10D1 :  COP of a Heat Pump Used for Home Heating  8 pts 

At steadystate, a house loses 10 kW of heat to the surroundings when the outdoor temperature is 0^{o}C and the indoor temperature is 22^{o}C. This energy must be replaced by a heat pump that uses R134a as the working fluid.  
Assume evaporator effluent
is a saturated vapor
and the condenser effluent is a saturated liquid. Assume the compressor has an isentropic efficiency of 85%. 

a.) Determine the operating pressures for the evaporator and condenser that provide at least a 10^{o}C driving force for heat exchange at the inlet and outlet of the evaporator and the condenser.  
b.) Calculate the mass flow rate of the R134a, the power requirement of the compressor and the COP of the heat pump.  
Read :  The operating pressures of the evaporator and condenser are governed by the saturation considerations. A rule of thumb in heat exchanger design is that you would like to have a DT of about 10^{o}C between two fluids entering the heat exchanger. In this case, since the outside temperature is 0^{o}C, assume that T_{1} is 10^{o}C. Since the inside temperature is 22^{o}C, the temperature of the R134a entering the condenser must be 32^{o}C to give us the desired DT of 10^{o}C. This leads to some messy interpolation that you can avoid by using the NIST Webbook. With these assumptions, there are really no tricks to this problem!  
Given:  T_{outside}  0  ^{o}C  Q_{out}  10  kW  
T_{inside}  22  ^{o}C  h_{S, comp}  0.85  
Find:  a.)  P_{1}  ???  kPa  P_{3}  ???  kPa  
b.)  m_{dot}  ???  kg/min  
W_{comp}  ???  kW  COP_{HP}  ???  
Diagram: 




Assumptions:  1   Each component is an open system, operating at steadystate.  
2   There are no pressure drops through the evaporator or condenser.  
3   The compressor operates adiabatically with an isentropic efficiency of 80%.  
4   The expansion through the valve is an isenthalpic throttling process.  
5   Kinetic and potential energy changes are negligible.  
6   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:  
DT_{evap} = ΔT_{cond} =  10  ^{o}C  
T_{1} = T_{2} =  10  ^{o}C  
T_{4} =  32  ^{o}C  
Equations / Data / Solve:  
Stream  T (^{o}C) 
P (kPa) 
H (kJ/kg) 
S (kJ/kgK) 
X (kg vap/kg) 
Phase  
1  10.0  200.60  201.10  1.0054  0.0699  VLE  
2  10.0  200.60  392.66  1.7334  1  Sat'd Vap.  
3  42.6  815.43  426.94  1.5485  N/A  Super. Vap.  
3S  37.7  815.43  421.80  1.7334  N/A  Super. Vap.  
4  32  815.43  201.10  1.0040  0  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 process. 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 steadystate, singleinlet, singleoutlet condenser with no shaft work and negligible kinetic and potential energy changes is:  

Eqn 1  
Solve Eqn 1 for the mass flow rate: 

Eqn 2  
We can look up H_{4} in the R134a Tables or in the NIST Webbook because it is saturated liquid at 32 °C.  
H_{4}  201.10  kJ/kg  
In order to determine H_{3}, we must use the isentropic efficiency of the compressor. 

Eqn 3  
We can solve Eqn 3 for H_{3} : 

Eqn 4  
Next, we need to determine H_{3S}. For an isentropic compressor, S_{3S} = S_{2} and we can lookup S_{2} in the NIST Webbook because stream 2 is a saturated vapor at 10^{o}C.  
S_{2}  1.7334  kJ/kgK  S_{3S}  1.7334  kJ/kgK  
H_{2}  392.66  kJ/kg  
Now, we know the value of two intensive properties at state 3S: S_{3S} and P_{3} (because the condenser is isobaric: P_{3} = P_{4} = P_{sat}(32^{o}C)  
P_{3} = P_{4} = P_{sat}(32^{o}C) =  815.43  kPa  
This involves some painful interpolation. I suggest you use the NIST Webbook to avoid the pain of interpolation.  
At 800 kPa:  T (^{o}C)  H (kJ/kg)  S (kJ/kgK)  
31.33  415.46  1.7140  
T_{3S,800}  H_{3S,800}  1.7334  T_{3S,800}  36.99  kJ/kg  
40  424.59  1.7436  H_{3S,800}  421.42  kJ/kg  
At 900 kPa:  T (^{o}C)  H (kJ/kg)  S (kJ/kgK)  
40  422.32  1.7283  
T_{3S,900}  H_{3S,900}  1.7334  T_{3S,900}  41.51  kJ/kg  
50  432.92  1.7616  H_{3S,900}  423.92  kJ/kg  
Finally, at 815.43 kPa :  T (^{o}C)  P (kPa)  H (kJ/kg)  
36.99  800  421.42  
T_{3S}  815.43  H_{3S}  T_{3S}  37.69  
41.51  900  423.92  H_{3S}  421.81  kJ/kg  
Now ,we can plug values into Eqn 4 to determine H_{3} :  H_{3}  426.94  kJ/kg  
Now, we can plug H_{3} back into Eqn 2 :  m_{dot}  0.04428  kg/s  
m_{dot}  2.657  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 steadystate, singleinlet, singleoutlet, adiabatic compressor with negligible kinetic and potential energy changes is:  

Eqn 5  
We already determined H_{2} and H_{3}, so all we need to do is plug numbers into Eqn 5.  
W_{comp}  1.518  kW  
Part c.)  We can determine the COP_{R} from its definition: 

Eqn 6  
COP_{R}  6.589  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  a.)  m_{dot}  2.66  kg/min  
b.)  W_{comp}  1.52  kW  c.)  COP_{R}  6.59 