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A reversible power cycle receives energy \(Q_{1}\) and \(Q_{2}\) from hot reservoirs at temperatures \(T_{1}\) and \(T_{2}\), respectively, and discharges energy \(Q_{3}\) to a cold reservoir at temperature \(T_{3}\). (a) Obtain an expression for the thermal efficiency in terms of the ratios \(T_{1} / T_{3}, T_{2} / T_{3}, q=Q_{2} / Q_{1}\). (b) Discuss the result of part (a) in each of these limits: lim \(q \rightarrow 0, \lim q \rightarrow \infty, \lim T_{1} \rightarrow \infty\)

Short Answer

Expert verified
\( \eta = 1 - \frac{ \frac{T_3}{T_1} + q \frac{T_3}{T_2}}{1+q} \) as derived. The limits give: \( \eta = 1 - \frac{T_3}{T_1} \) for \( q \to 0 \), \( \eta = 1 - \frac{T_3}{T_2} \) for \( q \to \infty \), and \( \eta = 1 \) for \( T_1 \to \infty \).

Step by step solution

01

- Understand the Problem

Given a reversible power cycle with energy inputs and discharges, identify temperatures and energy values: temperatures are given as \(T_1\), \(T_2\), \(T_3\) and energy values are \(Q_1\), \(Q_2\), \(Q_3\). The problem requires deriving an expression for thermal efficiency involving the provided ratios and analyzing specific limits.
02

- Define Thermal Efficiency

Thermal efficiency, \(\eta\), for a power cycle is defined as the net work output divided by the total energy input. For a reversible cycle, the thermal efficiency \( \eta = 1 - \frac{Q_3}{Q_1 + Q_2} \).
03

- Apply Energy Balance

For the cycle, energy input is \(Q_1 + Q_2\) and energy output is \(Q_3\). According to the first law of thermodynamics, the net work output is equal to the heat input minus the heat discharged, thus, \( W = Q_1 + Q_2 - Q_3 \).
04

- Use Carnot Efficiency Relations

For a Carnot engine, the efficiency is related to temperatures of the reservoirs: \( \eta_1 = 1 - \frac{T_3}{T_1} \) and \( \eta_2 = 1 - \frac{T_3}{T_2} \). The discharged heat, \( Q_3 \), can be expressed in terms of \(T_i\) and \(Q_i\) as: \( Q_3 = Q_1 \frac{T_3}{T_1} + Q_2 \frac{T_3}{T_2} \).
05

- Derive Expression for Thermal Efficiency

Substitute \( Q_3 \) into thermal efficiency formula to get: \[ \eta = 1 - \frac{Q_3}{Q_1 + Q_2} = 1 - \frac{Q_1 \frac{T_3}{T_1} + Q_2 \frac{T_3}{T_2}}{Q_1 + Q_2} \]. Simplify using the ratio \( q = \frac{Q_2}{Q_1} \) to obtain: \[ \eta = 1 - \frac{\frac{T_3}{T_1} + q \frac{T_3}{T_2}}{1+q} \].
06

- Analyze Limits

Evaluate the expression for different values of \( q \): 1) \( \lim_{ q \to 0 } \eta = 1 - \frac{\frac{T_3}{T_1}}{1} = 1 - \frac{T_3}{T_1} \). 2) \( \lim_{ q \to \infty } \eta = 1 - \frac{q \frac{T_3}{T_2}}{q} = 1 - \frac{T_3}{T_2} \). 3) \( \lim_{ T_1 \to \infty } \eta = 1 - 0 = 1 \).

