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Question: A thermal power plant produces electric power of \[600kW\] at \[4000V\], which is to be transported ...

A thermal power plant produces electric power of 600kW600kW at 4000V4000V, which is to be transported to a place 20km20km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with power factor unity. All the currents and voltages mentioned are rms values.

If the direct transmission method with a cable of resistance 0.4Ωkm10.4\Omega k{{m}^{-1}} is used, the power dissipation (in %) during transmission is?
A)20A)20
B)30B)30
C)40C)40
D)50D)50

Explanation

Solution

The power dissipated in a transmission line is proportional to square of the current. Here to find the power, we are given the values of electrical power produced and the voltage. Using the resistance per kilometer, we can find total resistance across 20km20km cable. By substituting these values in the formula for power dissipated, we can find the percentage loss in power.

Formula used:
P=I2RP={{I}^{2}}R
I=PVI=\dfrac{P}{V}

Complete answer:
Here,
Resistance, R=0.4Ω/kmR=0.4\Omega /km
Then, total resistance across 20km20km cable =0.4Ω×20=8Ω=0.4\Omega \times 20=8\Omega
We have,
P=I2RP={{I}^{2}}R ---------- 1
Where,
PP is the power dissipated
II is the rms current
RR is the resistance
We have,
Current, I=PVI=\dfrac{P}{V}
Where,
PP is the electrical power
VV is the voltage
Given that, P=600kWP=600kW and V=4000VV=4000V. Then
Current, I=600×1034000=150AI=\dfrac{600\times {{10}^{3}}}{4000}=150A
Substitute the value of II and RR in equation 1, we get,
P=1502×8=180000W=180kWP={{150}^{2}}\times 8=180000W=180kW
Percentage loss in power=180600×100=30%Percentage\text{ }loss\text{ }in\text{ }power=\dfrac{180}{600}\times 100=30\%

So, the correct answer is “Option B”.

Note:
The transmission over long distances creates power losses. The major part of the energy losses comes from power lines and Joule effect in transformers. The energy is lost as heat in the conductors. And the transmission lines have some amount of resistance. Hence power loss in transmission lines cannot be avoided.