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Two reversible power cycles are arranged in series. The first cycle receives energy by heat transfer from a hot reservoir at temperature \(T_{\mathrm{H}}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T

Short Answer

Expert verified
The combined efficiency is \( \eta_{\text{combined}} = \left( 1 - \frac{T}{T_{\text{H}}} \right) \left( 1 - \frac{T_{\text{C}}}{T} \right) \). The intermediate temperature is \( T = \sqrt{T_{\text{H}} T_{\text{C}}} \).

Step by step solution

01

- Thermal Efficiency of Single Reversible Cycle

The thermal efficiency of a single reversible cycle operating between two heat reservoirs at temperatures \( T_{\text{H}} \) and \( T_{\text{C}} \) is given by the Carnot efficiency:\[ \eta_{\text{single}} = 1 - \frac{T_{\text{C}}}{T_{\text{H}}} \]
02

- Thermal Efficiency of Each Cycle

Let \( \eta_1 \) be the efficiency of the first cycle and \( \eta_2 \) be the efficiency of the second cycle.\[ \eta_1 = 1 - \frac{T}{T_{\text{H}}} \]\[ \eta_2 = 1 - \frac{T_{\text{C}}}{T} \]
03

- Combined Efficiency

The combined efficiency \( \eta_{\text{combined}} \) of the two cycles arranged in series is the product of their individual efficiencies because the heat rejected by the first cycle is the heat received by the second cycle:\[ \eta_{\text{combined}} = \eta_1 * \eta_2 \]Substitute the expressions for \( \eta_1 \) and \( \eta_2 \):\[ \eta_{\text{combined}} = \left( 1 - \frac{T}{T_{\text{H}}} \right) \left( 1 - \frac{T_{\text{C}}}{T} \right) \]
04

- Equate Efficiencies for Special Case

For the special case where the thermal efficiencies of the two cycles are equal, set \( \eta_1 = \eta_2 \):\[ 1 - \frac{T}{T_{\text{H}}} = 1 - \frac{T_{\text{C}}}{T} \]
05

- Solve for Intermediate Temperature \(T\)

Solve the equation from Step 4 to find the intermediate temperature \( T \):\[ \frac{T}{T_{\text{H}}} = \frac{T_{\text{C}}}{T} \]\[ T^2 = T_{\text{H}} T_{\text{C}} \]\[ T = \sqrt{T_{\text{H}} T_{\text{C}}} \]

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Key Concepts

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

Carnot efficiency
The concept of Carnot efficiency is fundamental in thermodynamics.
It represents the maximum possible efficiency that any heat engine can achieve when operating between two temperature reservoirs.
This efficiency is determined solely by the temperatures of the hot and cold reservoirs.
The formula for Carnot efficiency is:
\( \eta_{\text{Carnot}} = 1 - \frac{T_{\text{C}}}{T_{\text{H}}}\) where \( T_{\text{C}}\) is the cold reservoir temperature and \( T_{\text{H}}\) is the hot reservoir temperature.

Notably, this equation tells us that the efficiency depends on the temperature difference between the two reservoirs.
The greater the difference, the higher the efficiency.
However, Carnot efficiency sets an idealized upper limit, meaning no real engine can be more efficient than a Carnot engine operating between the same temperatures.
Understanding this is crucial for analyzing more complex thermodynamic cycles.
Reversible cycles
A reversible cycle, like the Carnot cycle, is an idealized process that can proceed in either direction without increasing entropy.
In reality, all natural processes are irreversible to some extent.
However, reversible cycles are useful because they provide an upper bound on the efficiency of real engines.
In the context of the original problem, it involves two reversible cycles operating in series.

The first cycle receives heat from a hot reservoir and transfers some of this energy to a reservoir at an intermediate temperature.
The second cycle receives energy from this intermediate reservoir and transfers it to a cold reservoir.
For both these cycles to be reversible, there must be no entropy generation within the system.
Reversible cycles also help us understand how energy is conserved and how we can maximize work extraction.
Thermal efficiency
Thermal efficiency measures the effectiveness of a heat engine in converting the heat input into work output.
The general formula for thermal efficiency is given by:
\( \eta = \frac{W_{\text{out}}}{Q_{\text{in}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}}\)
where \( W_{\text{out}}\) is the work done and \( Q_{\text{in}}\) and \( Q_{\text{out}}\) are the heat input and output, respectively.

