Two machines 3.81∠0° and 3.81∠-18° kV

Let I flow from machine 1 to machine 2 through $Z = 3 + j10\ \Omega$. Then $S_1 = V_1I^*$ is the power delivered by machine 1 and $S_2 = V_2I^*$ is the power received by machine 2 (per phase; three-phase = 3 × per phase).

Current

$$\begin{aligned} V_1 &= 3810\angle0^\circ\ \text{V}\\ V_2 &= 3810\angle-18^\circ = 3623.53 - j1177.35\ \text{V}\\ I &= \frac{V_1 - V_2}{Z} = \frac{186.47 + j1177.35}{3 + j10}\\ &= 113.15 + j15.30 = 114.18\angle7.70^\circ\ \text{A} \end{aligned}$$

Powers per phase

$$\begin{aligned} S_1 &= V_1I^* = 3810(113.15 - j15.30) = 431.09 - j58.28\ \text{kVA}\\ S_2 &= V_2I^* = (3810\angle-18^\circ)(114.18\angle-7.70^\circ) = 391.98 - j188.64\ \text{kVA}\\ S_{line} &= |I|^2Z = 13036.1(3 + j10) = 39.11 + j130.36\ \text{kVA} \end{aligned}$$

Check: $431.09 - 391.98 = 39.11$ kW and $188.64 - 58.28 = 130.36$ kvar.

Results

Direction of flow

• Active power flows from machine 1 to machine 2, because $V_1$ leads $V_2$ by 18°.
• Reactive power flows from machine 2 towards machine 1. Machine 2 supplies 188.64 kvar: 130.36 kvar is used in the line reactance and 58.28 kvar reaches machine 1. Both voltage magnitudes are equal, so the flow of Q depends on the angle and on the line resistance.

Answer: Machine 1 is the generator (1293.3 kW, three-phase) and machine 2 the motor (1175.9 kW). The line loss is 117.3 kW. Q flows from machine 2 (565.9 kvar) to machine 1 (174.8 kvar received). The line absorbs 391.1 kvar.