Question

EXAMPLE 6 A household refrigeration system works with a vapor compression refrigeration system with two evaporators using Refrigerant 134a as the working fluid. This arrangement is used to achieve refrigeration at two different temperatures with a single compressor and a single condenser. The low temperature evaporator operates at -18°C with saturated vapor at its exit and has a refrigerating capacity of 10.5 kW (3 tons). The higher- temperature evaporator produces saturated vapor at 3.2 bar at its exit and has a refrigerating capacity of 7 kW (2 tons). Compression is isentropic to the condenser pressure of 10 bar. There are no significant pressure drops in the flows through the condenser and the two evaporators, and the refrigerant leaves the condenser as saturated liquid at 10 bar. Calculate (a) the mass flow rate of refrigerant through each evaporator, in kg/min. (b) the compressor power input, in kW. (c) the rate of heat transfer from the refrigerant passing through the condenser, in kW

Answer 1

P-h chart of R134 is used to solve the question as shown below.
Ph dringuam he sytom heun holow: 3 2 1o ban 3a P 3-2 bau Pint at-18'c 1-45 baa &6 JAR.A Prooy 3t4 amd 3 t ae isethalpit axpansibns 4 R13 4, the ounthalpiy at Fram phhau Vaudu points ane G05 605 h6 Smt salh hg -515 k7/ly al 1

Coanstral'zy bshas hm 46skt/l 585-455 130 kT ka 0.0g017 k/s 130 4.846 kg/mi'n phaG0s 455 15o leT/kg Han Maw rati thagugh g tmp ouaperdr 500.046667 kals at Mar barane 2d ft-8.44 x-6157+ (2TKL03) (4-84642-8)

59a.3 k Chh Cempuancon Poues inpub, p- o (h,-hi) D-4.4bra(635 -592-3) 5.44 kw Hrat trander at chdragm -i(h,-h) 4.&4 612-8635-45 aa.94 kw

10 20 P (bar) 350 000 4.0 450 550 009 009 000 4.2 14.55 0 09 S4.65 kJ kg-K 9800 0 003 4.725 4,750 4.775 Constant Specific Volume 0.005 Ot 4,825 4.850 Y4.875 Entropy 08 Temperature P40125 0060 4.925 Dryness Fraction 60 g96-0 li li 0 SL0O 10 80% HT 5/025 90 3010 GOO g0 3 0010 5125 0 11tT g.08 zyo 80 000 NA0.2m 09 20 O0 O.0 0.8 0.4 4.2 0.5 4.3 9 0 4.4 0,7 4.5 /0.9/ 4.6 0.3 9 0 0.5 X-0.11 S=3.9 4.1 000 009 040 000 350 000 h(kJ/kg)