Power by two generators, 1.1∠15° and 1.1∠0° pu

Total series impedance between the two emfs:

$$Z = j0.1 + (0.03 + j0.1) + j0.1 = 0.03 + j0.3\ \text{pu}$$

Let I flow from $E_1$ to $E_2$. Then $S_1 = E_1I^*$ is the power generated by G1 and $S_2 = E_2I^*$ is the power received by G2.

  j0.1     0.03+j0.1     j0.1
 +-/\/\/-----/\/\/-------/\/\/-+
 |       I -->                 |
(E1) 1.1/15 deg         1.1/0 deg (E2)
 |                             |
 +-----------------------------+

Current

$$\begin{aligned} E_1 &= 1.1\angle15^\circ = 1.0625 + j0.2847\\ E_2 &= 1.1\angle0^\circ = 1.1\\ I &= \frac{E_1 - E_2}{Z} = \frac{-0.0375 + j0.2847}{0.03 + j0.3}\\ &= 0.9272 + j0.2177 = 0.9524\angle13.21^\circ\ \text{pu} \end{aligned}$$

Powers

$$\begin{aligned} S_1 &= E_1I^* = (1.1\angle15^\circ)(0.9524\angle-13.21^\circ) = 1.0472 + j0.0327\ \text{pu}\\ S_2 &= E_2I^* = (1.1)(0.9524\angle-13.21^\circ) = 1.0200 - j0.2394\ \text{pu}\\ S_{loss} &= |I|^2Z = 0.9071(0.03 + j0.3) = 0.0272 + j0.2721\ \text{pu} \end{aligned}$$

Check: $1.0472 - 1.0200 = 0.0272$ pu and $0.0327 + 0.2394 = 0.2721$ pu.

Direction of power flow

• Real power flows from G1 to G2, because $E_1$ leads $E_2$ by 15°. G1 is generating and G2 is motoring.
• Both machines supply reactive power to the network, and it is all consumed in the series reactances ($I^2X = 0.2721$ pu).

Answer: G1 generates 1.047 pu of real power. G2 receives 1.020 pu. The line and reactances absorb 0.027 pu. Reactive power supplied is 0.033 pu by G1 and 0.239 pu by G2. P flows G1 → G2.