10C1  Analysis of a Dual Evaporator VC Refrigeration System  10 pts  
The figure shows the schematic diagram of a vaporcompression refrigeration system with two evaporators using Refrigerant 134a as the working fluid. This arrangement is used to achieve refrigeration at two different temperatures with a single compressor and a single condenser. The lowtemperature evaporator operates at 18°C with saturated vapor at its exit and has a refrigerating capacity of 3 tons. The highertemperature evaporator produces saturated vapor at 3.2 bar at its exit and has a refrigerating capacity of 2 tons. Compression is isentropic to the condenser pressure of 10 bar. There are no significant pressure drops in the flows through the condenser and the two evaporators, and the refrigerant leaves the condenser as saturated liquid at 10 bar. Calculate…  
a.)  the mass flow rate of refrigerant through each evaporator, in kg/min,  
b.)  the compressor power input, in kW,  
c.)  the rate of heat transfer from the refrigerant passing through the condenser, in kW.  
Read:  Don't let the diagram scare you. Analyze each unit or process in the cycle just as you would in an ordinary refrigeration cycle.  
Assume the cycle operates at steadystate and that each process is internally reversible, except for the expansions through each valve. These are throttling processes. The compressor and valves operate adiabatically. Changes in kinetic and potential energy are negligible.  
First determine the specific enthalpy at points 3 to 8. These are the easy ones. To determine the enthalpies at states 1 and 2, you will need to know the mass flow rates through each evaporator. You can determine the flow rates by using an energy balance on each evaporator.  
There are two stealth units on this flow diagram. A stream splitter where steam is divided before is enters the two evaporators and a mixer where streams 6 and 8 combine to form stream 1. The mixture is crucial to this problem. You can write mass and energy balances on the mixer. The mixer can be considered adiabatic. This will help you determine the specific enthalpy for stream 1. Then, because the compressor is isentropic, you can determine the specific enthalpy for stream 2.  
Given:  Q_{in,1}  3  tons  P_{2}  10  bar  
Q_{in,2}  2  tons  P_{3}  10  bar  
T_{6}  18  ^{o}C  P_{7}  3.2  bar  
Find :  Part (a)  m_{6}  ???  kg/min  
m_{8}  ???  kg/min  
Part (b)  W_{comp}  ???  kW  
Part (c)  Q_{out}  ???  kW  
Diagram:  Process flow diagram provided in the problem statement.  


Assumptions :  
 Each component is open system operating at steadystate.  
 All processes are internally reversible, except the expansion valves, which are isenthalpic throttling processes.  
 The compressor and valves operate adiabatically.  
 Kinetic and potential energy changes are negligible.  
Equations / Data / Solve :  
Stream  T (^{o}C) 
P (kPa) 
H (kJ/kg) 
S (kJ/kgK) 
Phase  
1  12.4 
144.6 
392.43 
1.7575 
Super. Vap.  
2  52.88 
1000 
433.9 
1.7575 
Super. Vap.  
3  1000 
255.50 
Sat'd Liq.  
4  320 
255.50 
VLE  
5  18.0 
144.6 
255.50 
VLE  
6  18.0 
144.6 
387.79 
Sat'd Vap.  
7  320 
400.04 
Sat'd Vap.  
8  144.6 
400.04 
Super. Vap.  
Part a.)  The key to determining m_{6} and m_{8} are the given values for Q_{in,1} and Q_{in,2}. We can use these values when we apply the 1st Law to each evaporator and determine m_{6} and m_{8}.  
Each evaporator operates at steadystate, involves no shaft work and has negligible changes in kinetic and internal energies. The appropriate forms of the 1st Law are:  

Eqn 1 

Eqn 2  
We can solve these equations for the unknown mass flow rates:  
Eqn 3 

Eqn 4  
Now, we need to determine the values of the H's to use in Eqns 3 & 4.  
The throttling valves are isenthalpic because they are adiabatic, have no shaft work interactions and changes in kinetic and potential energies are negligible. Therefore:  

