega</math>.m)<br />
:: <math>h_{s} \,</math> is the thickness of the surface layer (m)<br />
<br />
This derating factor will be used later in Step 5 when calculating the maximum allowable touch and step voltages.<br />
<br />
==== Step 3: Earthing Grid Resistance ====<br />
<br />
A good earthing grid has low resistance (with respect to remote earth) to minimise ground potential rise (GPR) and consequently avoid dangerous touch and step voltages. Calculating the earthing grid resistance usually goes hand in hand with earthing grid design - that is, you design the earthing grid to minimise grid resistance. The earthing grid resistance mainly depends on the area taken up by the earthing grid, the total length of buried earthing conductors and the number of earthing rods / electrodes. <br />
<br />
IEEE Std 80 offers two alternative options for calculating the earthing grid resistance (with respect to remote earth) - 1) the simplified method (Section 14.2) and 2) the Schwarz equations (Section 14.3), both of which are outlined briefly below. IEEE Std 80 also includes methods for reducing soil resistivity (in Section 14.5) and a treatment for concrete-encased earthing electrodes (in Section 14.6).<br />
<br />
===== Simplified Method =====<br />
<br />
IEEE Std 80 Equation 52 gives the simplified method as modified by Sverak to include the effect of earthing grid depth:<br />
<br />
: <math> R_{g} = \rho \left[ \frac{1}{L_{T}} + \frac{1}{\sqrt{20A}} \left( 1 + \frac{1}{1 + h \sqrt{20 / A}} \right) \right] \,</math><br />
<br />
Where <math> R_{g} \,</math> is the earthing grid resistance with respect to remote earth (<math>\Omega</math>)<br />
:: <math> \rho \,</math> is the soil resistivitiy (<math>\Omega</math>.m)<br />
:: <math> L_{T} \,</math> is the total length of buried conductors (m)<br />
:: <math> A \,</math> is the total area occupied by the earthing grid (<math>m^{2}</math>)<br />
:: <math> h \,</math> is the depth of the earthing grid (m)<br />
<br />
===== Schwarz Equations =====<br />
<br />
The Schwarz equations are a series of equations that are more accurate in modelling the effect of earthing rods / electrodes. The equations are found in IEEE Std 80 Equations 53, 54, 55<sup>[[Talk:Earthing_Calculation|''(footnote)'']]</sup> and 56, as follows:<br />
<br />
: <math> R_{g} = \frac{R_{1}R_{2} - R_{m}^{2}}{R_{1} + R_{2} - 2R_{m}} \,</math><br />
<br />
Where <math> R_{g} \, </math> is the earthing grid resistance with respect to remote earth (<math>\Omega</math>)<br />
:: <math> R_{1} \, </math> is the earth resistance of the grid conductors (<math>\Omega</math>)<br />
:: <math> R_{2} \, </math> is the earth resistance of the earthing electrodes (<math>\Omega</math>)<br />
:: <math> R_{m} \, </math> is the mutual earth resistance between the grid conductors and earthing electrodes (<math>\Omega</math>)<br />
<br />
And the grid, earthing electrode and mutual earth resistances are:<br />
<br />
: <math> R_{1} = \frac{\rho}{\pi L_{c}} \left[ \ln \left( \frac{2L_{c}}{a'} \right) + \frac{k_{1} L_{c}}{\sqrt{A}} - k_{2} \right] \, </math><br />
<br />
: <math> R_{2} = \frac{\rho}{2 \pi n_{r} L_{r}} \left[ \ln \left( \frac{4L_{r}}{b} \right) -1 + \frac{2k_{1} L_{r}}{\sqrt{A}} \left( \sqrt{n_{r}} - 1 \right)^{2} \right] \, </math><br />
<br />
: <math> R_{m} = \frac{\rho}{\pi L_{c}} \left[ \ln \left( \frac{2L_{c}}{L_{r}} \right) + \frac{k_{1} L_{c}}{\sqrt{A}} - k_{2} + 1 \right] \, </math><br />
<br />
Where <math> \rho \, </math> is the soil resistivity (<math>\Omega</math>.m)<br />
:: <math> L_{c} \, </math> is the total length of buried grid conductors (m)<br />
:: <math> a' \, </math> is <math> \sqrt{r . 2h} \,</math> for conductors buried at depth <math> h \, </math> metres and with cross-sectional radius <math> r \, </math> metres, or simply <math> r \, </math> for grid conductors on the surface<br />
:: <math> A \, </math> is the total area covered by the grid conductors (<math>m^{2}</math>)<br />
:: <math> L_{r} \, </math> is the length of each earthing electrode (m)<br />
:: <math> n_{r} \, </math> is number of earthing electrodes in area <math> A \, </math><br />
:: <math> b \, </math> is the cross-sectional radius of an earthing electrode (m)<br />
:: <math> k_{1} \, </math> and <math> k_{2} \, </math> are constant coefficients depending on the geometry of the grid<br />
<br />
<br />
The coefficient <math> k_{1} \, </math> can be approximated by the following:<br />
:* (1) For depth <math> h = 0 \, </math>: <math> k_{1} = -0.04L/R + 1.41 \, </math><br />
:* (2) For depth <math> h = \frac{1}{10} \sqrt{A} \, </math>: <math> k_{1} = -0.05L/R + 1.20 \, </math><br />
:* (3) For depth <math> h = \frac{1}{6} \sqrt{A} \, </math>: <math> k_{1} = -0.05L/R + 1.13 \, </math><br />
<br />
The coefficient <math> k_{2} \, </math> can be approximated by the following:<br />
:* (1) For depth <math> h = 0 \, </math>: <math> k_{2} = 0.15L/R + 5.50 \, </math><br />
:* (2) For depth <math> h = \frac{1}{10} \sqrt{A} \, </math>: <math> k_{2} = 0.10L/R + 4.68 \, </math><br />
:* (3) For depth <math> h = \frac{1}{6} \sqrt{A} \, </math>: <math> k_{2} = 0.05L/R + 4.40 \, </math><br />
<br />
Where in both cases, <math> L/R \, </math> is the length-to-width ratio of the earthing grid.<br />
<br />
==== Step 4: Maximum Grid Current ====<br />
<br />
The maximum grid current is the worst case earth fault current that would flow via the earthing grid back to remote earth. To calculate the maximum grid current, you firstly need to calculate the worst case symmetrical earth fault current at the facility that would have a return path through remote earth (call this <math>I_{k,e} \,</math>). This can be found from the power systems studies or from manual calculation. Generally speaking, the highest relevant earth fault level will be on the primary side of the largest distribution transformer (i.e. either the terminals or the delta windings).<br />
<br />
===== Current Division Factor =====<br />
<br />
Not all of the earth fault current will flow back through remote earth. A portion of the earth fault current may have local return paths (e.g. local generation) or there could be alternative return paths other than remote earth (e.g. overhead earth return cables, buried pipes and cables, etc). Therefore a current division factor <math>S_{f} \,</math> must be applied to account for the proportion of the fault current flowing back through remote earth. <br />
<br />
Computing the current division factor is a task that is specific to each project and the fault location and it may incorporate some subjectivity (i.e. "engineeing judgement"). In any case, IEEE Std 80 Section 15.9 has a good discussion on calculating the current division factor. In the most conservative case, a current division factor of <math>S_{f} = 1 \,</math> can be applied, meaning that 100% of earth fault current flows back through remote earth.<br />
<br />
The symmetrical grid current <math>I_{g} \,</math> is calculated by:<br />
<br />
: <math> I_{g} = I_{k,e} S_{f} \, </math><br />
<br />
===== Decrement Factor =====<br />
<br />
The symmetrical grid current is not the maximum grid current because of asymmetry in short circuits, namely a dc current offset. This is captured by the decrement factor, which can be calculated from IEEE Std 80 Equation 79:<br />
<br />
: <math> D_{f} = \sqrt{1 + \frac{T_{a}}{t_{f}} \left( 1 - e^{\frac{-2t_{f}}{T_{a}}} \right)} \, </math><br />
<br />
Where <math> D_{f} \, </math> is the decrement factor<br />
:: <math> t_{f} \, </math> is the duration of the fault (s)<br />
:: <math> T_{A} \, </math> is the dc time offset constant (see below)<br />
<br />
The dc time offset constant is derived from IEEE Std 80 Equation 74:<br />
<br />
: <math> T_{A} = \frac{X}{R} . \frac{1}{2 \pi f} \, </math><br />
<br />
Where <math> \frac{X}{R} \, </math> is the X/R ratio at the fault location<br />
:: <math> f \, </math> is the system frequency (Hz)<br />
<br />
The maximum grid current <math>I_{G} \,</math> is lastly calculated by:<br />
<br />
: <math> I_{G} = I_{g} D_{f} \, </math><br />
<br />
==== Step 5: Touch and Step Potential Criteria ====<br />
<br />
One of the goals of a safe earthing grid is to protect people against lethal electric shocks in the event of an earth fault. The magnitude of ac electric current (at 50Hz or 60Hz) that a human body can withstand is typically in the range of 60 to 100mA, when ventricular fibrillation and heart stoppage can occur. The duration of an electric shock also contributes to the risk of mortality, so the speed at which faults are cleared is also vital. Given this, we need to prescribe maximum tolerable limits for touch and step voltages that do not lead to lethal shocks. <br />
<br />
The maximum tolerable voltages for step and touch scenarios can be calculated empirically from IEEE Std Section 8.3 for body weights of 50kg and 70kg:<br />
<br />
Touch voltage limit - the maximum potential difference between the surface potential and the potential of an earthed conducting structure during a fault (due to ground potential rise):<br />
:* 50kg person: <math> E_{touch,50} = \left( 1000 + 1.5C_{s} \rho_{s} \right) \frac{0.116}{\sqrt{t_{s}}} \,</math><br />
:* 70kg person: <math> E_{touch,70} = \left( 1000 + 1.5C_{s} \rho_{s} \right) \frac{0.157}{\sqrt{t_{s}}} \,</math><br />
<br />
Step voltage limit - is the maximum difference in surface potential experience by a person bridging a distance of 1m with the feet without contact to any earthed object:<br />
:* 50kg person: <math> E_{step,50} = \left( 1000 + 6C_{s} \rho_{s} \right) \frac{0.116}{\sqrt{t_{s}}} \,</math><br />
:* 70kg person: <math> E_{step,70} = \left( 1000 + 6C_{s} \rho_{s} \right) \frac{0.157}{\sqrt{t_{s}}} \,</math><br />
<br />
Where <math> E_{touch} \,</math> is the touch voltage limit (V)<br />
:: <math> E_{step} \,</math> is the step voltage limit (V)<br />
:: <math> C_{s} \,</math> is the surface layer derating factor (as calculated in [[Earthing_Calculation#Step 2: Surface Layer Materials|Step 2]])<br />
:: <math> \rho_{s} \,</math> is the soil resistivity (<math>\Omega</math>.m)<br />
:: <math> t_{s} \,</math> is the maximum fault clearing time (s)<br />
<br />
The choice of body weight (50kg or 70kg) depends on the expected weight of the personnel at the site. Typically, where women are expected to be on site, the conservative option is to choose 50kg.<br />
<br />
==== Step 6: Ground Potential Rise (GPR) ====<br />
<br />
Normally, the potential difference between the local earth around the site and [[Remote_Earth|remote earth]] is considered to be zero (i.e. they are at the same potential). However an earth fault (where the fault current flows back through remote earth), the flow of current through the earth causes local potential gradients in and around the site. The maximum potential difference between the site and remote earth is known as the ground potential rise (GPR). It is important to note that this is a '''maximum''' potential potential difference and that earth potentials around the site will vary relative to the point of fault.<br />
<br />
The maximum GPR is calculated by:<br />
<br />
: <math> GPR = I_{G} R_{g}\, </math><br />
<br />
Where <math> GPR \, </math> is the maximum ground potential rise (V)<br />
:: <math> I_{G}\, </math> is the maximum grid current found earlier in [[Earthing_Calculation#Step 4: Maximum Grid Current|Step 4]] (A)<br />
:: <math> R_{g}\, </math> is the earthing grid resistance found earlier in [[Earthing_Calculation#Step 3: Earthing Grid Resistance|Step 3]] (<math>\Omega</math>)<br />
<br />
==== Step 7: Earthing Grid Design Verification ====<br />
<br />
Now we just need to verify that the earthing grid design is safe for touch and step potential. If the maximum GPR calculated above does not exceed either of the touch and step voltage limits (from [[Earthing_Calculation#Step 5: Touch and Step Potential Criteria|Step 5]]), then the grid design is safe. <br />
<br />
However if it '''does exceed''' the touch and step voltage limits, then some further analysis is required to verify the design, namely the calculation of the maximum mesh and step voltages as per IEEE Std 80 Section 16.5.<br />
<br />
===== Mesh Voltage Calculation =====<br />
<br />
The mesh voltage is the maximum touch voltage within a mesh of an earthing grid and is derived from IEEE Std 80 Equation 80:<br />
<br />
: <math> E_{m} = \frac{\rho_{s} K_{m} K_{i} I_{G}} {L_{M}} \, </math><br />
<br />
Where :: <math> \rho_{s} \, </math> is the soil resistivity (<math>\Omega</math>.m)<br />
:: <math> I_{G}\, </math> is the maximum grid current found earlier in [[Earthing_Calculation#Step 4: Maximum Grid Current|Step 4]] (A)<br />
:: <math> K_{m}\, </math> is the geometric spacing factor (see below)<br />
:: <math> K_{i}\, </math> is the irregularity factor (see below)<br />
:: <math> L_{M}\, </math> is the effective buried length of the grid (see below)<br />
<br />
====== Geometric Spacing Factor <math>K_{m}</math>======<br />
<br />
The geometric spacing factor <math> K_{m}\, </math> is calculated from IEEE Std 80 Equation 81:<br />
<br />
: <math> K_{m} = \frac{1}{2\pi} \left( \ln \left[ \frac{D^{2}}{16h \times d} + \frac{\left( D + 2h \right)^{2}}{8D \times d} - \frac{h}{4d} \right] + \frac{K_{ii}}{K_{h}} \ln \left[ \frac{8}{\pi (2n-1)} \right] \right) \,</math><br />
<br />
Where <math> D \, </math> is the spacing between parallel grid conductors (m)<br />
:: <math> h \, </math> is the depth of buried grid conductors (m)<br />
:: <math> d \, </math> is the cross-sectional diameter of a grid conductor (m)<br />
:: <math> K_{h} \, </math> is a weighting factor for depth of burial = <math> \sqrt{1+h} \, </math><br />
:: <math> K_{ii} \, </math> is a weighting factor for earth electrodes /rods on the corner mesh<br />
<br />
:::* <math> K_{ii} = 1 \, </math> for grids with earth electrodes along the grid perimeter or corners<br />
:::* <math> K_{ii} = \frac{1}{2n^{n/2}} \, </math> for grids with no earth electrodes on the corners or on the perimeter<br />
<br />
:: <math> n \, </math> is a geometric factor (see below)<br />
<br />
====== Geometric Factor <math>n</math>======<br />
<br />
The geometric factor <math>n \,</math> is calculated from IEEE Std 80 Equation 85:<br />
<br />
: <math> n = n_{a} \times n_{b} \times n_{c} \times n_{d} </math><br />
<br />
With <math> n_{a} = \frac{2L_{c}}{L_{p}} \, </math><br />
:: <math> n_{b} = 1 \, </math> for square grids, or otherwise <math> n_{b} = \sqrt{\frac{L_{p}}{4\sqrt{A}}} \, </math><br />
:: <math> n_{c} = 1 \, </math> for square and rectangular grids, or otherwise <math> n_{c} = \left[ \frac{L_{x}L_{y}}{A} \right]^{\frac{0.7A}{L_{x}L_{y}}} \, </math><br />
:: <math> n_{d} = 1 \, </math> for square, rectangular and L-shaped grids, or otherwise <math> n_{d} = \frac{D_{m}}{\sqrt{L_{x}^{2} + L_{y}^{2}}} \, </math><br />
<br />
Where <math> L_{c} \, </math> is the total length of horizontal grid conductors (m)<br />
:: <math> L_{p} \, </math> is the length of grid conductors on the perimeter (m)<br />
:: <math> A \, </math> is the total area of the grid (<math>m^{2}</math>)<br />
:: <math> L_{x} \, </math> and <math> L_{y} \, </math> are the maximum length of the grids in the x and y directions (m)<br />
:: <math> D_{m} \, </math> is the maximum distance between any two points on the grid (m)<br />
<br />
====== Irregularity Factor <math>K_{i}</math>======<br />
<br />
The irregularity factor <math> K_{i}\, </math> is calculated from IEEE Std 80 Equation 89:<br />
<br />
: <math> K_{i} = 0.644 + 0.148n \, </math><br />
<br />
Where <math> n \, </math> is the geometric factor derived above<br />
<br />
====== Effective Buried Length <math>L_{M}</math>======<br />
<br />
The effective buried length <math>L_{M} \,</math> is found as follows:<br />
<br />
* For grids with few or no earthing electrodes (and none on corners or along the perimeter):<br />
<br />
: <math>L_{M} = L_{c} + L_{R} \,</math><br />
<br />
Where <math> L_{c} \, </math> is the total length of horizontal grid conductors (m)<br />
:: <math> L_{R} \, </math> is the total length of earthing electrodes / rods (m)<br />
<br />
* For grids with earthing electrodes on the corners and along the perimeter:<br />
<br />
: <math>L_{M} = L_{c} + \left[ 1.55 + 1.22 \left( \frac{L_{r}}{\sqrt{L_{x}^{2} + L_{y}^{2}}} \right) \right] L_{R} \,</math><br />
<br />
Where <math> L_{c} \, </math> is the total length of horizontal grid conductors (m)<br />
:: <math> L_{R} \, </math> is the total length of earthing electrodes / rods (m)<br />
:: <math> L_{r} \, </math> is the length of each earthing electrode / rod (m)<br />
:: <math> L_{x} \, </math> and <math> L_{y} \, </math> are the maximum length of the grids in the x and y directions (m)<br />
<br />
===== Step Voltage Calculation =====<br />
<br />
The maximum allowable step voltage is calculated from IEEE Std 80 Equation 92:<br />
<br />
: <math> E_{s} = \frac{\rho_{s} K_{s} K_{i} I_{G}} {L_{S}} \, </math><br />
<br />
Where :: <math> \rho_{s} \, </math> is the soil resistivity (<math>\Omega</math>.m)<br />
:: <math> I_{G}\, </math> is the maximum grid current found earlier in [[Earthing_Calculation#Step 4: Maximum Grid Current|Step 4]] (A)<br />
:: <math> K_{s}\, </math> is the geometric spacing factor (see below)<br />
:: <math> K_{i}\, </math> is the irregularity factor (as [[Earthing_Calculation#Irregularity_Factor_Ki|derived above]] in the mesh voltage calculation)<br />
:: <math> L_{S}\, </math> is the effective buried length of the grid (see below)<br />
<br />
====== Geometric Spacing Factor <math>K_{s}</math> ======<br />
<br />
The geometric spacing factor <math> K_{s}\, </math> based on IEEE Std 80 Equation 81 is applicable for burial depths between 0.25m and 2.5m:<br />
<br />
: <math> K_{s} = \frac{1}{\pi} \left[ \frac{1}{2h} + \frac{1}{D+h} + \frac{1}{D} \left( 1 - 0.5^{n-2} \right) \right] \, </math><br />
<br />
Where <math> D \, </math> is the spacing between parallel grid conductors (m)<br />
:: <math> h \, </math> is the depth of buried grid conductors (m)<br />
:: <math> n \, </math> is a geometric factor (as [[Earthing_Calculation#Geometric_Factor_n|derived above]] in the mesh voltage calculation)<br />
<br />
====== Effective Buried Length <math>L_{S}</math> ======<br />
<br />
The effective buried length <math> L_{S}\, </math> for all cases can be calculated by IEEE Std 80 Equation 93:<br />
<br />
: <math> L_{S} = 0.75L_{c} + 0.85L_{R} \, </math><br />
<br />
Where <math> L_{c} \, </math> is the total length of horizontal grid conductors (m)<br />
:: <math> L_{R} \, </math> is the total length of earthing electrodes / rods (m)<br />
<br />
===== What Now? =====<br />
<br />
Now that the mesh and step voltages are calculated, compare them to the maximum tolerable touch and step voltages respectively. If:<br />
<br />
:* <math> E_{m} < E_{touch} \,</math>, and<br />
:* <math> E_{s} < E_{step} \,</math><br />
<br />
then the earthing grid design is safe. <br />
<br />
If not, however, then further work needs to be done. Some of the things that can be done to make the earthing grid design safe:<br />
<br />
:* Redesign the earthing grid to lower the grid resistance (e.g. more grid conductors, more earthing electrodes, increasing cross-sectional area of conductors, etc). Once this is done, re-compute the earthing grid resistance (see [[Earthing_Calculation#Step_3:_Earthing_Grid_Resistance|Step 3]]) and re-do the touch and step potential calculations.<br />
<br />
:* Limit the total earth fault current or create alternative earth fault return paths<br />
<br />
:* Consider soil treatments to lower the resistivity of the soil<br />
<br />
:* Greater use of high resistivity surface layer materials<br />
<br />
== Worked Example ==<br />
<br />
