Hi,
I have an example question however I am struggling to solve it. I have a solution (may not be correct) and my answer is completely different. Could anyone help me please?
A refrigerator uses carbon dioxide as a refrigerant. The refrigeration temperature is -20°C and the condenser temperature is 10°C.
Assume that the condenser and refrigerator operate at constant pressure and that the expansion (throttling) and the compressor are isentropic.
The carbon dioxide is a saturated liquid when it leaves the condenser and has a dryness fraction of 0.9 when it leaves the refrigerator.
a) Determine the circulation rate required to give 0.17 tons of refrigeration. Determine the COP.
1 ton of refrigeration = 3923W.
Solution
We first must evaluate T,P,S and H of the 4 streams of a refrigeration cycle.
The values can be obtained from refrigerant tables.
Remember in the 2-phase region, a temperature fixes a pressure.
Stream T(°C) P(atm) S(kJ/kgK) H(kJ/kg)
Entering the refrigerator -20 19.7 0.46 109.1***
Entering compressor -20 19.7 1.168* 295.3*
Leaving compressor 10 45.0 1.168 275**
Leaving condenser 10 45.0 0.407 109.1*** ([8] for correct enthalpies)
*Dryness fraction 0.9 means that S = 0.158 + 0.9(1.28-0.158) = 1.168kJ/kgK
and H = 39.7+ 0.9(323.7-39.7) = 295.3kJ/kg.
** We have to use a diagram for carbon dioxide – e.g. in Perry.
*** Isenthalpic process d-a
-20 and 19.7atm H is 109.1.
This is equivalent to a dryness fraction of
(109.7-39.7)/(323.7-39.7) = 0.27
This gives an entropy of 0.158+0.27(1.280-0.158) = 0.46kJ/kgK
Q in the refrigerator = hB - hA = 295.3-109.1 = 186.2kJ/kg
0.17 tons of refrigeration = 0.17*3923 = 667W = 0.667kW
Therefore mass flowrate required = 0.667/186.2 = 0.0036kg/s = 12.9kg/hr [2]
Q in the condenser = hC – hD = 275-109.1 = 265.9kJ/kg
COP = Qrefrigerator/(Qcondenser-Qrefrigerator) = 186.2/(265.9-186.2) = 2.34 [2]
My Solution
P1 = 1.97 MPa h1 = 154.45+(0.9×282.44) = 408.65 KJ/Kg
T1 = -20 °C S1 = 0.8328 + 0.9(1.1157) = 1.837 KJ/Kg
Ref: https://www.ohio.edu...rty_tables/CO2/
P2 = 45 MPa
T2 = 10 °C
S2 = s1
h2 = 422.8 KJ/Kg
P3 = 45 MPa h3 = hf =225.73 KJ/Kg
h4 = h3 = 225.73 KJ/Kg
0.17×3923 = 666.91 w
QL = m(h1-h4)
m = 0.0036 Kg/s
win = 0.0036 (422.8 – 408.65) = 0.051 kW
QH = 0.0036 (422.8 – 225.73) = 0.709 KW
COP = 0.6691/0.051 = 13.1