Eqn 5  Eqn 6  
Fortunately, we were given enough information to lookup the specific enthalpy of states 3, 6 and 7 in the saturated R134a tables or the NIST Webbook. Once we have these values, we can use Eqns 5 & 6 to evaluate the H's at states 4, 5 and 8 as well !  
H_{3}  255.50  kJ/kg  H_{6}  387.79  kJ/kg  
H_{4}  255.50  kJ/kg  H_{7}  400.04  kJ/kg  
H_{5}  255.50  kJ/kg  H_{8}  400.04  kJ/kg  
Now, we can plug values into Eqns 3 & 4 to determine the two unknown mass flow rates.  
1 ton =  211  kJ/min  m_{6}  4.7849  kg/min  
m_{8}  2.9196  kg/min  
Part b.)  We need to determine W_{S} for the compressor. We can accomplish this by applying the 1st Law to the compressor. The compressor operates at steadystate, is adiabatic and reversible and has negligible changes in kinetic and internal energies. The appropriate form of the 1st Law is:  

Eqn 7  
We know the mass flow rates from part (a), but we don't know either of the H's in Eqn 7 yet.  
We can determine H_{1} by applying the 1st Law to the mixing point where streams 6 and 8 combine to form stream 1. The mixer is a MIMO process that operates at steadystate, is adiabatic and has negligible changes in kinetic and internal energies. The appropriate form of the 1st Law is:  

Eqn 8  
A mass balance on the mixer tells us that:  Eqn 9  
Solve Eqn 8 for the only unknown in the equation: H_{1}.  

Eqn 10  
Now, we can plug values into Eqns 9 & 10 to evaluate m_{1} and H_{1} :  
m_{1}  7.7045  kg/min  H_{1}  392.43  kJ/kg  
Now, we need to work on evaluating H_{2}. The key to determining H_{2} is the fact that the compressor is both adiabatic and internally reversible, so it is isentropic. S_{2} = S1. We can lookup S_{1} because we know H_{1} and we know that:  

Eqn 11  
The saturated R134a tables or the NIST Webbook tell us:  
P_{sat}(18^{o}C) =  1.446  bar  
P_{1}  1.446  bar  
In evaluating S_{1}, we must first determine whether it is a superheated vapor or a saturated mixture.  
This is easier using the NIST Webbook than the R134a tables because no interpolation is required.  
At P = 1.446 bar :  
H_{sat liq}  176.23  kJ/kg  Since H_{1} > H_{sat vap}, state 1 is a superheated vapor.  
H_{sat vap}  387.79  kJ/kg  
T (^{o}C)  H (kJ/kg)  S (kJ/kgK)  
18  387.79  1.7396  T_{1}  12.36  ^{o}C  
T_{1}  392.43  S_{1}  S_{1}  1.7575  kJ/kgK  
10  394.37  1.7650  S_{2}  1.7575  kJ/kgK  
Now, we know the values of two intensive variables at state 2 (P_{2} & S_{2}), so we can go back to the steam tables or NIST Webbook and determine H_{2} by interpolation.  
At P = 10 bar :  
T (^{o}C)  H (kJ/kg)  S (kJ/kgK)  
50  430.88  1.7482  
T_{2}  H_{2}  1.7575  T_{2}  52.88  ^{o}C  
60  441.53  1.7806  H_{2}  433.94  kJ/kgK  
Finally, we can plug values back into Eqn 7 :  
W_{S}  319.83  kJ/min  W_{S}  5.3304  kW  
Part c.)  We can determine the heat transfer rate in the condenser by applying the 1st Law to it.  
The condenser operates at steadystate, involves no shaft work and has negligible changes in kinetic and internal energies. The appropriate form of the 1st Law is:  

Eqn 12  
In parts (a) & (b) we evaluated m_{1}, H_{2} and H_{3}, so we cannow plug values into Eqn 12:  
Q_{23}  1374.8  kJ/min  Q_{23}  22.914  kW  
Verify :  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers:  a.)  m_{6}  4.78  kg/min  
m_{8}  2.92  kg/min  
b.)  W_{S}  5.33  kW  
(The "" sign indicates that shaft work is done on the working fluid in the compressor.)  
c.)  Q_{23}  22.9  kW  
(The "" sign indicates that heat is transferred out of the working fluid in the condenser.)  