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Chapter 20: Problem 24

(II) An ideal (Carnot) engine has an efficiency of 38\(\% .\) If it were possible to run it backward as a heat pump, what would be its coefficient of performance?

Short Answer

Expert verified
The coefficient of performance is approximately 1.61.

Step by step solution

01

Understanding Carnot Efficiency

The efficiency (eta) of a Carnot engine is given by \(\eta = 1 - \frac{T_c}{T_h}\), where \(T_c\) is the temperature of the cold reservoir and \(T_h\) is the temperature of the hot reservoir. In this problem, we are given the efficiency as 38%, which means \(\eta = 0.38\).
02

Calculate Temperature Ratio

Using the efficiency, we can calculate the ratio \(\frac{T_c}{T_h}\). Rearranging the efficiency equation, we have: \(1 - \eta = \frac{T_c}{T_h}\). Substituting \(\eta = 0.38\), we get \(\frac{T_c}{T_h} = 1 - 0.38 = 0.62\).
03

Use Coefficient of Performance (COP) Formula

The coefficient of performance for a heat pump running backward (COP) is given by \( COP = \frac{1}{1 - \eta} \). Since \(1 - \eta = 0.62\), substitute this value into the formula to get \(COP = \frac{1}{0.62} \).
04

Calculate COP

Now compute the value of COP: \(COP = \frac{1}{0.62} \approx 1.61\). This represents the COP of the system if it operates as a heat pump.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Efficiency
Efficiency is a key parameter in determining how well an engine converts heat energy into useful work. For the Carnot engine, it is defined as the ratio of work output to the heat input. Mathematically, the efficiency (\( \eta \)) is expressed as:\[\eta = 1 - \frac{T_c}{T_h}\]where \( T_c \) is the temperature of the cold reservoir and \( T_h \) is the temperature of the hot reservoir. In practical terms, it indicates the fraction of heat that can be converted into work. Higher efficiency means more of the input energy is used effectively.
For instance, if an engine has an efficiency of 38%, this means that 38% of the heat energy is converted into work, while the remaining 62% is wasted as heat.
Coefficient of Performance
The Coefficient of Performance (COP) is crucial when analyzing how effectively a heat pump works. It measures the efficiency of a heat pump in adding heat to a system. For a Carnot engine operated in reverse鈥攁s a heat pump鈥攖he COP is calculated using:\[COP = \frac{1}{1 - \eta}\]Here, \( \eta \) is the efficiency of the original Carnot engine. The COP helps you understand how much heating effect you get for each unit of work input. A higher COP means the heat pump is extremely effective. In our example, with an efficiency of 38%, the calculation yields a COP of around 1.61.
This means the heat pump moves 1.61 units of heat for each unit of work input.
Heat Pump
A heat pump is a device that transfers heat from a colder area to a warmer one, often using mechanical work or electricity. When a Carnot engine functions in reverse, it acts as a heat pump.
The heat pump is particularly valuable in applications like heating homes, as it can provide more heating power than the electrical energy it consumes. - During the operation, it absorbs heat from a cold reservoir through work done on the system - Then releases it into a hot reservoir, providing efficient heating. The design of a heat pump takes advantage of the thermodynamic principles outlined by a Carnot cycle, ensuring maximum efficiency.
Temperature Ratio
The temperature ratio \( \frac{T_c}{T_h} \) is a pivotal factor in determining the efficiency and performance of both heat engines and pumps. It represents the relative difference between the cold and hot reservoir temperatures.
Within the context of a Carnot cycle, this ratio directly impacts the efficiency, as given by the equation:\[\eta = 1 - \frac{T_c}{T_h}\]A lower temperature ratio means the engine has the potential for higher efficiency. - By analyzing this ratio, one can infer how close the system operates to the ideal Carnot efficiency. - A heat pump operating effectively would rely on maintaining a strategic temperature ratio to optimize performance.
In practical applications, systems are often designed to exploit this ratio for improved energy utilization.

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Most popular questions from this chapter

(III) Consider an isolated gas-like system consisting of a box that contains \(N=10\) distinguishable atoms, each moving at the same speed \(v\). The number of unique ways that these atoms can be arranged so that \(N_{\mathrm{L}}\) atoms are within the left-hand half of the box and \(N_{\mathrm{R}}\) atoms are within the right-hand half of the box is given by \(N ! / N_{\mathrm{L}} ! N_{\mathrm{R}} !\), where, for example, the factorial \(4 !=4 \cdot 3 \cdot 2 \cdot 1\) (the only exception is that \(0 !=1\) ). Define each unique arrangement of atoms within the box to be a microstate of this system. Now imagine the following two possible macrostates: state \(\mathrm{A}\) where all of the atoms are within the left-hand half of the box and none are within the right-hand half; and state \(\mathrm{B}\) where the distribution is uniform (that is, there is the same number in each half). See Fig. \(20-21 .\) (a) Assume the system is initially in state \(\mathrm{A}\) and, at a later time, is found to be in state B. Determine the system's change in entropy. Can this process occur naturally? (b) Assume the system is initially in state \(\mathrm{B}\) and, at a later time, is found to be in state A. Determine the system's change in entropy. Can this process occur naturally?

(II) A very long solid nonconducting cylinder of radius \(R_{1}\) is uniformly charged with charge density \(\rho_{\mathrm{E}}\). It is surrounded by a cylindrical metal (conducting) tube of inner radius \(R_{2}\) and outer radius \(R_{3},\) which has no net charge (cross-sectional view shown in Fig. 37 ). If the axes of the two cylinders are parallel, but displaced from each other by a distance \(d\), determine the resulting electric field in the region electric field in the region \(R>R_{3}, \quad\) where the measured from the metal cylinder's axis. Assume \(d<\left(R_{2}-R_{1}\right)\)

(II) Energy may be stored for use during peak demand by pumping water to a high reservoir when demand is low and then releasing it to drive turbines when needed. Suppose water is pumped to a lake 135 \(\mathrm{m}\) above the turbines at a rate of \(1.35 \times 10^{5} \mathrm{kg} / \mathrm{s}\) for 10.0 \(\mathrm{h}\) at night. (a) How much energy \((\mathrm{kWh})\) is needed to do this each night? \((b)\) If all this energy is released during a 14 -h day, at 75\(\%\) efficiency, what is the average power output?

(II) When \(2.0 \mathrm{~kg}\) of water at \(12.0^{\circ} \mathrm{C}\) is mixed with \(3.0 \mathrm{~kg}\) of water at \(38.0^{\circ} \mathrm{C}\) in a well- insulated container, what is the change in entropy of the system? \((a)\) Make an estimate; (b) use the integral \(\Delta S=\int d Q / T\)

A 35\(\%\) efficient power plant puts out 920 \(\mathrm{MW}\) of electrical power. Cooling towers are used to take away the exhaust heat. \((a)\) If the air temperature \(\left(15^{\circ} \mathrm{C}\right)\) is allowed to rise 7.0 \(\mathrm{C}^{\circ}\) , estimate what volume of air \(\left(\mathrm{km}^{3}\right)\) is heated per day. Will the local climate be heated significantly? (b) If the heated air were to form a layer 150 \(\mathrm{m}\) thick, estimate how large an area it would cover for 24 \(\mathrm{h}\) of operation. Assume the air has density 1.2 \(\mathrm{kg} / \mathrm{m}^{3}\) and that its specific heat is about 1.0 \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{C}^{\circ}\) at constant pressure.
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