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 flame
Q_c= heat extracted from inside refrigerator
Q_r= waste heat expelled to room
T_f= temperature of flame
T_c= temperature inside refrigerator
T_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.

Q_f= heat input from flame
Q_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 is

So 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 flame
Q_c= heat extracted from inside refrigerator
Q_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 flame
T_c= temperature inside refrigerator
T_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 $\frac{Q_{\mathrm{r}}}{Q_{\mathrm{c}}}=\frac{T_{\mathrm{r}}\left(T_{\mathrm{c}}-T_{\mathrm{f}}\right)}{T_{\mathrm{c}}\left(T_{\mathrm{r}}-T_{\mathrm{f}}\right)}intheCOPformula\text{COP}=\frac{1}{\frac{Q_{\mathrm{r}}}{Q_{\mathrm{c}}}-1}$

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