Answer to Question #207658 in Electrical Engineering for Tumelo

Question #207658

4. Two power stations, A and B, which are in phase at 66 kV, supp!y respective

inductive loads of 455 MVA at an inductive power factor of 0.766 and 555

MVA at a lagging power factor of 0.88295 respectively. An inter-connector

with an impedance of 4.45∠75 0 ohms per line, links station A with station B.

By means of a phase regulator at A, the phase angle of the voltage at A is

advanced in order to increase the loading of station A to 620 MW.

4.1 Use standard formulae and calculate the angle of advance at power station A.


4.2 Calculate the transmission line active power loss.

(19.765 MW)

Expert's answer

4.1 The angle of advance at power station A.

"P=Re[V_A* \\frac{V_A-V_B}{Z}]"

"620*10^6=Re[66 \\angle 0* \\frac{66 \\angle 0-66}{4.45 \\angle 75}]*10^6"

"620*10^6= \\frac{66^2*10^6}{4.45^2+7.5^2}[4.45^2(1- cos \\theta )+7.5 sin \\theta]"

Solving the above equation we get "\\theta = 16.1^0"

4.2 The transmission line active power loss.

Power at B

"P=V_B(\\frac{V_A-V_B}{Z})=66 \\angle 0(\\frac{66 \\angle 16.1 -66}{4.45 \\angle 75}) = 600.2MW"

We can recall that Power at A = 620 MW

Therefore, Power loss "P_A-P_B=620-600.2=19.8 MW"

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