In this example, the touch and step potential calculations for an earthing grid design will be performed. The proposed site is a small industrial facility with a network connection via a transmission line and a delta-wye connected transformer.<br />
<br />
=== Step 1: Soil Resistivity ===<br />
<br />
The soil resistivity around the site was measured with a Wenner four-pin probe and found to be approximately 300 <math> \Omega </math>.m. <br />
<br />
=== Step 2: Surface Layer Materials ===<br />
<br />
A thin 100mm layer of blue metal (3,000 <math> \Omega </math>.m) is proposed to be installed on the site. The surface layer derating factor is:<br />
<br />
: <math> C_{s} = 1 - \frac{0.09 \left(1 - \frac{\rho}{\rho_{s}} \right)}{2h_{s} + 0.09} \,</math><br />
::: <math> = 1 - \frac{0.09 \left(1 - \frac{300}{3,000} \right)}{2 \times 0.1 + 0.09} \, </math><br />
::: <math> = 0.7207 \, </math><br />
<br />
=== Step 3: Earthing Grid Resistance ===<br />
<br />
[[Image:Earthing_grid_design.jpg|right|thumb|300px|Figure 1. Proposed rectangular earthing grid]]<br />
<br />
A rectangular earthing grid (see the figure right) with the following parameters is proposed:<br />
<br />
:* Length of 90m and a width of 50m<br />
:* 6 parallel rows and 7 parallel columns<br />
:* Grid conductors will be 120 <math>mm^{2}</math> and buried at a depth of 600mm<br />
:* 22 earthing rods will be installed on the corners and perimeter of the grid<br />
:* Each earthing rod will be 3m long<br />
<br />
Using the simplified equation, the resistance of the earthing grid with respect to remote earth is:<br />
<br />
: <math> R_{g} = \rho \left[ \frac{1}{L_{T}} + \frac{1}{\sqrt{20A}} \left( 1 + \frac{1}{1 + h \sqrt{20 / A}} \right) \right] \,</math><br />
::: <math> = 300 \left[ \frac{1}{956} + \frac{1}{\sqrt{20 \times 4,500}} \left( 1 + \frac{1}{1 + 0.6 \sqrt{20 / 4,500}} \right) \right] \,</math><br />
::: <math> = 2.2753 \Omega \, </math><br />
<br />
=== Step 4: Maximum Grid Current ===<br />
<br />
Suppose that the maximum single phase to earth fault at the HV winding of the transformer is 3.1kA and that the current division factor is 1 (all the fault current flows back to remote earth).<br />
<br />
The X/R ratio at the fault is approximately 15, the maximum fault duration 150ms and the system nominal frequency is 50Hz. The DC time offset is therefore:<br />
<br />
: <math> T_{A} = \frac{X}{R} . \frac{1}{2 \pi f} \, </math><br />
::: <math> = 15 \times \frac{1}{2 \pi 50} \, </math><br />
::: <math> = 0.04774 \, </math><br />
<br />
The decrement factor is then:<br />
<br />
: <math> D_{f} = \sqrt{1 + \frac{T_{a}}{t_{f}} \left( 1 - e^{\frac{-2t_{f}}{T_{a}}} \right)} \, </math><br />
::: <math> = \sqrt{1 + \frac{0.04774}{0.15} \left( 1 - e^{\frac{-2 \times 0.15}{0.04774}} \right)} \, </math><br />
::: <math> = 1.1479 \, </math><br />
<br />
Fianlly, the maximum grid current is:<br />
<br />
: <math> I_{G} = I_{g} D_{f} \, </math><br />
::: <math> = 3.1 \times 1.1479 \, </math><br />
::: <math> = 3.559 \, </math> kA<br />
<br />
=== Step 5: Touch and Step Potential Criteria ===<br />
<br />
Based on the average weight of the workers on the site, a body weight of 70kg is assumed for the maximum touch and step potential. A maximum fault clearing time of 150ms is also assumed.<br />
<br />
The maximum allowable touch potential is:<br />
<br />
: <math> E_{touch,70} = \left( 1000 + 1.5C_{s} \rho_{s} \right) \frac{0.157}{\sqrt{t_{s}}} \,</math><br />
::: <math> = \left( 1000 + 1.5 \times 0.7207 \times 3,000 \right) \frac{0.157}{\sqrt{0.15}} \,</math><br />
::: <math> = 1,720.04 \, </math> V<br />
<br />
The maximum allowable step potential is:<br />
<br />
: <math> E_{step,70} = \left( 1000 + 6C_{s} \rho_{s} \right) \frac{0.157}{\sqrt{t_{s}}} \,</math><br />
::: <math> = \left( 1000 + 6 \times 0.7207 \times 3,000 \right) \frac{0.157}{\sqrt{0.15}} \,</math><br />
::: <math> = 5,664.03 \, </math> V<br />
<br />
=== Step 6: Ground Potential Rise (GPR) ===<br />
<br />
The maximum ground potential rise is:<br />
<br />
: <math> GPR = I_{G} R_{g}\, </math><br />
::: <math> = 3,559 \times 2.2753 \, </math><br />
::: <math> = 8,097 \, </math> V<br />
<br />
The GPR far exceeds the maximum allowable touch and step potentials, and further analysis of mesh and step voltages need to be performed.<br />
<br />
=== Step 7: Earthing Grid Design Verification ===<br />
<br />
==== Mesh Voltage Calculation ====<br />
<br />
The components of the geometric factor <math>n_{a} \,</math>, <math>n_{b} \,</math>, <math>n_{c} \,</math> and <math>n_{d} \,</math> for the rectangular grid are:<br />
<br />
: <math> n_{a} = \frac{2L_{c}}{L_{p}} \, </math><br />
::: <math> = \frac{2 \times 890}{280} = 6.357 \, </math><br />
<br />
: <math> n_{b} = \sqrt{\frac{L_{p}}{4\sqrt{A}}} \, </math><br />
::: <math> = \sqrt{\frac{280}{4\sqrt{4500}}} = 1.022 \, </math><br />
<br />
: <math> n_{c} = n_{d} = 1 \, </math><br />
<br />
Therefore the geometric factor <math>n \,</math> is:<br />
<br />
: <math> n = n_{a} \times n_{b} \times n_{c} \times n_{d} \, </math><br />
::: <math> = 6.357 \times 1.022 \times 1 \times 1 \, </math><br />
::: <math> = 6.4939 \, </math><br />
<br />
The average spacing between parallel grid conductors <math> D \, </math> is:<br />
<br />
: <math> D = \frac{1}{2} \left( \frac{W_{g}}{n_{r} - 1} + \frac{L_{g}}{n_{c} - 1} \right) \, </math><br />
::: <math> = \frac{1}{2} \left( \frac{50}{6 - 1} + \frac{90}{7 - 1} \right) \, </math><br />
::: <math> = 12.5 \, </math><br />
<br />
where <math> W_{g} \, </math> and <math> L_{g} \, </math> are the width and length of the grid respectively (e.g. 50m and 90m)<br />
:: <math> n_{r} \, </math> and <math> n_{c} \, </math> is the number of parallel rows and columns respectively (e.g. 6 and 7)<br />
<br />
The geometric spacing factor <math> K_{m}\, </math> is:<br />
<br />
: <math> K_{m} = \frac{1}{2\pi} \left( \ln \left[ \frac{D^{2}}{16h \times d} + \frac{\left( D + 2h \right)^{2}}{8D \times d} - \frac{h}{4d} \right] + \frac{K_{ii}}{K_{h}} \ln \left[ \frac{8}{\pi (2n-1)} \right] \right) \,</math><br />
::: <math> K_{m} = \frac{1}{2\pi} \left( \ln \left[ \frac{12.5^{2}}{16 \times 0.6 \times 0.01236} + \frac{\left( 12.5 + 2 \times 0.6 \right)^{2}}{8 \times 12.5 \times 0.01236} - \frac{0.6}{4 \times 0.01236} \right] + \frac{1}{1.26} \ln \left[ \frac{8}{\pi (2 \times 6.4939 -1)} \right] \right) \,</math><br />
::: <math> = 0.964 \,</math><br />
<br />
The irregularity factor <math> K_{i}\, </math> is:<br />
<br />
: <math> K_{i} = 0.644 + 0.148n \, </math><br />
::: <math> = 0.644 + 0.148 \times 6.4939 \, </math><br />
::: <math> = 1.605 \,</math><br />
<br />
The effective buried length <math>L_{M} \,</math> is:<br />
<br />
: <math>L_{M} = L_{c} + \left[ 1.55 + 1.22 \left( \frac{L_{r}}{\sqrt{L_{x}^{2} + L_{y}^{2}}} \right) \right] L_{R} \,</math><br />
::: <math> = 890 + \left[ 1.55 + 1.22 \left( \frac{3}{\sqrt{90^{2} + 50^{2}}} \right) \right] \times 66 \,</math><br />
::: <math> = 994.65 \,</math>m<br />
<br />
Finally, the maximum mesh voltage is:<br />
<br />
: <math> E_{m} = \frac{\rho_{s} K_{m} K_{i} I_{G}} {L_{M}} \, </math><br />
::: <math> = \frac{300 \times 0.964 \times 1.605 \times 3,559} {994.65} \, </math><br />
::: <math> = 1,661 \,</math>V<br />
<br />
The maximum allowable touch potential is 1,720V, which exceeds the mesh voltage calculated above and the earthing system passes the touch potential criteria (although it is quite marginal).<br />
<br />
==== Step Voltage Calculation ====<br />
<br />
The geometric spacing factor <math> K_{s}\, </math> is:<br />
<br />
: <math> K_{s} = \frac{1}{\pi} \left[ \frac{1}{2h} + \frac{1}{D+h} + \frac{1}{D} \left( 1 - 0.5^{n-2} \right) \right] \, </math><br />
::: <math> = \frac{1}{\pi} \left[ \frac{1}{2 \times 0.6} + \frac{1}{12.5 + 0.6} + \frac{1}{12.5} \left( 1 - 0.5^{6.4939 - 2} \right) \right] \, </math><br />
::: <math> = 0.314 \,</math><br />
<br />
The effective buried length <math> L_{S}\, </math> is:<br />
<br />
: <math> L_{S} = 0.75L_{c} + 0.85L_{R} \, </math><br />
::: <math> = 0.75 \times 890 + 0.85 \times 66 \, </math><br />
::: <math> = 723.6 \,</math> m<br />
<br />
Finally, the maximum allowable step voltage is:<br />
<br />
: <math> E_{s} = \frac{\rho_{s} K_{s} K_{i} I_{G}} {L_{S}} \, </math><br />
::: <math> = \frac{300 \times 0.314 \times 1.605 \times 3,559} {723.6} \, </math><br />
::: <math> = 743.6 \,</math> V<br />
<br />
The maximum allowable step potential is 5,664V, which exceeds the step voltage calculated above and the earthing system passes the step potential criteria. Having passed both touch and step potential criteria, we can conclude that the earthing system is safe.<br />
<br />
== Computer Based Tools ==<br />
<br />
[[Image:SKM_Groundmat.jpg|right|thumb|300px|Figure 2. PTW GroundMat software output (courtesy of SKM Systems Analysis Inc)]]<br />
<br />
As can be seen from above, touch and step potential calculations can be quite a tedious and laborious task, and one that could conceivably be done much quicker by a computer. Even IEEE Std 80 recommends the use of computer software to calculate grid resistances, and mesh and step voltages, and also to create potential gradient visualisations of the site. <br />
<br />
Computer software packages can be used to assist in earthing grid design by modeling and simulation of different earthing grid configurations. The tools either come as standalone packages or plug-in modules to power system analysis software (such as PTW's [http://www.skm.com/products_grownmat.shtml GroundMat] or ETAP's [http://etap.com/ground-grid-systems/ground-grid-systems.htm Ground Grid Design Assessment]. Examples of standalone packages include [http://www.sestech.com/products/softpackages/autogridpro.htm SES Autogrid] and [http://www.elek.com.au/safegrid.htm SafeGrid].<br />
<br />
== What next? ==<br />
<br />
The minimum size for the earthing grid conductors can be used to specify the earthing grid conductor sizes in the material take-offs and earthing drawings. The touch and step potential calculations (where necessary) verify that the earthing grid design is safe for the worst earth faults to remote earth. The earthing drawings can therefore be approved for the next stage of reviews.<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=Category:Calculations&diff=184Category:Calculations2020-12-25T03:04:09Z<p>Jules: Created blank page</p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=Cable_Sizing_Calculation&diff=183Cable Sizing Calculation2020-12-25T03:03:49Z<p>Jules: /* What next? */</p>
<hr />
<div>== Introduction ==<br />
<br />
[[Image:Cables_header.JPG|right]]<br />
<br />
This article examines the sizing of electrical cables (i.e. cross-sectional area) and its implementation in various international standards. Cable sizing methods do differ across international standards (e.g. IEC, NEC, BS, etc) and some standards emphasise certain things over others. However the general principles underlying any cable sizing calculation do not change. In this article, a general methodology for sizing cables is first presented and then the specific international standards are introduced.<br />
<br />
=== Why do the calculation? ===<br />
<br />
The proper sizing of an electrical (load bearing) cable is important to ensure that the cable can:<br />
:* Operate continuously under full load without being damaged<br />
:* Withstand the worst short circuits currents flowing through the cable<br />
:* Provide the load with a suitable voltage (and avoid excessive voltage drops)<br />
:* (optional) Ensure operation of protective devices during an earth fault<br />
<br />
=== When to do the calculation? ===<br />
<br />
This calculation can be done individually for each power cable that needs to be sized, or alternatively, it can be used to produce cable sizing waterfall charts for groups of cables with similar characteristics (e.g. cables installed on ladder feeding induction motors).<br />
<br />
== General Methodology ==<br />
<br />
All cable sizing methods more or less follow the same basic six step process:<br />
<br />
: 1) Gathering data about the cable, its installation conditions, the load that it will carry, etc<br />
: 2) Determine the minimum cable size based on continuous current carrying capacity<br />
: 3) Determine the minimum cable size based on voltage drop considerations<br />
: 4) Determine the minimum cable size based on short circuit temperature rise<br />
: 5) Determine the minimum cable size based on earth fault loop impedance<br />
: 6) Select the cable based on the highest of the sizes calculated in step 2, 3, 4 and 5<br />
<br />
=== Step 1: Data Gathering ===<br />
<br />
The first step is to collate the relevant information that is required to perform the sizing calculation. Typically, you will need to obtain the following data:<br />
<br />
==== Load Details ====<br />
<br />
The characteristics of the load that the cable will supply, which includes:<br />
<br />
:* Load type: motor or feeder<br />
:* Three phase, single phase or DC<br />
:* System / source voltage<br />
:* Full load current (A) - or calculate this if the load is defined in terms of power (kW)<br />
:* Full load power factor (pu)<br />
:* Locked rotor or load starting current (A)<br />
:* Starting power factor (pu)<br />
:* Distance / length of cable run from source to load - this length should be as close as possible to the actual route of the cable and include enough contingency for vertical drops / rises and termination of the cable tails<br />
<br />
==== [[Cable_Construction|Cable Construction]] ====<br />
<br />
The basic characteristics of the cable's physical construction, which includes:<br />
<br />
:* [[Cable_Conductor_Materials|Conductor material]] - normally copper or aluminium<br />
:* Conductor shape - e.g. circular or shaped<br />
:* Conductor type - e.g. stranded or solid<br />
:* Conductor surface coating - e.g. plain (no coating), tinned, silver or nickel<br />
:* [[Cable_Insulation_Materials|Insulation type]] - e.g. PVC, XLPE, EPR<br />
:* Number of cores - single core or multicore (e.g. 2C, 3C or 4C)<br />
<br />
==== Installation Conditions ====<br />
<br />
How the cable will be installed, which includes:<br />
<br />
:* Above ground or underground<br />
:* Installation / arrangement - e.g. for underground cables, is it directly buried or buried in conduit? for above ground cables, is it installed on cable tray / ladder, against a wall, in air, etc.<br />
:* Ambient or soil temperature of the installation site<br />
:* Cable bunching, i.e. the number of cables that are bunched together<br />
:* Cable spacing, i.e. whether cables are installed touching or spaced<br />
:* Soil thermal resistivity (for underground cables)<br />
:* Depth of laying (for underground cables)<br />
:* For single core three-phase cables, are the cables installed in trefoil or laid flat?<br />
<br />
=== Step 2: Cable Selection Based on Current Rating ===<br />
<br />
Current flowing through a cable generates heat through the resistive losses in the conductors, dielectric losses through the insulation and resistive losses from current flowing through any cable screens / shields and armouring. <br />
<br />
The component parts that make up the cable (e.g. conductors, insulation, bedding, sheath, armour, etc) must be capable of withstanding the temperature rise and heat emanating from the cable. The current carrying capacity of a cable is the maximum current that can flow continuously through a cable without damaging the cable's insulation and other components (e.g. bedding, sheath, etc). It is sometimes also referred to as the continuous current rating or ampacity of a cable. <br />
<br />
Cables with larger conductor cross-sectional areas (i.e. more copper or aluminium) have lower resistive losses and are able to dissipate the heat better than smaller cables. Therefore a 16 <math>mm^{2}</math> cable will have a higher current carrying capacity than a 4 <math>mm^{2}</math> cable. <br />
<br />
==== Base Current Ratings ====<br />
<br />
[[Image:IEC_60364552_TableA5210.jpg|right|thumb|300px|Table 1. Example of base current rating table (Excerpt from IEC 60364-5-52)]]<br />
<br />
International standards and manufacturers of cables will quote base current ratings of different types of cables in tables such as the one shown on the right. Each of these tables pertain to a specific type of cable construction (e.g. copper conductor, PVC insulated, 0.6/1kV voltage grade, etc) and a base set of installation conditions (e.g. ambient temperature, installation method, etc). It is important to note that the current ratings are only valid for the quoted types of cables and base installation conditions.<br />
<br />
In the absence of any guidance, the following [[LV_Cable_Data|reference based current ratings]] may be used.<br />
<br />
==== Installed Current Ratings ====<br />
<br />
When the proposed installation conditions differ from the base conditions, derating (or correction) factors can be applied to the base current ratings to obtain the actual installed current ratings.<br />
<br />
International standards and cable manufacturers will provide derating factors for a range of installation conditions, for example ambient / soil temperature, grouping or bunching of cables, soil thermal resistivity, etc. The installed current rating is calculated by multiplying the base current rating with each of the derating factors, i.e.<br />
<br />
: <math>I_{c} = I_{b} . k_{d} \,</math><br />
<br />
where <math>I_{c}\,</math> is the installed current rating (A)<br />
:: <math>I_{b}\,</math> is the base current rating (A)<br />
:: <math>k_{d} \,</math> are the product of all the derating factors<br />
<br />
For example, suppose a cable had an ambient temperature derating factor of <math>k_{amb} = 0.94</math> and a grouping derating factor of <math>k_{g} = 0.85</math>, then the overall derating factor <math>k_{d} = 0.94 x 0.85 = 0.799</math>. For a cable with a base current rating of 42A, the installed current rating would be <math>I_{c} = 0.799 x 42 = 33.6A</math>.<br />
<br />
In the absence of any guidance, the following [[LV_Cable_Data|reference derating factors]] may be used.<br />
<br />
==== Cable Selection and Coordination with Protective Devices ====<br />
<br />
===== Feeders =====<br />
<br />
When sizing cables for non-motor loads, the upstream protective device ([[Electrical_Fuse|fuse]] or circuit breaker) is typically selected to also protect the cable against damage from [[thermal overload]]. The protective device must therefore be selected to exceed the full load current, but not exceed the cable's installed current rating, i.e. this inequality must be met:<br />
<br />
: <math>I_{l} \leq I_{p} \leq I_{c} \, </math><br />
<br />
Where <math>I_{l} \, </math> is the full load current (A)<br />
:: <math>I_{p} \, </math> is the protective device rating (A)<br />
:: <math>I_{c} \, </math> is the installed cable current rating (A)<br />
<br />
===== Motors =====<br />
<br />
Motors are normally protected by a separate thermal overload (TOL) relay and therefore the upstream protective device (e.g. [[Electrical_Fuse|fuse]] or circuit breaker) is not required to protect the cable against overloads. As a result, cables need only to be sized to cater for the full load current of the motor, i.e.<br />
<br />
: <math>I_{l} \leq I_{c} \, </math><br />
<br />
Where <math>I_{l} \, </math> is the full load current (A)<br />
:: <math>I_{c} \, </math> is the installed cable current rating (A)<br />
<br />
Of course, if there is no separate thermal overload protection on the motor, then the protective device needs to be taken into account as per the case for feeders above.<br />
<br />
=== Step 3: Voltage Drop ===<br />
<br />
