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Chapter 15: Problem 31

A coal-fired electrical generating station can use a higher \(T_{\mathrm{H}}\) than a nuclear plant; for safety reasons the core of a nuclear reactor is not allowed to get as hot as coal. Suppose that $T_{\mathrm{H}}=727^{\circ} \mathrm{C}\( for a coal station but \)T_{\mathrm{H}}=527^{\circ} \mathrm{C}$ for a nuclear station. Both power plants exhaust waste heat into a lake at \(T_{\mathrm{C}}=27^{\circ} \mathrm{C} .\) How much waste heat does each plant exhaust into the lake to produce \(1.00 \mathrm{MJ}\) of electricity? Assume both operate as reversible engines.

Short Answer

Expert verified
The coal-fired power plant exhausts 0.429 MJ of waste heat, and the nuclear power plant exhausts 0.6 MJ of waste heat into the lake to produce 1.00 MJ of electricity.

Step by step solution

01

Convert Celsius to Kelvin

First, we need to convert the given temperatures from degrees Celsius to Kelvin. To do this, we add 273 to the Celsius temperature. So, \(T_{\mathrm{H}}(coal) = 727 + 273 = 1000\mathrm{K}\) and \(T_{\mathrm{H}}(nuclear) = 527 + 273 = 800\mathrm{K}\). Similarly, \(T_{\mathrm{C}} = 27 + 273 = 300\mathrm{K}\).
02

Calculate the efficiency of both coal and nuclear power plants

Now, we need to find the efficiency of both power plants. Since the power plants are assumed to be reversible, we use the Carnot efficiency formula: Efficiency = \(1 - \frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\). For coal: Efficiency = \(1 - \frac{300\mathrm{K}}{1000\mathrm{K}} = 0.7\). For nuclear: Efficiency = \(1 - \frac{300\mathrm{K}}{800\mathrm{K}} = 0.625\).
03

Calculate the input heat for both power plants

We are given the amount of electricity produced by both power plants, which can be considered as their useful work output. Their efficiency multiplied by the heat input equals the useful work output. Therefore, the heat input can be found by dividing the useful work output (1.00 MJ) by the calculated efficiencies. For coal: \(Q_{\mathrm{in}}(coal) = \frac{1.00\mathrm{MJ}}{0.7} ≈ 1.43\mathrm{MJ}\). For nuclear: \(Q_{\mathrm{in}}(nuclear) = \frac{1.00\mathrm{MJ}}{0.625} ≈ 1.6\mathrm{MJ}\).
04

Calculate the waste heat for both coal and nuclear power plants

Finally, we can calculate the waste heat exhausted by both power plants using the input heat and their respective efficiencies. The waste heat equals \((1-\text{efficiency})\times Q_{\mathrm{in}}\). For coal: Waste heat = \((1-0.7) \times 1.43\mathrm{MJ} = 0.3 \times 1.43\mathrm{MJ} = 0.429\mathrm{MJ}\), and for nuclear: Waste heat = \((1-0.625) \times 1.6\mathrm{MJ} = 0.375 \times 1.6\mathrm{MJ} = 0.6\mathrm{MJ}\). So, the coal-fired power plant exhausts 0.429 MJ of waste heat, and the nuclear power plant exhausts 0.6 MJ of waste heat into the lake to produce 1.00 MJ of electricity.

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Most popular questions from this chapter

For a more realistic estimate of the maximum coefficient of performance of a heat pump, assume that a heat pump takes in heat from outdoors at $10^{\circ} \mathrm{C}$ below the ambient outdoor temperature, to account for the temperature difference across its heat exchanger. Similarly, assume that the output must be \(10^{\circ} \mathrm{C}\) hotter than the house (which itself might be kept at \(20^{\circ} \mathrm{C}\) ) to make the heat flow into the house. Make a graph of the coefficient of performance of a reversible heat pump under these conditions as a function of outdoor temperature (from $\left.-15^{\circ} \mathrm{C} \text { to }+15^{\circ} \mathrm{C} \text { in } 5^{\circ} \mathrm{C} \text { increments }\right)$
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