Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

In an absorption refrigerator, the energy driving the process is supplied not as work, but as heat from a gas flame. (Such refrigerators commonly use propane as fuel, and are used in locations where electricity is unavailable.* ) Let us define the following symbols, all taken to be positive by definition:
Q_f= heat input from flameQ_c= heat extracted from inside refrigeratorQ_r= waste heat expelled to roomT_f= temperature of flameT_c= temperature inside refrigeratorT_r= room temperature(a) Explain why the "coefficient of performance" (COP) for an absorption refrigerator should be defined as Q_c / Q_f.(b) What relation among Q_f, Q_c, and Q_r is implied by energy conservation alone? Will energy conservation permit the COP to be greater than 1 ?(c) Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures T_f, T_c, and T_r alone.

a) $COP = \frac{Q_{c}}{Q_{f}}$

b) $COP = \frac{Q_{c}}{(Q_{r} - Q_{c})}$ , COP can be greater than 1.

c) The value is $COP = \frac{T_{c} (T_{f} - T_{r})}{T_{f} (T_{r} - T_{c})}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part(a): Step 1: Given Information

Q_f= heat input from flameQ_c= heat extracted from inside refrigerator

Explain why COP for an absorption refrigerator should be defined as $Q_{c} / Q_{f}$

Part(a): Step 2: Explanation

In an absorption refrigerator electrical energy is transformed into the heat from a gas flame.
Since the heat extracted is the benefit and the fuel is the cost in operating the refrigerator therefore the coefficient of performance isSo input energy is Q_f and output is Q_c so COP is written as $COP = \frac{Q_{c}}{Q_{f}}$

Part(b): Step 1: Given information

Q_f= heat input from flameQ_c= heat extracted from inside refrigeratorQ_r= waste heat expelled to room

and $COP = \frac{Q_{c}}{Q_{f}}$

Find relationship among Q_f, Q_c, and Q_r ?

Will energy conservation permit the COP to be greater than 1 ?

Part (b): Step 2 : Explanation

In an absorption refrigerator heat from a gas flame is transformed into electrical energy

We have $COP = \frac{Q_{c}}{Q_{f}} . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Rewrite the expression to include the waste heat Q_R expelled to room, we get

$Q_{r} = Q_{f} + Q_{c}$

Rearrange the equation and Substitute this in the equation (1)

$COP = \frac{Q_{c}}{(Q_{r} - Q_{c})}$

Simplify this by dividing by Q_c

role="math" localid="1647272341714" $COP = \frac{1}{\frac{Q_{r}}{Q_{c}} - 1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

From the equation(2) $Q_{r} / Q_{c}$ is greater than 2, so COP can be greater than 1.

Part(C): Step 1:Given

T_f= temperature of flameT_c= temperature inside refrigeratorT_r= room temperature

$COP = \frac{Q_{c}}{(Q_{r} - Q_{c})}$

Find the upper limit of COP

Part(c): Step 2: Explanation

In an absorption refrigerator exchanges the electrical energy with the heat energy.

Use the second law of thermodynamics write the expression of the condition

$\frac{Q_{f}}{T_{f}} + \frac{Q_{c}}{T_{c}} = \frac{Q_{r}}{T_{r}}$

Substitute $Q_{f} = (Q_{r} - Q_{c})$ in the above equation

$\frac{(Q_{r} - Q_{c})}{T_{f}} + \frac{Q_{c}}{T_{c}} = \frac{Q_{r}}{T_{r}}$

Rearrange the above equation,

$\frac{Q_{r}}{Q_{c}} = \frac{T_{r} (T_{c} - T_{f})}{T_{c} (T_{r} - T_{f})}$

Substitute $\frac{Q_{r}}{Q_{c}} = \frac{T_{r} (T_{c} - T_{f})}{T_{c} (T_{r} - T_{f})} i n t h e C O P f o r m u l a COP = \frac{1}{\frac{Q_{r}}{Q_{c}} - 1}$

$COP = \frac{T_{c} (T_{f} - T_{r})}{T_{f} (T_{r} - T_{c})}$

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