Example Problem with Complete Solution

10B-1 : Ideal Ammonia Vapor-Compression Refrigerator 7 pts
A cold-storage warehouse uses a refrigeration system to keep groceries at 2oC while the temperature outside the warehouse is 30oC. The groceries and the outside air act as thermal reservoirs in this process. 
Although the warehouse is insulated, it absorbs heat from the surroundings at a rate of 775 kW. Determine the power requirement and COP for a Carnot refrigeration cycle and for an ideal ammonia 
vapor-compression refrigeration cycle that will maintain the temperature of the groceries under these conditions. The condenser operates at 1.6 MPa and the evaporator operates at 300 kPa.
 
Read : In Part (a), we can determine the COP of the Carnot refrigeration cycle directly from the [temperatures of the two thermal reservoirs. Then, we can use the definition of COP to evaluate Wcycle.
In Part (b), we can work our way around the cycle in the TS Diagram and evaluate H for every stream. Then, use the H values determined to evaluate Wcomp = Wcycle and finally evaluate COP from its definition.
The approach illustrated in this solution is to start from an equation that includes the variable that is your objective, in this case the definition of COP is the equation and the objective is the COP itself. Then, you must proceed to determine the values of all the variables needed to evaluate the COP.
These two approaches to solving cycle problems often turn out to require the same amount of effort. So, you can choose whichever method appeals to you.
Diagram:
Given: TC 2 oC QC 775 kW
275.15 K
TH 30 oC P1 = P2 = 300 kPa
303.15 K P3 = P4 = 1600 kPa
Find: a.) Carnot : Wcycle,rev ??? kW COPR,rev ???
b.) Ideal Ammonia VCR : Wcycle ??? kW COPR ???
Assumptions: 1 - Each component is an open system operating at steady-state.
2 - All processes are internally reversible, except the expansion valve, which is an  isenthalpic throttling processes.
3 - The compressor and valves operate adiabatically.
4 - Kinetic and potential energy changes are negligible.
5 - There are no pressure drops for flow through the heat exchangers.
Equations / Data / Solve:
Let's organize the data that we need to collect into a table.  This will make it easier to keep track of the values we have looked-up and the values we have calculated.
Stream T
(oC)		 P
(kPa)		 H
(kJ/kg)		 S
(kJ/kg-K)		 X
(kg vap/kg)		 Phase
1 -9.22 300 376.52 1.4582 0.1840 Two-Phase Mixture
2 -9.22 300 1432.5 5.4592 1 Saturated Vapor
3 113.53 1600 1680.1 5.4592 N/A Superheated Vapor
4 41.02 1600 376.52 1.3737 0 Saturated Liquid
Additional data that may be useful.
State T (oC) P (kPa) X H (kJ/kg) S (kJ/kg-K)
Sat Vap -9.22 300 1 1432.5 5.4592
Sat Liquid -9.22 300 0 138.39 0.55596
Sat Vap 41.02 1600 1 1471.0 4.8573
Sat Liquid 41.02 1600 0 376.5 1.3737
Part a.) Let's begin by determining the COP of a Carnot refrigeration cycle working between these two thermal reservoirs.
Eqn 1
Plugging in values for TH and TC yields : COPR,rev 9.83
Solve Eqn 1 for the unknown Wcycle :
Eqn 2
Plugging values into Eqn 2 yields : Wcycle,rev 78.87 kW
Part b.) Eqn 2 can be modified slightly to
apply to our ideal ammonia VCR cycle :
Eqn 3
We were given the value of QC, so we need to determine Wcycle before we can use Eqn 3 to evaluate COPR.
Only the compressor produces or consumes shaft work in the ideal VCR cycle, so let's begin by applying the 1st Law to the compressor.
The 1st Law for a steady-state, single-inlet, single-outlet, adiabatic compressor with negligible kinetic and potential energy changes is:
Eqn 4
We can immediately evaluate H2 because we know it is a saturated vapor at 300 kPa. Use the Saturated Ammonia Tables or the NIST Webbook.
H2 1432.5 kJ/kg
In order to determine H3, we need to determine the value of a 2nd intensive variable, because we only know the pressure, P3.
We can make use of the fact that an ideal VCR cycle is internally reversible and the compressor in an ideal VCR cycle is adiabatic. A device that is internally reversible and adiabatic is isentropic. S3 = S2.
We can evaluate S2 because we know it is a saturated vapor at 300 kPa. Use the Saturated Ammonia Tables or the NIST Webbook.
S2 = S3 5.4592 kJ/kg
Now, we can use P3 and S3 and the Superheated Ammonia Tables or the NIST Webbook to determine H3 by interpolation. This is much easier using the Isobaric Properties option in the NIST Webbook.
At 1600 kPa : T (oC) H (kJ/kg) S (kJ/kg-K)
110 1670.89 5.4354
T3 H3 5.4592 T3 113.54 kJ/kg
115 1683.89 5.4691 H3 1680.1 kJ/kg
We still need to determine the mass flow rate of the Ammonia through the cycle before we can use Eqn 4 to determine Wcycle.
The key to determining the mass flow rate is the value of QC that was given in the problem statement.
We need to apply the 1st Law for a steady-state, single-inlet, single-outlet, evaporator with negligible kinetic and potential energy changes. No shaft work crosses the boundary of the evaporator.
Eqn 5
Solve Eqn 5 for the mass flow rate :
Eqn 6
We already determined H2. The trouble is we don't know the value of H1.
The key to determining the H1 is the assumption that the Expansion Valve is isenthalpic. This assumption is based on the application of the 1st Law to the Valve.
We need to apply the 1st Law for a steady-state, single-inlet, single-outlet, expansion valve with negligible kinetic and potential energy changes. No shaft work crosses the boundary of the expansion valve. We assume the valve is adiabatic because it is small and there is little opportunity for heat exchange.
Eqn 7
Eqn 7 simplifies to :
Eqn 8 Or :
Eqn 9
Eqn 9 is helpful because we already have enough information to determine H4. We know stream 4 is a saturated liquid at 1600 kPa. So, we can use the Saturated Ammonia Tables or the NIST Webbook to evaluate H4.
H4 = H1 376.52 kJ/kg
We can finally put values back into Eqns 6, 4, & 3, in that order, to complete the solution to Part (b).
m 0.7339 kg/s Wcycle 181.7 kW
COPR 4.26
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : a.) Wcycle,rev 78.9 kW COPR,rev 9.83
b.) Wcycle 182 kW COPR 4.26