Power by two generators, 1.05∠5° and 1.05∠20° pu

The two internal reactances and the line form one series path between the two emfs:

$$Z = j0.08 + (0.025 + j0.1) + j0.08 = 0.025 + j0.26\ \text{pu}$$

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 absorbed by G2.

  j0.08    0.025+j0.1    j0.08
 +-/\/\/-----/\/\/-------/\/\/-+
 |       I -->                 |
(E1) 1.05/5 deg      1.05/20 deg (E2)
 |                             |
 +-----------------------------+

Current

$$\begin{aligned} E_1 &= 1.05\angle5^\circ = 1.0460 + j0.0915\\ E_2 &= 1.05\angle20^\circ = 0.9867 + j0.3591\\ I &= \frac{E_1 - E_2}{Z} = \frac{0.0593 - j0.2676}{0.025 + j0.26}\\ &= -0.9981 - j0.3242 = 1.0494\angle-162.01^\circ\ \text{pu} \end{aligned}$$

Powers

$$\begin{aligned} S_1 &= E_1I^* = -1.0737 + j0.2477\ \text{pu}\\ S_2 &= E_2I^* = -1.1012 - j0.0386\ \text{pu}\ \text{(absorbed by G2)}\\ S_{loss} &= |I|^2Z = 1.1013(0.025 + j0.26) = 0.0275 + j0.2863\ \text{pu} \end{aligned}$$

Check: $1.1012 - 1.0737 = 0.0275$ pu (I²R loss) and $0.2477 + 0.0386 = 0.2863$ pu (I²X).

Direction of power flow

• Real power flows from G2 to G1, since $E_2$ leads $E_1$ by 15°.
• Both generators supply reactive power to the network, and it is consumed in the series reactances. With equal emf magnitudes, each machine supplies part of the $I^2X$.

Answer: G2 generates 1.101 pu real power and G1 receives 1.074 pu. The loss is 0.0275 pu. G1 and G2 supply 0.248 pu and 0.039 pu reactive power. P flows G2 → G1.