A cable's conductor can be seen as an impedance and therefore whenever current flows through a cable, there will be a voltage drop across it, which can be derived by Ohm’s Law (i.e. V = IZ). The voltage drop will depend on two things:<br />
<br />
:* Current flow through the cable – the higher the current flow, the higher the voltage drop<br />
:* Impedance of the conductor – the larger the impedance, the higher the voltage drop<br />
<br />
==== Cable Impedances ====<br />
<br />
The impedance of the cable is a function of the cable size (cross-sectional area) and the length of the cable. Most cable manufacturers will quote a cable’s resistance and reactance in <math>\Omega</math>/km. The following [[LV_Cable_Data|typical cable impedances]] for low voltage AC and DC single core and multicore cables can be used in the absence of any other data.<br />
<br />
==== Calculating Voltage Drop ====<br />
<br />
For AC systems, the method of calculating voltage drops based on load power factor is commonly used. Full load currents are normally used, but if the load has high startup currents (e.g. motors), then voltage drops based on starting current (and power factor if applicable) should also be calculated.<br />
<br />
For a three phase system:<br />
<br />
: <math> V_{3\phi} = \frac{\sqrt{3} I (R_{c} \cos\phi + X_{c} \sin\phi) L}{1000} \, </math><br />
<br />
Where <math> V_{3\phi} \, </math> is the three phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
For a single phase system:<br />
<br />
: <math> V_{1\phi} = \frac{2 I (R_{c} \cos\phi + X_{c} \sin\phi) L}{1000} \, </math><br />
<br />
Where <math> V_{1\phi} \, </math> is the single phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
For a DC system:<br />
<br />
: <math> V_{dc} = \frac{2 I R_{c} L}{1000} \, </math><br />
<br />
Where <math> V_{dc} \, </math> is the dc voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the dc resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
==== Maximum Permissible Voltage Drop ====<br />
<br />
It is customary for standards (or clients) to specify maximum permissible voltage drops, which is the highest voltage drop that is allowed across a cable. Should your cable exceed this voltage drop, then a larger cable size should be selected.<br />
<br />
Maximum voltage drops across a cable are specified because load consumers (e.g. appliances) will have an input voltage tolerance range. This means that if the voltage at the appliance is lower than its rated minimum voltage, then the appliance may not operate correctly. <br />
<br />
In general, most electrical equipment will operate normally at a voltage as low as 80% nominal voltage. For example, if the nominal voltage is 230VAC, then most appliances will run at >184VAC. Cables are typically sized for a more conservative maximum voltage drop, in the range of 5 – 10% at full load.<br />
<br />
==== Calculating Maximum Cable Length due to Voltage Drop ====<br />
<br />
It may be more convenient to calculate the maximum length of a cable for a particular conductor size given a maximum permissible voltage drop (e.g. 5% of nominal voltage at full load) rather than the voltage drop itself. For example, by doing this it is possible to construct tables showing the maximum lengths corresponding to different cable sizes in order to speed up the selection of similar type cables. <br />
<br />
The maximum cable length that will achieve this can be calculated by re-arranging the voltage drop equations and substituting the maximum permissible voltage drop (e.g. 5% of 415V nominal voltage = 20.75V).<br />
For a three phase system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{3\phi}}{\sqrt{3} I (R_{c} \cos\phi + X_{c} \sin\phi)} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{3\phi} \, </math> is the maximum permissible three phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
<br />
For a single phase system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{1\phi}}{2 I (R_{c} \cos\phi + X_{c} \sin\phi)} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{1\phi} \, </math> is the maximum permissible single phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
<br />
For a DC system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{dc}}{2 I R_{c}} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{dc} \, </math> is the maximum permissible dc voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the dc resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
=== Step 4: Short Circuit Temperature Rise ===<br />
<br />
During a short circuit, a high amount of current can flow through a cable for a short time. This surge in current flow causes a temperature rise within the cable. High temperatures can trigger unwanted reactions in the cable insulation, sheath materials and other components, which can prematurely degrade the condition of the cable. As the cross-sectional area of the cable increases, it can dissipate higher fault currents for a given temperature rise. Therefore, cables should be sized to withstand the largest short circuit that it is expected to see. <br />
<br />
==== Minimum Cable Size Due to Short Circuit Temperature Rise ====<br />
<br />
The minimum cable size due to short circuit temperature rise is typically calculated with an equation of the form:<br />
<br />
: <math> A = \frac{\sqrt{i^{2}t}}{k} \, </math><br />
<br />
Where <math>A \, </math> is the minimum cross-sectional area of the cable (<math>mm^{2}</math>)<br />
:: <math>i \, </math> is the prospective short circuit current (A)<br />
:: <math>t \, </math> is the duration of the short circuit (s)<br />
:: <math>k \, </math> is a short circuit temperature rise constant<br />
<br />
The temperature rise constant is calculated based on the material properties of the conductor and the initial and final conductor temperatures (see the [[Short_Circuit_Temperature_Rise|derivation here]]). Different international standards have different treatments of the temperature rise constant, but by way of example, IEC 60364-5-54 calculates it as follows:<br />
<br />
: <math> k = 226 \sqrt{ \ln \left( 1 + \frac{\theta_{f}-\theta{i}}{234.5 + \theta_{i}} \right) } \, </math> (for copper conductors)<br />
<br />
: <math> k = 148 \sqrt{ \ln \left( 1 + \frac{\theta_{f}-\theta{i}}{228 + \theta_{i}} \right) } \, </math> (for aluminium conductors)<br />
<br />
Where <math>\theta_{i} \, </math> is the initial conductor temperature (deg C)<br />
:: <math>\theta_{f} \, </math> is the final conductor temperature (deg C)<br />
<br />
==== Initial and Final Conductor Temperatures ====<br />
<br />
The initial conductor temperature is typically chosen to be the maximum operating temperature of the cable. The final conductor temperature is typically chosen to be the limiting temperature of the insulation. In general, the cable's insulation will determine the maximum operating temperature and limiting temperatures. <br />
<br />
As a rough guide, the following temperatures are common for the different insulation materials:<br />
<br />
{| class="wikitable"<br />
|-<br />
!style="background:#efefef;" width="80px" | Material<br />
!style="background:#efefef;" width="110px" | Max Operating Temperature <sup>o</sup>C<br />
!style="background:#efefef;" width="100px" | Limiting Temperature <sup>o</sup>C<br />
|-<br />
| PVC || 75 || 160<br />
|-<br />
| EPR || 90 || 250<br />
|-<br />
| XLPE || 90 || 250<br />
|}<br />
<br />
==== Short Circuit Energy ====<br />
<br />
The short circuit energy <math>i^{2}t \, </math> is normally chosen as the maximum short circuit that the cable could potentially experience. However for circuits with current limiting devices (such as HRC fuses), then the short circuit energy chosen should be the maximum prospective let-through energy of the protective device, which can be found from manufacturer data. <br />
<br />
=== Step 5: Earth Fault Loop Impedance ===<br />
<br />
Sometimes it is desirable (or necessary) to consider the earth fault loop impedance of a circuit in the sizing of a cable. Suppose a bolted earth fault occurs between an active conductor and earth. During such an earth fault, it is desirable that the upstream protective device acts to interrupt the fault within a maximum disconnection time so as to protect against any inadvertent contact to exposed live parts. <br />
<br />
Ideally the circuit will have earth fault protection, in which case the protection will be fast acting and well within the maximum disconnection time. The maximum disconnection time is chosen so that a dangerous touch voltage does not persist for long enough to cause injury or death. For most circuits, a maximum disconnection time of 5s is sufficient, though for portable equipment and socket outlets, a faster disconnection time is desirable (i.e. <1s and will definitely require earth fault protection). <br />
<br />
However for circuits that do not have earth fault protection, the upstream protective device (i.e. [[Electrical_Fuse|fuse]] or circuit breaker) must trip within the maximum disconnection time. In order for the protective device to trip, the fault current due to a bolted short circuit must exceed the value that will cause the protective device to act within the maximum disconnection time. For example, suppose a circuit is protected by a [[Electrical_Fuse|fuse]] and the maximum disconnection time is 5s, then the fault current must exceed the fuse melting current at 5s (which can be found by cross-referencing the fuse time-current curves). <br />
<br />
By simple application of Ohm's law:<br />
<br />
: <math> I_{A} = \frac{V_{0}}{Z_{s}} \, </math><br />
<br />
Where <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> Z_{s} \, </math> is the impedance of the earth fault loop (<math>\Omega</math>)<br />
<br />
It can be seen from the equation above that the impedance of the earth fault loop must be sufficiently low to ensure that the earth fault current can trip the upstream protection.<br />
<br />
==== The Earth Fault Loop ====<br />
<br />
The earth fault loop can consist of various return paths other than the earth conductor, including the cable armour and the static earthing connection of the facility. However for practical reasons, the earth fault loop in this calculation consists only of the active conductor and the earth conductor.<br />
<br />
The earth fault loop impedance can be found by: <br />
<br />
: <math> Z_{s} = Z_{c} + Z_{e} \, </math><br />
<br />
Where <math> Z_{s} \,</math> is the earth fault loop impedance (<math>\Omega</math>)<br />
:: <math> Z_{c} \,</math> is the impedance of the active conductor (<math>\Omega</math>)<br />
:: <math> Z_{e} \,</math> is the impedance of the earth conductor (<math>\Omega</math>)<br />
<br />
Assuming that the active and earth conductors have identical lengths, the earth fault loop impedance can be calculated as follows:<br />
<br />
: <math> Z_{s} = \frac{L}{1000} \sqrt{(R_{c}+R_{e})^{2} + (X_{c}+X_{e})^{2}} \, </math><br />
<br />
Where <math> L \,</math> is the length of the cable (m)<br />
:: <math> R_{c} \,</math> and <math> R_{e} \,</math> are the ac resistances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
:: <math> X_{c} \,</math> and <math> X_{e} \,</math> are the reactances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
<br />
==== Maximum Cable Length ====<br />
<br />
The maximum earth fault loop impedance can be found by re-arranging the equation above:<br />
<br />
: <math> Z_{s,max} = \frac{V_{0}}{I_{A}} \, </math><br />
<br />
Where <math> Z_{s} \, </math> is the maximum earth fault loop impedance (<math>\Omega</math>)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
<br />
The maximum cable length can therefore be calculated by the following:<br />
<br />
: <math> L_{max} = \frac{1000 V_{0}}{I_{A} \sqrt{(R_{c}+R_{e})^{2} + (X_{c}+X_{e})^{2}}} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum cable length (m)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
:: <math> R_{c} \,</math> and <math> R_{e} \,</math> are the ac resistances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
:: <math> X_{c} \,</math> and <math> X_{e} \,</math> are the reactances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
<br />
Note that the voltage <math>V_{0}</math> at the protective device is not necessarily the nominal phase to earth voltage, but usually a lower value as it can be downstream of the main busbars. This voltage is commonly represented by applying some factor <math>c \,</math> to the nominal voltage. A conservative value of <math>c \,</math> = 0.8 can be used so that:<br />
<br />
: <math>V_{0} = c V_{n} = 0.8 V_{n} \,</math><br />
<br />
Where <math>V_{n}</math> is the nominal phase to earth voltage (V)<br />
<br />
== Worked Example ==<br />
<br />
In this example, we will size a cable for a 415V, 37kW three-phase motor from the MCC to the field.<br />
<br />
=== Step 1: Data Gathering ===<br />
<br />
The following data was collected for the cable to be sized:<br />
:* Cable type: Cu/PVC/GSWB/PVC, 3C+E, 0.6/1kV<br />
:* Operating temperature: 75C<br />
:* Cable installation: above ground on cable ladder bunched together with 3 other cables on a single layer and at 30C ambient temperature<br />
:* Cable run: 90m (including tails)<br />
:* Motor load: 37kW, 415V three phase, full load current = 61A, power factor = 0.85<br />
:* Protection: aM fuse of rating = 80A, max prospective fault <math>I^{2}t</math> = 90 <math>A^{2}s</math> , 5s melt time = 550A<br />
<br />
=== Step 2: Cable Selection Based on Current Rating ===<br />
<br />
Suppose the ambient temperature derating is 0.89 and the grouping derating for 3 bunched cables on a single layer is 0.82. The overall derating factor is 0.89 <math>\times</math> 0.82 = 0.7298. Given that a 16 <math>mm^{2}</math> and 25 <math>mm^{2}</math> have base current ratings of 80A and 101A respectively (based on [[PVC_Current_Ratings_(Copper)|Reference Method E]]), which cable should be selected based on current rating considerations?<br />
<br />
The installed current ratings for 16 <math>mm^{2}</math> and 25 <math>mm^{2}</math> is 0.7298 <math>\times</math> 80A = 58.38A and 0.7298 <math>\times</math> 101A = 73.71A respectively. Given that the full load current of the motor is 61A, then the installed current rating of the 16 <math>mm^{2}</math> cable is lower than the full load current and is not suitable for continuous use with the motor. The 25 <math>mm^{2}</math> cable on the other hand has an installed current rating that exceeds the motor full load current, and is therefore the cable that should be selected.<br />
<br />
=== Step 3: Voltage Drop ===<br />
<br />
Suppose a 25 <math>mm^{2}</math> cable is selected. If the maximum permissible voltage drop is 5%, is the cable suitable for a run length of 90m?<br />
<br />
A 25 <math>mm^{2}</math> cable has an [[AC_Resistance|ac resistance]] of 0.884 <math>\Omega</math>/km and an [[AC_Reactance|ac reactance]] of 0.0895 <math>\Omega</math>/km. The voltage drop across the cable is:<br />
<br />
: <math> V_{d} = \frac{90}{1000} \times \sqrt{3} \times 61 \times \left[ 0.884 \times 0.85 + 0.0895 \times \sin(\cos^{-1}(0.85) \right] = 7.593V \, </math><br />
<br />
A voltage drop of 7.593V is equivalent to <math>\frac{7.593}{415} = 1.83%</math>, which is lower than the maximum permissible voltage dorp of 5%. Therefore the cable is suitable for the motor based on voltage drop considerations.<br />
<br />
=== Step 4: Short Circuit Temperature Rise ===<br />
<br />
The cable is operating normally at 75C and has a prospective fault capacity (<math>I^{2}t</math>) of 90,000 <math>A^{2}s</math>. What is the minimum size of the cable based on short circuit temperature rise?<br />
<br />
PVC has a limiting temperature of 160C. Using the IEC formula, the short circuit temperature rise constant is 111.329. The minimum cable size due to short circuit temperature rise is therefore:<br />
<br />
: <math>A = \frac{\sqrt{90,000}}{111.329} = 2.695 mm^{2} \, </math><br />
<br />
In this example, we also use the fuse for earth fault protection and it needs to trip within 5s, which is at the upper end of the adiabatic period where the short circuit temperature rise equation is still valid. Therefore, it's a good idea to also check that the cable can withstand the short circuit temperature rise for for a 5s fault. The 80A motor fuse has a 5s melting current of 550A. The short circuit temperature rise is thus:<br />
<br />
: <math>A = \frac{\sqrt{550^{2} \times 5}}{111.329} = 11.047 mm^{2} \, </math><br />
<br />
Therefore, our 25 <math>mm^{2}</math> cable is still suitable for this application.<br />
<br />
=== Step 5: Earth Fault Loop Impedance ===<br />
<br />
Suppose there is no special earth fault protection for the motor and a bolted single phase to earth fault occurs at the motor terminals. Suppose that the earth conductor for our 25 <math>mm^{2}</math> cable is 10 <math>mm^{2}</math>. If the maximum disconnection time is 5s, is our 90m long cable suitable based on earth fault loop impedance?<br />
<br />
The 80A motor fuse has a 5s melting current of 550A. The ac resistances of the active and earth conductors are 0.884 <math>\Omega</math>/km and 2.33 <math>\Omega</math>/km) respectively. The reactances of the active and earth conductors are 0.0895 <math>\Omega</math>/km and 0.0967 <math>\Omega</math>/km) respectively.<br />
<br />
The maximum length of the cable allowed is calculated as:<br />
<br />
: <math> L_{max} = \frac{(1000)(0.8)(240)}{550 \sqrt{(0.884+2.33)^{2} + (0.0895+0.0967)^{2}}} = 108.43m \, </math><br />
<br />
The cable run is 90m and the maximum length allowed is 108m, therefore our cable is suitable based on earth fault loop impedance. In fact, our 25 <math>mm^{2}</math> cable has passed all the tests and is the size that should be selected.<br />
<br />
== Waterfall Charts ==<br />
<br />
[[Image:Cable_waterfall.jpg|right|thumb|400px|Table 2. Example of a cable waterfall chart]]<br />
<br />
Sometimes it is convenient to group together similar types of cables (for example, 415V PVC motor cables installed on cable ladder) so that instead of having to go through the laborious exercise of sizing each cable separately, one can select a cable from a pre-calculated chart. <br />
<br />
These charts are often called "waterfall charts" and typically show a list of load ratings and the maximum of length of cable permissible for each cable size. Where a particular cable size fails to meet the requirements for current carrying capacity or short circuit temperature rise, it is blacked out on the chart (i.e. meaning that you can't choose it).<br />
<br />
Preparing a waterfall chart is common practice when having to size many like cables and substantially cuts down the time required for cable selection.<br />
<br />
== International Standards ==<br />
<br />
=== IEC ===<br />
<br />
[https://webstore.iec.ch/publication/1878 IEC 60364-5-52 (2009)] "Electrical installations in buildings - Part 5-52: Selection and erection of electrical equipment - Wiring systems" is the IEC standard governing cable sizing.<br />
<br />
=== NEC ===<br />
<br />
[https://www.nfpa.org/nec/ NFPA 70 (2020)] "National Electricity Code" is the equivalent standard for IEC 60364 in North America and includes a section covering cable sizing in Article 300.<br />
<br />
=== BS ===<br />
<br />
[https://electrical.theiet.org/bs-7671/ BS 7671 (18th edition)] "Requirements for Electrical Installations - IEE Wiring Regulations" is the equivalent standard for IEC 60364 in the United Kingdom.<br />
<br />
=== AS/NZS ===<br />
<br />