In a single reversible cycle between two reservoirs at temperatures \( T_{\text{H}}\) and \( T_{\text{C}}\), thermal efficiency simplifies to Carnot efficiency:
\( \eta_{\text{single}} = 1 - \frac{T_{\text{C}}}{T_{\text{H}}}\).
For two cycles arranged in series, as in the given exercise, the combined efficiency is the product of individual efficiencies:
\( \eta_{\text{combined}} = \left( 1 - \frac{T}{T_{\text{H}}} \right) \left( 1 - \frac{T_{\text{C}}}{T} \right)\).
In special cases where the efficiencies are equal for both cycles, the intermediate temperature is the geometric mean of the hot and cold temperatures:
\( T = \sqrt{T_{\text{H}} T_{\text{C}}}\).
This demonstrates how intermediate temperatures can be derived based on known efficiencies and temperature conditions.

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Most popular questions from this chapter

A power cycle operating at steady state receives energy by heat transfer at a rate \(\dot{Q}_{\mathrm{H}}\) at \(T_{\mathrm{H}}=1000 \mathrm{~K}\) and rejects energy by heat transfer to a cold reservoir at a rate \(Q_{\mathrm{C}}\) at \(T_{C}=300 \mathrm{~K}\). For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible. (a) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{Q}_{c}=100 \mathrm{~kW}\) (b) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{W}_{\text {cycle }}=250 \mathrm{~kW}, \dot{Q}_{\mathrm{C}}=200 \mathrm{~kW}\) (c) \(\dot{W}_{\text {cyde }}=350 \mathrm{~kW}, \dot{Q}_{\mathrm{C}}=150 \mathrm{~kW}\) (d) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{Q}_{\mathrm{c}}=200 \mathrm{~kW}\)

A reversible power cycle receives energy \(Q_{\mathrm{H}}\) from a reservoir at temperature \(T_{\mathrm{H}}\) and rejects \(Q_{\mathrm{C}}\) to a reservoir at temperature \(T_{\mathrm{C}}\). The work developed by the power cycle is used to drive a reversible heat pump that removes energy \(Q_{\mathrm{C}}^{\prime}\) from a reservoir at temperature \(T_{\mathrm{C}}^{\top}\) and rejects energy \(Q_{\mathrm{H}}^{\prime}\) to a reservoir at temperature \(T_{\mathrm{H}}^{\prime}\) (a) Develop an expression for the ratio \(Q_{\mathrm{H}}^{\prime} / Q_{\mathrm{H}}\) in terms of the temperatures of the four reservoirs. (b) What must be the relationship of the temperatures \(T_{\mathrm{H}}, T_{\mathrm{C}}, T_{\mathrm{C}}^{\prime}\), and \(T^{\prime}{ }_{\mathrm{H}}\) for \(Q_{\mathrm{H}}^{\prime} / Q_{\mathrm{H}}\) to exceed a value of unity?

Describe a process that would satisfy the conservation of energy principle, but does not actually occur in nature.

A power cycle operating at steady state receives energy by heat transfer from the combustion of fuel at an average temperature of \(1000 \mathrm{~K}\). Owing to environmental considerations, the cycle discharges energy by heat transfer to the atmosphere at \(300 \mathrm{~K}\) at a rate no greater than \(60 \mathrm{MW}\). Based on the cost of fuel, the cost to supply the heat transfer is \(\$ 4.50\) per GJ. The power developed by the cycle is valued at \(\$ 0.08\) per \(\mathrm{kW}\) - h. For 8000 hours of operation annually, determine for any such cycle, in $\$$ per year, (a) the maximum value of the power generated and (b) the minimum fuel cost.

A reversible heat engine operates between two reservoirs at temperatures of \(650^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\). A reversible refrigerator operates between reservoirs at temperatures of \(35^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\). The refrigerator is driven by the heat engine. The heat transfer to the heat engine is \(1500 \mathrm{~kJ}\). A net work output of \(300 \mathrm{~kJ}\) is obtained from the combined enginerefrigerator plant. Determine the net heat transfer to the reservoir at \(35^{\circ} \mathrm{C}\) and the heat transfer to the refrigerant from cold reservoir.

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