Compensation capacity for 15 MW, 33 kV load

Treat it as a short line: $A = 1\angle0^\circ$, $B = Z = 5 + j25 = 25.495\angle78.69^\circ\ \Omega$. With line kV and three-phase MW/MVAr, the receiving-end power circle equations are:

$$\begin{aligned} P_R &= \frac{|V_S||V_R|}{|Z|}\cos(\theta-\delta) - \frac{|V_R|^2}{|Z|}\cos\theta \\ Q_R &= \frac{|V_S||V_R|}{|Z|}\sin(\theta-\delta) - \frac{|V_R|^2}{|Z|}\sin\theta \end{aligned}$$

with $\theta = 78.69^\circ$.

Part 1: Compensation for $|V_S| = |V_R| = 33$ kV

$$\frac{33\times33}{25.495} = 42.714,\quad \frac{33^2}{25.495}\cos\theta = 8.377,\quad \frac{33^2}{25.495}\sin\theta = 41.885$$

Active power condition:

$$\begin{aligned} 15 &= 42.714\cos(78.69^\circ-\delta) - 8.377 \\ \cos(78.69^\circ-\delta) &= 0.5473 \Rightarrow \delta = 21.87^\circ \end{aligned}$$

Reactive power the line can deliver at this condition:

$$Q_R = 42.714\sin(56.82^\circ) - 41.885 = -6.14\ \text{MVAr}$$

The load needs $Q_L = 15\tan(\cos^{-1}0.8) = 11.25$ MVAr (lagging), and the line itself needs 6.14 MVAr fed into its receiving end. So the compensator must supply:

$$Q_C = 11.25 - (-6.14) = 17.39\ \text{MVAr}$$

Part 2: Extra load with $|V_R| = 28$ kV and the same $Q_C$

$$\frac{33\times28}{25.495} = 36.242,\quad \frac{28^2}{25.495}\cos\theta = 6.031,\quad \frac{28^2}{25.495}\sin\theta = 30.154$$

For a new load $P$ at 0.8 pf lag, the line must deliver $Q_R = 0.75P - 17.39$. So:

$$\begin{aligned} P &= 36.242\cos(78.69^\circ-\delta) - 6.031 \\ 0.75P - 17.39 &= 36.242\sin(78.69^\circ-\delta) - 30.154 \end{aligned}$$

Solving these together (eliminating $P$):

$$\delta = 31.33^\circ,\quad P = 18.52\ \text{MW},\quad Q_R = -3.49\ \text{MVAr}$$

Check: $0.75 \times 18.52 - 17.39 = -3.49$ MVAr.

Extra load $= 18.52 - 15 = 3.52$ MW.

Answer: Compensator capacity ≈ 17.39 MVAr; extra load ≈ 3.52 MW at 0.8 pf lag (total 18.52 MW at 28 kV).