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An inventor claims to have developed a device that executes a power cycle while operating between reservoirs at 800 and \(350 \mathrm{~K}\) that has a thermal efficiency of (a) \(56 \%\), (b) \(40 \%\). Evaluate the claim for each case.

Short Answer

Expert verified
The claims are possible because both 56% and 40% are less than or equal to the Carnot efficiency of 56.25%.

Step by step solution

01

- Understand the Problem

An inventor's device operates between two reservoirs. The temperatures are given as 800 K for the hot reservoir and 350 K for the cold reservoir. The claim involves evaluating the device's thermal efficiency for both 56% and 40%.
02

- Recall the Formula for Carnot Efficiency

The maximum possible efficiency for any heat engine operating between two temperatures is given by the Carnot efficiency formula: \[ \eta_{carnot} = 1 - \frac{T_c}{T_h} \], where \( T_c \) is the temperature of the cold reservoir and \( T_h \) is the temperature of the hot reservoir.
03

- Calculate Carnot Efficiency

Using the given temperatures, calculate Carnot efficiency: \[ \eta_{carnot} = 1 - \frac{350}{800} = 1 - 0.4375 = 0.5625 \] or 56.25%.
04

- Evaluate the Claim for 56%

Compare the given efficiency of 56% to the Carnot efficiency of 56.25%. Since 56% is less than 56.25%, it is physically possible.
05

- Evaluate the Claim for 40%

Compare the given efficiency of 40% to the Carnot efficiency of 56.25%. Since 40% is significantly less than 56.25%, this claim is also physically possible.

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Key Concepts

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

thermal efficiency
Thermal efficiency is a measure of how well a heat engine converts heat into work. It's important because it tells us how much useful work we can get from a certain amount of heat energy. The higher the thermal efficiency, the better. For a heat engine, thermal efficiency \(\text{η}\) is given by the formula \( \text{η} = \frac{W_{net}}{Q_{in}} \), where \( W_{net} \) is the net work output and \( Q_{in} \) is the heat input.
  • The efficiency ranges from 0 to 1 (or 0% to 100%).
  • Perfect efficiency (100%) is not possible due to inherent energy losses such as friction and heat dissipation.

Therefore, thermal efficiency helps us understand the effectiveness of a heat engine, and serves as a benchmark when comparing different engines or cycles.
power cycle
A power cycle is a process where a working fluid (like steam or gas) goes through a series of state changes, transforming heat energy into mechanical work. This cycle is repeated over and over.
For example:
  • The Carnot cycle is a theoretical model of an ideal power cycle.
  • Real-life cycles include the Rankine cycle for steam engines and the Otto cycle for internal combustion engines.

The efficiency and effectiveness of a power cycle depend on factors like the type of working fluid, the temperature range, and the design of the engine. In our example, the power cycle involves reservoirs at 800 K and 350 K.
Carnot efficiency formula
The Carnot efficiency formula allows us to calculate the maximum possible efficiency a heat engine can achieve, operating between two temperature reservoirs. It's given by:
\[ \text{η}_{carnot} = 1 - \frac{T_c}{T_h} \]
where:
  • \( T_c \) is the temperature of the cold reservoir (in Kelvin).
  • \( T_h \) is the temperature of the hot reservoir (in Kelvin).

The Carnot efficiency sets the upper limit for the efficiency of any real engine. Any actual engine will have a lower efficiency than the Carnot efficiency because real processes have losses such as friction and non-reversible processes.
reservoir temperatures
Reservoir temperatures are crucial in determining the Carnot efficiency and thus the overall thermal efficiency of a heat engine. In thermodynamics, reservoirs are large thermal bodies that maintain a constant temperature even when heat is added or extracted.
In our example, the hot reservoir is at 800 K and the cold reservoir is at 350 K.
Higher temperature differences between the hot and cold reservoirs generally result in higher efficiencies:
  • Higher \( T_h \) relative to \( T_c \) means more potential work output.
  • Lower \( T_c \) relative to \( T_h \) also increases efficiency.

