Efficiency at 0.995 leading pf and zero regulation

Short line, $Z = 4 + j8\ \Omega$/phase, $V_R = 33$ kV, load 20 MW at 0.995 pf leading.

Efficiency

$$\begin{aligned} I &= \frac{20\times10^6}{\sqrt3\times33\,000\times0.995} = 351.67\ \text{A} \\ P_{loss} &= 3I^2R = 3(351.67)^2(4) = 1.484\ \text{MW} \\ \eta &= \frac{20}{20 + 1.484}\times100 = 93.09\% \end{aligned}$$

Present regulation (for reference): $\phi = 5.73^\circ$ leading,

$$V_S = 19\,052.6 + 351.67\angle5.73^\circ\,(4 + j8) = 20\,384.3\angle8.29^\circ\ \text{V}$$

i.e. 35.31 kV line, regulation $= 6.99\%$. It is still positive because the pf is only slightly leading.

Making regulation zero

For a short line,

$$\%VR \approx \frac{I(R\cos\phi - X\sin\phi)}{V_R}\times 100 \quad (\text{leading pf})$$

This is zero when

$$\tan\phi = \frac{R}{X} = \frac{4}{8} = 0.5 \Rightarrow \cos\phi = 0.894\ \text{leading}$$

So regulation is made zero by making the load (or load plus compensator) take more leading current:

• Connect a shunt capacitor bank / synchronous condenser at the receiving end.
• Required net leading reactive power: $Q = P\tan\phi = 20\times0.5 = 10$ MVAr.
• The load already supplies $20\tan(5.73^\circ) = 2.01$ MVAr leading, so the capacitor needed is

$$Q_C = 10 - 2.01 = 7.99\ \text{MVAr}$$ $$C_{ph} = \frac{Q_C}{\omega V_L^2} = \frac{7.99\times10^6}{314.16\times33\,000^2} = 23.36\ \mu\text{F (star)}$$

(An exact solution of $|V_S| = |V_R|$ gives net pf 0.847 leading and $Q_C$ ≈ 10.55 MVAr.)

Answer: Efficiency ≈ 93.09 %. Zero regulation needs a net pf of about 0.894 leading, i.e. a shunt capacitor of about 8 MVAr at the receiving end.