Consider a cold air-standard Diesel cycle. Operating data at principal states in

Question

Consider a cold air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.5. For k = 1.4, cv = 0.718 kJ/(kg K), and cp = 1.005 U/(kg K). determine the heat transfer per unit mass and work per unit mass for each process, in kJ/kg, and the cycle thermal efficiency. the exergetic transfers accompanying heat and work for each process, in kJ/kg. Devise and evaluate an exergetic efficiency for the cycle. Let T0 = 300 K and p0 = 100 kPa.

Explanation / Answer

a)

Process 1-2: Work done = (p2v2 - p1v1)/(1-k) = (4850.3*0.06098 - 100*0.9758)/(1-1.4) = -495.5 kJ/kg

Heat transfer = -495.5 + Cv(T2 - T1) = -495.5 + 0.718*(1030.7 - 340) = 0 kJ/kg

Process 2-3: Work done = p2*(v3 - v2) = 4850.3*(0.122 - 0.06098) = 296 kJ/kg

Heat transfer = 296 + Cv(T3-T2) = 296 + 0.718*(2061.4 - 1030.7) = 1036 kJ/kg

Process 3-4: Work done = (p4v4 - p3v3)/(1-k) = (263.9*0.9758 - 4850.3*0.122)/(1-1.4) = 835.5 kJ/kg

Heat transfer = 835.5 + Cv(T4-T3) = 835.5 + 0.718*(897.3-2061.4) = 0 kJ/kg

Process 4-1: Work done = 0 (since v4 = v1)

Heat trasfer = Cv(T1-T4) = 0.718*(340-897.3) = -400 kJ/kg

Thermal efficiency = W/Heat input = (-495.5 + 296 + 835.5)/1036 = 0.6139 = 61.39 %

b)

Exergy transfers:

Process 1-2: Exergy transfer = Cv(T2 - T1) + po(v2 - v1) - To(s2 - s1)

But s2 - s1 = Cp ln(T2/T1) - (Cp-Cv) ln (p2/p1) = 1.005*ln(1030.7/340) - (1.005-0.718)ln(4850.3/100) = 0

Exergy transfer = 0.718*(1030.7-340) + 100*(0.06098 - 0.9758) - 0 = 404.4 kJ/kg

Process 2-3: s3 - s2 = Cp ln(T3/T2) = 1.005*ln(2061.4/1030.7) = 0.6966 kJ/kg-K

Exergy transfer = Cv(T3 - T2) + po(v3 - v2) - To(s3 - s2) = 0.718*(2061.4 - 1030.7) + 100*(0.122-0.06098) - 300*0.6966 = 537.2 kJ/kg

Process 3-4: s4 - s3 = 0 (since heat transfer is zero.)

Exergy transfer = Cv(T4 - T3) + po(v4 - v3) - To(s4 - s3) = 0.718*(897.3-2061.4) + 100*(0.9758-0.122) - 0 = -750.4 kJ/kg

Process 4-1: s1 - s4 = Cp ln(T1/T4) - (Cp-Cv) ln(P1/P4) = 1.005 ln(340/897.3) - (1.005-0.718) ln(100/263.9) = -0.6966 kJ/kg-K

Exergy transfer = Cv(T1 - T4) + po(v1 - v4) - To(s1 - s4) = 0.718*(340-897.3) + 0 - 300*(-0.6966) = -191.2 kJ/kg

Exergtic efficiency = 750.4/(404.4+537.2) = 0.7969 = 79.69 %

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