Active power between machines, 200∠0° and 200∠15° V

For a purely reactive link the real power transfer is

$$P = \frac{|V_A||V_B|}{X}\sin(\delta_A - \delta_B)$$

and it flows from the machine with the leading angle to the one with the lagging angle.

Calculation (per phase)

$$\begin{aligned} P_{A\to B} &= \frac{200\times200}{10}\sin(0^\circ - 15^\circ)\\ &= 4000\times(-0.2588) = -1035.28\ \text{W} \end{aligned}$$

The negative sign means the power actually flows from B to A.

Check with phasors:

$$\begin{aligned} I_{A\to B} &= \frac{200\angle0^\circ - 200\angle15^\circ}{j10} = 5.221\angle-172.5^\circ\ \text{A}\\ S_A &= V_AI^* = 200\times5.221\angle172.5^\circ = -1035.28 + j136.30\ \text{VA}\\ S_B &= V_BI^* = -1035.28 - j136.30\ \text{VA} \end{aligned}$$

So A receives 1035.28 W and B delivers 1035.28 W. The link has no resistance, so there is no real-power loss. Each machine supplies 136.30 var to the link ($I^2X = 27.26\times10 = 272.59$ var).

Answer: About 1.035 kW per phase (3.106 kW three-phase) flows from machine B to machine A. B acts as the generator and A as the motor, because $V_B$ leads $V_A$ by 15°.