[https://infostore.saiglobal.com/en-us/standards/as-nzs-3008-1-1-2017-99685_saig_as_as_209556/ AS/NZS 3008.1 (2017)] "Electrical installations - Selection of cables - Cables for alternating voltages up to and including 0.6/1 kV" is the standard governing low voltage cable sizing in Australia and New Zealand. AS/NZS 3008.1.1 is for Australian conditions and AS/NZS 3008.1.2 is for New Zealand conditions.<br />
<br />
== What next? ==<br />
<br />
Having sized the power / load-bearing cables, the cable schedule can now be developed and then the cable material take-offs (MTO).<br />
<br />
[[Category:Calculations]]</div>Juleshttp://openelectrical.org/index.php?title=Cable_Sizing_Calculation&diff=182Cable Sizing Calculation2020-12-25T03:03:02Z<p>Jules: /* Initial and Final Conductor Temperatures */</p>
<hr />
<div>== Introduction ==<br />
<br />
[[Image:Cables_header.JPG|right]]<br />
<br />
This article examines the sizing of electrical cables (i.e. cross-sectional area) and its implementation in various international standards. Cable sizing methods do differ across international standards (e.g. IEC, NEC, BS, etc) and some standards emphasise certain things over others. However the general principles underlying any cable sizing calculation do not change. In this article, a general methodology for sizing cables is first presented and then the specific international standards are introduced.<br />
<br />
=== Why do the calculation? ===<br />
<br />
The proper sizing of an electrical (load bearing) cable is important to ensure that the cable can:<br />
:* Operate continuously under full load without being damaged<br />
:* Withstand the worst short circuits currents flowing through the cable<br />
:* Provide the load with a suitable voltage (and avoid excessive voltage drops)<br />
:* (optional) Ensure operation of protective devices during an earth fault<br />
<br />
=== When to do the calculation? ===<br />
<br />
This calculation can be done individually for each power cable that needs to be sized, or alternatively, it can be used to produce cable sizing waterfall charts for groups of cables with similar characteristics (e.g. cables installed on ladder feeding induction motors).<br />
<br />
== General Methodology ==<br />
<br />
All cable sizing methods more or less follow the same basic six step process:<br />
<br />
: 1) Gathering data about the cable, its installation conditions, the load that it will carry, etc<br />
: 2) Determine the minimum cable size based on continuous current carrying capacity<br />
: 3) Determine the minimum cable size based on voltage drop considerations<br />
: 4) Determine the minimum cable size based on short circuit temperature rise<br />
: 5) Determine the minimum cable size based on earth fault loop impedance<br />
: 6) Select the cable based on the highest of the sizes calculated in step 2, 3, 4 and 5<br />
<br />
=== Step 1: Data Gathering ===<br />
<br />
The first step is to collate the relevant information that is required to perform the sizing calculation. Typically, you will need to obtain the following data:<br />
<br />
==== Load Details ====<br />
<br />
The characteristics of the load that the cable will supply, which includes:<br />
<br />
:* Load type: motor or feeder<br />
:* Three phase, single phase or DC<br />
:* System / source voltage<br />
:* Full load current (A) - or calculate this if the load is defined in terms of power (kW)<br />
:* Full load power factor (pu)<br />
:* Locked rotor or load starting current (A)<br />
:* Starting power factor (pu)<br />
:* Distance / length of cable run from source to load - this length should be as close as possible to the actual route of the cable and include enough contingency for vertical drops / rises and termination of the cable tails<br />
<br />
==== [[Cable_Construction|Cable Construction]] ====<br />
<br />
The basic characteristics of the cable's physical construction, which includes:<br />
<br />
:* [[Cable_Conductor_Materials|Conductor material]] - normally copper or aluminium<br />
:* Conductor shape - e.g. circular or shaped<br />
:* Conductor type - e.g. stranded or solid<br />
:* Conductor surface coating - e.g. plain (no coating), tinned, silver or nickel<br />
:* [[Cable_Insulation_Materials|Insulation type]] - e.g. PVC, XLPE, EPR<br />
:* Number of cores - single core or multicore (e.g. 2C, 3C or 4C)<br />
<br />
==== Installation Conditions ====<br />
<br />
How the cable will be installed, which includes:<br />
<br />
:* Above ground or underground<br />
:* Installation / arrangement - e.g. for underground cables, is it directly buried or buried in conduit? for above ground cables, is it installed on cable tray / ladder, against a wall, in air, etc.<br />
:* Ambient or soil temperature of the installation site<br />
:* Cable bunching, i.e. the number of cables that are bunched together<br />
:* Cable spacing, i.e. whether cables are installed touching or spaced<br />
:* Soil thermal resistivity (for underground cables)<br />
:* Depth of laying (for underground cables)<br />
:* For single core three-phase cables, are the cables installed in trefoil or laid flat?<br />
<br />
=== Step 2: Cable Selection Based on Current Rating ===<br />
<br />
Current flowing through a cable generates heat through the resistive losses in the conductors, dielectric losses through the insulation and resistive losses from current flowing through any cable screens / shields and armouring. <br />
<br />
The component parts that make up the cable (e.g. conductors, insulation, bedding, sheath, armour, etc) must be capable of withstanding the temperature rise and heat emanating from the cable. The current carrying capacity of a cable is the maximum current that can flow continuously through a cable without damaging the cable's insulation and other components (e.g. bedding, sheath, etc). It is sometimes also referred to as the continuous current rating or ampacity of a cable. <br />
<br />
Cables with larger conductor cross-sectional areas (i.e. more copper or aluminium) have lower resistive losses and are able to dissipate the heat better than smaller cables. Therefore a 16 <math>mm^{2}</math> cable will have a higher current carrying capacity than a 4 <math>mm^{2}</math> cable. <br />
<br />
==== Base Current Ratings ====<br />
<br />
[[Image:IEC_60364552_TableA5210.jpg|right|thumb|300px|Table 1. Example of base current rating table (Excerpt from IEC 60364-5-52)]]<br />
<br />
International standards and manufacturers of cables will quote base current ratings of different types of cables in tables such as the one shown on the right. Each of these tables pertain to a specific type of cable construction (e.g. copper conductor, PVC insulated, 0.6/1kV voltage grade, etc) and a base set of installation conditions (e.g. ambient temperature, installation method, etc). It is important to note that the current ratings are only valid for the quoted types of cables and base installation conditions.<br />
<br />
In the absence of any guidance, the following [[LV_Cable_Data|reference based current ratings]] may be used.<br />
<br />
==== Installed Current Ratings ====<br />
<br />
When the proposed installation conditions differ from the base conditions, derating (or correction) factors can be applied to the base current ratings to obtain the actual installed current ratings.<br />
<br />
International standards and cable manufacturers will provide derating factors for a range of installation conditions, for example ambient / soil temperature, grouping or bunching of cables, soil thermal resistivity, etc. The installed current rating is calculated by multiplying the base current rating with each of the derating factors, i.e.<br />
<br />
: <math>I_{c} = I_{b} . k_{d} \,</math><br />
<br />
where <math>I_{c}\,</math> is the installed current rating (A)<br />
:: <math>I_{b}\,</math> is the base current rating (A)<br />
:: <math>k_{d} \,</math> are the product of all the derating factors<br />
<br />
For example, suppose a cable had an ambient temperature derating factor of <math>k_{amb} = 0.94</math> and a grouping derating factor of <math>k_{g} = 0.85</math>, then the overall derating factor <math>k_{d} = 0.94 x 0.85 = 0.799</math>. For a cable with a base current rating of 42A, the installed current rating would be <math>I_{c} = 0.799 x 42 = 33.6A</math>.<br />
<br />
In the absence of any guidance, the following [[LV_Cable_Data|reference derating factors]] may be used.<br />
<br />
==== Cable Selection and Coordination with Protective Devices ====<br />
<br />
===== Feeders =====<br />
<br />
When sizing cables for non-motor loads, the upstream protective device ([[Electrical_Fuse|fuse]] or circuit breaker) is typically selected to also protect the cable against damage from [[thermal overload]]. The protective device must therefore be selected to exceed the full load current, but not exceed the cable's installed current rating, i.e. this inequality must be met:<br />
<br />
: <math>I_{l} \leq I_{p} \leq I_{c} \, </math><br />
<br />
Where <math>I_{l} \, </math> is the full load current (A)<br />
:: <math>I_{p} \, </math> is the protective device rating (A)<br />
:: <math>I_{c} \, </math> is the installed cable current rating (A)<br />
<br />
===== Motors =====<br />
<br />
Motors are normally protected by a separate thermal overload (TOL) relay and therefore the upstream protective device (e.g. [[Electrical_Fuse|fuse]] or circuit breaker) is not required to protect the cable against overloads. As a result, cables need only to be sized to cater for the full load current of the motor, i.e.<br />
<br />
: <math>I_{l} \leq I_{c} \, </math><br />
<br />
Where <math>I_{l} \, </math> is the full load current (A)<br />
:: <math>I_{c} \, </math> is the installed cable current rating (A)<br />
<br />
Of course, if there is no separate thermal overload protection on the motor, then the protective device needs to be taken into account as per the case for feeders above.<br />
<br />
=== Step 3: Voltage Drop ===<br />
<br />
A cable's conductor can be seen as an impedance and therefore whenever current flows through a cable, there will be a voltage drop across it, which can be derived by Ohm’s Law (i.e. V = IZ). The voltage drop will depend on two things:<br />
<br />
:* Current flow through the cable – the higher the current flow, the higher the voltage drop<br />
:* Impedance of the conductor – the larger the impedance, the higher the voltage drop<br />
<br />
==== Cable Impedances ====<br />
<br />
The impedance of the cable is a function of the cable size (cross-sectional area) and the length of the cable. Most cable manufacturers will quote a cable’s resistance and reactance in <math>\Omega</math>/km. The following [[LV_Cable_Data|typical cable impedances]] for low voltage AC and DC single core and multicore cables can be used in the absence of any other data.<br />
<br />
==== Calculating Voltage Drop ====<br />
<br />
For AC systems, the method of calculating voltage drops based on load power factor is commonly used. Full load currents are normally used, but if the load has high startup currents (e.g. motors), then voltage drops based on starting current (and power factor if applicable) should also be calculated.<br />
<br />
For a three phase system:<br />
<br />
: <math> V_{3\phi} = \frac{\sqrt{3} I (R_{c} \cos\phi + X_{c} \sin\phi) L}{1000} \, </math><br />
<br />
Where <math> V_{3\phi} \, </math> is the three phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
For a single phase system:<br />
<br />
: <math> V_{1\phi} = \frac{2 I (R_{c} \cos\phi + X_{c} \sin\phi) L}{1000} \, </math><br />
<br />
Where <math> V_{1\phi} \, </math> is the single phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
For a DC system:<br />
<br />
: <math> V_{dc} = \frac{2 I R_{c} L}{1000} \, </math><br />
<br />
Where <math> V_{dc} \, </math> is the dc voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the dc resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
==== Maximum Permissible Voltage Drop ====<br />
<br />
It is customary for standards (or clients) to specify maximum permissible voltage drops, which is the highest voltage drop that is allowed across a cable. Should your cable exceed this voltage drop, then a larger cable size should be selected.<br />
<br />
Maximum voltage drops across a cable are specified because load consumers (e.g. appliances) will have an input voltage tolerance range. This means that if the voltage at the appliance is lower than its rated minimum voltage, then the appliance may not operate correctly. <br />
<br />
In general, most electrical equipment will operate normally at a voltage as low as 80% nominal voltage. For example, if the nominal voltage is 230VAC, then most appliances will run at >184VAC. Cables are typically sized for a more conservative maximum voltage drop, in the range of 5 – 10% at full load.<br />
<br />
==== Calculating Maximum Cable Length due to Voltage Drop ====<br />
<br />
It may be more convenient to calculate the maximum length of a cable for a particular conductor size given a maximum permissible voltage drop (e.g. 5% of nominal voltage at full load) rather than the voltage drop itself. For example, by doing this it is possible to construct tables showing the maximum lengths corresponding to different cable sizes in order to speed up the selection of similar type cables. <br />
<br />
The maximum cable length that will achieve this can be calculated by re-arranging the voltage drop equations and substituting the maximum permissible voltage drop (e.g. 5% of 415V nominal voltage = 20.75V).<br />
For a three phase system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{3\phi}}{\sqrt{3} I (R_{c} \cos\phi + X_{c} \sin\phi)} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{3\phi} \, </math> is the maximum permissible three phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
<br />
For a single phase system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{1\phi}}{2 I (R_{c} \cos\phi + X_{c} \sin\phi)} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{1\phi} \, </math> is the maximum permissible single phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
<br />
For a DC system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{dc}}{2 I R_{c}} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{dc} \, </math> is the maximum permissible dc voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the dc resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
=== Step 4: Short Circuit Temperature Rise ===<br />
<br />
During a short circuit, a high amount of current can flow through a cable for a short time. This surge in current flow causes a temperature rise within the cable. High temperatures can trigger unwanted reactions in the cable insulation, sheath materials and other components, which can prematurely degrade the condition of the cable. As the cross-sectional area of the cable increases, it can dissipate higher fault currents for a given temperature rise. Therefore, cables should be sized to withstand the largest short circuit that it is expected to see. <br />
<br />
==== Minimum Cable Size Due to Short Circuit Temperature Rise ====<br />
<br />
The minimum cable size due to short circuit temperature rise is typically calculated with an equation of the form:<br />
<br />
: <math> A = \frac{\sqrt{i^{2}t}}{k} \, </math><br />
<br />
Where <math>A \, </math> is the minimum cross-sectional area of the cable (<math>mm^{2}</math>)<br />
:: <math>i \, </math> is the prospective short circuit current (A)<br />
:: <math>t \, </math> is the duration of the short circuit (s)<br />
:: <math>k \, </math> is a short circuit temperature rise constant<br />
<br />
The temperature rise constant is calculated based on the material properties of the conductor and the initial and final conductor temperatures (see the [[Short_Circuit_Temperature_Rise|derivation here]]). Different international standards have different treatments of the temperature rise constant, but by way of example, IEC 60364-5-54 calculates it as follows:<br />
<br />
: <math> k = 226 \sqrt{ \ln \left( 1 + \frac{\theta_{f}-\theta{i}}{234.5 + \theta_{i}} \right) } \, </math> (for copper conductors)<br />
<br />
: <math> k = 148 \sqrt{ \ln \left( 1 + \frac{\theta_{f}-\theta{i}}{228 + \theta_{i}} \right) } \, </math> (for aluminium conductors)<br />
<br />
Where <math>\theta_{i} \, </math> is the initial conductor temperature (deg C)<br />
:: <math>\theta_{f} \, </math> is the final conductor temperature (deg C)<br />
<br />
==== Initial and Final Conductor Temperatures ====<br />
<br />
The initial conductor temperature is typically chosen to be the maximum operating temperature of the cable. The final conductor temperature is typically chosen to be the limiting temperature of the insulation. In general, the cable's insulation will determine the maximum operating temperature and limiting temperatures. <br />
<br />
As a rough guide, the following temperatures are common for the different insulation materials:<br />
<br />
{| class="wikitable"<br />
|-<br />
!style="background:#efefef;" width="80px" | Material<br />
!style="background:#efefef;" width="110px" | Max Operating Temperature <sup>o</sup>C<br />
!style="background:#efefef;" width="100px" | Limiting Temperature <sup>o</sup>C<br />
|-<br />
| PVC || 75 || 160<br />
|-<br />
| EPR || 90 || 250<br />
|-<br />
| XLPE || 90 || 250<br />
|}<br />
<br />
==== Short Circuit Energy ====<br />
<br />
The short circuit energy <math>i^{2}t \, </math> is normally chosen as the maximum short circuit that the cable could potentially experience. However for circuits with current limiting devices (such as HRC fuses), then the short circuit energy chosen should be the maximum prospective let-through energy of the protective device, which can be found from manufacturer data. <br />
<br />
=== Step 5: Earth Fault Loop Impedance ===<br />
<br />
Sometimes it is desirable (or necessary) to consider the earth fault loop impedance of a circuit in the sizing of a cable. Suppose a bolted earth fault occurs between an active conductor and earth. During such an earth fault, it is desirable that the upstream protective device acts to interrupt the fault within a maximum disconnection time so as to protect against any inadvertent contact to exposed live parts. <br />
<br />
Ideally the circuit will have earth fault protection, in which case the protection will be fast acting and well within the maximum disconnection time. The maximum disconnection time is chosen so that a dangerous touch voltage does not persist for long enough to cause injury or death. For most circuits, a maximum disconnection time of 5s is sufficient, though for portable equipment and socket outlets, a faster disconnection time is desirable (i.e. <1s and will definitely require earth fault protection). <br />
<br />
However for circuits that do not have earth fault protection, the upstream protective device (i.e. [[Electrical_Fuse|fuse]] or circuit breaker) must trip within the maximum disconnection time. In order for the protective device to trip, the fault current due to a bolted short circuit must exceed the value that will cause the protective device to act within the maximum disconnection time. For example, suppose a circuit is protected by a [[Electrical_Fuse|fuse]] and the maximum disconnection time is 5s, then the fault current must exceed the fuse melting current at 5s (which can be found by cross-referencing the fuse time-current curves). <br />
<br />
By simple application of Ohm's law:<br />
<br />
: <math> I_{A} = \frac{V_{0}}{Z_{s}} \, </math><br />
<br />
Where <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> Z_{s} \, </math> is the impedance of the earth fault loop (<math>\Omega</math>)<br />
<br />
It can be seen from the equation above that the impedance of the earth fault loop must be sufficiently low to ensure that the earth fault current can trip the upstream protection.<br />
