Please Show the Full Steps!!! Thanks!!!!! Refrigerant 22 enters an adiabatic ste

Question

Please Show the Full Steps!!! Thanks!!!!!

Refrigerant 22 enters an adiabatic steady state compressor as saturated vapor at - 4 degree C. The refrigerant exits the compressor at a pressure of 16 bars. The actual work input required to operate the compressor is 47.48 kJ/kg. If the effects of kinetie and potential energy on the compressor are negligible, determine The isentropic efficiency of the compressor (etac) The actual compressor rate of entropy production (sigmacv/m) in kJ/kg.K The isentropic compressor rate of entropy production (sigmacv/m). m kJ/kg.K

Explanation / Answer

From R-22 properties at T1 = -4 deg C and quality x1 = 1 (sat. vapor) we get P1 = 4.36 bar, h1 = 249 kJ/kg and s1 = 0.934 kJ/kg-K

For isentropic compression, From R-22 properties at P2 = 16 bars and s2_is = s1 = 0.934 kJ/kg-K we get h2_is = 281 kJ/kg, T2_is = 62.7 deg C

Ideal work done = h2 - h1 = 281 - 249 = 32 kJ/kg

a) Isntropic eff = Ideal work done / actual work done

= 32 / 47.48

= 0.674 or 67.4 %

b)

Actual work done = h2 - h1

47.48 = h2 - 249

h2 = 296.48 kJ/kg

From R-22 properties at P2 = 16 bars and h2 = 296.48 kJ/kg we get s2 = 0.978 kJ/kg-K

Actual entropy production = s2 - s1

= 0.978 - 0.934

= 0.044 kJ/kg-K

c)

Isentropic entropy production = 0

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