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Chapter 20: Problem 45

A Carnot engine operates between two heat reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}} .\) An inventor proposes to increase the efficiency by running one engine between \(T_{\mathrm{H}}\) and an intermediate temperature \(T^{\prime}\) and a second engine between \(T^{\prime}\) and \(T_{C}\), using as input the heat expelled by the first engine. Compute the efficiency of this composite system. and compare it to that of the original engine.

Short Answer

Expert verified
The efficiency of the composite Carnot system is equal to \(\eta_{1}(1 + \eta_{2})\). This efficiency will only exceed that of a single Carnot engine, which is \(1 - \frac{T_C}{T_H}\), if the intermediate temperature \(T'\) lies between \(T_H\) and \(T_C\).

Step by step solution

01

Understanding Carnot engines and efficiency

Carnot engines operate in a Carnot cycle with high temperature reservoir T_H, and a low temperature reservoir T_C. The efficiency of a Carnot engine operating between these two temperatures, according to the second law of thermodynamics, is given by: \( \eta = 1 - \frac{T_C}{T_H} \)
02

Formulate efficiency of engines

According to the inventor's proposition, let's have two engines. Engine 1 operating between temperatures T_H and T', and Engine 2 operating between temperatures T' and T_C. The efficiencies of these two engines, using the formula from step 1, are: \(\eta_{1} = 1 - \frac{T'}{T_H}\) for engine 1, and \(\eta_{2} = 1 - \frac{T_C}{T'}\) for engine 2.
03

Calculate composite system efficiency

The total work obtained from the 2 engines, W_total, is the sum of the work from engine 1 and engine 2, which is: \( W_{total} = Q_H . \eta_{1} + Q_H . \eta_{1} . \eta_{2} = Q_H . \eta_{1} (1 + \eta_{2})\). Where \(Q_H\), is the heat absorbed from the high temperature reservoir by engine 1. The total heat absorbed from the reservoir, \(Q_H\), minus the total work done, \(W_{total}\), is equal to the heat expelled, \(Q_C\), to the cold reservoir. Therefore, we have \(Q_C = Q_H - W_{total}\). Substitute \(W_{total}\) from above gives, \(Q_C = Q_H - Q_H . \eta_{1} (1 + \eta_{2}) = Q_H(1 - \eta_{1} (1 + \eta_{2}))\). The composite engine's efficiency, \(\eta_{comp}\), which is the ratio of the total work done to the total heat input, gives us \(\eta_{comp} =1 - \frac{Q_C}{Q_H} = \eta_{1}(1 + \eta_{2})\).
04

Comparison with original engine

Now, we can compare this efficiency to the original Carnot engine's efficiency, which is \(\eta = 1 - \frac{T_C}{T_H}\). For the inventor's proposition to be advantageous, the efficiency of the composite system should be greater than the original engine's efficiency. But comparing the two directly would give a complex inequality involving temperature and efficiency. A better way is to compare the temperatures. So, \(T' > T_C, T' < T_H\). This shows that the composite system will only be more efficient if the intermediate temperature \(T'\) lies between \(T_H\) and \(T_C\). If this isn't the case, the composite system's efficiency would be less than or equal to the original Carnot engine's efficiency.

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Key Concepts

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

Carnot cycle
The Carnot cycle is a theoretical construct used to understand the efficiency of heat engines. It is based on reversible processes, which means it operates in a manner that can be reversed without any loss of energy to friction or other dissipative forces. This cycle involves four main stages:
  • Isothermal expansion: The engine absorbs heat from a high-temperature reservoir while the gas expands, doing work on the surroundings.
  • Adiabatic expansion: The gas expands further without any heat exchange, resulting in a decrease in its temperature.
  • Isothermal compression: The engine releases heat to a low-temperature reservoir while the gas is compressed.
  • Adiabatic compression: The cycle is completed by compressing the gas further, increasing its temperature back to the starting point.
These processes form a closed loop in the state of the system, ensuring maximum theoretical efficiency. However, it is important to note that real engines can't reach the efficiency of a Carnot cycle due to irreversibilities such as friction and thermal losses. Understanding the Carnot cycle helps in designing more efficient real-world engines by approaching the conditions of a theoretically efficient cycle.
Second law of thermodynamics
The second law of thermodynamics is a key principle in understanding why certain processes occur spontaneously and the limitations of energy conversion in thermodynamic systems. It introduces the concept of entropy, which is often associated with the level of disorder or randomness in a system. In the context of heat engines, the second law implies that not all heat absorbed by the engine can be converted into work. Some heat must be released to a cooler reservoir, ensuring the spontaneous direction of heat flow.
  • This law can be expressed in terms of efficiency for heat engines through the formula: o The maximum possible efficiency (\(\eta\)) is is given by: \[\eta = 1 - \frac{T_C}{T_H} \]where \(T_H\) is the absolute temperature of the hot reservoir and \(T_C\) is that of the cold reservoir.
  • It suggests that, for any engine, reaching 100% efficiency is impossible, as this would mean all heat is converted into work, violating the law.
  • The second law also highlights the necessity of temperature differences for engines to do work: the greater the temperature difference, the higher the potential efficiency.
By understanding the second law, engineers seek to optimize systems by minimizing losses and irreversibilities, moving closer towards ideal efficiency.
Temperature reservoirs
Temperature reservoirs are essential components in thermodynamic cycles such as the Carnot cycle. They are idealized bodies with a large thermal energy capacity, allowing them to maintain a constant temperature even when heat is added or removed. This consistency is crucial for the cyclic operation of engines.
  • The high-temperature reservoir (\(T_H\)) is the source of heat energy for the engine, providing energy that is partially converted into work.
  • The low-temperature reservoir (\(T_C\)) acts as a sink, receiving the waste heat produced after the work has been done.
  • The efficiency of an engine depends significantly on the difference between these temperatures. Greater differences result in higher maximum efficiencies.
  • These reservoirs are often large bodies of water or the atmosphere in practical applications, providing a stable temperature reference point.
In multi-stage engines, like the one proposed by the inventor in the exercise, an intermediate reservoir can be used. This intermediate temperature (\(T'\)) allows successive stages to increase overall efficiency by exploiting additional temperature gradients. Ultimately, these reservoirs and the temperatures they maintain are pivotal in determining an engine's performance and efficiency.

