Question

A transmission line that has a resistance per unit length of 2.40 10-4 Ω/m is to be used to transmit 5.00 MW across 400 mi (6.44 105 m). The output voltage of the source is 4.50 kV. (a) What is the line loss if a transformer is used to step up the voltage to 506 kV? (b) What fraction of the input power is lost to the line under these circumstances? (c) What difficulties would be encountered in attempting to transmit the 5.00 MW at the source voltage of 4.50 kV?

Answer 1

Answer:

(a) 30.174 kW

(b) $6.035\times 10^{- 3}\ N$

(c) I = 1.112 kA

Solution:

As per the question:

Resistance per unit length, $\sigma = 2.40\times 10^{- 4}\Omega/ m$

Power transmitted, P = 5.00 MW = $5\times 10^{6}\ W$

Length, L = $6.44\times 10^5\ m$

Output voltage, V = 4.50 kV

Stepped-up Voltage, V' = 506 kV

Now,

(a) The line loss can be calculated as:

Current, I = $\frac{P}{V'} = \frac{5\times 10^{6}}{506\times 10^{3}} = 9.88\ A$

To calculate the power loss:

$P_{L} = I^{2}R$

where

R = Resistance in the transmission line

R = $2\sigma L = 2\times 2.40\times 10^{- 4}\times 6.44\times 10^5 = 309.12\Omega$

Thus

$P_{L} = 9.88^{2}\times 309.12 = 30.174\ kW$

(b) Fraction of the input power lost is given by:

$f = \frac{P_{L}}{P} = \frac{30.174\times 10^{3}}{5\times 10^{6}} = 6.035\times 10^{- 3}\ W$

(c) To calculate the current at the sending end at ther source voltage of 4.50 kV:

$I = \frac{P}{V} = \frac{5\times 10^{6}}{4.50\times 10^{3}} = 1.112\ kA$

This value of current is very high and sending such a high watt power is impossible, even if the circuit at the other end is open then also this current can be destructive.