Example Problem with Complete Solution

10D-1 COP of a Heat Pump Used for Home Heating 8 pts
On a particular day when the outside temperature is 5°C, a house requires a heat transfer rate of 12 kW to maintain the inside temperature at 20°C. A vapor-compression heat pump with Refrigerant 22 as the working fluid is to be used to provide the necessary heating. Specify appropriate evaporator and condenser pressures of a cycle for this purpose. Let the refrigerant be saturated vapor at the evaporator exit and saturated liquid at the condenser exit. Calculate…
               
a.) the mass flow rate of refrigerant, in kg/min,
b.) the compressor power, in kW,
c.) the coefficient of performance.
               
Read : The operating pressure of the evaporator and condenser are governed by the saturation considerations discussed in the problem 1 of this homework. A rule of thumb in heat exchanger design is that you would like to have a ΔT of about 10oC between two fluid exchanging heat. In this case, since the outdoor temperature is 5oC. assume that T1 is -5oC. Conveniently, at Pcond = 12 bar, the saturation temperature for R-22 is about 30oC. This will give us the desired ΔT of 10oC between the R-22 in the condenser and the air inside the house. So let's use PC = 12 bar for this problem. With these assumptions, there are really no tricks to this problem!
Given: Toutside 5 oC Qout 12 kW
Tinside 20 oC
Find : a.) mdot ? kg/min Part (c) COPHP
b.) Wcomp ? kW
Diagram : Figure 1
Figure 2
Assumptions :
- Each component is an open system, operating at steady-state.
- There are no pressure drops through the evaporator or condenser.
- The compressor operates adiabatically with an isentropic efficiency of 80%.
- The expansion through the valve is an isenthalpic throttling process.
- Kinetic and potential energy changes are negligible.
The evaporator and condenser pressures must be chosen to allow for sufficient DT's to avoid excessive heat exchanger sizes (surface area). For a preliminary design assume:
ΔTevap 10 oC
T1 = T2 -5 oC
Pcond = P3 = P4 12 bar
Also, assume : ηS, comp 0.80 %
Equations / Data / Solve :
Stream
T
(oC)
P
(kPa)
H
(kJ/kg)
S
(kJ/kg-K)
Phase
1
-5.0
VLE
2
-5.0
403.2
1.7581
Sat'd Vap.
3
12
435.57
Super. Vap.
3S
47.2
12
429.08
1.7581
Super. Vap.
4
12
236.96
Sat'd Liq.
Part a.) The mass flow rate is the same through each piece of equipment.  Therefore, we can determine the mass flow rate by applying the 1st Law to any one device.  Because we are given the heat transfer rate at the condenser, it is the device where we will have the fewest unknowns in the 1st Law so we are most likely to be able to determine the mass flow rate of the working fluid.
The 1st Law for a steady-state, single-inlet, single outlet condenser with no shaft work and negligible kinetic and potential energy changes is:
Equation 1
Eqn 1
Solve Eqn 1 for the mass flow rate: Equation 2
Eqn 2
We can look up H4 in the R-22 tables in the NIST Webbook because it is saturated liquid at 12 bar.
H4 236.96 kJ/kg
In order to determine H3, we must use the isentropic efficiency of the compressor.
Equation 3
Eqn 3
We can solve Eqn 3 for H3 : Equation 4
Eqn 4
Next, we need to determin H3S.  For an isentropic compressor, S3S = S2 and we can look up S2 in the NIST Webbook because stream 2 is a saturated vapor at -5oC.
S2 1.7581 kJ/kg-K S3S 1.7581 kJ/kg-K
H2 403.16 kJ/kg
Now, we know the value of two intensive properties at state 3S: S3S and P3 (because the condenser is isobaric:  P3 = P4 = 12 bar.
At 12 bar:
T (oC) H (kJ/kg) S (kJ/kg-K)
40 422.94 1.7388
T3S H3S 1.7581 T3S 47.23 K
50 431.44 1.7655 H3S 429.08 kJ/kg
Now ,we can plug values into Eqn 4 to determine H3 :
H3 435.57 kJ/kg
Now, we can plug H3 back into Eqn 2 : mdot 0.060421 kg/s
3.625 kg/min
Part b.) In order to determine the shaft work requirement for the compressor, we must apply the 1st Law to it.  The 1st Law for a steady-state, single-inlet, single outlet, adiabatic compressor with negligible kinetic and potential energy changes is:
Equation 5
Eqn 5
We already determined H2 and H3, so all we need to do is plug numbers into Eqn 5.
Wcomp 1.958 kW
Part c.) We can determine the COPR from its definition:
Equation 6
Eqn 6 COPR 6.129
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answers: a.) mdot 3.63 kg/min
b.) Wcomp 1.96 kW
c.) COPR 6.13