Jules at 08:23, 22 November 2020

2020-11-22T08:23:24Z

Jules: Created page with "== Introduction == The RL switching / closing transient is one of the most common electrical transients that is encountered in practice, and is also the basis for the computa..."

2020-11-22T08:21:52Z

Created page with "== Introduction == The RL switching / closing transient is one of the most common electrical transients that is encountered in practice, and is also the basis for the computa..."

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== Introduction ==

The RL switching / closing transient is one of the most common electrical transients that is encountered in practice, and is also the basis for the computation of [[Short_Circuit_Calculation|short circuit currents]].

== Derivation ==

[[Image:RL_Switching.PNG|right|thumb|400px|Figure 1. Basic RL switching circuit]]

Consider the basic switching circuit in the figure to the right, consisting of an AC voltage source V, a switch S, a resistance R and an inductance L (all ideal circuit elements). At time <math>t=0 \,</math>, the switch S will close and complete the circuit. Suppose the voltage source can be characterised as a sinusoid (as a function of time):

: <math> V(t) = V_{m} \sin(\omega t + \theta) \,</math>

Where <math>\theta \,</math> is an arbitrary phase angle to capture the time of switching.

At the point of switching, the voltage is given by [http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws Kirchhoff's voltage law]:

: <math> V_{m} \sin(\omega t + \theta) = R I(t) + L \frac{dI(t)}{dt} \,</math>

The current <math>I(t) \,</math> must reach a steady state current of:

: <math>I_{s} = \frac{V}{Z} = \frac{V}{R + j \omega L} \,</math> ... Equ. (1)

And have a steady state power factor:

: <math> \cos{\phi_{s}} = \frac{R}{|Z|} = \frac{R}{\sqrt{R^{2} + \omega^{2} L^{2}}} \,</math>

However, at <math>t=0 \,</math>, <math>I(0) = 0 \,</math> and the inductance <math>L \,</math> will prevent the circuit from reaching the steady-state current instantaneously. Therefore, there must be some transient that will provide a continuous transition path from <math>I(0) = 0 \,</math> to the steady-state current <math>I(t_{s}) = I_{s} \,</math>.

Equation (1) can be re-written as follows:

: <math> V_{m} \left[ \sin(\omega t)\cos \theta + \cos(\omega t)\sin \theta \right] = R I(t) + L \frac{dI(t)}{dt} \,</math>

Taking the Laplace transform of both sides, we get:

: <math> V_{m} \left( \frac{\omega \cos \theta}{s^{2} + \omega^{2}} + \frac{s \sin \theta}{s^{2} + \omega^{2}} \right) = R i(s) + s L i(s) - L I(0) \,</math>

We assume that the initial current <math>I(0) = 0 \,</math>, so therefore re-arranging the equation above we get:

: <math>i(s) = \frac{V_{m}}{L} \left( \frac{1}{\frac{R}{L} + s} \right) \left( \frac{\omega \cos \theta}{s^{2} + \omega^{2}} + \frac{s \sin \theta}{s^{2} + \omega^{2}} \right) \,</math>

:: <math> = \frac{V_{m}}{L} \left[ \frac{\omega \cos \theta}{(s^{2} + \omega^{2})(\frac{R}{L} + s)} + \frac{s \sin \theta}{(s^{2} + \omega^{2})(\frac{R}{L} + s)} \right] \,</math> ... Equ. (2)

It can be shown that the expression <math> \frac{1}{(s + \alpha)(s^{2} + \omega^{2})} </math> can be simplified as follows:

: <math> \frac{1}{(s + \alpha)(s^{2} + \omega^{2})} = \frac{1}{\alpha^{2} + \omega^{2}} \left( \frac{1}{s + \alpha} - \frac{s}{s^{2} + \omega^{2}} + \frac{\alpha}{s^{2} + \omega^{2}} \right) \, </math>

Therefore, the [http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms inverse Laplace transforms] of the terms in Equation (2) can be evaluated in a fairly straightforward manner:

'''1st term:''' <math> \mathcal{L}^{-1} \left[ \frac{\omega \cos \theta}{(s^{2} + \omega^{2})(\frac{R}{L} + s)} \right] = \frac{\omega \cos \theta}{\left( \frac{R}{L} \right)^{2} + \omega^{2}} \left[ e^{-\frac{R}{L}t} - \cos (\omega t) + \frac{R}{\omega L} \sin (\omega t) \right] \,</math>

