TABLE A-16 Properties of saturated refrigerant-134a (CF H2):Temperature table o,
ID: 1717578 • Letter: T
Question
TABLE A-16 Properties of saturated refrigerant-134a (CF H2):Temperature table o, m/kg; u, kJ/kg; h, kJ/kg; s, kJ/(kg. K) Internal energy Enthalpy 0.5164 | 0.7055 | 0.3569 | -0.04 | 204.45 | 0.00 | 222.88 | 222.88 | 0.0000 | 0.9560 7 1 0.93050.72330.205214.31211.29 14.37216.01230.380.0600 0.9411 1.01990.7265 0.188216.75 212.4316.82214.80231.620.06990.9390 1.11600.7296 0.1728 19.21213.5719.29 213.57232.850.0798 0.9370 1.21920.73280.1590 21.68214.7021.77212.32234.080.0897 0.9351 1.32990.736 0.146424.17215.8424.26211.0s 235.310.0996 0.9332 14483 0.73950.130 26.67 216.9726.77 209.76 236.53 0.10940.9315 157480.74282729.1218.1029.30208.45 237.74 0.1192 0.9298 85400.7498106834.25 220.3634.39205.77 240.15 o.13880.9267 2.1704 0.75690.0919 39.38 222.6039.54203.00 242.540.1583 0.9239 .52740.76440.079444.56 224.8444.75 200.1 244.900.17770.9213 2.9282 0.7720.0689 49.79 227.0650.02197.21 247.23 0.1970 0.9190 .3760.7801 0.0655.08 229.27 535194.19 249.53 02162 0.9169 3.87560.78840.052560.43231.46 60.73191.07 251.800.23540.91s0 42940.797 0.0460 65.83 233.63 6618 187.85254.03 0.25450.9132 0416 0.80620.0405 71.29235.787169184.52256.220.2735 0.9116 5.7160 0.81570.0358 76.80 237.77.26 181.09 258.36 0.29240.9102 .45660.8257031782.37240.01 82.90 175 260.450.31130.9089Explanation / Answer
(a)
Work done = Change in enthalpy between states
Enthalpy at state 1 is the enthalpy of saturated vapor at -90 C, h1 = 241.94 kJ/kg
Entropy at state 1, s1 = 0.9246 kJ/kg K
Entropy at state 2, s2 = 0.78 x s1 (adiabatic efficiency)
= 0.7211
At 9 bars, the entropy of saturated vapor is 0.9054 kJ/kgK which is greater than the entropy of calculated state 2.
So state 2 lies in two phase state. It's dryness fraction is calculated by
0.7211 = 0.3656 + a x (0.9054 - 0.3656)
a = 0.6585
Enthalpy at state 2, h2 = 99.56 + 0.6585 x 166.62 = 209.28 kJ/kg
Work done = h1 - h2 = 241.92 - 209.28 = 32.64 kJ/kg
(b)
As state 2 is found to be in two phase mixture its temperature will be same as that of saturated temperature at 9 bars which is 35.530 C
(d)
Irreversibility = Tsurr x Change in enthalpy = 293 x (0.9246 - 0.7211) = 59.62 kJ/kg
(e)
Effectiveness = Actual work / (Actual work + Irreversibility) = 32.64 / (32.64 + 59.62) = 0.3537
(f) If the process is assumed to be reversible and adiabatic, entropy remains constant between two processes.
Then state 2 will be in superheated state. Its enthalpy in this state will be 271.3 kJ/kg.
Work done = 241.94 - 271.3 = - 29.36 kJ/kg
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