- 2070 Magh · 8 marks
Saturated mixture at 150°C: pressure, quality, u, h
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.
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.
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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