Therefore, to achieve higher efficiencies, heat engines strive to operate between a very high temperature source and a very low temperature sink.
heat engine
A heat engine is a device that converts thermal energy into mechanical work. It does this by exploiting the differences in temperatures between two reservoirs. The main components of heat engines include a working fluid, a source of heat (hot reservoir), and a heat sink (cold reservoir).
In this example:
  • The working fluid absorbs heat from the hot reservoir (800 K).
  • It then does work (like moving a piston or turning a turbine).
  • Finally, it releases unused heat to the cold reservoir (350 K).

The effectiveness of a heat engine is measured by its thermal efficiency. By understanding and applying the Carnot efficiency formula, we can determine how close a given heat engine's performance is to the theoretical maximum efficiency.

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Most popular questions from this chapter

For each \(\mathrm{kW}\) of power input to an ice maker at steady state, determine the maximum rate that ice can be produced, in \(\mathrm{kg} / \mathrm{h}\), from liquid water at \(0^{\circ} \mathrm{C}\). Assume that \(333 \mathrm{~kJ} / \mathrm{kg}\) of energy must be removed by heat transfer to freeze water at \(0^{\circ} \mathrm{C}\), and that the surroundings are at \(20^{\circ} \mathrm{C}\).

To maintain the passenger compartment of an automobile traveling at \(13.4 \mathrm{~m} / \mathrm{s}\) at \(21^{\circ} \mathrm{C}\) when the surrounding air temperature is \(32^{\circ} \mathrm{C}\), the vehicle's air conditioner removes \(5.275 \mathrm{~kW}\), by heat transfer. Estimate the amount of engine horsepower required to drive the air conditioner. Referring to typical manufacturer's data, compare your estimate with the actual horsepower requirement. Discuss the relationship between the initial unit cost of an automobile air-conditioning system and its operating cost.

The preliminary design of a space station calls for a power cycle that at steady state receives energy by heat transfer at \(T_{\mathrm{H}}=600 \mathrm{~K}\) from a nuclear source and rejects energy to space by thermal radiation according to Eq. 2.33. For the radiative surface, the temperature is \(T_{\mathrm{C}}\), the emissivity is \(0.6\), and the surface receives no radiation from any source. The thermal efficiency of the power cycle is one- half that of a reversible power cycle operating between reservoirs at \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}}\) - (a) For \(T_{\mathrm{C}}=400 \mathrm{~K}\), determine \(\dot{W}_{\text {cycle }} / \mathrm{A}\), the net power developed per unit of radiator surface area, in \(\mathrm{kW} / \mathrm{m}^{2}\), and the thermal efficiency. (b) Plot \(\dot{W}_{\text {cycle }} / \mathrm{A}\) and the thermal efficiency versus \(T_{\mathrm{C}}\), and determine the maximum value of \(\dot{W}_{\text {cycle }} / \mathrm{A}\). (c) Determine the range of temperatures \(T_{\mathrm{C}}\), in \(\mathrm{K}\), for which \(\dot{W}_{\text {cycle }} / \mathrm{A}\) is within 2 percent of the maximum value obtained in part (b). The Stefan-Boltzmann constant is \(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\).

To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{C}\), would it be better to increase \(T_{\mathrm{H}}\) or decrease \(T_{\mathrm{C}}\) by equal amounts?

Using the Kelvin-Planck statement of the second law of thermodynamics, demonstrate the following corollaries: (a) The coefficient of performance of an irreversible refrigeration cycle is always less than the coefficient of performance of a reversible refrigeration cycle when both exchange energy by heat transfer with the same two reservoirs. (b) All reversible refrigeration cycles operating between the same two reservoirs have the same coefficient of performance. (c) The coefficient of performance of an irreversible heat pump cycle is always less than the coefficient of performance of a reversible heat pump cycle when both exchange energy by heat transfer with the same two reservoirs. (d) All reversible heat pump cycles operating between the same two reservoirs have the same coefficient of performance.

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