Answer:
A) The quality of the refrigerant at the evaporator inlet = 0.48
B) The refrigeration load = 5.39 kW
C) COP = 2.14
Explanation:
A) From the refrigerant R-144 table I attached,
At P=60kpa and interpolating at - 34°C,we obtain enthalpy;
h1 = 230.03 Kj/kg
Also at P= 1.2MPa which is 1200kpa and interpolating at 65°C,we obtain enthalpy ;
h2 = 295.16 Kj/Kg
Also at P= 1.2MPa which is 1200kpa and interpolating at 42°C,we obtain enthalpy ;
h3 = 111.23 Kj/Kg
h4 is equal to h3 and thus h4 = 111.23 Kj/kg
We want to find the refrigerant quality at the evaporation inlet which is state 4 and P= 60 Kpa.
Thus, from the table attached, we see that hf = 3.84 at that pressure and hg = 227.8
Now, to find the quality of the refrigerant, we'll use the formula,
x4 = (h4 - hg) /(hf - hg)
Where x4 is the quality of the refrigerant. Thus;
x4 = (111.23 - 3.84)/(227.8 - 3.84) = 0.48
B) The mass flow rate of the refrigerant can be determined by applying a 1st law energy balance across the
condenser. Thus, the water properties can be obtained by using a saturated liquid at the given temperatures;
So using the first table in the image i attached; interpolating at 18°C; hw1 = hf = 75.54 kJ/kg
Also interpolating at 26°C; hw2 = hf = 109.01 kJ/kg
Now;
(mr) (h2 − h3)= (mw) (hw2 − hw1)
mr is mass flow rate
Making mr the subject, we get;
mr = [(mw) (hw2 − hw1)] /(h2 − h3)
mr = [(0.25 kg/s)(109.01 − 75.54) kJ/kg
] /(295.13 − 111.37) kJ/kg
mr = 8.3675/183.76
mr = 0.0455 kg/s
Formula for refrigeration load is;
QL = mr(h1 − h4)
Thus,
QL = (0.0455 kg/s)(230.03 − 111.37) kJ/kg = 5.39 kW
C) The formula for specific work into the compressor is;
W(in) = [(h2 − h1)] − (Q(in
)/mr)
= (295.13 − 230.03) kJ/kg − (0.450kJ/s
/0.0455 kg/s)
= 65.10 Kj/kg - 9.89 Kj/kg
55.21 kJ/kg
Formula for COP is;
COP = qL
/W(in)
Thus; COP = (h1 − h4)/W(in
)
= [(230.03 − 111.37) kJ/kg
] /55.24 kJ/kg
= 2.14
