Mumbai University > Mechanical Engineering > Sem8 > Refrigeration and Air Conditioning

Marks: 5M

Year: Dec 2012

The Carnot refrigerator works on the reversed Carnot cycle. A reversed Carnot cycle using air as working medium is shown on p-v and t-s diagrams. The processes involved during the cycle are:

Fig. Reversed Carnot Cycle

1. Isentropic compression process- Air is compressed is entropically as shown by curve 1-2 on diagrams. No heat is absorbed or rejected by the air.

2. Isothermal compression process- Air is now compressed isothermally (T2=T3) and the heat rejected per kg of air during the process is given by

$$q_{2-3}= \ \ \ area \ \ \ 2-3-3'-2' =T_3 (s_2-s_3 )=T_2 (s_2-s_3 )$$

3. Isentropic expansion process- Air is now expanded is entropically as shown by curve 3-4.

4. Isothermal expansion process- Air is expanded isothermally and heat absorbed by air (or heat extracted from cold body) during this process per kg of air is given by

$$q_{4-1}=area \ \ \ 4-1-2'-3'=T_4 (s_1-s_4 )=T_4 (s_2-s_3 )=T_1 (s_2-s_3 )$$

Therefore, work done during cycle per kg of air

$$=\text{heat rejected-heat absorbed}=(q_{2-3})-(q_{4-1}) \\ =T_2 (s_2-s_3 )-T_1 (s_2-s_3 )=(T_2-T_1 )(s_2-s_3 )$$

C.O.P. of refrigerator working on reversed Carnot cycle,

$$C.O.P_R=\frac{(\text{Heat absorbed})}{(\text{Work done})}=\frac{q_{4-1}}{(q_{2-3}-q_{4-1})}=\frac{T_1 (s_2-s_3 )}{(T_2-T_1 )(s_2-s_3 )} =\frac{T_1}{(T_2-T_1 )}$$

Factors which affect the COP of the cycle:

• C.O.P. of the reversed Carnot cycle may be improved by decreasing higher temperature i.e. temperature of hot body T2
• Increasing lower temperature i.e. temperature of cold body T1 also improves C.O.P.
• Although this cycle gives maximum C.O.P. yet it is impossible to achieve this because the isentropic processes of the cycle require high speed while the isothermal processes require an extremely low speed which is impracticable for air or gas.

Practical difficulties with Carnot refrigeration system:

• Although this cycle is theoretically most efficient between the fixed temperature limits, yet it is impossible to achieve this because the isentropic processes of the cycle require high speed while the isothermal processes require an extremely low speed. Such variation of speed in a stroke is not possible and makes it impracticable.
• With finite specific heats of a medium, it is not possible to carry out isothermal heat transfer process in practice.
• Enclosed area of p-v diagram which represents net work done is small. Any irreversibilities of the system would considerably increase the work required to operate the cycle.
• Extreme pressures and large volumes are developed since pressure takes place in both adiabatic and isothermal compression processes.