Question

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.

Short answer

The net work developed per cycle is given by \( W_{\text{cycle}} = Q_{\text{H}} \left( 1 - \frac{T_{\text{C}}^{\prime}}{T_{\text{H}}^{\prime}} \right) - T_{\text{C}}^{\prime} \sigma \). If there are no irreversibilities, \( W_{\text{cycle}} = Q_{\text{H}} \left( 1 - \frac{T_{\text{C}}}{T_{\text{H}}} \right) \).

Step-by-step solution

Relate the Net Work to Heat Transfers

According to the first law of thermodynamics, the net work done per cycle for a thermodynamic system is the difference between the heat added to the system and the heat removed from the system. Thus, we have \[ W_{\text{cycle}} = Q_{\text{in}} - Q_{\text{out}} \] In this context, the heat input is \( Q_{\text{H}} \) and the heat output is \( Q_{\text{C}} \).

Express Entropy Change

The total change in entropy for a complete cycle is the sum of the entropy changes due to heat transfer and the entropy produced due to irreversibilities, \( \sigma \). For the heat transfers at temperatures \( T_{\text{H}}^{\prime} \) and \( T_{\text{C}}^{\prime} \), we get: \[ \frac{Q_{\text{H}}}{T_{\text{H}}^{\prime}} - \frac{Q_{\text{C}}}{T_{\text{C}}^{\prime}} + \sigma = 0 \] Rearranging, we find: \[ Q_{\text{H}} = Q_{\text{C}} \left( \frac{T_{\text{H}}^{\prime}}{T_{\text{C}}^{\prime}} \right) - T_{\text{C}}^{\prime} \sigma \]

Develop the Expression for Work

Substitute the expression for \( Q_{\text{H}} \) into the first law equation:: \[ W_{\text{cycle}} = Q_{\text{H}} - Q_{\text{C}} \] Plug in \( Q_{\text{C}} \left( \frac{T_{\text{H}}^{\prime}}{T_{\text{C}}^{\prime}} \right) - T_{\text{C}}^{\prime} \sigma \) and simplify: \[ W_{\text{cycle}} = Q_{\text{H}} \left( 1 - \frac{T_{\text{C}}^{\prime}}{T_{\text{H}}^{\prime}} \right) - T_{\text{C}}^{\prime} \sigma \]

Relate Reservoir Temperatures

For the heat transfers involving reservoirs, the temperatures of the system segments doing the heat transfer must be equal to the reservoir temperatures under reversible conditions. Hence, if the system is in contact with a hot reservoir at temperature \( T_{\text{H}} \), then \( T_{\text{H}}^{\prime} = T_{\text{H}} \). Similarly, for the cold reservoir at temperature \( T_{\text{C}} \), \( T_{\text{C}}^{\prime} = T_{\text{C}} \).

Solve for Work with No Internal Irreversibilities

For no internal irreversibilities, \( \sigma = 0 \). In this case, the expression for net work simplifies to: \[ W_{\text{cycle}} = Q_{\text{H}} \left( 1 - \frac{T_{\text{C}}^{\prime}}{T_{\text{H}}^{\prime}} \right) \]

Solve for Work with No Internal or External Irreversibilities

With no internal or external irreversibilities, the process is fully reversible. In such cases, \( \sigma = 0 \) and there's no additional entropy exchange except due to reversible heat transfers, thus \( T_{\text{H}}^{\prime} = T_{\text{H}}, T_{\text{C}}^{\prime} = T_{\text{C}} \). Hence, \[ W_{\text{cycle}} = Q_{\text{H}} \left( 1 - \frac{T_{\text{C}}}{T_{\text{H}}} \right) \]

Key concepts

First Law of Thermodynamics

The first law of thermodynamics is pivotal in understanding energy conservation within a thermodynamic cycle. It states that the energy of an isolated system is constant.
In simpler terms, energy cannot be created or destroyed, only transformed from one form to another. For a thermodynamic cycle, the net work done, \( W_{\text{cycle}} \), is the difference between the total heat added to the system, \( Q_{\text{H}} \), and the heat removed from the system, \( Q_{\text{C}} \). We can represent this as:
\( W_{\text{cycle}} = Q_{\text{H}} - Q_{\text{C}} \).
This equation highlights how energy input into the system (heat) is converted into work and waste heat.
By understanding the energy changes, we can further assess system efficiency and potential areas for optimization.

Entropy Change

Entropy, often associated with disorder, plays a crucial role in thermodynamic processes. It measures the energy distribution among particles in a system. In a complete cycle, the change in entropy includes contributions from heat transfers and irreversibilities. For heat transfers at temperatures \( T_{\text{H}}^{\text{prime}} \) and \( T_{\text{C}}^{\text{prime}} \), the entropy change is expressed as:
\( \frac{Q_{\text{H}}}{T_{\text{H}}^{\text{prime}}} - \frac{Q_{\text{C}}}{T_{\text{C}}^{\text{prime}}} + \sigma = 0 \)
Here, \( \sigma \) accounts for the entropy produced due to system irreversibilities. This equation helps balance the entropy equations by introducing the concept of irreversibilities, further analyzed in other sections.

Heat Transfer

Heat transfer is central to thermodynamic cycles. It involves movement of thermal energy from one part of the system to another. In our cycle, heat is received from a high-temperature source \( T_{\text{H}}^{\text{prime}} \) and discharged to a low-temperature sink \( T_{\text{C}}^{\text{prime}} \). The efficiency of the cycle hinges on maximizing heat input and minimizing heat losses. Understanding how heat is transferred allows us to determine the net work performed, represented as:
\( W_{\text{cycle}} = Q_{\text{H}} \left( 1 - \frac{T_{\text{C}}^{\text{prime}}}{T_{\text{H}}^{\text{prime}}} \right) - T_{\text{C}}^{\text{prime}} \sigma \).
Proper management of heat transfer can reduce losses and improve system performance.

Irreversibilities

Irreversibilities account for inefficiencies in thermodynamic processes. These are deviations from ideal conditions and lead to entropy production represented by \( \sigma \). This can be due to friction, unrestrained expansions, and other non-idealities.
Understanding irreversibilities helps in calculating the net work done:
\( W_{\text{cycle}} = Q_{\text{H}} \left( 1 - \frac{T_{\text{C}}^{\text{prime}}}{T_{\text{H}}^{\text{prime}}} \right) - T_{\text{C}}^{\text{prime}} \sigma \).
If irreversibilities are minimized ( \( \sigma = 0 \)), we achieve better efficiency. Hence, identifying and reducing sources of irreversibilities is key to optimizing thermodynamic cycles.

Reservoir Temperatures

Thermodynamic cycles rely on heat transfers between reservoirs at different temperatures. The efficiency of the cycle depends on the temperature difference between the high-temperature \( T_{\text{H}} \) and low-temperature \( T_{\text{C}} \) reservoirs.
In an ideal reversible process, the system's heat transfer segments match the reservoir temperatures: \( T_{\text{H}}^{\text{prime}} = T_{\text{H}} \) and \( T_{\text{C}}^{\text{prime}} = T_{\text{C}} \).
For no irreversibilities within the system ( \( \sigma = 0 \)), the net work output is:
\( W_{\text{cycle}} = Q_{\text{H}} \left( 1 - \frac{T_{\text{C}}}{T_{\text{H}}} \right) \).
This illustrates how significant the reservoir temperatures and the degree of irreversibilities influence system efficiency and work output.