A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at −30°C by rejecting its waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C. The inlet state of the compressor is 60 kPa and −34°C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator

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

answer: 500 wph.

explanation: