Mumbai university > MECH > SEM 3 > THERMO

Marks: 6M

Year: May 2014

Stirling Cycle Processes:

1. The air is compressed isothermally from state 1 to 2 (TL to TH).

2. The air at state-2 is passed into the regenerator from the top at a temperature T1. The air passing through the regenerator matrix gets heated from TL to TH.

3. The air at state-3 expands isothermally in the cylinder until it reaches state-4.

4. The air coming out of the engine at temperature TH (condition 4) enters into regenerator from the bottom and gets cooled while passing through the regenerator matrix at constant volume and it comes out at a temperature TL, at condition 1 and the cycle is repeated.

It can be shown that the heat absorbed by the air from the regenerator matrix during the process 2-3 is equal to the heat given by the air to the regenerator matrix during the process 4-1, then the exchange of heat with external source will be only during the isothermal processes.

Net work done, W, is given as,

$$W = Q_s - Q_r$$

Heat supplied = Qs = heat supplied during the isothermal process 3-4

$$Q_s = P_3V_3ln(\frac{V_4}{V_3}) = mRT_H ln(r)$$

Where $r = \frac{V_4}{V_3}$ is the compressin ratio

Heat rejected = QR = Heat rejected during the isothermal compression process, 1-2.

$$Q_s = P_1V_1ln(\frac{V_1}{V_2}) = mRT_L ln(r)$$

Therefore , $W = mRln(r)[T_H - T_L]$

The efficiency of the cycle is given as,

$$η_{th} = \frac{W}{Q_s} = \frac{mRln(r)[T_H - T_L]}{mRln(r)[T_H]} = \frac{T_H - T _L}{T_H}$$

i.e $$η_{th} = 1 - \frac{T_L}{T_H}$$

Thus the efficiency of Stirling cycle is equal to that of Carnot cycle efficiency when both are working with the same temperature limits.