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It is well known that the thermal efficiency of heat engines increases as the temperature of the energy source increases. In an attempt to improve the efficiency of a power plant, somebody suggests transferring heat from the available energy source to a higher-temperature medium by a heat pump before energy is supplied to the power plant. What do you think of this suggestion? Explain.

Short Answer

Expert verified
Answer: Using a heat pump to raise the temperature of the medium can potentially increase the power plant's thermal efficiency. However, the heat pump requires additional work input, which may or may not result in a favorable overall energy balance. A detailed energy and economic analysis is needed to determine the feasibility and efficiency of such a system in practice.

Step by step solution

01

Understanding the concept of thermal efficiency of heat engines

Thermal efficiency is a measure of how well a heat engine converts heat energy into useful work. It is a dimensionless performance metric defined as the ratio of the output work done to the input heat energy. An ideal heat engine would convert all the heat energy it receives into useful work, but in reality, some portion of the heat is always lost to the environment.
02

Explaining the temperature effect on efficiency

According to the second law of thermodynamics, a heat engine cannot have 100% efficiency. To attain a higher thermal efficiency, the difference between the energy source temperature and the temperature at which energy is rejected must be increased. This is because when operating between two fixed temperatures (i.e., source and sink), the heat engine's efficiency increases with the source temperature.
03

Understanding the concept of a heat pump

A heat pump is a device that transfers heat energy from a lower-temperature source to a higher-temperature medium. This is usually accomplished by using external work done on the heat pump system. The performance of a heat pump is expressed as the Coefficient of Performance (COP), which is defined as the ratio of heat energy transferred to a higher-temperature medium to the work done on the system.
04

Evaluating the suggestion to use a heat pump

The suggestion to use a heat pump to raise the medium's temperature before supplying energy to the power plant aims to increase the power plant's thermal efficiency. However, this process will require additional work input to operate the heat pump. According to the first law of thermodynamics, the energy input to a system has to be equal to the sum of its internal energy change, work done by the system, and energy rejected as heat. In this case, the increased thermal efficiency of the power plant will come at the cost of additional work to operate the heat pump, which means that the overall energy balance may or may not be favorable.
05

Conclusion

While using a heat pump to increase the energy source's temperature might improve the thermal efficiency of the power plant, it also requires additional work to be done on the heat pump system. Therefore, the overall advantage of this suggestion depends on whether the benefits of increased efficiency outweigh the costs of operating the heat pump. It is crucial to perform a detailed energy and economic analysis to determine the feasibility and efficiency of such a system in practice.

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Most popular questions from this chapter

Using EES (or other) software, determine the maximum work that can be extracted from a pond containing \(10^{5} \mathrm{kg}\) of water at \(350 \mathrm{K}\) when the temperature of the surroundings is \(300 \mathrm{K}\). Notice that the temperature of water in the pond will be gradually decreasing as energy is extracted from it; therefore, the efficiency of the engine will be decreasing. Use temperature intervals of \((a) 5 \mathrm{K},(b) 2 \mathrm{K}\) and \((c) 1 \mathrm{K}\) until the pond temperature drops to \(300 \mathrm{K}\). Also solve this problem exactly by integration and compare the results.

A household refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of \(800 \mathrm{kJ} / \mathrm{h}\). If the COP of the refrigerator is \(2.2,\) determine the power the refrigerator draws when running.

A heat pump supplies heat energy to a house at the rate of \(140,000 \mathrm{kJ} / \mathrm{h}\) when the house is maintained at \(25^{\circ} \mathrm{C} .\) Over a period of one month, the heat pump operates for 100 hours to transfer energy from a heat source outside the house to inside the house. Consider a heat pump receiving heat from two different outside energy sources. In one application the heat pump receives heat from the outside air at \(0^{\circ} \mathrm{C} .\) In a second application the heat pump receives heat from a lake having a water temperature of \(10^{\circ} \mathrm{C}\). If electricity costs \(\$ 0.105 / \mathrm{kWh}\), determine the maximum money saved by using the lake water rather than the outside air as the outside energy source.

An old gas turbine has an efficiency of 21 percent and develops a power output of \(6000 \mathrm{kW}\). Determine the fuel consumption rate of this gas turbine, in L/min, if the fuel has a heating value of \(42,000 \mathrm{kJ} / \mathrm{kg}\) and a density of \(0.8 \mathrm{g} / \mathrm{cm}^{3}\)

A heat pump is absorbing heat from the cold outdoors at \(5^{\circ} \mathrm{C}\) and supplying heat to a house at \(25^{\circ} \mathrm{C}\) at a rate of \(18,000 \mathrm{kJ} / \mathrm{h}\). If the power consumed by the heat pump is \(1.9 \mathrm{kW},\) the coefficient of performance of the heat pump is \((a) 1.3\) (b) 2.6 \((c) 3.0\) \((d) 3.8\) \((e) 13.9\)

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