Question

in a 275KV transmission line with line constants A = 0.8585 o and B = 200Đ75 o , if the voltage profile at each end is to be maintained at 275kV, the power at Unity Power Factor
(UPF) will be nearly
98 MW
118 MW
144 MW
184 MW​

Answer 1

Answer:

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Answer 2

Answer:

b 118

Explanation:

Concept:

The reactive power flow in a transmission line is given by

\(Q = \frac{{{V_S}{V_R}}}{B}\sin \left( {\beta - \delta } \right) - \frac{{AV_R^2}}{B}\sin \left( {\beta - \alpha } \right) = 0\)

Where VS and VR are the sending end and receiving end voltages of transmission line respectively.

For unity power factor load, reactive power is zero i.e. Q = 0

The active power flow in a transmission line is given by,

\(P = \frac{{{V_S}{V_R}}}{B}\cos \left( {\beta - \delta } \right) - \frac{{AV_R^2}}{B}\cos \left( {\beta - \alpha } \right)\)

Calculation:

Given that, VS = 275 kV, VR = 275 kV

B = 200 ∠75° ⇒ β = 75°

A = 0.85 ∠5° ⇒ α = 5°

\(Q = \frac{{275\; \times \;275}}{{200}}\sin \left( {75^\circ - 5} \right) - \frac{{0.85\; \times \;{{275}^2}}}{{200}}\sin \left( {75^\circ - 5^\circ } \right) = 0\)

⇒ δ = 22°

Power delivered at unity power factor is,

\(P = \frac{{275\; \times \;25}}{{200}}\cos \left( {75^\circ - 22} \right) - \frac{{0.85\; \times \;{{275}^2}}}{{200}}\cos \left( {75^\circ - 5^\circ } \right)\)

⇒ P = 117.6 kW ≈ 118 MW