33 kV, 12 MW leading load: efficiency, zero regulation

Short line: $Z = 6 + j16\ \Omega$, $V_R$ = 33 kV, load 12 MW at 0.9 pf leading.

Efficiency

$$\begin{aligned} I &= \frac{12\times10^6}{\sqrt3\times33\,000\times0.9} = 233.27\ \text{A} \\ P_{loss} &= 3I^2R = 3(233.27)^2(6) = 0.979\ \text{MW} \\ \eta &= \frac{12}{12 + 0.979}\times100 = 92.45\% \end{aligned}$$

Present regulation

$$\begin{aligned} V_S &= 19\,052.6 + 233.27\angle25.84^\circ\,(6 + j16) = 19\,102.3\angle11.99^\circ\ \text{V} \\ &\Rightarrow 33.09\ \text{kV (line)},\quad \%VR = 0.26\% \end{aligned}$$

The regulation is already very small, because the leading load current nearly cancels the drop.

Making regulation zero

Zero regulation needs $|V_S| = |V_R|$. With line kV, MW, MVAr:

$$(V_R^2 + RP + XQ)^2 + (XP - RQ)^2 = V_R^4$$ $$(1089 + 72 + 16Q)^2 + (192 - 6Q)^2 = 1089^2$$

The root nearer zero is

$$Q = -6.01\ \text{MVAr} \quad (\text{pf} = 0.894\ \text{leading})$$

The load supplies $-12\tan25.84^\circ = -5.81$ MVAr, so a little more leading VAr is needed:

$$Q_C = 6.01 - 5.81 = 0.20\ \text{MVAr}$$

So connect a small shunt capacitor (about 0.2 MVAr, $C = Q_C/\omega V_L^2 \approx 0.58\ \mu$F per phase in star) at the receiving end, or raise the leading current of a synchronous load slightly. In general, regulation is made zero by adjusting receiving-end reactive power (capacitor or reactor) until the net pf is near $\cos\phi = X/|Z|$ leading.

Note: the approximate rule $\tan\phi = R/X$ gives pf 0.936 lead here, but at such small regulation the neglected quadrature term matters, so the exact value above is used.

Answer: Efficiency ≈ 92.45 %. Regulation (now 0.26 %) becomes zero with about 0.2 MVAr of extra shunt capacitance at the load (net pf ≈ 0.894 leading).