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Key Concepts

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

Carnot efficiency
The Carnot efficiency is a theoretical maximum efficiency for a heat engine operating between two temperature reservoirs. It is based on an idealized reversible process. The efficiency is given by the formula: \[\text{η}_c = 1 - \frac{T_c}{T_h} \] where \( T_c \) is the temperature of the cold reservoir and \( T_h \) is the temperature of the hot reservoir. This means no real engine can be more efficient than a Carnot engine working between the same temperatures. It's essential to understand that Carnot efficiency depends solely on the temperatures of the heat reservoirs, not on the working substance or mechanism of the engine. This principle helps set benchmarks for the best possible performance of thermal cycles.
Energy balance
Energy balance in a thermodynamic cycle involves accounting for the energy entering and leaving the system. In the context of the first law of thermodynamics, this means the total energy input must equal the total energy output plus any work done by the cycle. For the given problem: \[\text{Work output, } W = Q_1 + Q_2 - Q_3 \] where \ Q_1\ and \ Q_2\ are the heat inputs, and \ Q_3\ is the heat discharged to the cold reservoir. \ This relationship ensures that all the energy flows are accounted for and is fundamental to understanding how thermal cycles function. Maintaining an accurate energy balance is crucial for analyzing any thermodynamic process and determining efficiency.
First law of thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transformed from one form to another. Mathematically, it's expressed as: \[\text{ΔU} = Q - W \] where \ ΔU\ is the change in internal energy, \ Q\ is the heat added to the system, and \ W\ is the work done by the system. In the context of a power cycle, it implies that the net work output is equal to the total heat input minus the rejected heat: \[\text{Net work output} = \text{Heat input} - \text{Heat rejected} \] This principle is instrumental in solving problems related to thermal efficiency as it ensures that all energy transfers are precisely accounted for.
Thermodynamic cycles
Thermodynamic cycles are processes in which a working fluid undergoes a series of state changes, eventually returning to its original state. The purpose of these cycles is to convert heat energy into mechanical work. In a typical power cycle, such as the one described in the exercise, energy is received from high-temperature reservoirs and partially converted into work, with the remaining energy discharged to a lower temperature reservoir. The efficiency of such cycles is heavily influenced by how closely they can approximate an ideal Carnot cycle. Understanding thermodynamic cycles is essential for designing engines, refrigerators, and other systems that operate on thermal energy.
Reversible processes
A reversible process is an idealized process that happens infinitely slowly, ensuring the system remains in thermodynamic equilibrium at all times. No real process is completely reversible, but many can be approximated as such to simplify analysis. The significance of reversible processes is that they represent the most efficient path for energy conversions. In the exercise problem, using a reversible cycle helps derive the maximum possible efficiency (Carnot efficiency). This is because reversible processes involve no entropy generation, leading to no irreversible losses. Thus, they provide an ideal benchmark for evaluating real-world systems and their efficiencies.

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Most popular questions from this chapter

A system undergoes a thermodynamic cycle while receiving energy by heat transfer from a tank of liquid water initially at \(90^{\circ} \mathrm{C}\) and rejecting energy by heat transfer at \(15^{\circ} \mathrm{C}\) to the surroundings. If the final water temperature is \(15^{\circ} \mathrm{C}\), determine the minimum theoretical volume of water in the tank, \(\mathrm{m}^{3}\), for the cycle to produce net work equal to \(1.6 \times 10^{5} \mathrm{~kJ}\).

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A system undergoing a thermodynamic cycle receives \(Q_{\mathrm{H}}\) at temperature \(T_{\mathrm{H}}^{\prime}\) and discharges \(Q_{\mathrm{C}}\) at temperature \(T_{\mathrm{C}}^{\prime}\). There are no other heat transfers. (a) Show that the net work developed per cycle is given by $$ W_{\text {cycle }}=Q_{\mathrm{H}}\left(1-\frac{T_{\mathrm{C}}^{\prime}}{T_{\mathrm{H}}^{\prime}}\right)-T_{\mathrm{C}}^{\prime} \sigma $$ where \(\sigma\) is the amount of entropy produced per cycle owing to irreversibilities within the system. (b) If the heat transfers \(Q_{\mathrm{H}}\) and \(Q_{\mathrm{C}}\) are with hot and cold reservoirs, respectively, what is the relationship of \(T_{\mathrm{H}}^{\prime}\) to the temperature of the hot reservoir \(T_{\mathrm{H}}\) and the relationship of \(T_{\mathrm{C}}^{\prime}\) to the temperature of the cold reservoir \(T_{\mathrm{C}}\) ? (c) Obtain an expression for \(W_{\text {cycle }}\) if there are (i) no internal irreversibilities, (ii) no internal or external irreversibilities.

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