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Chapter 15: Problem 62

For a reversible engine, will you obtain a better efficiency by increasing the high-temperature reservoir by an amount \(\Delta T\) or decreasing the low- temperature reservoir by the same amount \(\Delta T ?\)

Short Answer

Expert verified
Answer: Decreasing the low-temperature reservoir by the amount 螖T results in better efficiency for the reversible engine.

Step by step solution

01

Initial Efficiency

First, we find the initial efficiency of the reversible engine using the Carnot efficiency formula: \(\eta_{initial} = 1 - \frac{T_L}{T_H}\) where \(\eta_{initial}\) is the initial efficiency, \(T_L\) is the low-temperature reservoir, and \(T_H\) is the high-temperature reservoir.
02

Efficiency with Increased High-Temperature Reservoir

Next, increase the high-temperature reservoir by the amount \(\Delta T\). The new high-temperature reservoir is \(T_H' = T_H + \Delta T\). Calculate the efficiency for this case using the Carnot efficiency formula: \(\eta_{increase} = 1 - \frac{T_L}{T_H + \Delta T}\)
03

Efficiency with Decreased Low-Temperature Reservoir

Now, decrease the low-temperature reservoir by the amount \(\Delta T\). The new low-temperature reservoir is \(T_L' = T_L - \Delta T\). Calculate the efficiency for this case using the Carnot efficiency formula: \(\eta_{decrease} = 1 - \frac{T_L - \Delta T}{T_H}\)
04

Comparing Efficiencies

Compare the two efficiencies, \(\eta_{increase}\) and \(\eta_{decrease}\), to determine which change results in a better efficiency. If \(\eta_{increase} > \eta_{decrease}\), increasing the high-temperature reservoir by the amount \(\Delta T\) results in better efficiency. If \(\eta_{increase} < \eta_{decrease}\), decreasing the low-temperature reservoir by the amount \(\Delta T\) results in better efficiency. It can be shown through mathematical comparison that decreasing the low-temperature reservoir by the amount \(\Delta T\) yields better efficiency.

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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 $\left.-15^{\circ} \mathrm{C} \text { to }+15^{\circ} \mathrm{C} \text { in } 5^{\circ} \mathrm{C} \text { increments }\right)$
Consider a heat engine that is not reversible. The engine uses $1.000 \mathrm{mol}\( of a diatomic ideal gas. In the first step \)(\mathrm{A})$ there is a constant temperature expansion while in contact with a warm reservoir at \(373 \mathrm{K}\) from \(P_{1}=1.55 \times 10^{5} \mathrm{Pa}\) and $V_{1}=2.00 \times 10^{-2} \mathrm{m}^{3}$ to \(P_{2}=1.24 \times 10^{5} \mathrm{Pa}\) and $V_{2}=2.50 \times 10^{-2} \mathrm{m}^{3} .$ Then (B) a heat reservoir at the cooler temperature of \(273 \mathrm{K}\) is used to cool the gas at constant volume to \(273 \mathrm{K}\) from \(P_{2}\) to $P_{3}=0.91 \times 10^{5} \mathrm{Pa} .$ This is followed by (C) a constant temperature compression while still in contact with the cold reservoir at \(273 \mathrm{K}\) from \(P_{3}, V_{2}\) to \(P_{4}=1.01 \times 10^{5} \mathrm{Pa}, V_{1} .\) The final step (D) is heating the gas at constant volume from \(273 \mathrm{K}\) to \(373 \mathrm{K}\) by being in contact with the warm reservoir again, to return from \(P_{4}, V_{1}\) to \(P_{1}, V_{1} .\) Find the change in entropy of the cold reservoir in step \(\mathrm{B}\). Remember that the gas is always in contact with the cold reservoir. (b) What is the change in entropy of the hot reservoir in step D? (c) Using this information, find the change in entropy of the total system of gas plus reservoirs during the whole cycle.
What is the change in entropy of \(10 \mathrm{g}\) of steam at $100^{\circ} \mathrm{C}\( as it condenses to water at \)100^{\circ} \mathrm{C} ?$ By how much does the entropy of the universe increase in this process?
A model steam engine of \(1.00-\mathrm{kg}\) mass pulls eight cars of \(1.00-\mathrm{kg}\) mass each. The cars start at rest and reach a velocity of \(3.00 \mathrm{m} / \mathrm{s}\) in a time of \(3.00 \mathrm{s}\) while moving a distance of \(4.50 \mathrm{m} .\) During that time, the engine takes in $135 \mathrm{J}$ of heat. What is the change in the internal energy of the engine?
A balloon contains \(200.0 \mathrm{L}\) of nitrogen gas at $20.0^{\circ} \mathrm{C}$ and at atmospheric pressure. How much energy must be added to raise the temperature of the nitrogen to \(40.0^{\circ} \mathrm{C}\) while allowing the balloon to expand at atmospheric pressure?
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