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Most popular questions from this chapter

Human Entropy. A person with skin of surface area \(1.85 \mathrm{~m}^{2}\) and temperature \(30.0^{\circ} \mathrm{C}\) is resting in an insulated room where the ambient air temperature is \(20.0^{\circ} \mathrm{C}\). Assume that this person gcts rid of excess heat by radiation only. By how much does the person change the entropy of the air in this room cach second? (Recall that the room radiates back into the person and that the emissivity of the skin is \(1.00 .\) )

In your summer job with a venture capital firm, you are given funding requests from four inventors of heat cngines. The inventors claim the following data for their operating prototypes: for a Carnot device that operates between \(95 \mathrm{~F}\) and \(80^{\circ} \mathrm{F}\). (c) You have an air conditioner with an EER of 10.9 . Your home on average requires a total cooling output of \(\left|Q_{\mathrm{C}}\right|=1.9 \times 10^{10} \mathrm{~J}\) per year. If electricity costs you 15.3 cents per \(\mathrm{kW} \cdot \mathrm{h}\), how much do you spend per year, on average, to operate your air conditioner? (Assume that the unit's EER accurately represcits the operation of your air conditioner. A seasonal energy efficiency ratio (SEFR) is often used. The SFFR is calculated over a range of outside temperatures to get a more accurate seasonal average.) (d) You are considering replacing your air conditioner with a more efficient one with an EER of 14.6 . Based on the EER, how much would that save you on electricity costs in an average year?

A heat engine uses a large insulated tank of ice water as its cold reservoir. In 100 cycles the engine takes in \(8000 \mathrm{~J}\) of heat energy from the hot reservoir and the rejected heat melts \(0.0180 \mathrm{~kg}\) of ice in the tank. During these 100 cycles, how much work is performed by the engine?

As a budding mechanical engineer, you are called upon to design a Carnot engine that has \(2.00 \mathrm{~mol}\) of a monatomic ideal gas as its working substance and operates from a high-temperature reservoir at \(500^{\circ} \mathrm{C}\). The engine is to lift a \(15.0 \mathrm{~kg}\) weight \(2.00 \mathrm{~m}\) per cycle, using \(500 \mathrm{~J}\) of heat input. The gas in the cngine chamber can have a minimum volume of \(5.00 \mathrm{~L}\) during the cycle. (a) Draw a \(p V\) -diagram for this cycle. Show in your diagram where heat enters and leaves the gas. (b) What must be the temperature of the cold reservoir? (c) What is the thermal efficicncy of the cngine? (d) How much heat cnergy does this engine waste per cycle? (e) What is the maximum pressure that the gas chamher will have to withstand?

You are conducting experiments to study prototype heat engines. In one test, \(4.00 \mathrm{~mol}\) of argon gas are taken around the cycle shown in Fig. \(\mathrm{P} 20.55 .\) The pressure is low cnough for the gas to be treated as ideal. You measure the gas temperature in states \(a, b, c,\) and \(d\) and find \(T_{a}=250.0 \mathrm{~K}\) \(T_{h}=300.0 \mathrm{~K}, \quad T_{c}=380.0 \mathrm{~K}, \quad\) and \(T_{d}=316.7 \mathrm{~K}\) (a) Calculate the efficiency \(e\) of the cycle. (b) Disappointed by the cycle's low efficicncy, you consider doubling the number of moles of gas while kecping the pressure and volume the same. What would \(e\) be then? (c) You remember that the efficiency of a Carnot cycle increases if the temperature of the hot reservoir is increased. So, you return to using \(4.00 \mathrm{~mol}\) of gas but double the volume in states \(c\) and \(d\) while keeping the pressures the same. The resulting temperatures in these states are \(T_{c}=760.0 \mathrm{~K}\) and \(T_{d}=633.4 \mathrm{~K}\). \(T_{a}\) and \(T_{b}\) remain the same as in part (a). Calculate \(e\) for this cycle with the new \(T_{c}\) and \(T_{d}\) values. (d) Encouraged by the increase in efficiency, you raise \(T_{c}\) and \(T_{d}\) still further. But \(e\) doesn't increase very much; it scems to be approaching a limiting valuc. If \(T_{a}=250.0 \mathrm{~K}\) and \(T_{b}=300.0 \mathrm{~K}\) and you keep volumes \(V_{a}\) and \(V_{b}\) the same as in part (a), then \(T_{d} T_{d}=T_{b} / T_{a}\) and \(T_{c}=1.20 T_{d}\). Derive an expression for \(e\) as a function of \(T_{d}\) for this cycle. What value docs \(e\) approach as \(T_{d}\) becomes very large?
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