'''2nd term:''' <math> \mathcal{L}^{-1} \left[ \frac{s \sin \theta}{(s^{2} + \omega^{2})(\frac{R}{L} + s)} \right] = \frac{\sin \theta}{\left( \frac{R}{L} \right)^{2} + \omega^{2}} \left[ -\frac{R}{L} e^{-\frac{R}{L}t} + \omega \sin (\omega t) + \frac{R}{L} \cos (\omega t) \right] \,</math>

Combining the two terms together (and including the constants), we get the transient current:

: <math> I(t) = \frac{V_{m}}{L \left( \frac{R}{L} \right)^{2} + \omega^{2}} \left[ (\omega \cos \theta -\frac{R}{L} \sin \theta) e^{-\frac{R}{L}t} + (\frac{R}{L} \cos \theta + \omega \sin \theta) \sin (\omega t) - (\omega \cos \theta - \frac{R}{L} \sin \theta) \cos (\omega t) \right] \,</math> ... Equ. (3)

Earlier, we found that the steady state power factor is:

: <math> \cos{\phi_{s}} = \frac{R}{\sqrt{R^{2} + \omega^{2} L^{2}}} \,</math>

This can be re-arranged as follows:

: <math> \cos{\phi_{s}} = \frac{R}{L} \frac{1}{\sqrt{(\frac{R}{L})^{2} + \omega^{2}}} \,</math>

Likewise, the sine of the power angle is:

: <math> \sin{\phi_{s}} = \frac{\omega}{\sqrt{(\frac{R}{L})^{2} + \omega^{2}}} \,</math>

Using these two equations above, we can simplify Equation (3) even further:

: <math> I(t) = \frac{V_{m} \sqrt{(\frac{R}{L})^{2} + \omega^{2}}}{L \left[ \left( \frac{R}{L} \right)^{2} + \omega^{2} \right]} \left[ ( \cos \theta \sin \phi_{s} - \sin \theta \cos \phi_{s}) e^{-\frac{R}{L}t} + (\cos \theta \cos \phi_{s} + \sin \theta \sin \phi_{s}) \sin (\omega t) - (\cos \theta \sin \phi_{s} - \sin \theta \cos \phi_{s}) \cos (\omega t) \right] \,</math>

Using some [http://en.wikipedia.org/wiki/Trigonometric_identities#Angle_sum_and_difference_identities angle sum and difference trigonometric identities], we get:

: <math> I(t) = \frac{V_{m}} {\sqrt{(R^{2} + \omega^{2} L^{2}}} \left[ ( - \sin (\theta - \phi_{s}) e^{-\frac{R}{L}t} + (\cos \theta \cos \phi_{s} + \sin (\theta - \phi_{s}) \sin (\omega t) + (\cos (\theta - \phi_{s}) \cos (\omega t) \right] \,</math>

Simplifying again with the same trig identities, we get the final equation:

: <math> I(t) = \frac{V_{m}} {\sqrt{(R^{2} + \omega^{2} L^{2}}} \left[ ( \sin (\omega t + \theta - \phi_{s}) - \sin (\theta - \phi_{s}) e^{-\frac{R}{L}t} \right] \,</math> ... Equ. (4)

== Interpretation ==

[[Image:RL_Transient.png|right|thumb|400px|Figure 2. RL switching transient current]]

The figure right depicts a plot of the transient current in Equation (4) for the parameters R/L = 40 and switching angle <math>\theta</math> = 0<sup>o</sup>. Here we see the classic transient current waveform for an RL switching (closing) circuit with the time constant R/L.

[[Category:Modelling / Analysis]]

@@ Line 1: / Line 1: @@
 == Introduction ==
-The RL switching / closing transient is one of the most common electrical transients that is encountered in practice, and is also the basis for the computation of [[Short_Circuit_Calculation|short circuit currents]].
+The RL switching / closing transient is one of the most common electrical transients that is encountered in practice, and is also the basis for the computation of [[Short Circuit|short circuit currents]].
 == Derivation ==
@@ Line 88: / Line 88: @@
 The figure right depicts a plot of the transient current in Equation (4) for the parameters R/L = 40 and switching angle <math>\theta</math> = 0<sup>o</sup>. Here we see the classic transient current waveform for an RL switching (closing) circuit with the time constant R/L.
 [[Category:Modelling / Analysis]]

RL Circuit Switching - Revision history

Jules at 08:23, 22 November 2020

Jules: Created page with "== Introduction == The RL switching / closing transient is one of the most common electrical transients that is encountered in practice, and is also the basis for the computa..."