/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A system undergoes a refrigerati... [FREE SOLUTION] | 91Ó°ÊÓ

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A system undergoes a refrigeration cycle while receiving \(Q_{\mathrm{C}}\) by heat transfer at temperature \(T_{\mathrm{C}}\) and discharging energy \(Q_{\mathrm{H}}\) by heat transfer at a higher temperature \(T_{\mathrm{H}}\) There are no other heat transfers. (a) Using energy and exergy balances, show that the net work input to the cycle cannot be zero. (b) Show that the coefficient of performance of the cycle can be expressed as $$ \beta=\left(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}\right)\left(1-\frac{T_{\mathrm{H}} \mathrm{E}_{\mathrm{d}}}{T_{0}\left(Q_{\mathrm{H}}-Q_{\mathrm{C}}\right)}\right) $$ where \(\mathrm{E}_{\mathrm{d}}\) is the exergy destruction and \(T_{0}\) is the temperature of the exergy reference environment. (c) Using the result of part (b), obtain an expression for the maximum theoretical value for the coefficient of performance.

Short Answer

Expert verified
The net work cannot be zero by energy and exergy balances. The coefficient of performance \(\beta \) is given by \(\beta = \frac{T_{\text{C}}}{T_{\text{H}} - T_{\text{C}} \right(1 - \frac{\text{T}_{\text{H}} \text{E}_{\text{d}}}{T_{0} (\text{Q}_{\text{H}} - \text{Q}_{\text{C}})} \). Maximum \(\beta_{\text{max}}\) is \(\frac{T_{\text{C}}}{T_{\text{H}} - T_{\text{C}}}\.

Step by step solution

01

Energy Balance

Begin by applying the First Law of Thermodynamics (energy balance) to the refrigeration cycle. The net work input, denoted by \(\text{W}_{\text{net}}\), can be expressed as: \[\text{W}_{\text{net}} = Q_{\text{H}} - Q_{\text{C}} \] where \Q_{\text{H}}\ is the heat rejected to the high temperature reservoir and \Q_{\text{C}}\ is the heat absorbed from the low temperature reservoir.
02

Exergy Balance

Next, apply the exergy balance for the cycle. The exergy destruction, \(E_{\text{d}}\), accounts for irreversibilities in the cycle. The exergy balance can be written as: \[\text{W}_{\text{net}} = \text{W}_{\text{rev}} - E_{\text{d}} \] where \text{W}_{\text{rev}}\ is the reversible work. This shows that \text{W}_{\text{net}}\ cannot be zero unless \E_{\text{d}}\ is zero, which is impossible in real cycles.
03

Coefficient of Performance

From part (a), using the relation \(\text{W}_{\text{net}} = Q_{\text{H}} - Q_{\text{C}} \), the coefficient of performance \(\beta \) is given by: \[ \beta = \frac{Q_{\text{C}}}{\text{W}_{\text{net}}} = \frac{Q_{\text{C}}}{Q_{\text{H}} - Q_{\text{C}}} \]
04

Express \beta \ in Terms of Exergy

Using exergy balance, substitute \(\text{W}_{\text{net}} = T_{0} \frac{E_{\text{d}}}{Q_{\text{H}} - Q_{\text{C}}} \) into the coefficient of performance equation: \[ \beta = \frac{Q_{\text{C}}}{T_{\text{H}} - T_{\text{C}}}\right(1 - \frac{\text{T}_{\text{H}} \text{E}_{\text{d}}}{T_{0} \text{(Q}_{\text{H}} - \text{Q}_{\text{C}})} \]
05

Maximum Theoretical Coefficient of Performance

To find the maximum theoretical coefficient of performance, set the exergy destruction \(E_{\text{d}} = 0\): \[ \beta_{\text{max}} = \frac{T_{\text{C}}}{T_{\text{H}} - T_{\text{C}}} \]

