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Question 7.4: Problem: What is the temperature and specific volume of a sa...

Problem: What is the temperature and specific volume of a saturated water‐steam mixture with 1% quality at a pressure of 50 kPa?

Find: Saturation temperature T_{sat} and specific volume v.

Known: Mixture is at saturation pressure P_{sat} = 50 kPa, quality x = 0.4.

Governing Equation:

Mixture property (y)                      y=xy_g + (1-x)y_f

Properties: From Table 7.3 at P_{sat} = 50 kPa, saturation temperature T_{sat} = 81.33 °C, specific volume of liquid v_f = 0.001030 m³ / kg and vapour v_g = 3.240 m³ / kg.

Table 7.3 Properties of saturated water as a function of pressure.

Pressure (MPa) Temp. (°C) Spec. Vol. (m³/kg) Int. Energy (kJ/kg) Enthalpy (kJ/kg) Entropy (kJ/kgK)
p T_{sat} v_f v_g u_f u_g h_f h_g s_f s_g
0.000 611 3 0.01 0.001 000 206.14 0 2375.3 0.00 2501.4 0.0000 9.1562
0.0010 6.98 0.001 000 129.21 29.3 2385.0 29.30 2514.2 0.1059 8.9756
0.0015 13.03 0.001 001 87.98 54.71 2393.3 54.71 2525.3 0.1957 8.8279
0.0020 17.50 0.001 001 67.00 73.48 2399.5 73.48 2533.5 0.2607 8.7237
0.0025 21.08 0.001 002 54.25 88.48 2404.4 88.49 2540.0 0.3120 8.6432
0.0030 24.08 0.001 003 45.67 101.04 2408.5 101.05 2545.5 0.3545 8.5776
0.0040 28.96 0.001 004 34.80 121.45 2415.2 121.46 2554.4 0.4226 8.4746
0.0050 32.88 0.001 005 28.19 137.81 2420.5 137.82 2561.5 0.4764 8.3951
0.0075 40.29 0.001 008 19.24 168.78 2430.5 168.79 2574.8 0.5764 8.2515
0.010 45.81 0.001 010 14.67 191.82 2437.9 191.83 2584.7 0.6493 8.1502
0.015 53.97 0.001 014 10.02 225.92 2448.7 225.94 2599.1 0.7549 8.0085
0.020 60.06 0.001 017 7.649 251.38 2456.7 251.40 2609.7 0.8320 7.9085
0.025 64.97 0.001 020 6.204 271.90 2463.1 271.93 2618.2 0.8931 7.8314
0.030 69.10 0.001 022 5.229 289.20 2468.4 289.23 2625.3 0.9439 7.7686
0.040 75.87 0.001 027 3.993 317.53 2477.0 317.58 2636.8 1.0259 7.6700
0.050 81.33 0.001 030 3.240 340.44 2483.9 340.49 2645.9 1.0910 7.5939
0.075 91.78 0.001 037 2.217 384.31 2496.7 384.39 2663.0 1.2130 7.4564
0.100 99.63 0.001 043 1.694 417.36 2506.1 417.46 2675.5 1.3026 7.3594
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v=xv_g+(1-x)v_f=0.01 \times 3.240 \ m^3/kg +(1-0.01)\times 0.001 \ 030 \ m^3/kg=0.033 \ 420 \ m^3/kg

Answer: The mixture temperature is 81 °C and its specific volume is 0.0334 m³ / kg.

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