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Answers without the blur. Sign up and see all textbooks for free! Q.4.31

Expert-verified Found in: Page 141 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Suppose that the throttling valve in the refrigerator of the previous problem is replaced with a small turbine-generator in which the fluid expands adiabatically, doing work that contributes to powering the compressor. Will this change affect the COP of the refrigerator? If so, by how much? Why do you suppose real refrigerators use a throttle instead of a turbine?

The COP is enhanced by 1.08, and practical refrigerators use throttle rather than the turbine to reduce cost and complexity.

See the step by step solution

## Step 1: To find

The argument of whether and by how much, if any, the change will affect the COP of a refrigerator.and the reason why real refrigerators use a throttle instead of a turbine.

## Step 2: Explanation

The refrigerator runs on HFC-234a at pressures ranging from 1 to 10 bar and temperatures ranging from $-26.4°$ to $39.4°$. Along the throttling path, entropy is conserved.

The expression of the entropy at point 4

$S=x{S}_{1}+\left(1-x\right){S}_{\mathrm{g}}$

Where, x is the fraction of liquid,

${S}_{1}$ is the entropy of liquid and

${S}_{\mathrm{g}}$ is the entropy of the gas.

Rearrange the above expression

$x=\frac{S-{S}_{2}}{{S}_{1}-{S}_{2}}$

Substitute role="math" localid="1648669923161" $S=0.384\mathrm{kJ}/\mathrm{K}·\mathrm{kg}$

${S}_{1}=0.068\mathrm{kJ}/\mathrm{K}·\mathrm{kg}$

${S}_{\mathrm{g}}=0.940\mathrm{kJ}/\mathrm{K}·\mathrm{kg}$ from table 4.3 in the above expression.

$x=\frac{\left(0.384\mathrm{kJ}/\mathrm{K}·\mathrm{kg}\right)-\left(0.940\mathrm{kJ}/\mathrm{K}·\mathrm{kg}\right)}{\left(0.068\mathrm{kJ}/\mathrm{K}·\mathrm{kg}\right)-\left(0.940\mathrm{kJ}/\mathrm{K}·\mathrm{kg}\right)}\phantom{\rule{0ex}{0ex}}=0.6376$

## Step 3: Continuation for the explanation

Now,

The expression for the enthalpy ${\mathrm{H}}_{4}$at point 4

${H}_{4}=x{H}_{1}+\left(1-x\right){H}_{g}$

Here, ${H}_{1}$ = is enthalpy of liquid and

${H}_{\mathrm{g}}$= is enthalpy of gas.

$\text{Substitute}{H}_{1}=16\mathrm{kJ}/\mathrm{kg}\phantom{\rule{0ex}{0ex}}{H}_{\mathrm{g}}=231\mathrm{kJ}/\mathrm{kg}\phantom{\rule{0ex}{0ex}}x=0.6376\text{from table in the above expression.}$

${H}_{4}=\left(0.6376\right)\left(16\mathrm{kJ}/\mathrm{kg}\right)+\left(1-0.6376\right)\left(231\mathrm{kJ}/\mathrm{kg}\right)\phantom{\rule{0ex}{0ex}}=93.92\mathrm{kJ}/\mathrm{kg}$

The expression of the COP

$\mathrm{COP}=\frac{{H}_{1}-{H}_{4}}{{H}_{2}-{H}_{3}-\left({H}_{1}-{H}_{4}\right)}$

Here,${H}_{1}=$enthalpy at point 1,

${H}_{2}$= is enthalpy at point 2 and

localid="1648670937721" ${H}_{3}=$is enthalpy at point 3 .

$\text{Substitute}{H}_{1}=231\mathrm{kJ}/\mathrm{kg}\phantom{\rule{0ex}{0ex}}{\mathrm{H}}_{2}=279.08\mathrm{kJ}/\mathrm{kg},\phantom{\rule{0ex}{0ex}}{\mathrm{H}}_{3}=105\mathrm{kJ}/\mathrm{kg}\phantom{\rule{0ex}{0ex}}{\mathrm{H}}_{4}=93.92\mathrm{kJ}/kg\text{in above expression}$

$\mathrm{COP}=\frac{\left(231\mathrm{kJ}/\mathrm{kg}\right)-\left(93.92\mathrm{kJ}/\mathrm{kg}\right)}{\left(279.08\mathrm{kJ}/\mathrm{kg}\right)-\left(105\mathrm{kJ}/\mathrm{kg}\right)-\left(\left(231\mathrm{kJ}/\mathrm{kg}\right)-\left(93.92\mathrm{kJ}/\mathrm{kg}\right)\right)}\phantom{\rule{0ex}{0ex}}=3.70$

$\text{The value of increment in COP is}$

$=\left(3.70-2.62\right)\phantom{\rule{0ex}{0ex}}=1.08$

Hence The COP is enhanced by 1.08, and practical refrigerators use throttle rather than the turbine to reduce cost and complexity. ### Want to see more solutions like these? 