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Key Concepts

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

Energy Balance
The energy balance is rooted in the First Law of Thermodynamics. This law emphasizes conservation of energy, stating that energy cannot be created or destroyed, only transferred or transformed. For a refrigeration cycle, the energy balance can be represented as:
\( W_{\text{net}} = Q_{\text{H}} - Q_{\text{C}} \)
Here, \( W_{\text{net}} \) is the net work input to the cycle. \( Q_{\text{H}} \) represents the heat ejected to the high-temperature reservoir, and \( Q_{\text{C}} \) is the heat absorbed from the low-temperature reservoir. This equation tells us that the work supplied to the cycle is equal to the energy difference between heat rejected and heat absorbed.
It's important to note that for the refrigeration cycle to operate, the net work input, \( W_{\text{net}} \), cannot be zero. If it were, it would imply no energy input, making the heat transfer process impossible.
Exergy Balance
Exergy balance considers the quality and usability of energy, incorporating the concept of irreversibility within a thermodynamic process. Exergy destruction, indicated by \( E_{\text{d}} \), reflects the energy lost due to inefficiencies. The exergy balance equation is:
\( W_{\text{net}} = W_{\text{rev}} - E_{\text{d}} \)
Here, \( W_{\text{rev}} \) is the reversible work, work that would occur in an ideal, lossless system. With real systems, however, irreversibilities (such as friction and unrestrained expansion) mean that \( E_{\text{d}} \) is always greater than zero. Therefore, \( W_{\text{net}} \) must also be greater than zero for any real processes. This assures us that net work input can't be zero unless there are no losses, which is unrealistic.
Coefficient of Performance (COP)
The coefficient of performance (COP) is a measure of a refrigeration cycle's efficiency. For a refrigeration cycle, the COP is defined by the ratio:
\( \beta = \frac{ Q_{\text{C}} }{ W_{\text{net}} } = \frac{ Q_{\text{C}} }{ Q_{\text{H}} - Q_{\text{C}} } \)
This equation indicates how effectively the system uses work to transfer heat from the cold reservoir.
In a practical scenario, the COP gives insights into how much cooling you get per unit of work input. Higher COP values are desirable as they imply more efficient systems, translating to lower operational costs.
Exergy Destruction
Exergy destruction refers to the portion of energy that is irrevocably lost due to irreversibilities within the system. It signifies inefficiencies such as friction, turbulence, and other dissipative effects.
In the exergy balance equation, it was seen that:
\( W_{\text{net}} = W_{\text{rev}} - E_{\text{d}} \)
Here, \( E_{\text{d}} \) reflects the real system's deviation from an ideal system. Decreasing exergy destruction leads to enhanced cycle performance. This relationship highlights the importance of minimizing inefficiencies to improve the system's overall performance and efficiency.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, is crucial in understanding energy interactions. It states that energy can neither be created nor destroyed in an isolated system. For the refrigeration cycle, it lays the foundation for energy balance:
\( W_{\text{net}} = Q_{\text{H}} - Q_{\text{C}} \)
This principle ensures that the energy transferred as work is balanced by the heat flow into and out of the system. By applying this law, we make sure that all energy within the system is accounted for, allowing accurate calculation and analysis of the system's performance.

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Most popular questions from this chapter

\(7.92\) Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas enters a turbine operating at steady state at 50 bar, \(500 \mathrm{~K}\) with a velocity of \(50 \mathrm{~m} / \mathrm{s}\). The inlet area is \(0.02 \mathrm{~m}^{2}\). At the exit, the pressure is 20 bar, the temperature is \(440 \mathrm{~K}\), and the velocity is \(10 \mathrm{~m} / \mathrm{s}\) The power developed by the turbine is \(3 \mathrm{MW}\), and heat transfer occurs across a portion of the surface where the average temperature is \(462 \mathrm{~K}\). Assume ideal gas behavior for the carbon dioxide and neglect the effect of gravity. Let \(T_{0}=298 \mathrm{~K}, p_{0}=1\) bar. (a) Determine the rate of heat transfer, in \(\mathrm{kW}\). (b) Perform a full exergy accounting, in \(\mathrm{kW}\), based on the net rate exergy is carried into the turbine by the carbon dioxide.

A steam turbine operating at steady state develops \(9750 \mathrm{hp}\). The turbine receives 100,000 pounds of steam per hour at \(400 \mathrm{lbf} / \mathrm{in} .^{2}\) and \(600^{\circ} \mathrm{F}\). At a point in the turbine where the pressure is \(60 \mathrm{lbf} / \mathrm{in}^{2}\). and the temperature is \(300^{\circ} \mathrm{F}\), steam is bled off at the rate of \(25,000 \mathrm{lb} / \mathrm{h}\). The remaining steam continues to expand through the turbine, exiting at \(2 \mathrm{lbf} / \mathrm{in}^{2}\) and \(90 \%\) quality. (a) Determine the rate of heat transfer between the turbine and its surroundings, in Btu/h. (b) Devise and evaluate an exergetic efficiency for the turbine. Kinetic and potential energy effects can be ignored. Let \(T_{0}=\) \(77^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).

A system consists of \(2 \mathrm{~kg}\) of water at \(100^{\circ} \mathrm{C}\) and 1 bar. Determine the exergy, in kJ, if the system is at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1 \mathrm{bar}\).

Refrigerant \(134 \mathrm{a}\) at \(100 \mathrm{lb} / \mathrm{in} .^{2}, 200^{\circ} \mathrm{F}\) enters a valve operating at steady state and undergoes a throttling process. (a) Determine the exit temperature, in \({ }^{\circ} F\), and the exergy destruction rate, in Btu per lb of Refrigerant 134a flowing, for an exit pressure of \(50 \mathrm{lbf} / \mathrm{in}^{2}\). \(^{2}\) (b) Plot the exit temperature, in \({ }^{\circ} \mathrm{F}\), and the exergy destruction rate, in Btu per lb of Refrigerant 134a flowing, each versus exit pressure ranging from 50 to \(100 \mathrm{lbf} /\) in. \({ }^{2}\) Let \(T_{0}=70^{\circ} \mathrm{F}, p_{0}=14.7 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\)

A pump operating at steady state takes in saturated liquid water at \(65 \mathrm{lbf} / \mathrm{in}^{2}\) at a rate of \(10 \mathrm{lb} / \mathrm{s}\) and discharges water at \(1000 \mathrm{lbf} / \mathrm{in}^{2}\). The isentropic pump efficiency is \(80.22 \%\). Heat transfer with the surroundings and the effects of motion and gravity can be neglected. If \(T_{0}=75^{\circ} \mathrm{F}\), determine for the pump (a) the exergy destruction rate, in Btu/s (b) the exergetic efficiency.

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