Ammonia is compressed from T_1 = 298 K and P_1 = 2 bar to P_2 = 20 bar. The ammo

Question

Ammonia is compressed from T_1 = 298 K and P_1 = 2 bar to P_2 = 20 bar. The ammonia is ideal at the inlet but not at the outlet. The compressor efficiency was assumed to be 1 and that gave a work requirement of 3300 J/mol. The actual measured outlet temperature during operation was determined to be 423 K. Calculate the efficiency of the compressor using this data and the virial equation of state. T_c = 405.7 K omega = 0.254 P_C = 112.8 bar C_p/R = 4.27 + 0.04 (T - 298) (T in degree K) R = 8.31 J/mol-K

Explanation / Answer

HR/RTC= OR*(Bo- Tr*(dB/dTr+w*(B1-Tr*dB1/dTr)

At the outlet, it is not ideal. Hence

Tr=T/TC = 423/405.7 =1.042, Pr= P/PC = 20/112.8=0.177

Bo= 0.083-0.422/Tr 1.6=1.069

B1=0.139-0.172/Tr 4.2 = 0.139- 0.172/(1.042)4.2 = -0.00571

dB/dT = 0.722/(1.042)5.2 = 0.583

HR/RTC = 0.177[1.069- 1.042+0.233(1.069+1.042*0.585)]1=1.36

HR= 8.314*405.7*1.36=4587 J/mol

We know that HR= H-Hideal

Hideal = = 8.314[0.427*(425-298)+ 0.004*[(4252-2982)/2-298*(425-298)]==719 J/mole.

HR= H-719

We know that from 1st law of thermodynamics for a flow process (neglecting kinetic and potential energy changes)

H= Q+W

Hideal = Q+Wideal

Hreal = Q+Wreal

HR= Hideal- Hreal = Wideal-Wreal = 719

Wreal = 3300-719 = 2581 J/mole

Efficiency= Wreal/Wideal =100* (2581/3300)= 78%

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