Question

Two reversible power cycles are arranged in series. The first cycle receives energy by heat transfer from a reservoir at temperature \(T_{\mathrm{H}}\) and rejects energy to a reservoir at an intermediate temperature \(T\). The second cycle receives the energy rejected by the first cycle from the reservoir at temperature \(T\). and rejects energy to a reservoir at temperature \(T_{\mathrm{C}}\) lower than \(T\). Derive an expression for the intermediate temperature \(T\) in terms of \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}}\) when (a) the net work of the two power cycles is equal. (b) the thermal efficiencies of the two power cycles are equal.

Short answer

In both cases, the intermediate temperature \( T = \sqrt{T_H T_C} \).

Step-by-step solution

Understand the Problem

Two reversible power cycles operate in series. The first cycle receives heat from a hot reservoir at temperature \( T_H \) and rejects heat to an intermediate reservoir at temperature \( T \). The second cycle takes this heat from temperature \( T \) and rejects it to a cold reservoir at temperature \( T_C \). The task is to find an expression for the intermediate temperature \( T \) in two different cases: when the net work of the cycles is equal, and when the thermal efficiencies of the cycles are equal.

Define Thermal Efficiency

The thermal efficiency of a reversible power cycle operating between two temperatures is given by \( \eta = 1 - \frac{T_{ \text{cold}}}{T_{ \text{hot}}} \). For the first cycle, the efficiency is \( \eta_1 = 1 - \frac{T}{T_H} \). For the second cycle, the efficiency is \( \eta_2 = 1 - \frac{T_C}{T} \).

Case (a) - Net Work Equality

In this case, the net work done by both cycles is equal. Let the heat input to the first cycle be \( Q_H \). The work done by the first cycle is \( W_1 = \eta_1 Q_H = \left( 1 - \frac{T}{T_H} \right) Q_H \). The heat rejected by the first cycle and taken by the second cycle is \( Q = (1 - \eta_1) Q_H = \frac{T}{T_H} Q_H \). The work done by the second cycle is \( W_2 = \eta_2 Q = \left( 1 - \frac{T_C}{T} \right) \frac{T}{T_H} Q_H \). Equate \( W_1 \) and \( W_2 \) and solve for \( T \): \[ \left( 1 - \frac{T}{T_H} \right) Q_H = \left( 1 - \frac{T_C}{T} \right) \frac{T}{T_H} Q_H \] \[ 1 - \frac{T}{T_H} = \frac{T - T_C}{T_H} \] \[ T = \sqrt{T_H T_C} \]

Case (b) - Thermal Efficiency Equality

In this case, the thermal efficiencies of both cycles are equal. Equate the efficiencies \( \eta_1 \) and \( \eta_2 \): \[ \eta_1 = \eta_2 \] \[ 1 - \frac{T}{T_H} = 1 - \frac{T_C}{T} \] Simplifying, we obtain \[ \frac{T}{T_H} = \frac{T_C}{T} \] \[ T^2 = T_H T_C \] \[ T = \sqrt{T_H T_C} \]

Key concepts

thermodynamic cycles

Thermodynamic cycles are fundamental concepts in the study of thermodynamics. These cycles consist of a series of processes that transfer heat and work between a system and its surroundings, ultimately returning to the initial state. Examples of thermodynamic cycles include the Carnot cycle, the Rankine cycle, and the Brayton cycle.

In this exercise, we deal with two reversible power cycles, which are idealized processes that operate with maximum efficiency. Reversible cycles are theoretical because they assume no energy losses. These cycles are arranged in series, meaning the heat rejected by the first cycle is utilized by the second cycle.

intermediate temperature

The intermediate temperature plays a critical role in linking the two reversible cycles. The first cycle receives heat from a high-temperature reservoir at temperature, denoted as \( T_H \). It then rejects heat to a lower-temperature reservoir at an intermediate temperature, denoted as \( T \).

This intermediate temperature is crucial because it dictates how much heat is retained and transferred to the second cycle. The second cycle begins by receiving this rejected heat at temperature \( T \) and then rejects heat to a colder reservoir at temperature \( T_C \). Finding the correct expression for this intermediate temperature based on different conditions helps ensure that the system operates efficiently.

thermal efficiency equality

Thermal efficiency is a measure of how effectively a thermodynamic cycle converts heat into work. The efficiency of a reversible cycle operating between a hot and a cold reservoir is given by \[ \eta = 1 - \frac{T_{cold}}{T_{hot}} \].

For our two cycles:

• The first cycle's efficiency: \ \eta_1 = 1 - \frac{T}{T_H} \
• The second cycle's efficiency: \ \eta_2 = 1 - \frac{T_C}{T} \

If these efficiencies are equal, meaning \ \eta_1 = \eta_2 \, we can derive that the intermediate temperature \( T \) must satisfy the equality: \ \frac{T}{T_H} = \frac{T_C}{T} \.

Solving this expression provides us with \ T = \sqrt{T_H T_C} \. This means that when the thermal efficiencies of both cycles are the same, the intermediate temperature \( T \) is the geometric mean of the high and low temperatures.

net work equality

Net work equality refers to the scenario where the total work output of the two power cycles is the same. The work done by each cycle is related to the heat input and the cycle's efficiency.

The first cycle receives heat \( Q_H \) and does work calculated as \ W_1 = \eta_1 Q_H = (1 - \frac{T}{T_H}) Q_H\. The heat rejected by the first cycle is \ Q = \frac{T}{T_H} Q_H \, which is then received by the second cycle.

The second cycle performs work \ W_2 = \eta_2 Q = (1 - \frac{T_C}{ T}) \frac{T}{T_H} Q_H \. To achieve net work equality, we set \ \displaystyle W_1 = W_2 \ and simplify to find \ T = \sqrt{T_H T_C} \.

This shows net work equality also leads to the intermediate temperature being the geometric mean of \( T_H \) and \( T_C \).

heat transfer

Heat transfer between reservoirs and cycles is a fundamental concept in understanding thermodynamic systems. In this problem, it is essential to see how heat is transferred from a high-temperature reservoir to a sequence of cycles and eventually to a lower-temperature reservoir.

• The first cycle receives heat \( Q_H \) from the high-temperature reservoir \( T_H \)
• The cycle then rejects an amount of heat \( Q \) to the intermediate temperature reservoir \( T \)
• The second cycle absorbs this heat \( Q \) at \( T \) and subsequently rejects heat \( Q_C \) to the low-temperature reservoir \( T_C \)

Understanding this sequence helps analyze the efficiency and work output of the cycles as well as the determination of critical temperatures that influence the overall system's performance.