Two machines 11∠0° and 11∠40° kV via j15 Ω

Assumption: 11 kV is the line-to-line EMF. The per-phase EMFs are therefore $E_1 = 6.351\angle0^\circ$ kV and $E_2 = 6.351\angle40^\circ$ kV. The current $I$ is taken as flowing from machine 1 to machine 2 through $Z = j15\ \Omega$.

Current

$$\begin{aligned} I &= \frac{E_1 - E_2}{jX} = \frac{6351\angle0^\circ - 6351\angle40^\circ}{j15}\\ &= 289.6\angle-160^\circ\ \text{A} \end{aligned}$$

Complex power (three-phase)

Machine 1, with power taken out of its terminal:

$$S_1 = 3E_1I^* = -5.185 + j1.887\ \text{MVA}$$

Machine 2, with power taken into it (since $I$ enters it):

$$S_2 = 3E_2I^* = -5.185 - j1.887\ \text{MVA}$$

Check with the standard formulas:

$$\begin{aligned} P &= \frac{E_1E_2\sin\delta}{X} = \frac{11\times11\times\sin40^\circ}{15} = 5.185\ \text{MW}\\ Q &= \frac{E^2(1-\cos\delta)}{X} = \frac{121(1-\cos40^\circ)}{15} = 1.887\ \text{Mvar} \end{aligned}$$

(i) Real power

• Machine 1: $P_1 = -5.185$ MW out of it, so it consumes 5.185 MW and acts as a motor.
• Machine 2: $P_{2,in} = -5.185$ MW, so it generates 5.185 MW and acts as a generator.

Machine 2's EMF leads by 40°, so real power flows from 2 to 1. The line is lossless ($R = 0$), so the two powers are equal.

(ii) Reactive power

• Machine 1: $Q_1 = +1.887$ Mvar out of it, so it delivers 1.887 Mvar.
• Machine 2: $Q_{2,in} = -1.887$ Mvar, so it also delivers 1.887 Mvar.

Both machines supply reactive power to the line, which absorbs:

$$3|I|^2X = 3\times289.6^2\times15 = 3.774\ \text{Mvar}$$

Answer: Machine 1 is a motor consuming 5.185 MW, and machine 2 is a generator producing 5.185 MW. Each machine supplies 1.887 Mvar, a total of 3.774 Mvar absorbed by the line reactance.