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PhyMedAdv 2013 • Paper 2

Direct Transmission Power Dissipation Percentage Calculation

Comprehension Passage

A thermal power plant produces electric power of 600 kW600\text{ kW} at 4000 V4000\text{ V}, which needs to be delivered to a facility 20 km20\text{ km} away for consumer usage. Power can be transported either directly using a high current carrying capacity cable, or through a system involving step-up and step-down transformers at opposite ends. Direct transmission leads to significant energy loss through dissipation, whereas the transformer method reduces dissipation drastically. In this process, a step-up transformer at the power station reduces the transmission current. At the consumer end, a step-down transformer reduces the voltage to the required lower value. Assume the transmission cable is strictly resistive, all transformers are ideal with a power factor of unity, and all specified values of voltage and current are root-mean-square (rms) values.

If direct transmission is implemented using a cable having a resistance of 0.4 Ω km−10.4\ \Omega\text{ km}^{-1}, calculate the percentage power dissipation during power transmission.

Options

A

2020

B

3030

Correct
C

4040

D

5050

Topics & Concepts

Alternating CurrentTransformersAC Circuits

Step-by-Step Solution

To find the percentage power dissipation during direct power transmission, we follow these steps:

1. Given Data:

  • Power produced by plant, P=600 kW=600×103 WP = 600 \text{ kW} = 600 \times 10^3 \text{ W}
  • Transmission voltage, V=4000 VV = 4000 \text{ V}
  • Transmission distance, d=20 kmd = 20 \text{ km}
  • Resistance per unit length of cable, r=0.4 Ω km−1r = 0.4 \ \Omega \text{ km}^{-1}

2. Resistance of the Transmission Cable (RR): R=r×d=0.4 Ω km−1×20 km=8 ΩR = r \times d = 0.4 \ \Omega \text{ km}^{-1} \times 20 \text{ km} = 8 \ \Omega

3. Transmission Current (II): Since the power factor is unity (cos⁡ϕ=1\cos\phi = 1), the rms current in direct transmission is given by: I=PV=600×103 W4000 V=150 AI = \frac{P}{V} = \frac{600 \times 10^3 \text{ W}}{4000 \text{ V}} = 150 \text{ A}

4. Power Dissipated in Cable (PlossP_{\text{loss}}): The power dissipated as heat in the resistive cable is: Ploss=I2R=(150 A)2×8 Ω=22500×8=180000 W=180 kWP_{\text{loss}} = I^2 R = (150 \text{ A})^2 \times 8 \ \Omega = 22500 \times 8 = 180000 \text{ W} = 180 \text{ kW}

5. Percentage Power Dissipation: Percentage Dissipation=(PlossP)×100=(180 kW600 kW)×100=30%\text{Percentage Dissipation} = \left( \frac{P_{\text{loss}}}{P} \right) \times 100 = \left( \frac{180 \text{ kW}}{600 \text{ kW}} \right) \times 100 = 30\%

Hence, the correct option is B.

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