# Reverse Carnot Cycle

Reverse Carnot Cycle is a reversible cycle, it is used as an example of a refrigeration cycle operating between a constant temperature heat source and sink. On this basis, various refrigeration cycles are being compared.

Similarly, refrigeration working on, has a maximum COP. It is not possible to make the refrigerating machines working on the principle of reversed Carnot cycle.

###### In the first video, we have explained the working and in the second video, one solved example related to it.

Auses air as a working medium as shown in Fig. 1 and Fig 2 as a Reverse Carnot Cycle working and T-S diagram. Fig 1 Reverse Carnot cycle

## Process of Reverse Carnot Cycle

Four processes of the cycle are as follows.

• Process 1-2: Isentropic Compression.
• Process 2-3: Isothermal Compression.
• Process 3-4: Isentropic Expansion.
• Process 4-1: Isothermal Expansion. Fig. 2 T-S diagram of Reverse Carnot Cycle

Let p1, v1, and T1 be the pressure, volume, and temperature of the air.

1. Isentropic Compression: The air is compressed isentropically as shown by curve1-2 in the T-S diagram. During this process, the pressure increases from p1 to p2, and the specific volume decreases from v1 to v2 and the temperature increases from T1 to T2. During this process Entropy ‘s’ remains constant (s1=s2). No heat is observed or rejected by the air.

1. Isothermal Compression: Air is compressed isothermally at a constant temp T2=T3 as shown by the curve 2-3 in the T-s diagram. During this process, the pressure of the air increases from p2 to p3, specific volume decreases from 2 to v3 and the temperature remains constant T2=T3.

Heat rejected by the air: qr = T3(S2-S3) = T2(S2-S3)

where qr denotes heat rejection.

1. Isentropic Expansion: Air is expanded isentropically as shown by the curve 3-4 in the T-s diagram. The pressure of the air decreases from p3 to p4, specific volume increases from v3 to v4, and the temperature decreases from T3 to T4. No heat is rejected or absorbed by the air.

1. Isothermal Expansion: Air is compressed isothermally at a constant temp T4=T1 as shown by the curve 4-1 in the T-s diagram. The pressure of the air decreases from p4 to p1, specific volume increases from v4 to v1.

Heat absorbed by the air: qa = T4(S1-S4) = T1(S2-S3)

where qa denotes the heat absorption.

## Work Done

Work done during the cycle per kg of air

work done  = Heat rejected – Heat absorbed  = qr – qa

= T2(s2-s3) – T1(s2-s3)

= (T2 – T1)(s2 – s3)

## Reverse Carnot Cycle C.O.P

Coefficient of performance of the system working on the Reverse Carnot cycle.

C.O.P  =   Heat absorbed/ work done

=  qa/qa – qr

=  T1( S1 – S3 )/ ( T2 – T1 ) (S1 – S3 )

= T1/ T2 – T1

Reverse Carnot cycle is most efficient between the fixed temperature limits yet no refrigerator could be made using this cycle. This is due to the fact that the isentropic process requires high speed and the isothermal process requires extremely low speed.

### Reverse Carnot Cycle Example

Let us take an example related to it.

Question.

A Reverse Carnot cycle for refrigeration absorbs heat at 270K and rejects at 300K. Calculate the Coefficient of performance of this cycle. If the cycle is absorbing 1130kj/min at 270 K. How many kJ work is required per second?

T1 = 270 K and T2 = 300 K

As we have discussed above the C.O.P od the cycle is given by,

C.O.P = T1/ T2 – T1

C.O.P = 270 / 300 – 270

C.O.P = 9

Work Require per second

Heat absorbed at 270 K, Q1 = 1130 kJ/min = 18.83 kJ/s.

As we have discussed above

C.O.P = Heat Absorbed  / Work Done

9 = 18.83 / Work Done

Work Done = 2.1 kJ / s

Categories: Refrigeration