Question

An 800 kV transmission line has a maximum power transfer capacity of P. When operated at 400 kV with the series reactance unchanged, the new maximum power transfer capacity is approximately
1. P
2. 2 P
3. P / 2
4. P / 4

Answer 1

Correct Answer - Option 4 : P / 4

Concept:

Power transfer capability of the transmission line is the indication of how much power transfers can be increased without compromising system security.

\(P\propto \frac{{{V}_{S}}\cdot {{V}_{R}}}{{{X}_{S}}}\)

\(or~{{P}_{max}}=\frac{V_{S}^{2}}{{{X}_{S}}}\)

It can be increased either by increasing voltage or by decreasing the reactance of the line.

Calculation:

It is given that X_S = constant

\({{P}_{max}}=\frac{V_{S}^{2}}{{{X}_{S}}}~or~P\propto V_{S}^{2}\)

At V_S = 800 V, P_m1 ∝ (800)²

And at V_S = 400 V, P_m2 ∝ (400)²

\(\Rightarrow \frac{{{P}_{m1}}}{{{P}_{m2}}}=\frac{{{\left( 800 \right)}^{2}}}{{{\left( 400 \right)}^{2}}}=4\)

⇒ P₂ = P₁ / 4

And power transfer capability at V_S = 800 V is P.

⇒ Power transfer capability at V_S = 400 V

\(\Rightarrow {{P}_{2}}=\frac{P}{4}\)