Question

3. An air-conditioning system is shown in Fig.3, in which air flows over tubes carrying Refrigerant 134a. Air enters with a volumetric flow rate of 50 m3/min at 32°C, 1 bar, and exits at 22°C, 0.95 bar. Refrigerant enters the tubes at 5 bar with a quality of 20% and exits at 5 bar, 20°C. Ignoring heat transfer at the outer surface of the air conditioner, and neglecting kinetic and potential energy effects, determine at steady state (a) the mass flow rate of the refrigerant, in kg/min. (b) the rate of heat transfer, in kJ/min, between the air and refrigerant.

Answer 1

Solution : it Py=1 bord TIF 32°C - 305k CAV.) = 50 m3/min - --- -- -- --- R = 1349 1349 P4 = sbar 74=20°C Refrigerant R=134a P3=5 bar 0+> 13=0.2 - - - - - Air - P2= 0.95 bar T2= 22°C = 295k Because the air and refrigerant are speperate streams, the steady state mass balance reduces to mi = m2 = mair m3 = ma = MR.

By 5.F. E- E, equestion esteady flow energy equation) 0 = cev - wiev + main [ Chrhe) + ( "izvz?) 1 + g (3,-32) mr [Cha-ha) + ( ug?<V?) +9 (33-34)] for assumption based and solve for mir 1. Chrha) me= mair - Cha-ha) obtain mair, using the ideal gas equation (AV) RTI 50 X 100 =57-12 kg/min 18-314 29 mair = 57.12 kg/min

properties from table A-22 hi = 305.22 kJ/kg ha= 295.17 kJ/kg further, using data from table A-ll ha= hf3 + 13 hfg3 has 71-33 + 0.2 (184.74) = 108. 28 kJ/kg from teeble . A-12 h4 = 260.34 kJ/kg MR = (305:22 - 295:17) [260.34 - 108-28) 57.12 x mr= 3.775 kg/min transfer due to heat cecair denotes transfer the energy for air only

dair = mair (hz-he) Qu'r = 57.12 ( 295-17-305.22) @air = -574 kJ/min for a control volume enclosing only the refrigerant Stream : i ar = mache-ha) cir = 3.775 (260-34-108.28) Tor= 574 kJ/min