Answer to Question #207660 in Electrical Engineering for Tumelo

Question

Two power stations, A and B, work at 88 kV and deliver power to their separate

consumers as follows:

Power station B: 540 MVA at a power factor of 0.809 lagging

Power station A: 480 MVA at a power factor of 0.8387 lagging

An inter-connector with an impedance of Z = 7.25∠820 ohms per line joins power

station A with power station B. The voltage at A is advanced with an angle of 7.80

with respect to the voltage of B.

1.1 Use the complex power method and calculate the active power and reactive power

received and transferred.

(144.885 MW; - 10.3855 MVAr; 142.135 MW; - 29.9525 MVAr)

1.2 Write down the final distribution of active and reactive power between the tvo power

Stations.

(A sends 547.447 MW and sends 251.041 MVA'r ; B receives 579.0045 MW and receives 287.451 MVA'r)

1.3 Calculate the final power factor at the generators of each power station.

(0.909 lagging ; 0.8957 lagging)

Answer 1

1.1

$I_1=I_i \implies \frac{V_A-V_B}{2}= \frac{88 \angle 7.8 -88\angle 0}{7.25 \angle 82}= 1.668 \angle 11.56^0$

At power station A

$I_1 =\frac{480*10^6}{\sqrt{3}*88}=3.49183 \angle -32.997$

$I= I_1+I_2=1.668 \angle 11.56 +3.14918 \angle -32.997 = 4.82 \angle -21.437 ^0$

Complex power at station A

$P_A= V_AI= 88 \angle 7.8 * 4.82 \angle 21.44 = 144.885-j10.3835$

Active power = 387.412 MW

Reactive power = 10.3835

At power station B

$I_2 =\frac{540*10^6}{\sqrt{3}*88}=3.542831kA$

$I= I_1+I_2=4.81718 \angle -21.437 +3.54283\angle -36.002 = 142.135 \angle -29.9525$

Active power = 142.135 MW

Reactive power = 29.9525

1.2

From the above calculations, A sends 547.447 MW and sends 251.041 MVA'r ; B receives 579.0045 MW and receives 287.451 MVA'

1.3

$cos \phi = 0.6145 \implies \phi = 0.909 lagging$

$cos \phi = 0.62497 \implies \phi = 0.8957 lagging$