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5.43 A refrigeration cycle operating between two reserv
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Concepts and reason

Heat:

Heat is the amount of energy flowing from one body to another due to the difference in their temperatures.

Energy:

Energy is the capacity to do work. It is an extensive property. It includes the kinetic energy and potential energies. For a closed system, the energy transfer takes place across system boundary by means of only work and heat.

Refrigerator:

It is a device used to transfer the heat from low temperature medium to high temperature medium.

Fundamentals

First law of thermodynamics:

It states that energy cannot be created nor destroyed but can be changed from one form to another. Thus, energy is conserved. For an energy imbalance in a system, the energy change inside the system is difference between energy entering into the system and the energy leaving the system.

Energy balance neglecting the kinetic and potential energies is given as

E: - Eow = AE

The schematic diagram of a refrigeration system is shown in figure (1).

Hot Reservoir
ΤΗ
----->2H
Wcycle
Refrigeration
Cycle
-
-
-
-
-
-
-
-
1
Cold Reservoir
Figure 1

Coefficient of performance for a refrigerator (COP):

The efficiency of a refrigerator is indicated as the coefficient of performance . It is given by

B
Worce

Here, heat removed from refrigerated space is and work input to the system is .

From the first law of thermodynamics, the energy conversion in refrigeration system is expressed as

Wycle = QH-Qc
…… (1)

Here, the heat rejected to warm environment is .

The relation between temperatures and heat transfers in a refrigeration system can be expressed as

о.
Т
…… (2)

Here, the temperature at warm environment is and the temperature at the refrigerated space is .

From the relations (1) and (2), the COP of refrigerator can be expressed in different forms as:

Belc
Q-Qc
Qu/Qc-1
- Tu To
T,TC -1

Calculate the maximum possible coefficient of performance between given temperatures.

B=T, T.

Substitute 275 K
for and 315 K
for .

275 K
B=
315 K-275 K
= 6.875

(a)

Calculate the new COP for given conditions.

B.
Qc
W cyck

Substitute 1000 kJ
for and for .

1000 kJ
B.
80 kJ
= 12.5

Compare the and .

B. = 12.5> B =6.875

(b)

Calculate the new COP for given conditions.

B =
le
Qu-Qc

Substitute 1200 kJ
for and 2000 kJ
for .

1200 kJ
B =
2000 kJ -1200 kJ
= 1.5

Compare the and .

B. = 1.5<B=6.875

(c)

Calculate the new COP for given conditions.

B.
Woyce

Substitute equation (1) for .

B-911-Woede
W
ycle

Substitute 1575 kJ
for and 200 kJ
for .

1575 kJ - 200 kJ
200 kJ
= 6.875

Compare the and .

B. = 6.875 = B =6.875

(d)

Compare the and .

Substitute 6 for and 6.875 for .

Be=6<B=6.875

Ans: Part a

The cycle is impossible.

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