Line 65:

: <math>||Z_{s}|| = \frac{V^{2}}{S_{SC}} = \frac{33kV^{2}}{800MVA} = 1.361 \Omega \, </math>

: <math>||Z_{s}|| = \frac{V^{2}}{S_{SC}} = \frac{33kV^{2}}{800MVA} = 1.361 \Omega \, </math>

−

Converted to per-unit values (on a 100MVA base):

+

Converted to per-unit values (on a 100 MVA base):

−

: <math>||Z_{s,pu}|| = \frac{||Z_{s}||}{Z_{base}} = \frac{1.361}{4.84} =  0.281 pu \, </math>

+

: <math>||Z_{s,pu}|| = \frac{||Z_{s}||}{Z_{base}} = \frac{1.361}{10.89} =  0.125 pu \, </math>

For an X/R ratio of 10, the source impedance is therefore:

For an X/R ratio of 10, the source impedance is therefore:

Line 77:

: <math>\boldsymbol{S_{pu}} = 0.5 - j0.375 pu = 0.625 \angle -0.6435 \, </math> (angle in radians)

: <math>\boldsymbol{S_{pu}} = 0.5 - j0.375 pu = 0.625 \angle -0.6435 \, </math> (angle in radians)

−

Plugging the parameters <math>S = 0.625 \, </math>, <math>Z = 0.281 \, </math>, <math>\phi = -0.6435 \, </math> and <math>\theta = 1.471 \, </math> into Equation (1), we get the load bus voltage:

+

Plugging the parameters <math>S = 0.625 \, </math>, <math>Z = 0.125\, </math>, <math>\phi = -0.6435 \, </math> and <math>\theta = 1.471 \, </math> into Equation (1), we get the load bus voltage:

−

: <math>V_r = 0.8480 pu \, </math>

+

: <math>V_r = 0.9433 pu \, </math>

== Intuition for General Power Flow Solutions ==

== Intuition for General Power Flow Solutions ==

Open Electrical - Recent changes [en]

Simple Power Flow Example