<br />
==== The Earth Fault Loop ====<br />
<br />
The earth fault loop can consist of various return paths other than the earth conductor, including the cable armour and the static earthing connection of the facility. However for practical reasons, the earth fault loop in this calculation consists only of the active conductor and the earth conductor.<br />
<br />
The earth fault loop impedance can be found by: <br />
<br />
: <math> Z_{s} = Z_{c} + Z_{e} \, </math><br />
<br />
Where <math> Z_{s} \,</math> is the earth fault loop impedance (<math>\Omega</math>)<br />
:: <math> Z_{c} \,</math> is the impedance of the active conductor (<math>\Omega</math>)<br />
:: <math> Z_{e} \,</math> is the impedance of the earth conductor (<math>\Omega</math>)<br />
<br />
Assuming that the active and earth conductors have identical lengths, the earth fault loop impedance can be calculated as follows:<br />
<br />
: <math> Z_{s} = \frac{L}{1000} \sqrt{(R_{c}+R_{e})^{2} + (X_{c}+X_{e})^{2}} \, </math><br />
<br />
Where <math> L \,</math> is the length of the cable (m)<br />
:: <math> R_{c} \,</math> and <math> R_{e} \,</math> are the ac resistances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
:: <math> X_{c} \,</math> and <math> X_{e} \,</math> are the reactances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
<br />
==== Maximum Cable Length ====<br />
<br />
The maximum earth fault loop impedance can be found by re-arranging the equation above:<br />
<br />
: <math> Z_{s,max} = \frac{V_{0}}{I_{A}} \, </math><br />
<br />
Where <math> Z_{s} \, </math> is the maximum earth fault loop impedance (<math>\Omega</math>)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
<br />
The maximum cable length can therefore be calculated by the following:<br />
<br />
: <math> L_{max} = \frac{1000 V_{0}}{I_{A} \sqrt{(R_{c}+R_{e})^{2} + (X_{c}+X_{e})^{2}}} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum cable length (m)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
:: <math> R_{c} \,</math> and <math> R_{e} \,</math> are the ac resistances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
:: <math> X_{c} \,</math> and <math> X_{e} \,</math> are the reactances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
<br />
Note that the voltage <math>V_{0}</math> at the protective device is not necessarily the nominal phase to earth voltage, but usually a lower value as it can be downstream of the main busbars. This voltage is commonly represented by applying some factor <math>c \,</math> to the nominal voltage. A conservative value of <math>c \,</math> = 0.8 can be used so that:<br />
<br />
: <math>V_{0} = c V_{n} = 0.8 V_{n} \,</math><br />
<br />
Where <math>V_{n}</math> is the nominal phase to earth voltage (V)<br />
<br />
== Worked Example ==<br />
<br />
In this example, we will size a cable for a 415V, 37kW three-phase motor from the MCC to the field.<br />
<br />
=== Step 1: Data Gathering ===<br />
<br />
The following data was collected for the cable to be sized:<br />
:* Cable type: Cu/PVC/GSWB/PVC, 3C+E, 0.6/1kV<br />
:* Operating temperature: 75C<br />
:* Cable installation: above ground on cable ladder bunched together with 3 other cables on a single layer and at 30C ambient temperature<br />
:* Cable run: 90m (including tails)<br />
:* Motor load: 37kW, 415V three phase, full load current = 61A, power factor = 0.85<br />
:* Protection: aM fuse of rating = 80A, max prospective fault <math>I^{2}t</math> = 90 <math>A^{2}s</math> , 5s melt time = 550A<br />
<br />
=== Step 2: Cable Selection Based on Current Rating ===<br />
<br />
Suppose the ambient temperature derating is 0.89 and the grouping derating for 3 bunched cables on a single layer is 0.82. The overall derating factor is 0.89 <math>\times</math> 0.82 = 0.7298. Given that a 16 <math>mm^{2}</math> and 25 <math>mm^{2}</math> have base current ratings of 80A and 101A respectively (based on [[PVC_Current_Ratings_(Copper)|Reference Method E]]), which cable should be selected based on current rating considerations?<br />
<br />
The installed current ratings for 16 <math>mm^{2}</math> and 25 <math>mm^{2}</math> is 0.7298 <math>\times</math> 80A = 58.38A and 0.7298 <math>\times</math> 101A = 73.71A respectively. Given that the full load current of the motor is 61A, then the installed current rating of the 16 <math>mm^{2}</math> cable is lower than the full load current and is not suitable for continuous use with the motor. The 25 <math>mm^{2}</math> cable on the other hand has an installed current rating that exceeds the motor full load current, and is therefore the cable that should be selected.<br />
<br />
=== Step 3: Voltage Drop ===<br />
<br />
Suppose a 25 <math>mm^{2}</math> cable is selected. If the maximum permissible voltage drop is 5%, is the cable suitable for a run length of 90m?<br />
<br />
A 25 <math>mm^{2}</math> cable has an [[AC_Resistance|ac resistance]] of 0.884 <math>\Omega</math>/km and an [[AC_Reactance|ac reactance]] of 0.0895 <math>\Omega</math>/km. The voltage drop across the cable is:<br />
<br />
: <math> V_{d} = \frac{90}{1000} \times \sqrt{3} \times 61 \times \left[ 0.884 \times 0.85 + 0.0895 \times \sin(\cos^{-1}(0.85) \right] = 7.593V \, </math><br />
<br />
A voltage drop of 7.593V is equivalent to <math>\frac{7.593}{415} = 1.83%</math>, which is lower than the maximum permissible voltage dorp of 5%. Therefore the cable is suitable for the motor based on voltage drop considerations.<br />
<br />
=== Step 4: Short Circuit Temperature Rise ===<br />
<br />
The cable is operating normally at 75C and has a prospective fault capacity (<math>I^{2}t</math>) of 90,000 <math>A^{2}s</math>. What is the minimum size of the cable based on short circuit temperature rise?<br />
<br />
PVC has a limiting temperature of 160C. Using the IEC formula, the short circuit temperature rise constant is 111.329. The minimum cable size due to short circuit temperature rise is therefore:<br />
<br />
: <math>A = \frac{\sqrt{90,000}}{111.329} = 2.695 mm^{2} \, </math><br />
<br />
In this example, we also use the fuse for earth fault protection and it needs to trip within 5s, which is at the upper end of the adiabatic period where the short circuit temperature rise equation is still valid. Therefore, it's a good idea to also check that the cable can withstand the short circuit temperature rise for for a 5s fault. The 80A motor fuse has a 5s melting current of 550A. The short circuit temperature rise is thus:<br />
<br />
: <math>A = \frac{\sqrt{550^{2} \times 5}}{111.329} = 11.047 mm^{2} \, </math><br />
<br />
Therefore, our 25 <math>mm^{2}</math> cable is still suitable for this application.<br />
<br />
=== Step 5: Earth Fault Loop Impedance ===<br />
<br />
Suppose there is no special earth fault protection for the motor and a bolted single phase to earth fault occurs at the motor terminals. Suppose that the earth conductor for our 25 <math>mm^{2}</math> cable is 10 <math>mm^{2}</math>. If the maximum disconnection time is 5s, is our 90m long cable suitable based on earth fault loop impedance?<br />
<br />
The 80A motor fuse has a 5s melting current of 550A. The ac resistances of the active and earth conductors are 0.884 <math>\Omega</math>/km and 2.33 <math>\Omega</math>/km) respectively. The reactances of the active and earth conductors are 0.0895 <math>\Omega</math>/km and 0.0967 <math>\Omega</math>/km) respectively.<br />
<br />
The maximum length of the cable allowed is calculated as:<br />
<br />
: <math> L_{max} = \frac{(1000)(0.8)(240)}{550 \sqrt{(0.884+2.33)^{2} + (0.0895+0.0967)^{2}}} = 108.43m \, </math><br />
<br />
The cable run is 90m and the maximum length allowed is 108m, therefore our cable is suitable based on earth fault loop impedance. In fact, our 25 <math>mm^{2}</math> cable has passed all the tests and is the size that should be selected.<br />
<br />
== Waterfall Charts ==<br />
<br />
[[Image:Cable_waterfall.jpg|right|thumb|400px|Table 2. Example of a cable waterfall chart]]<br />
<br />
Sometimes it is convenient to group together similar types of cables (for example, 415V PVC motor cables installed on cable ladder) so that instead of having to go through the laborious exercise of sizing each cable separately, one can select a cable from a pre-calculated chart. <br />
<br />
These charts are often called "waterfall charts" and typically show a list of load ratings and the maximum of length of cable permissible for each cable size. Where a particular cable size fails to meet the requirements for current carrying capacity or short circuit temperature rise, it is blacked out on the chart (i.e. meaning that you can't choose it).<br />
<br />
Preparing a waterfall chart is common practice when having to size many like cables and substantially cuts down the time required for cable selection.<br />
<br />
== International Standards ==<br />
<br />
=== IEC ===<br />
<br />
[https://webstore.iec.ch/publication/1878 IEC 60364-5-52 (2009)] "Electrical installations in buildings - Part 5-52: Selection and erection of electrical equipment - Wiring systems" is the IEC standard governing cable sizing.<br />
<br />
=== NEC ===<br />
<br />
[https://www.nfpa.org/nec/ NFPA 70 (2020)] "National Electricity Code" is the equivalent standard for IEC 60364 in North America and includes a section covering cable sizing in Article 300.<br />
<br />
=== BS ===<br />
<br />
[https://electrical.theiet.org/bs-7671/ BS 7671 (18th edition)] "Requirements for Electrical Installations - IEE Wiring Regulations" is the equivalent standard for IEC 60364 in the United Kingdom.<br />
<br />
=== AS/NZS ===<br />
<br />
[https://infostore.saiglobal.com/en-us/standards/as-nzs-3008-1-1-2017-99685_saig_as_as_209556/ AS/NZS 3008.1 (2017)] "Electrical installations - Selection of cables - Cables for alternating voltages up to and including 0.6/1 kV" is the standard governing low voltage cable sizing in Australia and New Zealand. AS/NZS 3008.1.1 is for Australian conditions and AS/NZS 3008.1.2 is for New Zealand conditions.<br />
<br />
== What next? ==<br />
<br />
Having sized the power / load-bearing cables, the cable schedule can now be developed and then the cable material take-offs (MTO).<br />
<br />
[[Category:Calculations]</div>Juleshttp://openelectrical.org/index.php?title=File:IEC_60364552_TableA5210.jpg&diff=181File:IEC 60364552 TableA5210.jpg2020-12-25T03:01:39Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=File:Cable_waterfall.jpg&diff=180File:Cable waterfall.jpg2020-12-25T03:01:24Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=File:Cables_header.JPG&diff=179File:Cables header.JPG2020-12-25T03:01:07Z<p>Jules: </p>
<hr />
<div></div>Juleshttp://openelectrical.org/index.php?title=Cable_Sizing_Calculation&diff=178Cable Sizing Calculation2020-12-25T02:59:52Z<p>Jules: Created page with "== Introduction == right This article examines the sizing of electrical cables (i.e. cross-sectional area) and its implementation in various inte..."</p>
<hr />
<div>== Introduction ==<br />
<br />
[[Image:Cables_header.JPG|right]]<br />
<br />
This article examines the sizing of electrical cables (i.e. cross-sectional area) and its implementation in various international standards. Cable sizing methods do differ across international standards (e.g. IEC, NEC, BS, etc) and some standards emphasise certain things over others. However the general principles underlying any cable sizing calculation do not change. In this article, a general methodology for sizing cables is first presented and then the specific international standards are introduced.<br />
<br />
=== Why do the calculation? ===<br />
<br />
The proper sizing of an electrical (load bearing) cable is important to ensure that the cable can:<br />
:* Operate continuously under full load without being damaged<br />
:* Withstand the worst short circuits currents flowing through the cable<br />
:* Provide the load with a suitable voltage (and avoid excessive voltage drops)<br />
:* (optional) Ensure operation of protective devices during an earth fault<br />
<br />
=== When to do the calculation? ===<br />
<br />
This calculation can be done individually for each power cable that needs to be sized, or alternatively, it can be used to produce cable sizing waterfall charts for groups of cables with similar characteristics (e.g. cables installed on ladder feeding induction motors).<br />
<br />
== General Methodology ==<br />
<br />
All cable sizing methods more or less follow the same basic six step process:<br />
<br />
: 1) Gathering data about the cable, its installation conditions, the load that it will carry, etc<br />
: 2) Determine the minimum cable size based on continuous current carrying capacity<br />
: 3) Determine the minimum cable size based on voltage drop considerations<br />
: 4) Determine the minimum cable size based on short circuit temperature rise<br />
: 5) Determine the minimum cable size based on earth fault loop impedance<br />
: 6) Select the cable based on the highest of the sizes calculated in step 2, 3, 4 and 5<br />
<br />
=== Step 1: Data Gathering ===<br />
<br />
The first step is to collate the relevant information that is required to perform the sizing calculation. Typically, you will need to obtain the following data:<br />
<br />
==== Load Details ====<br />
<br />
The characteristics of the load that the cable will supply, which includes:<br />
<br />
:* Load type: motor or feeder<br />
:* Three phase, single phase or DC<br />
:* System / source voltage<br />
:* Full load current (A) - or calculate this if the load is defined in terms of power (kW)<br />
:* Full load power factor (pu)<br />
:* Locked rotor or load starting current (A)<br />
:* Starting power factor (pu)<br />
:* Distance / length of cable run from source to load - this length should be as close as possible to the actual route of the cable and include enough contingency for vertical drops / rises and termination of the cable tails<br />
<br />
==== [[Cable_Construction|Cable Construction]] ====<br />
<br />
The basic characteristics of the cable's physical construction, which includes:<br />
<br />
:* [[Cable_Conductor_Materials|Conductor material]] - normally copper or aluminium<br />
:* Conductor shape - e.g. circular or shaped<br />
:* Conductor type - e.g. stranded or solid<br />
:* Conductor surface coating - e.g. plain (no coating), tinned, silver or nickel<br />
:* [[Cable_Insulation_Materials|Insulation type]] - e.g. PVC, XLPE, EPR<br />
:* Number of cores - single core or multicore (e.g. 2C, 3C or 4C)<br />
<br />
==== Installation Conditions ====<br />
<br />
How the cable will be installed, which includes:<br />
<br />
:* Above ground or underground<br />
:* Installation / arrangement - e.g. for underground cables, is it directly buried or buried in conduit? for above ground cables, is it installed on cable tray / ladder, against a wall, in air, etc.<br />
:* Ambient or soil temperature of the installation site<br />
:* Cable bunching, i.e. the number of cables that are bunched together<br />
:* Cable spacing, i.e. whether cables are installed touching or spaced<br />
:* Soil thermal resistivity (for underground cables)<br />
:* Depth of laying (for underground cables)<br />
:* For single core three-phase cables, are the cables installed in trefoil or laid flat?<br />
<br />
=== Step 2: Cable Selection Based on Current Rating ===<br />
<br />
Current flowing through a cable generates heat through the resistive losses in the conductors, dielectric losses through the insulation and resistive losses from current flowing through any cable screens / shields and armouring. <br />
<br />
The component parts that make up the cable (e.g. conductors, insulation, bedding, sheath, armour, etc) must be capable of withstanding the temperature rise and heat emanating from the cable. The current carrying capacity of a cable is the maximum current that can flow continuously through a cable without damaging the cable's insulation and other components (e.g. bedding, sheath, etc). It is sometimes also referred to as the continuous current rating or ampacity of a cable. <br />
<br />
Cables with larger conductor cross-sectional areas (i.e. more copper or aluminium) have lower resistive losses and are able to dissipate the heat better than smaller cables. Therefore a 16 <math>mm^{2}</math> cable will have a higher current carrying capacity than a 4 <math>mm^{2}</math> cable. <br />
<br />
==== Base Current Ratings ====<br />
<br />
[[Image:IEC_60364552_TableA5210.jpg|right|thumb|300px|Table 1. Example of base current rating table (Excerpt from IEC 60364-5-52)]]<br />
<br />
International standards and manufacturers of cables will quote base current ratings of different types of cables in tables such as the one shown on the right. Each of these tables pertain to a specific type of cable construction (e.g. copper conductor, PVC insulated, 0.6/1kV voltage grade, etc) and a base set of installation conditions (e.g. ambient temperature, installation method, etc). It is important to note that the current ratings are only valid for the quoted types of cables and base installation conditions.<br />
<br />
In the absence of any guidance, the following [[LV_Cable_Data|reference based current ratings]] may be used.<br />
<br />
==== Installed Current Ratings ====<br />
<br />
When the proposed installation conditions differ from the base conditions, derating (or correction) factors can be applied to the base current ratings to obtain the actual installed current ratings.<br />
<br />
International standards and cable manufacturers will provide derating factors for a range of installation conditions, for example ambient / soil temperature, grouping or bunching of cables, soil thermal resistivity, etc. The installed current rating is calculated by multiplying the base current rating with each of the derating factors, i.e.<br />
<br />
: <math>I_{c} = I_{b} . k_{d} \,</math><br />
<br />
where <math>I_{c}\,</math> is the installed current rating (A)<br />
:: <math>I_{b}\,</math> is the base current rating (A)<br />
:: <math>k_{d} \,</math> are the product of all the derating factors<br />
<br />
For example, suppose a cable had an ambient temperature derating factor of <math>k_{amb} = 0.94</math> and a grouping derating factor of <math>k_{g} = 0.85</math>, then the overall derating factor <math>k_{d} = 0.94 x 0.85 = 0.799</math>. For a cable with a base current rating of 42A, the installed current rating would be <math>I_{c} = 0.799 x 42 = 33.6A</math>.<br />
<br />
In the absence of any guidance, the following [[LV_Cable_Data|reference derating factors]] may be used.<br />
<br />
==== Cable Selection and Coordination with Protective Devices ====<br />
<br />
===== Feeders =====<br />
<br />
When sizing cables for non-motor loads, the upstream protective device ([[Electrical_Fuse|fuse]] or circuit breaker) is typically selected to also protect the cable against damage from [[thermal overload]]. The protective device must therefore be selected to exceed the full load current, but not exceed the cable's installed current rating, i.e. this inequality must be met:<br />
<br />
: <math>I_{l} \leq I_{p} \leq I_{c} \, </math><br />
<br />
Where <math>I_{l} \, </math> is the full load current (A)<br />
:: <math>I_{p} \, </math> is the protective device rating (A)<br />
:: <math>I_{c} \, </math> is the installed cable current rating (A)<br />
<br />
===== Motors =====<br />
<br />
Motors are normally protected by a separate thermal overload (TOL) relay and therefore the upstream protective device (e.g. [[Electrical_Fuse|fuse]] or circuit breaker) is not required to protect the cable against overloads. As a result, cables need only to be sized to cater for the full load current of the motor, i.e.<br />
<br />
