Air at the start of a Diesel cycle compression process is at 90 kPa, 25 oC. The

Question

Air at the start of a Diesel cycle compression process is at 90 kPa, 25 oC. The compression ratio for this cycle is 12.6 and the maximum cycle temperature is 1370 K. Answer the following questions using the cold-standard assumprions.

a.)What is the temperature at the end of the compression process? ______ K

b.)What is the temperature at the end of the expansion process? ______ K

c.)How much heat is rejected by the cycle? ______ kJ/kg

d.) What is the thermal efficiency of this cycle? _______

Please show work neatly!

Explanation / Answer

P1= 90KPa ; T1= 298.15K; r= 12.6; T3=1370K
From table:k=1.4;Cv=0.718;Cp=1.005

Pv=RT, solving for v1, v1=RT1/P1; v= 0.950767 m3/kg

v1/v2=12.6, solving for v2, v2=v1/12.6; v2=0.07545 m3/kg
For Isentropic process (1 to 2):
T2= T1* (v1/v2)^k-1
T2= 298.15*(12.6^0.4) = 821.4535K

Process between 2 to 3: Pressure Constant
T2/v2=T3/v3, solving for v3
v3 = 1370*0.07545/821.4535 = 0.12583 m3/kg

Process 3 to 4 is also isentropic:
T4= T3* (v3/v4)^k-1
v1=v4
T4= 1370*(0.12583/0.950767)^0.4 = 610.1121K

Heat in in P constant
qin = Cp*(T3-T2) = 1.005*(1370-821.4535) = 551.2892 kJ/kg

Heat ou in v constant
qout = Cv*(T1-T4) = 0.718*(298.15-610.1121) = -223.9887 kJ/kg

Efficiency = wnet/qin, however, wnet = qin qout; then
(qin+qout)/qin = (551.2892-223.9887)/551.2892 = 0.5937 or 59.37%

Answering the qustions:

a) T2 = 821.4535K

b)T4 = 610.1121K

c)qout = -223.9887 kJ/kg

d) Efficiency =0.5937

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