Question

Calculate the change in entropy and heat transfer through cylinder walls, if 0.4m3 of a gas at a pressure of 10 bar and 200°C expands by the law PV 1.35 = Constant. During the process there is loss of 380 KJ of internal energy. (Take CP = 1.05KJ/kg k and CV = 0.75KJ/kgK)

Answer 1

ds = ?

dQ = ?

V1 = 0.4m3

P1 = 10bar = 1000KN/m2

T1 = 2000C = 473K

PV1.35 = C

dU = 380KJ

CP = 1.05, CV = 0.75

Since P1V1 = mRT

m = 1000 × 0.4/[(1.005 – 0.75) × 473]

= 2.82kg ...(i)

dU = mCV(T2 – T1)

– 380 = 2.82 × 0.75 (T2 – 473)

T2 = 292K ...(ii)

W1-2 = mR(T1 – T2)/(n – 1)

= [2.82 × 0.3 (473 – 292)]/(1.35 – 1)

= 437.5KJ ...(iii)

y = CP/CV = 1.05/0.75 = 1.4

Q1–2 = [(g – n)W1-2]/(g – 1)

= [(1.4 – 1.35) x 437.5]/(1.4 – 1)

= 54.69KJ

S2 – S1 = [(g – n)mR ln V2/V1]/(g –1) ...(iv)

Since in isentropic process T1/T2 = (V2/V1)n–1

473/292 = (V2/V1) 1.35 – 1

V2/V1 = 3.96; Putting in equation 4

S2 – S1 = [(1.4 – 1.35) × 2.82 × 0.3 × ln 3.96]/( 1.4–1)

S2 – S1 = 0.145 KJ/K