10B1 :  Ideal Ammonia VaporCompression Refrigerator  7 pts 

A coldstorage warehouse uses a refrigeration system to keep groceries at 2^{o}C while the temperature outside the warehouse is 30^{o}C. 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  
vaporcompression 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 W_{cycle}.  
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 W_{comp} = W_{cycle} 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:  T_{C}  2  ^{o}C  Q_{C}  775  kW  
275.15  K  
T_{H}  30  ^{o}C  P_{1} = P_{2} =  300  kPa  
303.15  K  P_{3} = P_{4} =  1600  kPa  
Find:  a.)  Carnot :  W_{cycle,rev}  ???  kW  COP_{R,rev}  ???  
b.)  Ideal Ammonia VCR :  W_{cycle}  ???  kW  COP_{R}  ???  
Assumptions:  1   Each component is an open system operating at steadystate.  
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 lookedup and the values we have calculated.  
Stream  T (^{o}C) 
P (kPa) 
H (kJ/kg) 
S (kJ/kgK) 
X (kg vap/kg) 
Phase  
1  9.22  300  376.52  1.4582  0.1840  TwoPhase 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 (^{o}C)  P (kPa)  X  H (kJ/kg)  S (kJ/kgK)  
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 T_{H} and T_{C} yields :  COP_{R,rev}  9.83  
Solve Eqn 1 for the unknown W_{cycle} : 

Eqn 2  
Plugging values into Eqn 2 yields :  W_{cycle,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 Q_{C}, so we need to determine W_{cycle} before we can use Eqn 3 to evaluate COP_{R}.  
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 steadystate, singleinlet, singleoutlet, adiabatic compressor with negligible kinetic and potential energy changes is:  

Eqn 4  
We can immediately evaluate H_{2} because we know it is a saturated vapor at 300 kPa. Use the Saturated Ammonia Tables or the NIST Webbook.  
H_{2}  1432.5  kJ/kg  
In order to determine H_{3}, we need to determine the value of a 2nd intensive variable, because we only know the pressure, P_{3}.  
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. S_{3} = S_{2}.  
We can evaluate S_{2} because we know it is a saturated vapor at 300 kPa. Use the Saturated Ammonia Tables or the NIST Webbook.  
S_{2} = S_{3}  5.4592  kJ/kg  
Now, we can use P_{3} and S_{3} and the Superheated Ammonia Tables or the NIST Webbook to determine H_{3} by interpolation. This is much easier using the Isobaric Properties option in the NIST Webbook.  
At 1600 kPa :  T (^{o}C)  H (kJ/kg)  S (kJ/kgK)  
110  1670.89  5.4354  
T_{3}  H_{3}  5.4592  T_{3}  113.54  kJ/kg  
115  1683.89  5.4691  H_{3}  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 W_{cycle}.  
The key to determining the mass flow rate is the value of Q_{C} that was given in the problem statement.  
We need to apply the 1st Law for a steadystate, singleinlet, singleoutlet, 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 H_{2}. The trouble is we don't know the value of H_{1}.  
The key to determining the H_{1} 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 steadystate, singleinlet, singleoutlet, 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 H_{4}. 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 H_{4}.  
H_{4} = H_{1}  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  W_{cycle}  181.7  kW  
COP_{R}  4.26  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  a.)  W_{cycle,rev}  78.9  kW  COP_{R,rev}  9.83  
b.)  W_{cycle}  182  kW  COP_{R}  4.26 