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# 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

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

(16.060)

4.2 Calculate the transmission line active power loss.

(19.765 MW)

1
2021-06-18T07:54:18-0400

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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