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 C_P = 1.05KJ/kg k and C_V = 0.75KJ/kgK)

Answer 1

ds = ?

dQ = ?

V₁ = 0.4m³

P₁ = 10bar = 1000KN/m²

T₁ = 2000C = 473K

PV^1.35 = C

dU = 380KJ

C_P = 1.05, CV = 0.75

Since P₁V₁ = mRT

m = 1000 × 0.4/[(1.005 – 0.75) × 473]

= 2.82kg ...(i)

dU = mC_V(T₂ – T₁)

– 380 = 2.82 × 0.75 (T₂ – 473)

T₂ = 292K ...(ii)

W_1-2 = mR(T₁ – T₂)/(n – 1)

= [2.82 × 0.3 (473 – 292)]/(1.35 – 1)

= 437.5KJ ...(iii)

y = C_P/C_V = 1.05/0.75 = 1.4

Q_1–2 = [(g – n)W_1-2]/(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