: <math>I_{l} \leq I_{c} \, </math><br />
<br />
Where <math>I_{l} \, </math> is the full load current (A)<br />
:: <math>I_{c} \, </math> is the installed cable current rating (A)<br />
<br />
Of course, if there is no separate thermal overload protection on the motor, then the protective device needs to be taken into account as per the case for feeders above.<br />
<br />
=== Step 3: Voltage Drop ===<br />
<br />
A cable's conductor can be seen as an impedance and therefore whenever current flows through a cable, there will be a voltage drop across it, which can be derived by Ohm’s Law (i.e. V = IZ). The voltage drop will depend on two things:<br />
<br />
:* Current flow through the cable – the higher the current flow, the higher the voltage drop<br />
:* Impedance of the conductor – the larger the impedance, the higher the voltage drop<br />
<br />
==== Cable Impedances ====<br />
<br />
The impedance of the cable is a function of the cable size (cross-sectional area) and the length of the cable. Most cable manufacturers will quote a cable’s resistance and reactance in <math>\Omega</math>/km. The following [[LV_Cable_Data|typical cable impedances]] for low voltage AC and DC single core and multicore cables can be used in the absence of any other data.<br />
<br />
==== Calculating Voltage Drop ====<br />
<br />
For AC systems, the method of calculating voltage drops based on load power factor is commonly used. Full load currents are normally used, but if the load has high startup currents (e.g. motors), then voltage drops based on starting current (and power factor if applicable) should also be calculated.<br />
<br />
For a three phase system:<br />
<br />
: <math> V_{3\phi} = \frac{\sqrt{3} I (R_{c} \cos\phi + X_{c} \sin\phi) L}{1000} \, </math><br />
<br />
Where <math> V_{3\phi} \, </math> is the three phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
For a single phase system:<br />
<br />
: <math> V_{1\phi} = \frac{2 I (R_{c} \cos\phi + X_{c} \sin\phi) L}{1000} \, </math><br />
<br />
Where <math> V_{1\phi} \, </math> is the single phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
For a DC system:<br />
<br />
: <math> V_{dc} = \frac{2 I R_{c} L}{1000} \, </math><br />
<br />
Where <math> V_{dc} \, </math> is the dc voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the dc resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
==== Maximum Permissible Voltage Drop ====<br />
<br />
It is customary for standards (or clients) to specify maximum permissible voltage drops, which is the highest voltage drop that is allowed across a cable. Should your cable exceed this voltage drop, then a larger cable size should be selected.<br />
<br />
Maximum voltage drops across a cable are specified because load consumers (e.g. appliances) will have an input voltage tolerance range. This means that if the voltage at the appliance is lower than its rated minimum voltage, then the appliance may not operate correctly. <br />
<br />
In general, most electrical equipment will operate normally at a voltage as low as 80% nominal voltage. For example, if the nominal voltage is 230VAC, then most appliances will run at >184VAC. Cables are typically sized for a more conservative maximum voltage drop, in the range of 5 – 10% at full load.<br />
<br />
==== Calculating Maximum Cable Length due to Voltage Drop ====<br />
<br />
It may be more convenient to calculate the maximum length of a cable for a particular conductor size given a maximum permissible voltage drop (e.g. 5% of nominal voltage at full load) rather than the voltage drop itself. For example, by doing this it is possible to construct tables showing the maximum lengths corresponding to different cable sizes in order to speed up the selection of similar type cables. <br />
<br />
The maximum cable length that will achieve this can be calculated by re-arranging the voltage drop equations and substituting the maximum permissible voltage drop (e.g. 5% of 415V nominal voltage = 20.75V).<br />
For a three phase system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{3\phi}}{\sqrt{3} I (R_{c} \cos\phi + X_{c} \sin\phi)} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{3\phi} \, </math> is the maximum permissible three phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
<br />
For a single phase system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{1\phi}}{2 I (R_{c} \cos\phi + X_{c} \sin\phi)} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{1\phi} \, </math> is the maximum permissible single phase voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the ac resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> X_{c} \, </math> is the ac reactance of the cable (<math>\Omega</math>/km)<br />
:: <math> \cos\phi \, </math> is the load power factor (pu)<br />
<br />
For a DC system:<br />
<br />
: <math> L_{max} = \frac{1000 V_{dc}}{2 I R_{c}} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum length of the cable (m)<br />
:: <math> V_{dc} \, </math> is the maximum permissible dc voltage drop (V)<br />
:: <math> I \, </math> is the nominal full load or starting current as applicable (A)<br />
:: <math> R_{c} \, </math> is the dc resistance of the cable (<math>\Omega</math>/km)<br />
:: <math> L \, </math> is the length of the cable (m)<br />
<br />
=== Step 4: Short Circuit Temperature Rise ===<br />
<br />
During a short circuit, a high amount of current can flow through a cable for a short time. This surge in current flow causes a temperature rise within the cable. High temperatures can trigger unwanted reactions in the cable insulation, sheath materials and other components, which can prematurely degrade the condition of the cable. As the cross-sectional area of the cable increases, it can dissipate higher fault currents for a given temperature rise. Therefore, cables should be sized to withstand the largest short circuit that it is expected to see. <br />
<br />
==== Minimum Cable Size Due to Short Circuit Temperature Rise ====<br />
<br />
The minimum cable size due to short circuit temperature rise is typically calculated with an equation of the form:<br />
<br />
: <math> A = \frac{\sqrt{i^{2}t}}{k} \, </math><br />
<br />
Where <math>A \, </math> is the minimum cross-sectional area of the cable (<math>mm^{2}</math>)<br />
:: <math>i \, </math> is the prospective short circuit current (A)<br />
:: <math>t \, </math> is the duration of the short circuit (s)<br />
:: <math>k \, </math> is a short circuit temperature rise constant<br />
<br />
The temperature rise constant is calculated based on the material properties of the conductor and the initial and final conductor temperatures (see the [[Short_Circuit_Temperature_Rise|derivation here]]). Different international standards have different treatments of the temperature rise constant, but by way of example, IEC 60364-5-54 calculates it as follows:<br />
<br />
: <math> k = 226 \sqrt{ \ln \left( 1 + \frac{\theta_{f}-\theta{i}}{234.5 + \theta_{i}} \right) } \, </math> (for copper conductors)<br />
<br />
: <math> k = 148 \sqrt{ \ln \left( 1 + \frac{\theta_{f}-\theta{i}}{228 + \theta_{i}} \right) } \, </math> (for aluminium conductors)<br />
<br />
Where <math>\theta_{i} \, </math> is the initial conductor temperature (deg C)<br />
:: <math>\theta_{f} \, </math> is the final conductor temperature (deg C)<br />
<br />
==== Initial and Final Conductor Temperatures ====<br />
<br />
The initial conductor temperature is typically chosen to be the maximum operating temperature of the cable. The final conductor temperature is typically chosen to be the limiting temperature of the insulation. In general, the cable's insulation will determine the maximum operating temperature and limiting temperatures. <br />
<br />
As a rough guide, the following temperatures are common for the different insulation materials:<br />
<br />
{| border = "1" cellpadding="5" cellspacing="0" style = "text-align:center;"<br />
|-<br />
!style="background:#efefef;" width="80px" | Material<br />
!style="background:#efefef;" width="110px" | Max Operating Temperature <sup>o</sup>C<br />
!style="background:#efefef;" width="100px" | Limiting Temperature <sup>o</sup>C<br />
|-<br />
| PVC || 75 || 160<br />
|-<br />
| EPR || 90 || 250<br />
|-<br />
| XLPE || 90 || 250<br />
|}<br />
<br />
==== Short Circuit Energy ====<br />
<br />
The short circuit energy <math>i^{2}t \, </math> is normally chosen as the maximum short circuit that the cable could potentially experience. However for circuits with current limiting devices (such as HRC fuses), then the short circuit energy chosen should be the maximum prospective let-through energy of the protective device, which can be found from manufacturer data. <br />
<br />
=== Step 5: Earth Fault Loop Impedance ===<br />
<br />
Sometimes it is desirable (or necessary) to consider the earth fault loop impedance of a circuit in the sizing of a cable. Suppose a bolted earth fault occurs between an active conductor and earth. During such an earth fault, it is desirable that the upstream protective device acts to interrupt the fault within a maximum disconnection time so as to protect against any inadvertent contact to exposed live parts. <br />
<br />
Ideally the circuit will have earth fault protection, in which case the protection will be fast acting and well within the maximum disconnection time. The maximum disconnection time is chosen so that a dangerous touch voltage does not persist for long enough to cause injury or death. For most circuits, a maximum disconnection time of 5s is sufficient, though for portable equipment and socket outlets, a faster disconnection time is desirable (i.e. <1s and will definitely require earth fault protection). <br />
<br />
However for circuits that do not have earth fault protection, the upstream protective device (i.e. [[Electrical_Fuse|fuse]] or circuit breaker) must trip within the maximum disconnection time. In order for the protective device to trip, the fault current due to a bolted short circuit must exceed the value that will cause the protective device to act within the maximum disconnection time. For example, suppose a circuit is protected by a [[Electrical_Fuse|fuse]] and the maximum disconnection time is 5s, then the fault current must exceed the fuse melting current at 5s (which can be found by cross-referencing the fuse time-current curves). <br />
<br />
By simple application of Ohm's law:<br />
<br />
: <math> I_{A} = \frac{V_{0}}{Z_{s}} \, </math><br />
<br />
Where <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> Z_{s} \, </math> is the impedance of the earth fault loop (<math>\Omega</math>)<br />
<br />
It can be seen from the equation above that the impedance of the earth fault loop must be sufficiently low to ensure that the earth fault current can trip the upstream protection.<br />
<br />
==== The Earth Fault Loop ====<br />
<br />
The earth fault loop can consist of various return paths other than the earth conductor, including the cable armour and the static earthing connection of the facility. However for practical reasons, the earth fault loop in this calculation consists only of the active conductor and the earth conductor.<br />
<br />
The earth fault loop impedance can be found by: <br />
<br />
: <math> Z_{s} = Z_{c} + Z_{e} \, </math><br />
<br />
Where <math> Z_{s} \,</math> is the earth fault loop impedance (<math>\Omega</math>)<br />
:: <math> Z_{c} \,</math> is the impedance of the active conductor (<math>\Omega</math>)<br />
:: <math> Z_{e} \,</math> is the impedance of the earth conductor (<math>\Omega</math>)<br />
<br />
Assuming that the active and earth conductors have identical lengths, the earth fault loop impedance can be calculated as follows:<br />
<br />
: <math> Z_{s} = \frac{L}{1000} \sqrt{(R_{c}+R_{e})^{2} + (X_{c}+X_{e})^{2}} \, </math><br />
<br />
Where <math> L \,</math> is the length of the cable (m)<br />
:: <math> R_{c} \,</math> and <math> R_{e} \,</math> are the ac resistances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
:: <math> X_{c} \,</math> and <math> X_{e} \,</math> are the reactances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
<br />
==== Maximum Cable Length ====<br />
<br />
The maximum earth fault loop impedance can be found by re-arranging the equation above:<br />
<br />
: <math> Z_{s,max} = \frac{V_{0}}{I_{A}} \, </math><br />
<br />
Where <math> Z_{s} \, </math> is the maximum earth fault loop impedance (<math>\Omega</math>)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
<br />
The maximum cable length can therefore be calculated by the following:<br />
<br />
: <math> L_{max} = \frac{1000 V_{0}}{I_{A} \sqrt{(R_{c}+R_{e})^{2} + (X_{c}+X_{e})^{2}}} \, </math><br />
<br />
Where <math> L_{max} \, </math> is the maximum cable length (m)<br />
:: <math> V_{0} \, </math> is the phase to earth voltage at the protective device (V)<br />
:: <math> I_{A} \, </math> is the earth fault current required to trip the protective device within the minimum disconnection time (A)<br />
:: <math> R_{c} \,</math> and <math> R_{e} \,</math> are the ac resistances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
:: <math> X_{c} \,</math> and <math> X_{e} \,</math> are the reactances of the active and earth conductors respectively (<math>\Omega</math>/km)<br />
<br />
Note that the voltage <math>V_{0}</math> at the protective device is not necessarily the nominal phase to earth voltage, but usually a lower value as it can be downstream of the main busbars. This voltage is commonly represented by applying some factor <math>c \,</math> to the nominal voltage. A conservative value of <math>c \,</math> = 0.8 can be used so that:<br />
<br />
: <math>V_{0} = c V_{n} = 0.8 V_{n} \,</math><br />
<br />
Where <math>V_{n}</math> is the nominal phase to earth voltage (V)<br />
<br />
== Worked Example ==<br />
<br />
In this example, we will size a cable for a 415V, 37kW three-phase motor from the MCC to the field.<br />
<br />
=== Step 1: Data Gathering ===<br />
<br />
The following data was collected for the cable to be sized:<br />
:* Cable type: Cu/PVC/GSWB/PVC, 3C+E, 0.6/1kV<br />
:* Operating temperature: 75C<br />
:* Cable installation: above ground on cable ladder bunched together with 3 other cables on a single layer and at 30C ambient temperature<br />
:* Cable run: 90m (including tails)<br />
:* Motor load: 37kW, 415V three phase, full load current = 61A, power factor = 0.85<br />
:* Protection: aM fuse of rating = 80A, max prospective fault <math>I^{2}t</math> = 90 <math>A^{2}s</math> , 5s melt time = 550A<br />
<br />
=== Step 2: Cable Selection Based on Current Rating ===<br />
<br />
Suppose the ambient temperature derating is 0.89 and the grouping derating for 3 bunched cables on a single layer is 0.82. The overall derating factor is 0.89 <math>\times</math> 0.82 = 0.7298. Given that a 16 <math>mm^{2}</math> and 25 <math>mm^{2}</math> have base current ratings of 80A and 101A respectively (based on [[PVC_Current_Ratings_(Copper)|Reference Method E]]), which cable should be selected based on current rating considerations?<br />
<br />
The installed current ratings for 16 <math>mm^{2}</math> and 25 <math>mm^{2}</math> is 0.7298 <math>\times</math> 80A = 58.38A and 0.7298 <math>\times</math> 101A = 73.71A respectively. Given that the full load current of the motor is 61A, then the installed current rating of the 16 <math>mm^{2}</math> cable is lower than the full load current and is not suitable for continuous use with the motor. The 25 <math>mm^{2}</math> cable on the other hand has an installed current rating that exceeds the motor full load current, and is therefore the cable that should be selected.<br />
<br />
=== Step 3: Voltage Drop ===<br />
<br />
Suppose a 25 <math>mm^{2}</math> cable is selected. If the maximum permissible voltage drop is 5%, is the cable suitable for a run length of 90m?<br />
<br />
A 25 <math>mm^{2}</math> cable has an [[AC_Resistance|ac resistance]] of 0.884 <math>\Omega</math>/km and an [[AC_Reactance|ac reactance]] of 0.0895 <math>\Omega</math>/km. The voltage drop across the cable is:<br />
<br />
: <math> V_{d} = \frac{90}{1000} \times \sqrt{3} \times 61 \times \left[ 0.884 \times 0.85 + 0.0895 \times \sin(\cos^{-1}(0.85) \right] = 7.593V \, </math><br />
<br />
A voltage drop of 7.593V is equivalent to <math>\frac{7.593}{415} = 1.83%</math>, which is lower than the maximum permissible voltage dorp of 5%. Therefore the cable is suitable for the motor based on voltage drop considerations.<br />
<br />
=== Step 4: Short Circuit Temperature Rise ===<br />
<br />
The cable is operating normally at 75C and has a prospective fault capacity (<math>I^{2}t</math>) of 90,000 <math>A^{2}s</math>. What is the minimum size of the cable based on short circuit temperature rise?<br />
<br />
PVC has a limiting temperature of 160C. Using the IEC formula, the short circuit temperature rise constant is 111.329. The minimum cable size due to short circuit temperature rise is therefore:<br />
<br />
: <math>A = \frac{\sqrt{90,000}}{111.329} = 2.695 mm^{2} \, </math><br />
<br />
In this example, we also use the fuse for earth fault protection and it needs to trip within 5s, which is at the upper end of the adiabatic period where the short circuit temperature rise equation is still valid. Therefore, it's a good idea to also check that the cable can withstand the short circuit temperature rise for for a 5s fault. The 80A motor fuse has a 5s melting current of 550A. The short circuit temperature rise is thus:<br />
<br />
: <math>A = \frac{\sqrt{550^{2} \times 5}}{111.329} = 11.047 mm^{2} \, </math><br />
<br />
Therefore, our 25 <math>mm^{2}</math> cable is still suitable for this application.<br />
<br />
=== Step 5: Earth Fault Loop Impedance ===<br />
<br />
Suppose there is no special earth fault protection for the motor and a bolted single phase to earth fault occurs at the motor terminals. Suppose that the earth conductor for our 25 <math>mm^{2}</math> cable is 10 <math>mm^{2}</math>. If the maximum disconnection time is 5s, is our 90m long cable suitable based on earth fault loop impedance?<br />
<br />
The 80A motor fuse has a 5s melting current of 550A. The ac resistances of the active and earth conductors are 0.884 <math>\Omega</math>/km and 2.33 <math>\Omega</math>/km) respectively. The reactances of the active and earth conductors are 0.0895 <math>\Omega</math>/km and 0.0967 <math>\Omega</math>/km) respectively.<br />
<br />
The maximum length of the cable allowed is calculated as:<br />
<br />
: <math> L_{max} = \frac{(1000)(0.8)(240)}{550 \sqrt{(0.884+2.33)^{2} + (0.0895+0.0967)^{2}}} = 108.43m \, </math><br />
<br />
The cable run is 90m and the maximum length allowed is 108m, therefore our cable is suitable based on earth fault loop impedance. In fact, our 25 <math>mm^{2}</math> cable has passed all the tests and is the size that should be selected.<br />
<br />
== Waterfall Charts ==<br />
<br />
[[Image:Cable_waterfall.jpg|right|thumb|400px|Table 2. Example of a cable waterfall chart]]<br />
<br />
Sometimes it is convenient to group together similar types of cables (for example, 415V PVC motor cables installed on cable ladder) so that instead of having to go through the laborious exercise of sizing each cable separately, one can select a cable from a pre-calculated chart. <br />
<br />
These charts are often called "waterfall charts" and typically show a list of load ratings and the maximum of length of cable permissible for each cable size. Where a particular cable size fails to meet the requirements for current carrying capacity or short circuit temperature rise, it is blacked out on the chart (i.e. meaning that you can't choose it).<br />
<br />
