- 2067 Mangsir · 6 marks
Steam throttled 700 to 350 kPa: entropy change
Short answer
Throttling holds enthalpy constant. From 700 kPa, x=0.96: s₁ = 6.5184 kJ/kg·K. At 350 kPa, h constant gives x = 0.9759, s₂ = 6.8143 kJ/kg·K. Entropy change per unit mass is +0.296 kJ/kg·K.
Steam at 700kPa with a quality of 0.96, is throttled down to 350kpa. Calculate the change of entropy per unit mass of steam. [Refer the attached table for properties of steam.]
Throttling through a valve is a steady, adiabatic process with no work and negligible change in kinetic and potential energy. So the enthalpy is constant: h₂ = h₁.
State 1: 700 kPa, x₁ = 0.96 (from saturated steam table)
h_f = 697.14, h_fg = 2065.6 kJ/kg; s_f = 1.9921, s_fg = 4.7149 kJ/kg·K
h₁ = 697.14 + 0.96 × 2065.6 = 2680.1 kJ/kg s₁ = 1.9921 + 0.96 × 4.7149 = 6.5184 kJ/kg·K
State 2: 350 kPa, h₂ = 2680.1 kJ/kg
At 350 kPa: T_sat = 138.9 °C, h_f = 584.31, h_fg = 2147.7, h_g = 2732.0 kJ/kg; s_f = 1.7275, s_fg = 5.2126 kJ/kg·K.
Since h_f < h₂ < h_g, the steam is wet:
x₂ = (2680.1 − 584.31)/2147.7 = 0.9759 s₂ = 1.7275 + 0.9759 × 5.2126 = 6.8143 kJ/kg·K
Entropy change
Δs = s₂ − s₁ = 6.8143 − 6.5184
Answer: Δs ≈ 0.296 kJ/kg·K (positive, as throttling is irreversible; the steam becomes slightly drier, x = 0.976).
✓ Answer checked by Claude Opus 5.5 (high effort)
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