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Question 7.7: Problem: Superheated water vapour at 0.1 MPa and 250 °C is e...

Problem: Superheated water vapour at 0.1 MPa and 250 °C is expanded in an isentropic process to a temperature of 35 °C. Show that the final state is a saturated mixture and find its quality.

Find: The quality x2 x_2 of the final saturated mixture.

Known: Initial saturation pressure P1,sat P_{1,sat} = 0.1 MPa, initial saturation temperature T1,sat T_{1,sat} = 250 °C, final saturation temperature T2,sat T_{2,sat} = 35 °C, isentropic process.

Properties: From Table 7.4 at P1,sat P_{1,sat} = 0.1 MPa and T1,sat T_{1,sat} = 250 °C, specific entropy s1 s_1 = 8.0333 kJ / kgK. From Table 7.2 at T2,sat T_{2,sat} = 35 °C, specific entropy of liquid sf s_f = 0.5053 kJ / kg and vapour sg s_g = 8.3531 kJ / kg.

Table 7.4 Properties of superheated water vapour at 0.1 MPa pressure.

P = 0.10 MPa (TsatT_{sat} = 99.63 °C)
T (°C) v (m³ /kg) u (kJ/kg) h (kJ/kg) s (kJ/kgK)
Tsat T_{sat} 1.694 2506.1 2675.5 7.3594
100 1.6958 2506.7 2676.2 7.3614
150 1.9364 2582.8 2776.4 7.6143
200 2.172 2658.1 2875.3 7.8343
250 2.406 2733.7 2974.3 8.0333
300 2.639 2810.4 3074.3 8.2158
400 3.103 2967.9 3278.2 8.5435
500 3.565 3131.6 3488.1 8.8342
600 4.028 3301.9 3704.4 9.0976
700 4.490 3479.2 3928.2 9.3398
800 4.952 3663.5 4158.6 9.5652
900 5.414 3854.8 4396.1 9.7767
1000 5.875 4052.8 4640.3 9.9764
1100 6.337 4257.3 4891.0 10.1659
1200 6.799 4467.7 5147.6 10.3463
1300 7.260 4683.5 5409.5 10.5183

Table 7.2 Properties of saturated water as a function of temperature.

Temp. (°C) Pressure (MPa) Spec. Vol. (m³/kg) Int. Energy (kJ/kg) Enthalpy (kJ/kg) Entropy (kJ/kgK)
Tsat T_{sat} P vf v_f vg v_g uf u_f ug u_g hf h_f hg h_g sf s_f sg s_g
0.01 0.000 611 3 0.001 000 206.14 0.00 2375.3 0.00 2501.4 0.0000 9.1562
5 0.000 872 1 0.001 000 147.12 20.97 2382.3 20.98 2510.6 0.0761 9.0257
10 0.001 227 6 0.001 000 106.38 42.00 2389.2 42.01 2519.8 0.1510 8.9008
15 0.001 705 1 0.001 001 77.93 62.99 2396.1 62.99 2528.9 0.2245 8.7814
20 0.002 339 0.001 002 57.79 83.95 2402.9 83.96 2538.1 0.2966 8.6672
25 0.003 169 0.001 003 43.36 104.88 2409.8 104.89 2547.2 0.3674 8.5580
30 0.004 246 0.001 004 32.89 125.78 2416.6 125.79 2556.3 0.4369 8.4533
35 0.005 628 0.001 006 25.22 146.67 2423.4 146.68 2565.3 0.5053 8.3531
40 0.007 384 0.001 008 19.52 167.56 2430.1 167.57 2574.3 0.5725 8.2570
45 0.009 593 0.001 010 15.26 188.44 2436.8 188.45 2583.2 0.6387 8.1648
50 0.012 349 0.001 012 12.03 209.32 2443.5 209.33 2592.1 0.7038 8.0763
55 0.015 758 0.001 015 9.568 230.21 2450.1 230.23 2600.9 0.7679 7.9913
60 0.019 940 0.001 017 7.671 251.11 2456.6 251.13 2609.6 0.8312 7.9096
65 0.025 03 0.001 020 6.197 272.02 2463.1 272.06 2618.3 0.8935 7.8310
70 0.031 19 0.001 023 5.042 292.95 2469.6 292.98 2626.8 0.9549 7.7553
75 0.038 58 0.001 026 4.131 313.90 2475.9 313.93 2643.7 1.0155 7.6824
80 0.047 39 0.001 029 3.407 334.86 2482.2 334.91 2635.3 1.0753 7.6122
85 0.057 83 0.001 033 2.828 355.84 2488.4 355.90 2651.9 1.1343 7.5445
90 0.070 14 0.001 036 2.361 376.85 2494.5 376.92 2660.1 1.1925 7.4791
95 0.084 55 0.001 040 1.982 397.88 2500.6 397.96 2668.1 1.2500 7.4159
100 0.101 35 0.001 044 1.6729 418.94 2506.5 419.04 2676.1 1.3069 7.3549
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Since the water undergoes an isentropic process, s2=s1 s_2 = s_1 = 8.0333 kJ / kgK. Since s2,f<s2<s2,g s_{2,f} < s_2 < s_{2,g} , the final state is a saturated mixture.

s2=x2sg+(1−x2)sf=sf+x2(sg–sf) s_2=x_2s_g + (1-x_2)s_f=s_f + x_2(s_g – s_f)

x=s2–sfsg−sf=8.0333 kJ/kgK–0.5053 kJ/kgK8.3531 kJ/kgK–0.5053 kJ/kgK=0.95925 x=\frac{s_2 – s_f}{s_g-s_f} =\frac{8.0333 \ kJ/kgK – 0.5053 \ kJ/kgK}{8.3531 \ kJ/kgK – 0.5053 \ kJ/kgK} =0.95925

Answer: The quality of the final saturated mixture is 95.9%.

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