Preparing a waterfall chart is common practice when having to size many like cables and substantially cuts down the time required for cable selection.<br />
<br />
== International Standards ==<br />
<br />
=== IEC ===<br />
<br />
[https://webstore.iec.ch/publication/1878 IEC 60364-5-52 (2009)] "Electrical installations in buildings - Part 5-52: Selection and erection of electrical equipment - Wiring systems" is the IEC standard governing cable sizing.<br />
<br />
=== NEC ===<br />
<br />
[https://www.nfpa.org/nec/ NFPA 70 (2020)] "National Electricity Code" is the equivalent standard for IEC 60364 in North America and includes a section covering cable sizing in Article 300.<br />
<br />
=== BS ===<br />
<br />
[https://electrical.theiet.org/bs-7671/ BS 7671 (18th edition)] "Requirements for Electrical Installations - IEE Wiring Regulations" is the equivalent standard for IEC 60364 in the United Kingdom.<br />
<br />
=== AS/NZS ===<br />
<br />
[https://infostore.saiglobal.com/en-us/standards/as-nzs-3008-1-1-2017-99685_saig_as_as_209556/ AS/NZS 3008.1 (2017)] "Electrical installations - Selection of cables - Cables for alternating voltages up to and including 0.6/1 kV" is the standard governing low voltage cable sizing in Australia and New Zealand. AS/NZS 3008.1.1 is for Australian conditions and AS/NZS 3008.1.2 is for New Zealand conditions.<br />
<br />
== What next? ==<br />
<br />
Having sized the power / load-bearing cables, the cable schedule can now be developed and then the cable material take-offs (MTO).<br />
<br />
[[Category:Calculations]</div>Juleshttp://openelectrical.org/index.php?title=Category:Fundamentals&diff=177Category:Fundamentals2020-11-23T23:20:41Z<p>Jules: Created blank page</p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=Category:Modelling_/_Analysis&diff=176Category:Modelling / Analysis2020-11-23T23:20:19Z<p>Jules: Created blank page</p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=Short_Circuit&diff=175Short Circuit2020-11-22T13:53:28Z<p>Jules: /* Far-from-Generator Short Circuit */</p>
<hr />
<div>A short circuit is an electrical fault where a conductive path (usually of low impedance) is formed between two or more conductive parts of an electrical system (e.g. phase-phase, phase-earth, phase-neutral, etc). This article looks at the nature of short circuits and tries to break down and explain the constituent parts of fault currents. Note that the terms "short circuit" and "fault" are often used interchangeably.<br />
<br />
In most networks, a short circuit is similar to the [[RL_Circuit_Switching|closing transient of an RL circuit]], where the R and L components are the impedances of the source(s). The transient characteristics of short circuit currents vary depending on whether they are near or far from synchronous generators. The sections below describe the two general types of short circuits:<br />
<br />
== Near-to-Generator Short Circuit ==<br />
<br />
A fault close to a synchronous generator has the following maximum short circuit current <math>i_{sc}(t)</math>:<br />
<br />
: <math>i_{sc}(t) = E \sqrt{2} \left[ \left( \frac{1}{X_{d}''} - \frac{1}{X_{d}'} \right) e^{-t/t_{d}''} + \left( \frac{1}{X_{d}'} - \frac{1}{X_{d}} \right) e^{-t/t_{d}'} + \frac{1}{X_{d}} \right] \sin (\omega t) + \frac{E \sqrt{2}}{X_{d}''} e^{-t/t_{a}} \, </math><br />
<br />
Where <math>E \, </math> is the phase-to-neutral rms voltage at the generator terminals (V)<br />
:: <math>X_{d}'' \, </math> is the generator direct-axis subtransient reactance (<math>\Omega</math>)<br />
:: <math>X_{d}' \, </math> is the generator direct-axis transient reactance (<math>\Omega</math>)<br />
:: <math>X_{d} \, </math> is the generator synchronous reactance (<math>\Omega</math>)<br />
:: <math>T_{d}'' \, </math> is the generator subtransient time constant (s)<br />
:: <math>T_{d}' \, </math> is the generator transient time constant (s)<br />
:: <math>T_{a} \, </math> is the aperiodic time constant (s)<br />
<br />
From the above equation, it can be seen that the short circuit current can be broken up into an aperiodic current (dc component of the short circuit):<br />
<br />
: <math> \frac{E \sqrt{2}}{X_{d}''} e^{-t/t_{a}} \, </math><br />
<br />
[[Image:NTG_dc.PNG]]<br />
<br />
And a series of three damped sinusoidal waveforms corresponding to the following distinct stages:<br />
<br />
(1) Subtransient component: <math> E \sqrt{2} \left( \frac{1}{X_{d}''} - \frac{1}{X_{d}'} \right) e^{-t/t_{d}''} \sin (\omega t) \, </math><br />
<br />
[[Image:NTG_subtransient.PNG]]<br />
<br />
This period typically lasts 10 to 20ms from the start of the fault. The subtransient reactance is due to the flux casued by the stator currents crossing the air gap and reaching the rotor surface or amortisseur / damper windings. <br />
<br />
(2) Transient component: <math> E \sqrt{2} \left( \frac{1}{X_{d}'} - \frac{1}{X_{d}} \right) e^{-t/t_{d}'} \sin (\omega t) \, </math><br />
<br />
[[Image:NTG_transient.PNG]]<br />
<br />
This period typically lasts 100 to 400ms after the subtransient period. The transient reactance occurs when all the damping currents in the rotor surface or amortisseur / damper windings have decayed, but while the damping currents in the field winding are still in action. <br />
<br />
(3) Steady-state component: <math> E \sqrt{2} \frac{1}{X_{d}} \sin (\omega t) \, </math><br />
<br />
The steady-state occurs after the transient period when all the damping currents in the field windings have decayed, and essentially remains until the fault is cleared.<br />
<br />
[[Image:NTG_steady.PNG]]<br />
<br />
Putting these all together, we get the familiar near-to-generator short circuit waveform:<br />
<br />
[[Image:NTG_overall.PNG]]<br />
<br />
== Far-from-Generator Short Circuit ==<br />
<br />
In short circuits occurring far from synchronous generators, we can ignore the effects of the generator subtransient behaviour. It can be shown through [[RL_Circuit_Switching|transient circuit analysis]] that the maximum far-from-generator short circuit is as follows:<br />
<br />
: <math> i_{sc}(t) = \frac {E \sqrt{2}}{Z_{sc}} \left[ \sin \left( \omega t + \phi \right) - \sin(\phi) e^{-\frac{R}{X} \omega t} \right] \, </math><br />
<br />
Where <math>E \, </math> is the rms voltage of the circuit (V)<br />
:: <math> Z_{sc} \, </math> is the fault impedance (<math>\Omega</math>)<br />
:: <math> \frac{R}{X} \, </math> is the R/X ratio at the point of fault (pu)<br />
:: <math> \phi </math> is a phase angle comprising the angle at the time of fault and the steady-state power angle (rad)<br />
<br />
We can see that there are two components: <br />
<br />
(1) A decaying aperiodic component: <math> - \frac {E \sqrt{2}}{Z_{sc}} \sin(\phi) e^{-\frac{R}{X} \omega t } \, </math><br />
<br />
[[Image:FFG_dc.PNG]]<br />
<br />
(note here that the arbitrary phase angle <math> \phi </math> yields a negative magnitude)<br />
<br />
(2) A steady state component: <math> \frac {E \sqrt{2}}{Z_{sc}} \sin \left( \omega t + \phi \right) \, </math><br />
<br />
[[Image:FFG_steady.PNG]]<br />
<br />
Putting these together, we get the total far-from-generator fault current:<br />
<br />
[[Image:FFG_overall.PNG]]<br />
<br />
During the transient period, the peak transient current is typically 1.5 to 2.5 times higher than the peak steady state current.<br />
<br />
[[Category:Fundamentals]]<br />
[[Category:Modelling / Analysis]]</div>Juleshttp://openelectrical.org/index.php?title=File:FFG_steady.PNG&diff=174File:FFG steady.PNG2020-11-22T13:49:54Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:FFG_overall.PNG&diff=173File:FFG overall.PNG2020-11-22T13:49:40Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:FFG_dc.PNG&diff=172File:FFG dc.PNG2020-11-22T13:49:23Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=Short_Circuit&diff=171Short Circuit2020-11-22T13:48:38Z<p>Jules: /* Far-from-Generator Short Circuit */</p>
<hr />
<div>A short circuit is an electrical fault where a conductive path (usually of low impedance) is formed between two or more conductive parts of an electrical system (e.g. phase-phase, phase-earth, phase-neutral, etc). This article looks at the nature of short circuits and tries to break down and explain the constituent parts of fault currents. Note that the terms "short circuit" and "fault" are often used interchangeably.<br />
<br />
In most networks, a short circuit is similar to the [[RL_Circuit_Switching|closing transient of an RL circuit]], where the R and L components are the impedances of the source(s). The transient characteristics of short circuit currents vary depending on whether they are near or far from synchronous generators. The sections below describe the two general types of short circuits:<br />
<br />
== Near-to-Generator Short Circuit ==<br />
<br />
A fault close to a synchronous generator has the following maximum short circuit current <math>i_{sc}(t)</math>:<br />
<br />
: <math>i_{sc}(t) = E \sqrt{2} \left[ \left( \frac{1}{X_{d}''} - \frac{1}{X_{d}'} \right) e^{-t/t_{d}''} + \left( \frac{1}{X_{d}'} - \frac{1}{X_{d}} \right) e^{-t/t_{d}'} + \frac{1}{X_{d}} \right] \sin (\omega t) + \frac{E \sqrt{2}}{X_{d}''} e^{-t/t_{a}} \, </math><br />
<br />
Where <math>E \, </math> is the phase-to-neutral rms voltage at the generator terminals (V)<br />
:: <math>X_{d}'' \, </math> is the generator direct-axis subtransient reactance (<math>\Omega</math>)<br />
:: <math>X_{d}' \, </math> is the generator direct-axis transient reactance (<math>\Omega</math>)<br />
:: <math>X_{d} \, </math> is the generator synchronous reactance (<math>\Omega</math>)<br />
:: <math>T_{d}'' \, </math> is the generator subtransient time constant (s)<br />
:: <math>T_{d}' \, </math> is the generator transient time constant (s)<br />
:: <math>T_{a} \, </math> is the aperiodic time constant (s)<br />
<br />
From the above equation, it can be seen that the short circuit current can be broken up into an aperiodic current (dc component of the short circuit):<br />
<br />
: <math> \frac{E \sqrt{2}}{X_{d}''} e^{-t/t_{a}} \, </math><br />
<br />
[[Image:NTG_dc.PNG]]<br />
<br />
And a series of three damped sinusoidal waveforms corresponding to the following distinct stages:<br />
<br />
(1) Subtransient component: <math> E \sqrt{2} \left( \frac{1}{X_{d}''} - \frac{1}{X_{d}'} \right) e^{-t/t_{d}''} \sin (\omega t) \, </math><br />
<br />
[[Image:NTG_subtransient.PNG]]<br />
<br />
This period typically lasts 10 to 20ms from the start of the fault. The subtransient reactance is due to the flux casued by the stator currents crossing the air gap and reaching the rotor surface or amortisseur / damper windings. <br />
<br />
(2) Transient component: <math> E \sqrt{2} \left( \frac{1}{X_{d}'} - \frac{1}{X_{d}} \right) e^{-t/t_{d}'} \sin (\omega t) \, </math><br />
<br />
[[Image:NTG_transient.PNG]]<br />
<br />
This period typically lasts 100 to 400ms after the subtransient period. The transient reactance occurs when all the damping currents in the rotor surface or amortisseur / damper windings have decayed, but while the damping currents in the field winding are still in action. <br />
<br />
(3) Steady-state component: <math> E \sqrt{2} \frac{1}{X_{d}} \sin (\omega t) \, </math><br />
<br />
The steady-state occurs after the transient period when all the damping currents in the field windings have decayed, and essentially remains until the fault is cleared.<br />
<br />
[[Image:NTG_steady.PNG]]<br />
<br />
Putting these all together, we get the familiar near-to-generator short circuit waveform:<br />
<br />
[[Image:NTG_overall.PNG]]<br />
<br />
== Far-from-Generator Short Circuit ==<br />
<br />
In short circuits occurring far from synchronous generators, we can ignore the effects of the generator subtransient behaviour. It can be shown through [[RL_Circuit_Switching|transient circuit analysis]] that the maximum far-from-generator short circuit is as follows:<br />
<br />
: <math> i_{sc}(t) = \frac {E \sqrt{2}}{Z_{sc}} \left[ \sin \left( \omega t + \phi \right) - \sin(\phi) e^{-\frac{R}{X} \omega t} \right] \, </math><br />
<br />
Where <math>E \, </math> is the rms voltage of the circuit (V)<br />
:: <math> Z_{sc} \, </math> is the fault impedance (<math>\Omega</math>)<br />
:: <math> \frac{R}{X} \, </math> is the R/X ratio at the point of fault (pu)<br />
:: <math> \phi </math> is a phase angle comprising the angle at the time of fault and the steady-state power angle (rad)<br />
<br />
We can see that there are two components: <br />
<br />
(1) A decaying aperiodic component: <math> - \frac {E \sqrt{2}}{Z_{sc}} \sin(\phi) e^{-\frac{R}{X} \omega t } \, </math><br />
<br />
[[Image:FFG_dc.PNG]]<br />
<br />
(2) A steady state component: <math> \frac {E \sqrt{2}}{Z_{sc}} \sin \left( \omega t + \phi \right) \, </math><br />
<br />
[[Image:FFG_steady.PNG]]<br />
<br />
Putting these together, we get the total far-from-generator fault current:<br />
<br />
[[Image:FFG_overall.PNG]]<br />
<br />
During the transient period, the peak transient current is typically 1.5 to 2.5 times higher than the peak steady state current.<br />
<br />
[[Category:Fundamentals]]<br />
[[Category:Modelling / Analysis]]</div>Juleshttp://openelectrical.org/index.php?title=Short_Circuit&diff=170Short Circuit2020-11-22T13:47:04Z<p>Jules: /* Far-from-Generator Short Circuit */</p>
<hr />
<div>A short circuit is an electrical fault where a conductive path (usually of low impedance) is formed between two or more conductive parts of an electrical system (e.g. phase-phase, phase-earth, phase-neutral, etc). This article looks at the nature of short circuits and tries to break down and explain the constituent parts of fault currents. Note that the terms "short circuit" and "fault" are often used interchangeably.<br />
<br />
In most networks, a short circuit is similar to the [[RL_Circuit_Switching|closing transient of an RL circuit]], where the R and L components are the impedances of the source(s). The transient characteristics of short circuit currents vary depending on whether they are near or far from synchronous generators. The sections below describe the two general types of short circuits:<br />
<br />
== Near-to-Generator Short Circuit ==<br />
<br />
A fault close to a synchronous generator has the following maximum short circuit current <math>i_{sc}(t)</math>:<br />
<br />
: <math>i_{sc}(t) = E \sqrt{2} \left[ \left( \frac{1}{X_{d}''} - \frac{1}{X_{d}'} \right) e^{-t/t_{d}''} + \left( \frac{1}{X_{d}'} - \frac{1}{X_{d}} \right) e^{-t/t_{d}'} + \frac{1}{X_{d}} \right] \sin (\omega t) + \frac{E \sqrt{2}}{X_{d}''} e^{-t/t_{a}} \, </math><br />
<br />
Where <math>E \, </math> is the phase-to-neutral rms voltage at the generator terminals (V)<br />
:: <math>X_{d}'' \, </math> is the generator direct-axis subtransient reactance (<math>\Omega</math>)<br />
:: <math>X_{d}' \, </math> is the generator direct-axis transient reactance (<math>\Omega</math>)<br />
:: <math>X_{d} \, </math> is the generator synchronous reactance (<math>\Omega</math>)<br />
:: <math>T_{d}'' \, </math> is the generator subtransient time constant (s)<br />
:: <math>T_{d}' \, </math> is the generator transient time constant (s)<br />
:: <math>T_{a} \, </math> is the aperiodic time constant (s)<br />
<br />
From the above equation, it can be seen that the short circuit current can be broken up into an aperiodic current (dc component of the short circuit):<br />
<br />
: <math> \frac{E \sqrt{2}}{X_{d}''} e^{-t/t_{a}} \, </math><br />
<br />
[[Image:NTG_dc.PNG]]<br />
<br />
And a series of three damped sinusoidal waveforms corresponding to the following distinct stages:<br />
<br />
(1) Subtransient component: <math> E \sqrt{2} \left( \frac{1}{X_{d}''} - \frac{1}{X_{d}'} \right) e^{-t/t_{d}''} \sin (\omega t) \, </math><br />
<br />
[[Image:NTG_subtransient.PNG]]<br />
<br />
This period typically lasts 10 to 20ms from the start of the fault. The subtransient reactance is due to the flux casued by the stator currents crossing the air gap and reaching the rotor surface or amortisseur / damper windings. <br />
<br />
(2) Transient component: <math> E \sqrt{2} \left( \frac{1}{X_{d}'} - \frac{1}{X_{d}} \right) e^{-t/t_{d}'} \sin (\omega t) \, </math><br />
<br />
[[Image:NTG_transient.PNG]]<br />
<br />
This period typically lasts 100 to 400ms after the subtransient period. The transient reactance occurs when all the damping currents in the rotor surface or amortisseur / damper windings have decayed, but while the damping currents in the field winding are still in action. <br />
<br />
(3) Steady-state component: <math> E \sqrt{2} \frac{1}{X_{d}} \sin (\omega t) \, </math><br />
<br />
The steady-state occurs after the transient period when all the damping currents in the field windings have decayed, and essentially remains until the fault is cleared.<br />
<br />
[[Image:NTG_steady.PNG]]<br />
<br />
Putting these all together, we get the familiar near-to-generator short circuit waveform:<br />
<br />
[[Image:NTG_overall.PNG]]<br />
<br />
== Far-from-Generator Short Circuit ==<br />
<br />
In short circuits occurring far from synchronous generators, we can ignore the effects of the generator subtransient behaviour. It can be shown through [[RL_Circuit_Switching|transient circuit analysis]] that the maximum far-from-generator short circuit is as follows:<br />
<br />
: <math> i_{sc}(t) = \frac {E \sqrt{2}}{Z_{sc}} \left[ \sin \left( \omega t + \phi \right) - \sin \phi e^{-\frac{R}{X} \omega t} \right] \, </math><br />
<br />
Where <math>E \, </math> is the rms voltage of the circuit (V)<br />
:: <math> Z_{sc} \, </math> is the fault impedance (<math>\Omega</math>)<br />
:: <math> \frac{R}{X} \, </math> is the R/X ratio at the point of fault (pu)<br />
:: <math> \phi </math> is a phase angle comprising the angle at the time of fault and the steady-state power angle (rad)<br />
<br />
We can see that there are two components: <br />
<br />
(1) A decaying aperiodic component: <math> - \frac {E \sqrt{2}}{Z_{sc}} \sin + \phi e^{-\frac{R}{X} \omega t } \, </math><br />
<br />
[[Image:FFG_dc.PNG]]<br />
<br />
(2) A steady state component: <math> \frac {E \sqrt{2}}{Z_{sc}} \sin \left( \omega t + + \phi \right) \, </math><br />
<br />
[[Image:FFG_steady.PNG]]<br />
<br />
Putting these together, we get the total far-from-generator fault current:<br />
<br />
[[Image:FFG_overall.PNG]]<br />
<br />
During the transient period, the peak transient current is typically 1.5 to 2.5 times higher than the peak steady state current.<br />
<br />
[[Category:Fundamentals]]<br />
[[Category:Modelling / Analysis]]</div>Juleshttp://openelectrical.org/index.php?title=Short_Circuit&diff=169Short Circuit2020-11-22T13:39:26Z<p>Jules: Created page with "A short circuit is an electrical fault where a conductive path (usually of low impedance) is formed between two or more conductive parts of an electrical system (e.g. phase-ph..."</p>
<hr />
<div>A short circuit is an electrical fault where a conductive path (usually of low impedance) is formed between two or more conductive parts of an electrical system (e.g. phase-phase, phase-earth, phase-neutral, etc). This article looks at the nature of short circuits and tries to break down and explain the constituent parts of fault currents. Note that the terms "short circuit" and "fault" are often used interchangeably.<br />
<br />
In most networks, a short circuit is similar to the [[RL_Circuit_Switching|closing transient of an RL circuit]], where the R and L components are the impedances of the source(s). The transient characteristics of short circuit currents vary depending on whether they are near or far from synchronous generators. The sections below describe the two general types of short circuits:<br />
<br />
== Near-to-Generator Short Circuit ==<br />
<br />
