--- title: "Saturated mixture at 150°C: pressure, quality, u, h" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 3 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:saturated-mixture-150c-one-third-liquid" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/saturated-mixture-150c-one-third-liquid site: BCT Engineering --- # Saturated mixture at 150°C: pressure, quality, u, h Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances - Asked in: 2070 Magh, 8 marks - 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/saturated-mixture-150c-one-third-liquid - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3 ## Question A vessel contains 2 kg of saturated liquid water and saturated water vapor mixture at a temperature of 150°C. One third of the volume is saturated liquid and two third is saturated vapor. Determine the pressure, quality, volume, internal energy and enthalpy of the mixture. ## Short answer At 150°C, Psat=476.1 kPa. From volume fractions, V=0.00651 m³, quality x=0.0055. Internal energy U=m(u_f+x u_fg)=1285 kJ, enthalpy H=m(h_f+x h_fg)=1288 kJ. ## Answer **Given:** m = 2 kg of wet steam at 150 °C; liquid volume = V/3, vapour volume = 2V/3. **Saturated water at 150 °C (tables):** P_sat = 476.1 kPa; v_f = 0.001091, v_g = 0.3925 m³/kg; u_f = 631.7, u_fg = 1927.3 kJ/kg; h_f = 632.2, h_fg = 2113.7 kJ/kg. **Step 1: Pressure** P = P_sat(150 °C) = **476.1 kPa (≈ 0.476 MPa)** **Step 2: Volume and masses** m_f = (V/3)/v_f, m_g = (2V/3)/v_g, and m_f + m_g = 2: V [1/(3 × 0.001091) + 2/(3 × 0.3925)] = 2 V [305.5 + 1.6985] = 2 ⇒ V = 2/307.2 = **0.00651 m³** m_f = (0.00651/3)/0.001091 = 1.989 kg m_g = (2 × 0.00651/3)/0.3925 = 0.01105 kg **Step 3: Quality** x = m_g/m = 0.01105/2 = **0.00553** **Step 4: Internal energy and enthalpy** u = u_f + x u_fg = 631.7 + 0.00553 × 1927.3 = 642.4 kJ/kg U = m u = 2 × 642.4 = **1284.7 kJ** h = h_f + x h_fg = 632.2 + 0.00553 × 2113.7 = 643.9 kJ/kg H = m h = **1287.8 kJ** **Answer: P = 476.1 kPa, x = 0.0055, V = 0.00651 m³, U ≈ 1285 kJ, H ≈ 1288 kJ.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. P = A. 101.3 kPa B. Psat(150 C) = 476.1 kPa C. 1 MPa D. 0.0055 kPa Answer: B. Saturated mixture at T fixes P. (Key point: Pressure) 2. Volume equations: A. mf = mg B. V = 2 vf C. mf = (V/3)/vf, mg = (2V/3)/vg, sum = 2 kg D. V = 2 vg Answer: C. V = 0.00651 m3. (Key point: Volume) 3. x = A. 0.667 B. 0.333 C. 0.5 D. mg/m = 0.0055 (small since vapour is light) Answer: D. Vapour volume is large but its mass tiny. (Key point: Quality) 4. U = A. m(uf + x ufg) = 1285 kJ B. 2 x ug = 4800 kJ C. 2 x uf = 1263 kJ D. 2 x hf = 1265 kJ Answer: A. u = 642.4 kJ/kg. 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