Final temperature of the steam = 304.29 K = 31.14°C
Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T = 0 J/K.
But 0.01 J/k.s was obtained though, which is approximately 0.
Explanation:
For an adiabatic system, the Pressure and temperature are related thus
P¹⁻ʸ Tʸ = constant
where γ = ratio of specific heats. For steam, γ = 1.33
P₁¹⁻ʸ T₁ʸ = P₂¹⁻ʸ T₂ʸ
P₁ = 700 bar
P₂ = 10 bar
T₁ = 600°C = 873.15 K
T₂ = ?
(700⁻⁰•³³)(873.15¹•³³) = (10⁻⁰•³³)(T₂¹•³³)
T₂ = 304.29 K = 31.14°C
b) Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T
To prove this
Entropy of the process
dQ - dW = dU
dQ = dU + dW
dU = mCv dT
dW = PdV
dQ = TdS
TdS = mCv dT + PdV
dS = (mCv dT/T) + (PdV/T) =
PV = mRT; P/T = mR/V
dS = (mCv dT/T) + (mRdV/V)
On integrating, we obtain
ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)
To obtain V₁ and V₂, we use PV = mRT
V/m = specific volume
Pv = RT
R for steam = 461.52 J/kg.K
For V₁
P = 700 bar = 700 × 10⁵ Pa, T = 873.15 K
v₁ = (461.52 × 873.15)/(700 × 10⁵) = 0.00570 m³/kg
For V₂
P = 10 bar = 10 × 10⁵ Pa, T = 304.29 K
v₂ = (461.52 × 304.29)/(10 × 10⁵) = 0.143 m³/kg
ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)
m = 2 kg/min = 0.0333 kg/s, Cv = 1410.8 J/kg.K
ΔS = [0.03333 × 1410.8 × In (304.29/873.15)] + (0.03333 × 461.52 × In (0.143/0.00570)
ΔS = - 49.56 + 49.57 = 0.01 J/K.s ≈ 0 J/K.s