A fault close to a synchronous generator has the following maximum short circuit current <math>i_{sc}(t)</math>:<br />
<br />
: <math>i_{sc}(t) = E \sqrt{2} \left[ \left( \frac{1}{X_{d}''} - \frac{1}{X_{d}'} \right) e^{-t/t_{d}''} + \left( \frac{1}{X_{d}'} - \frac{1}{X_{d}} \right) e^{-t/t_{d}'} + \frac{1}{X_{d}} \right] \sin (\omega t) + \frac{E \sqrt{2}}{X_{d}''} e^{-t/t_{a}} \, </math><br />
<br />
Where <math>E \, </math> is the phase-to-neutral rms voltage at the generator terminals (V)<br />
:: <math>X_{d}'' \, </math> is the generator direct-axis subtransient reactance (<math>\Omega</math>)<br />
:: <math>X_{d}' \, </math> is the generator direct-axis transient reactance (<math>\Omega</math>)<br />
:: <math>X_{d} \, </math> is the generator synchronous reactance (<math>\Omega</math>)<br />
:: <math>T_{d}'' \, </math> is the generator subtransient time constant (s)<br />
:: <math>T_{d}' \, </math> is the generator transient time constant (s)<br />
:: <math>T_{a} \, </math> is the aperiodic time constant (s)<br />
<br />
From the above equation, it can be seen that the short circuit current can be broken up into an aperiodic current (dc component of the short circuit):<br />
<br />
: <math> \frac{E \sqrt{2}}{X_{d}''} e^{-t/t_{a}} \, </math><br />
<br />
[[Image:NTG_dc.PNG]]<br />
<br />
And a series of three damped sinusoidal waveforms corresponding to the following distinct stages:<br />
<br />
(1) Subtransient component: <math> E \sqrt{2} \left( \frac{1}{X_{d}''} - \frac{1}{X_{d}'} \right) e^{-t/t_{d}''} \sin (\omega t) \, </math><br />
<br />
[[Image:NTG_subtransient.PNG]]<br />
<br />
This period typically lasts 10 to 20ms from the start of the fault. The subtransient reactance is due to the flux casued by the stator currents crossing the air gap and reaching the rotor surface or amortisseur / damper windings. <br />
<br />
(2) Transient component: <math> E \sqrt{2} \left( \frac{1}{X_{d}'} - \frac{1}{X_{d}} \right) e^{-t/t_{d}'} \sin (\omega t) \, </math><br />
<br />
[[Image:NTG_transient.PNG]]<br />
<br />
This period typically lasts 100 to 400ms after the subtransient period. The transient reactance occurs when all the damping currents in the rotor surface or amortisseur / damper windings have decayed, but while the damping currents in the field winding are still in action. <br />
<br />
(3) Steady-state component: <math> E \sqrt{2} \frac{1}{X_{d}} \sin (\omega t) \, </math><br />
<br />
The steady-state occurs after the transient period when all the damping currents in the field windings have decayed, and essentially remains until the fault is cleared.<br />
<br />
[[Image:NTG_steady.PNG]]<br />
<br />
Putting these all together, we get the familiar near-to-generator short circuit waveform:<br />
<br />
[[Image:NTG_overall.PNG]]<br />
<br />
== Far-from-Generator Short Circuit ==<br />
<br />
In short circuits occurring far from synchronous generators, we can ignore the effects of the generator subtransient behaviour. It can be shown through [[RL_Circuit_Switching|transient circuit analysis]] that the maximum far-from-generator short circuit is as follows:<br />
<br />
: <math> i_{sc}(t) = \frac {E \sqrt{2}}{Z_{sc}} \left[ \sin \left( \omega t + \frac{\pi}{2} \right) - e^{-\frac{R}{X} \omega t} \right] \, </math><br />
<br />
Where <math>E \, </math> is the rms voltage of the circuit (V)<br />
:: <math> Z_{sc} \, </math> is the fault impedance (<math>\Omega</math>)<br />
:: <math> \frac{R}{X} \, </math> is the R/X ratio at the point of fault (pu)<br />
<br />
We can see that there are two components: <br />
<br />
(1) A decaying aperiodic component: <math> - \frac {E \sqrt{2}}{Z_{sc}} e^{-\frac{R}{X} \omega t} \, </math><br />
<br />
[[Image:FFG_dc.PNG]]<br />
<br />
(2) A steady state component: <math> \frac {E \sqrt{2}}{Z_{sc}} \sin \left( \omega t + \frac{\pi}{2} \right) \, </math><br />
<br />
[[Image:FFG_steady.PNG]]<br />
<br />
Putting these together, we get the total far-from-generator fault current:<br />
<br />
[[Image:FFG_overall.PNG]]<br />
<br />
During the transient period, the peak transient current is typically 1.5 to 2.5 times higher than the peak steady state current.<br />
<br />
[[Category:Fundamentals]]<br />
[[Category:Modelling / Analysis]]</div>Juleshttp://openelectrical.org/index.php?title=File:NTG_transient.PNG&diff=168File:NTG transient.PNG2020-11-22T13:38:38Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:NTG_subtransient.PNG&diff=167File:NTG subtransient.PNG2020-11-22T13:38:13Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:NTG_steady.PNG&diff=166File:NTG steady.PNG2020-11-22T13:38:00Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:NTG_overall.PNG&diff=165File:NTG overall.PNG2020-11-22T13:37:46Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:NTG_dc.PNG&diff=164File:NTG dc.PNG2020-11-22T13:37:33Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=Induction_Motor_Performance&diff=163Induction Motor Performance2020-11-22T13:35:57Z<p>Jules: </p>
<hr />
<div>Induction motors typically run at highest efficiency and power factor at full-load. However when motors are run at lower operating points, there is generally a corresponding decline in performance. This drop in performance tends to vary non-linearly (with respect to the load factor), with larger drops in performance at low load factors (0% to 40%). For example, the figure below shows the performance test data of a 155kW, 2000V, 50Hz induction motor at load factors from 0% to 100%:<br />
<br />
[[Image:Motor_performance.png|700px]]<br />
<br />
== Related Topics ==<br />
<br />
:* [[Induction Motor Model]]<br />
:* [[Induction Motor Torque]]<br />
<br />
[[Category:Modelling / Analysis]]</div>Juleshttp://openelectrical.org/index.php?title=Main_Page&diff=162Main Page2020-11-22T08:31:10Z<p>Jules: </p>
<hr />
<div><strong>The Open Electrical wiki is currently being migrated</strong><br />
<br />
Note that not all pages have been migrated yet.<br />
<br />
* [[:Category:Fundamentals|Fundamentals]]<br />
* [[:Category:Calculations|Calculations]]<br />
* [[:Category:Modelling / Analysis|Modelling and Analysis]]</div>Juleshttp://openelectrical.org/index.php?title=Cable_Voltage_Grade&diff=161Cable Voltage Grade2020-11-22T08:28:09Z<p>Jules: </p>
<hr />
<div>== Definition ==<br />
[[Image:Cable_Voltage_Grade.png|right|thumb|400px|Figure 1. Meaning of U and U0]]<br />
<br />
AC cables are designed to be suitable for specific design voltages, which is called the "Voltage Grade" (or "Voltage Designation", "Voltage Class" or "Voltage Rating") of the cable. The voltage grade is commonly expressed in the following form: <math> U_{0} / U \, </math><br />
<br />
Where <math> U_{0} \, </math> is the power frequency voltage between phase and earth (V rms)<br />
:: <math> U \, </math> is the power frequency voltage between two phase conductors (V rms)<br />
<br />
For example, some standard IEC voltage grades are 0.6/1kV, 1.9/3.3kV, 3.8/6.6kV, 6.35/11kV, 12.7/22kV, 19/33kV, etc.<br />
<br />
For standard insulated cables, <math>U</math> is always higher than <math>U_{0}</math> because as shown in Figure 1, the measurement of <math>U</math> is across two layers of insulation (one per phase conductor), whereas the measurement of <math>U_{0}</math> is only across one layer of insulation to earth. Obviously, the voltage rating is higher the more insulation there is. <br />
<br />
Some manufacturers design cables with voltage grades where <math>U_{0}=U</math>, for example 1.0/1.0kV. In such cases, the <math>U</math> rating can technically be higher, but the manufacturer has decided not to certify the cable for a higher voltage grade.<br />
<br />
The manufacturer designs the cable (i.e. insulation, bedding, sheaths, etc) for the specified voltage grade. A cable can therefore be operated at voltages that do not exceed the voltage grade, e.g. a 0.6/1kV cable can be operated for any phase-to-earth and phase-to-phase voltages not exceeding 0.6kV and 1kV respectively. You may notice that LV cables are mainly specified to 0.6/1kV cables even though they are operated are much lower voltages (e.g. 240/415V, 220/380V, etc). This is due to the fact that the mechanical requirements of the insulation thickness are greater than the electrical requirements.<br />
<br />
== Neutral Earthing Considerations ==<br />
<br />
For MV and HV cables (>1kV), some consideration must be given to the neutral earthing arrangements of the system when specifying a voltage grade. In the US, the [http://www.aeic.org/ Association of Edison Illuminating Companies (AEIC)] has the following specifications depending on the neutral earthing arrangements and clearing time of the protective device:<br />
<br />
:* 100% insulation level<br />
:* 133% insulation level<br />
:* 173% insulation level<br />
<br />
The insulation levels above refer to the nominal phase-to-phase voltages. For example, a cable with a 133% insulation level on a 33kV system is rated for 133% the nominal phase-to-phase system voltage, i.e. 133% x 33kV = 43.89kV. <br />
<br />
=== 100% Insulation Level ===<br />
<br />
The 100% is the insulation level normally used for cables on solidly earthed systems, or on any system where the protective device will clear earth faults within 1 minute.<br />
<br />
=== 133% Insulation Level ===<br />
<br />
The 133% insulation level is specified for systems where the protective device is expected to clear earth faults within 1 hour, and is typically specified for high impedance earthed or unearthed systems.<br />
<br />
=== 173% Insulation Level ===<br />
<br />
The 173% insulation level is specified for systems where the time to clear an earth fault is indefinite. This is typically recommended for unearthed or resonant earthed systems.<br />
<br />
[[Category:Fundamentals]]</div>Juleshttp://openelectrical.org/index.php?title=Time_Domain_Simulations_of_Power_Systems&diff=160Time Domain Simulations of Power Systems2020-11-22T08:27:08Z<p>Jules: Created page with "In time domain simulations of power systems, a dynamic model is defined as any model that can be described by a set of differential-algebraic equations (DAE), commonly present..."</p>
<hr />
<div>In time domain simulations of power systems, a dynamic model is defined as any model that can be described by a set of differential-algebraic equations (DAE), commonly presented in the following form:<br />
<br />
: <math> \dot{\boldsymbol{x}} = \boldsymbol{f}(\boldsymbol{x},\boldsymbol{y},\boldsymbol{p},t) \,</math><br />
<br />
: <math> \boldsymbol{0} = \boldsymbol{g}(\boldsymbol{x},\boldsymbol{y},\boldsymbol{p},t) \,</math><br />
<br />
where <math>\boldsymbol{x} \,</math> is a vector of state variables<br />
:<math>\boldsymbol{y} \,</math> is a vector of algebraic variables<br />
:<math>\boldsymbol{p} \,</math> is a vector of constant or controllable parameters<br />
:<math>t \,</math> is time (an independent variable)<br />
:<math>\boldsymbol{f}(.) \,</math> are the differential equations<br />
:<math>\boldsymbol{g}(.) \,</math> are the algebraic equations<br />
<br />
== Related Topics ==<br />
<br />
:* [[Explicit Numerical Integrators]]<br />
:* [[Modified Admittance Matrix]]<br />
:* [[Network Interfacing]]<br />
:* [[Synchronous Machine Models]]<br />
<br />
[[Category:Modelling / Analysis]]</div>Juleshttp://openelectrical.org/index.php?title=Transformer_Inrush&diff=159Transformer Inrush2020-11-22T08:26:20Z<p>Jules: Created page with "Transformer (magnetising) inrush currents are caused by sudden step-changes in the transformer magnetising voltage and can be divided into three general types: :* '''Energisa..."</p>
<hr />
<div>Transformer (magnetising) inrush currents are caused by sudden step-changes in the transformer magnetising voltage and can be divided into three general types:<br />
<br />
:* '''Energisation inrush''' - occurs when a transformer is switched on from a de-energised state<br />
:* '''Recovery inrush''' - occurs when voltage is restored to the transformer after a brief voltage dip or interruption<br />
:* '''Sympathetic inrush''' - occurs when a transformer is switched on and is connected in parallel to an already energised transformer. The abrupt voltage drop caused by energising the first transformer can cause inrush currents (in the opposite direction) in the already energised transformer.<br />
<br />
Transformer energisation is considered to cause the highest inrush currents and will be the focus of this article. When energised, an initial magnetising inrush current flows into the transformer. Typically, the inrush current lasts in the order of 0.1s and has the following magnitude:<br />
<br />
:* For transformers up to 2500kVA: 8x nominal current<br />
:* For transformers greater than 2500kVA: 10x nominal current<br />
<br />
Transformer inrush currents are predominantly due to saturation of the transformer core and can be modelled as an [[RL Circuit Switching|RL switching transient]] with a saturable inductance L. <br />
<br />
== Technical Background ==<br />
<br />
[[Image:TX_Inrush_Simple.png|right|thumb|350px|Figure 1. Simple model for transformer inrush]]<br />
<br />
Consider the application of a sinusoidal voltage to a transformer that is modelled very simply as a coil of N turns. By Faraday's law, we get:<br />
<br />
: <math> V_{p} \sin(\omega t + \theta) = N \frac{d \phi}{dt} \, </math><br />
<br />
Where <math> V_{p} \, </math> is the voltage amplitude (V)<br />
:: <math> \theta \, </math> is an arbitrary switching angle (radians)<br />
:: <math> N \, </math> is the number of turns in the transformer coil<br />
:: <math> \phi \, </math> is the transformer magnetising flux<br />
<br />
To solve this differential equation for flux given some initial (residual) flux <math> \phi (0) \, </math>, we take the [http://en.wikipedia.org/wiki/Laplace_transform Laplace transform] of it:<br />
<br />
: <math> V_{p} \left[ \frac{s \sin \theta + \omega \cos \theta}{s^{2} + \omega^{2}} \right] = N \left[ s \phi (s) - \phi (0) \right] \, </math><br />
<br />
Solving for <math>\phi (s) \, </math>:<br />
<br />
: <math> \phi (s) = \frac{V_{p}}{N} \left[ \frac{s \sin \theta + \omega \cos \theta}{s(s^{2} + \omega^{2})} \right] + \frac{\phi (0)}{s} \, </math><br />
<br />
Simplifying further:<br />
<br />
: <math> \phi (s) = \frac{V_{p}}{N} \left[ \frac{\sin \theta}{s^{2} + \omega^{2}} + \frac{\cos \theta}{\omega} \left( \frac{1}{s} - \frac{s}{s^{2} + \omega^{2}} \right) \right] + \frac{\phi (0)}{s} \, </math><br />
<br />
Taking the inverse Laplace transform, we finally get:<br />
<br />
: <math> \phi (t) = \frac{V_{p}}{\omega N} \left[ \sin \theta \sin (\omega t) + \cos \theta (1 - \cos \omega t) \right] + \phi (0) \, </math> (1)<br />
<br />
While this is a gross simplification of reality, the equation above can give us some intuition about the nature of the transformer flux during a transient switching on (energisation) event. <br />
<br />
=== Effects of Switching Angle ===<br />
<br />
From equation (1) above, it can be shown that the switching angle results in a dc offset of the flux waveform. The figure below shows the effects of the switching angle on the flux waveform at three angles (relative to the ac supply voltage): 1) a positive zero crossing (<math>\theta = 0</math>), 2) a voltage peak (<math>\theta = \frac{\pi}{2}</math>) and 3) a negative zero crossing (<math>\theta = \pi</math>).<br />
<br />
[[Image:Effects_Switch_Angle.png|left|frame|Figure 2. The effects of switching angle on flux]]<br />
<br clear=all><br />
<br />
Here it can be seen that switching at a positive zero crossing shifts the flux waveform up to a peak flux of 2pu (and vice versa for a negative zero crossing, i.e. peak flux down to -2pu). There is no dc offset when switching at a voltage peak. Any other switching angle will result in a dc offset in between the zero crossing waveforms. <br />
<br />
=== Effects of Residual Flux ===<br />
<br />
It can be readily seen from transient flux waveform in equation (1) that the residual flux causes a dc offset in the flux waveform, i.e. a positive residual flux offsets the waveform up and vice versa for a negative residual flux. <br />
<br />
Therefore, depending on the switching angle, a residual flux can actually be beneficial to keeping inrush currents low. For example, if the switching angle was a negative zero crossing (e.g. <math>\theta = \pi</math>, this would offset the flux waveform down. But a positive residual flux would have the opposite effect by offsetting the flux waveform up.<br />
<br />
On the other hand, the residual flux can reinforce the dc offset caused by the switching angle and can result in peak flux values (absolute) well over 2pu. <br />
<br />
=== Transient Flux in Practical Transformers ===<br />
<br />
In our idealised model above, the transient flux waveform on switching has constant dc offsets depending on the switching angle and residual flux. However in a practical transformer, the effect of leakage impedances (i.e. winding resistances and leakage reactances) will lead to an exponentially decaying dc offset component (like in the [[RL Circuit Switching|switching of RL circuits]]), with the rate of decay depending on the time constant <math>\frac{R}{L}</math>.<br />
<br />
The figure below presents a more realistic transient flux waveform on switching, displaying a decaying dc component:<br />
<br />
[[Image:TX_Inrush_Flux.png|left|frame|Figure 3. Example of transient flux waveform in practical transformers]]<br />
<br clear=all><br />
<br />
=== Non-Linear Relationship between Flux and Current ===<br />
<br />
[[Image:Hysteresis.PNG|right|thumb|220px|Figure 4. Flux-current hysteresis loop]]<br />
<br />
We saw above that during transient conditions, the magnetising flux can rise up over 2pu (depending on the switching angle and residual flux). The other key factor that leads to high inrush currents is the non-linear relationship between flux and current. <br />
<br />
The transformer core exhibits saturation characteristics like the hysteresis loop in the figure shown on the right. Here it can be seen that at 1 pu flux, the core is already saturating and the slope of the curve begins to approache a horizontal line. Therefore, if the flux is increased further above 1 pu flux, the current drawn can be several orders of magnitude higher. <br />
<br />
=== Summary ===<br />
<br />
Putting all the pieces together, the general intuition explaining transformer energisation inrush can be put as follows:<br />
<br />
:1) On the sudden application of a voltage to the transformer (i.e. circuit breaker is closed), a transient flux waveform is generated that, depending on the switching angle and residual flux, can reach a peak of over two times nominal flux.<br />
<br />
:2) The relationship between flux and current is highly non-linear. At higher than nominal flux, the core will begin to saturate and the related magnetising currents can get very high. This is equivalent to saying that as the core saturates, the magnetising reactance decreases significantly.<br />
<br />
== References ==<br />
<br />
:*[1] Greenwood, A., [http://www.amazon.com/gp/product/0471620580/ref=as_li_ss_tl?ie=UTF8&tag=openelect-20&linkCode=as2&camp=1789&creative=390957&creativeASIN=0471620580 "Electrical Transients in Power Systems"], Wiley, 1991 <br />
<br />
== Related Topics ==<br />
<br />
:* [[RL Circuit Switching]]<br />
<br />
[[Category:Modelling / Analysis]]</div>Juleshttp://openelectrical.org/index.php?title=File:Hysteresis.PNG&diff=158File:Hysteresis.PNG2020-11-22T08:26:15Z<p>Jules: </p>
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<div></div>Juleshttp://openelectrical.org/index.php?title=File:Effects_Switch_Angle.png&diff=157File:Effects Switch Angle.png2020-11-22T08:26:01Z<p>Jules: </p>
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<div></div>Jules