Answer to Question #207658 in Electrical Engineering for Tumelo

Question

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.

(16.060)

4.2 Calculate the transmission line active power loss.

(19.765 MW)

Answer 1

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 $P_A-P_B=620-600.2=19.8 MW$