Example Problem with Complete Solution

10D-1 : COP of a Heat Pump Used for Home Heating 8 pts
At steady-state, a house loses 10 kW of heat to the surroundings when the outdoor temperature is 0oC and the indoor temperature is 22oC. This energy must be replaced by a heat pump that uses R-134a 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 10oC driving force for heat exchange at the inlet and outlet of the evaporator and the condenser.
b.) Calculate the mass flow rate of the R-134a, 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 10oC between two fluids entering the heat exchanger. In this case, since the outside temperature is 0oC, assume that T1 is -10oC. Since the inside temperature is 22oC, the temperature of the R-134a entering the condenser must be 32oC to give us the desired DT of 10oC. 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: Toutside 0 oC Qout 10 kW
Tinside 22 oC hS, comp 0.85
Find: a.) P1 ??? kPa P3 ??? kPa
b.) mdot ??? kg/min
Wcomp ??? kW COPHP ???
Diagram:
Assumptions: 1 - Each component is an open system, operating at steady-state.
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:
DTevap = ΔTcond = 10 oC
T1 = T2 = -10 oC
T4 = 32 oC
Equations / Data / Solve:
Stream T
(oC)
P
(kPa)
H
(kJ/kg)
S
(kJ/kg-K)
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 steady-state, single-inlet, single-outlet 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 H4 in the R-134a Tables or in the NIST Webbook because it is saturated liquid at 32 C.
H4 201.10 kJ/kg
In order to determine H3, we must use the isentropic efficiency of the compressor.
Eqn 3
We can solve Eqn 3 for H3 :
Eqn 4
Next, we need to determine 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 -10oC.
S2 1.7334 kJ/kg-K S3S 1.7334 kJ/kg-K
H2 392.66 kJ/kg
Now, we know the value of two intensive properties at state 3S: S3S and P3 (because the condenser is isobaric: P3 = P4 = Psat(32oC)
P3 = P4 = Psat(32oC) = 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 (oC) H (kJ/kg) S (kJ/kg-K)
31.33 415.46 1.7140
T3S,800 H3S,800 1.7334 T3S,800 36.99 kJ/kg
40 424.59 1.7436 H3S,800 421.42 kJ/kg
At 900 kPa: T (oC) H (kJ/kg) S (kJ/kg-K)
40 422.32 1.7283
T3S,900 H3S,900 1.7334 T3S,900 41.51 kJ/kg
50 432.92 1.7616 H3S,900 423.92 kJ/kg
Finally, at 815.43 kPa : T (oC) P (kPa) H (kJ/kg)
36.99 800 421.42
T3S 815.43 H3S T3S 37.69
41.51 900 423.92 H3S 421.81 kJ/kg
Now ,we can plug values into Eqn 4 to determine H3 : H3 426.94 kJ/kg
Now, we can plug H3 back into Eqn 2 : mdot 0.04428 kg/s
mdot 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 steady-state, single-inlet, single-outlet, adiabatic compressor with negligible kinetic and potential energy changes is:
Eqn 5
We already determined H2 and H3, so all we need to do is plug numbers into Eqn 5.
Wcomp 1.518 kW
Part c.) We can determine the COPR from its definition:
Eqn 6
COPR 6.589
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : a.) mdot 2.66 kg/min
b.) Wcomp 1.52 kW c.) COPR 6.59