Zero regulation for 40 MW leading load

Short line: $Z = 12 + j36\ \Omega$, $V_R$ = 132 kV, load 40 MW at 0.9 pf leading.

Present regulation

$$\begin{aligned} I &= \frac{40\times10^6}{\sqrt3\times132\,000\times0.9} = 194.39\ \text{A},\quad \phi = 25.84^\circ\ \text{lead} \\ V_S &= 76\,210.2 + 194.39\angle25.84^\circ\,(12 + j36) \\ &= 75\,613.9\angle5.55^\circ\ \text{V} \Rightarrow 130.97\ \text{kV (line)} \\ \%VR &= \frac{130.97 - 132}{132}\times100 = -0.78\% \end{aligned}$$

The regulation is negative: the load is too leading, so the receiving-end voltage is higher than the sending-end voltage.

Condition for zero regulation

Approximate condition: $R\cos\phi - X\sin\phi = 0$, i.e. $\tan\phi = R/X = 1/3$, giving pf $= 0.949$ leading.

Exact condition $|V_S| = |V_R| = 132$ kV with $P$ = 40 MW (line kV, MW, MVAr):

$$(V_R^2 + RP + XQ)^2 + (XP - RQ)^2 = V_R^4$$ $$(17\,424 + 480 + 36Q)^2 + (1440 - 12Q)^2 = 17\,424^2$$

Taking the root nearer zero:

$$Q = -15.44\ \text{MVAr}\quad(\text{pf} = 0.933\ \text{leading})$$

How to make it zero

The load now supplies $Q_L = -40\tan25.84^\circ = -19.37$ MVAr (leading). The leading VAr must be reduced to 15.44 MVAr, so connect a shunt reactor at the receiving end of

$$Q_{reactor} = 19.37 - 15.44 = 3.93\ \text{MVAr}$$ $$L_{ph} = \frac{V_L^2}{\omega Q} = \frac{132\,000^2}{314.16\times3.93\times10^6} = 14.1\ \text{H (star)}$$

(Alternatively, reduce the excitation of a synchronous load/condenser so that it draws less leading current.)

Answer: Regulation is now −0.78 %. It is made zero by absorbing ≈ 3.9 MVAr with a shunt reactor at the receiving end, bringing the net pf to ≈ 0.933 leading.