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4.15

Expert-verifiedFound in: Page 130

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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)$\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{{Q}_{\mathrm{f}}}$

b) $\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)}$, COP can be greater than 1.

c) The value is $\mathrm{COP}=\frac{{T}_{\mathrm{c}}\left({T}_{\mathrm{f}}-{T}_{\mathrm{r}}\right)}{{T}_{\mathrm{f}}\left({T}_{\mathrm{r}}-{T}_{\mathrm{c}}\right)}$

Q_{f}= heat input from flameQ_{c}= heat extracted from inside refrigerator

Explain why COP for an absorption refrigerator should be defined as ${Q}_{\mathrm{c}}/{Q}_{\mathrm{f}}$

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$\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{{Q}_{\mathrm{f}}}$

Q_{f}= heat input from flameQ_{c}= heat extracted from inside refrigeratorQ_{r}= waste heat expelled to room

and $\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{{Q}_{\mathrm{f}}}$

Find relationship among Q_{f}, Q_{c}, and Q_{r} ?

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

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

We have $\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{{Q}_{\mathrm{f}}}..........................\left(1\right)$

Rewrite the expression to include the waste heat Q_{R} expelled to room, we get

${Q}_{\mathrm{r}}={Q}_{\mathrm{f}}+{Q}_{\mathrm{c}}$

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

$\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)}$

Simplify this by dividing by Q_{c}

role="math" localid="1647272341714" $\text{COP}=\frac{1}{\frac{{Q}_{\mathrm{r}}}{{Q}_{\mathrm{c}}}-1}.............................\left(2\right)$

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

T_{f}= temperature of flameT_{c}= temperature inside refrigeratorT_{r}= room temperature

$\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)}$

Find the upper limit of COP

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

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

$\frac{{Q}_{\mathrm{f}}}{{T}_{\mathrm{f}}}+\frac{{Q}_{\mathrm{c}}}{{T}_{\mathrm{c}}}=\frac{{Q}_{\mathrm{r}}}{{T}_{\mathrm{r}}}$

Substitute ${Q}_{\mathrm{f}}=\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)$in the above equation

$\frac{\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)}{{T}_{\mathrm{f}}}+\frac{{Q}_{\mathrm{c}}}{{T}_{\mathrm{c}}}=\frac{{Q}_{\mathrm{r}}}{{T}_{\mathrm{r}}}$

Rearrange the above equation,

$\frac{{Q}_{\mathrm{r}}}{{Q}_{\mathrm{c}}}=\frac{{T}_{\mathrm{r}}\left({T}_{\mathrm{c}}-{T}_{\mathrm{f}}\right)}{{T}_{\mathrm{c}}\left({T}_{\mathrm{r}}-{T}_{\mathrm{f}}\right)}$

Substitute $\frac{{Q}_{\mathrm{r}}}{{Q}_{\mathrm{c}}}=\frac{{T}_{\mathrm{r}}\left({T}_{\mathrm{c}}-{T}_{\mathrm{f}}\right)}{{T}_{\mathrm{c}}\left({T}_{\mathrm{r}}-{T}_{\mathrm{f}}\right)}intheCOPformula\text{COP}=\frac{1}{\frac{{Q}_{\mathrm{r}}}{{Q}_{\mathrm{c}}}-1}$

$\mathrm{COP}=\frac{{T}_{\mathrm{c}}\left({T}_{\mathrm{f}}-{T}_{\mathrm{r}}\right)}{{T}_{\mathrm{f}}\left({T}_{\mathrm{r}}-{T}_{\mathrm{c}}\right)}$

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