--- title: "Steam throttled 700 to 350 kPa: entropy change" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 5 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-throttled-700kpa-entropy-change" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-throttled-700kpa-entropy-change site: BCT Engineering --- # Steam throttled 700 to 350 kPa: entropy change Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics - Asked in: 2067 Mangsir, 6 marks - Video: [Thermodynamics: Open System Energy and Entropy Balance on a Throttling Device](https://www.youtube.com/watch?v=_2uyfxoEeSM) — Purdue OXE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-throttled-700kpa-entropy-change - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5 ## Question 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.] ## 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. ## Answer 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What stays constant during throttling? A. Temperature B. Pressure C. Entropy D. Enthalpy Answer: D. A throttle is steady, adiabatic with no work and negligible KE/PE change, so h2 = h1. (Key point: Throttling property) 2. Why is the steam at 350 kPa a wet mixture? A. The pressure is below atmospheric B. h2 equals hf C. hf < h2 < hg at 350 kPa D. h2 is greater than hg Answer: C. The value h2 = 2680.1 lies between hf = 584.31 and hg = 2732.0, so x2 = 0.9759. (Key point: State 2 condition) 3. Why is the entropy change positive? A. Work is done B. Heat is added C. Throttling is irreversible D. Pressure rises Answer: C. Throttling is an irreversible adiabatic process, so entropy increases: about 0.296 kJ/kg K. (Key point: Sign of entropy change) 4. How is s2 found? A. s2 = hf/T B. s2 = sg at 350 kPa C. s2 = s1 D. s2 = sf + x2 sfg at 350 kPa, with x2 from h2 Answer: D. Use h2 = h1 to get x2 = (h2 - hf)/hfg, then s2 = sf + x2 sfg = 6.8143. 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