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Chapter 4: Problem 28

A coal power plant consumes 100,000 kg of coal per hour and produces 500 MW of power. If the heat of combustion of coal is \(30 \mathrm{MJ} / \mathrm{kg}\), what is the efficiency of the power plant?

Short Answer

Expert verified
The efficiency of the coal power plant is approximately \(16.67 \%\).

Step by step solution

01

Find the input power

To find the input power, first, we need to convert the heat of combustion of coal to the same unit as the output power, which is measured in MW. We know that 1 MJ = 1e-3 MWs: Heat of combustion of coal = \(30 \, \mathrm{MJ/kg} \times 1e-3 \, \mathrm{MWs/MJ} = 0.03 \, \mathrm{MWs/kg}\) Now, we can find the input power by multiplying the coal consumption rate with the heat of combustion of coal: Input power = Coal consumption per hour 脳 Heat of combustion of coal = \(100,000 \, \mathrm{kg/hour} \times 0.03 \, \mathrm{MWs/kg}\) = \(3,000 \, \mathrm{MWs/hour}\) Remember that 1 MW = 1 MJs, so we could also express input power as 3,000,000 MJs.
02

Calculate the efficiency

Efficiency is defined as the ratio of output power to input power, expressed as a percentage: Efficiency = \(\frac{\mathrm{(output \, power)}}{\mathrm{(input \, power)}} \times 100\) Now, we can plug in the output power (500 MW) and the input power (3,000 MWs/hour) into the equation: Efficiency = \(\frac{500 \, \mathrm{MW}}{3,000 \, \mathrm{MWs/hour}} \times 100\) To be consistent with the unit, we could express the output power in terms of MJs: So, 500 MW = 500,000 MJs Efficiency = \(\frac{500,000 \, \mathrm{MJs}}{3,000,000 \, \mathrm{MJs}} \times 100\) = \(\frac{1}{6} \times 100\) = \(16.67 \%\) The efficiency of the coal power plant is approximately 16.67%.

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