Empirical formula; voltage for 150 MW over 100 km

Appropriateness of the empirical formula

$$V_{econ} = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{\cos\phi \times N_c \times 150}}\ \text{kV}\quad (L\ \text{in km},\ P\ \text{in MW})$$

• It is derived from the balance of costs: conductor and loss cost fall with voltage, insulation and tower cost rise with voltage. The square root reflects $V_{econ} = \sqrt{A/B}$ from minimising $C = A/V + BV$.
• Length term $L/1.6$ ($L$ converted to miles): longer lines need higher voltage to keep drop and loss low.
• Power term $1000P/(\cos\phi\,N_c\,150)$: proportional to kVA per circuit; more power per circuit needs higher voltage. Increasing $N_c$ shares the power and lowers the voltage required.
• Merits: quick, needs only $P$, $L$, pf, $N_c$; gives a good starting point; matches common practice.
• Limitations: it is an old empirical (Still's type) formula; it ignores actual costs of conductor, energy and land, does not check stability, regulation, corona or thermal limits, and gives non-standard values. So the result must be rounded to a standard voltage and checked with the SIL-based capability curve.

Selection for 150 MW over 100 km

$\cos\phi$ not given, assumed 0.9.

$N_c = 1$:

$$\begin{aligned} V &= 5.5\sqrt{\frac{100}{1.6} + \frac{150000}{0.9 \times 150}} = 5.5\sqrt{62.5 + 1111.11} \\ &= 5.5 \times 34.258 = 188.42\ \text{kV} \end{aligned}$$

$N_c = 2$:

$$V = 5.5\sqrt{62.5 + 555.56} = 5.5 \times 24.861 = 136.73\ \text{kV}$$

Capability: m.f. at 100 km $= 2.75 - \frac{20}{80}(0.5) = 2.625$.

• 132 kV D/C: economical voltage 136.7 kV is very close to 132 kV; capability 228.7 MW (52% margin); if one circuit trips the other still carries 114 MW (76% of load).
• 220 kV S/C: economical voltage 188.4 kV lies between 132 and 220 kV; capability is more than twice the need and there is no backup circuit.

Answer: 132 kV, double circuit (closest to economical voltage, adequate capability, better reliability).