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Q.4.31

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An Introduction to Thermal Physics
Found in: Page 141
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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=xS1+(1-x)Sg

Where, x is the fraction of liquid,

S1 is the entropy of liquid and

Sg is the entropy of the gas.

Rearrange the above expression

x=S-S2S1-S2

Substitute role="math" localid="1648669923161" S=0.384 kJ/K·kg

S1=0.068 kJ/K·kg

Sg=0.940 kJ/K·kg from table 4.3 in the above expression.

x=(0.384 kJ/K·kg)-(0.940 kJ/K·kg)(0.068 kJ/K·kg)-(0.940 kJ/K·kg) =0.6376

Step 3: Continuation for the explanation

Now,

The expression for the enthalpy H4at point 4

H4=xH1+(1-x)Hg

Here, H1 = is enthalpy of liquid and

Hg= is enthalpy of gas.

Substitute H1=16 kJ/kg Hg= 231 kJ/kg x=0.6376 from table in the above expression.

H4=(0.6376)(16 kJ/kg)+(1-0.6376)(231 kJ/kg)= 93.92 kJ/kg

The expression of the COP

COP=H1-H4H2-H3-H1-H4

Here,H1=enthalpy at point 1,

H2= is enthalpy at point 2 and

localid="1648670937721" H3 =is enthalpy at point 3 .

Substitute H1=231 kJ/kg H2=279.08 kJ/kg, H3=105 kJ/kg H4= 93.92 kJ/kg in above expression

COP=(231 kJ/kg)-(93.92 kJ/kg)(279.08 kJ/kg)-(105 kJ/kg)-((231 kJ/kg)-(93.92 kJ/kg)) = 3.70

The value of increment in COP is

=(3.70-2.62)= 1.08

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

Most popular questions for Physics Textbooks

Under many conditions, the rate at which heat enters an air conditioned building on a hot summer day is proportional to the difference in temperature between inside and outside, Th-Tc. (If the heat enters entirely by conduction, this statement will certainly be true. Radiation from direct sunlight would be an exception.) Show that, under these conditions, the cost of air conditioning should be roughly proportional to the square of the temperature difference. Discuss the implications, giving a numerical example.

Consider a household refrigerator that uses HFC-134a as the refrigerant, operating between the pressures of 1.0bar and 10bars.

(a) The compression stage of the cycle begins with saturated vapor at 1 bar and ends at 10 bars. Assuming that the entropy is constant during compression, find the approximate temperature of the vapor after it is compressed. (You'll have to do an interpolation between the values given in Table 4.4.)

(b) Determine the enthalpy at each of the points 1,2,3 and 4 , and calculate the coefficient of performance. Compare to the COP of a Carnot refrigerator operating between the same extreme temperatures. Does this temperature range seem reasonable for a household refrigerator? Explain briefly.

(c) What fraction of the liquid vaporizes during the throttling step?

It has been proposed to use the thermal gradient of the ocean to drive a heat engine. Suppose that at a certain location the water temperature is 22°Cat the ocean surface and 4°C at the ocean floor.

(a) What is the maximum possible efficiency of an engine operating between these two temperatures?

(b) If the engine is to produce 1GW of electrical power, what minimum volume of water must be processed (to suck out the heat) in every second?

Derive equation 4.10 for the efficiency of the Otto cycle.

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:
Qf= heat input from flameQc= heat extracted from inside refrigeratorQr= waste heat expelled to roomTf= temperature of flameTc= temperature inside refrigeratorTr= room temperature(a) Explain why the "coefficient of performance" (COP) for an absorption refrigerator should be defined as Qc / Qf.(b) What relation among Qf, Qc, and Qr 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 Tf, Tc, and Tr alone.
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