10C1 :  Analysis of a Dual Evaporator VC Refrigeration System  10 pts 

The special R134a refrigeration system with two evaporators, shown below, is used to cool both a refrigerator, with evaporator 2, and a freezer, with evaporator 1.  


Evaporators 1 and 2 have refrigeration capacities of 3 tons and 5 tons, respectively. A ton of refrigeration is defined as the heat of fusion absorbed by melting 1 short ton of pure ice at 0^{o}C in 24 hours.  
The key here is that one ton of refrigeration is equivalent to 211 kJ/min. The condenser operates at 800 kPa. Evaporators 1 and 2 operate at 15^{o}C and 250 kPa, respectively.  
The R134a leaves each evaporator as a saturated vapor and it leaves the condenser as a saturated liquid. Calculate... a.) m_{6} and m_{7} in kg/min b.) W_{S,comp} in kW c.) Q_{cond} 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 energies are negligible.  
First, determine the specific enthalpy at states 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 mass flow rates by applying the 1st Law to each evaporator.  
There are two stealth units on this flow diagram. A stream splitter where steam is divided before it enters Evaporator #2 or Valve #2 and a Mixer where streams 6 and 8 combine to form stream 1. The Mixer 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}  5  tons  P_{3}  800  kPa  
Q_{in,2}  3  tons  x_{3}  0  kg vap/kg  
T_{6}  15  ^{o}C  x_{6}  1  kg vap/kg  
P_{2}  800  kPa  P_{7}  250  kPa  
x_{7}  1  kg vap/kg  
Find:  Part (a)  m_{6}  ???  kg/min  Part (b)  W_{comp}  ???  kW  
m_{8}  ???  kg/min  Part (c)  Q_{out}  ???  kW  
Diagram:  The process flow diagram was provided in the problem statement.  


Assumptions:  1   Each component is an open system operating at steadystate.  
2   All processes are internally reversible, except the expansion valves, which are isenthalpic throttling processes.  
3   The compressor and valves operate adiabatically.  
4   Kinetic and potential energy changes are negligible.  
Equations / Data / Solve:  
Stream  T (oC) 
P (kPa) 
H (kJ/kg) 
S (kJ/kgK) 
X (kg vap/kg) 
Phase  
1  12.2  163.94  391.98  1.7462  N/A  Super. Vap.  
2  40.77  800  425.39  1.7462  N/A  Super. Vap.  
3  31.3  800  243.65  1.1497  0  Sat'd Liq.  
4  4.28  250  243.65  1.1626  0.245  VLE  
5  15  163.94  243.65  1.1716  0.303  VLE  
6  15  163.94  389.63  1.7371  1  Sat'd Vap.  
7  4.28  250  396.08  1.7296  1  Sat'd Vap.  
8  7.3  163.94  396.08  1.7617  N/A  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 to determine m_{6} and m_{8}.  
Each evaporator operates at steadystate, involves no shaft work and has negligible changes in kinetic and potential 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}  243.65  kJ/kg  H_{6}  389.63  kJ/kg  
H_{4}  243.65  kJ/kg  H_{7}  396.08  kJ/kg  
H_{5}  243.65  kJ/kg  H_{8}  396.08  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}  7.227  kg/min  
m_{8}  4.152  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 potential 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 mixer 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 potential 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}  11.38  kg/min  H_{1}  391.98  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} = S_{1}. We can lookup S_{1} because we know H_{1} and we know that:  

Eqn 11  
The Saturated R134a Tables or the NIST Webbook tells us:  P_{sat}(18^{o}C) =  163.94  kPa  
P_{1}  163.94  kPa  
In evaluating S_{1}, we must first determine whether stream 1 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 = 163.94 kPa :  H_{sat liq}  180.14  kJ/kg  
H_{sat vap}  389.63  kJ/kg  
Since H_{1} > H_{sat vap}, state 1 is a superheated vapor.  
T (^{o}C)  H (kJ/kg)  S (kJ/kgK)  
13  391.30  1.7435  Interpolation yields :  T_{1}  12.17  ^{o}C  
T_{1}  391.98  S_{1}  S_{1}  1.7462  kJ/kgK  
10  393.80  1.7531  S_{2}  1.7462  kJ/kgK  
Be careful with this interpolation. Use a narrow temperature range because the answer is very sensitive to this result.  
Now, we know the values of two intensive variables at state 2 (P_{2} & S_{2}), so we can go back to the R134a Tables or NIST Webbook and determine H_{2} by interpolation.  
At P = 800 kPa :  
T (^{o}C)  H (kJ/kg)  S (kJ/kgK)  
40  424.59  1.7436  
T_{2}  H_{2}  1.7462  Interpolation yields :  T_{2}  40.78  ^{o}C  
45  429.74  1.7599  H_{2}  425.40  kJ/kgK  
Finally, we can plug values back into Eqn 7 :  
W_{S}  380.20  kJ/min  W_{S}  6.337  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 potential 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}  2068.2  kJ/min  Q_{23}  34.47  kW  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  a.)  m_{6}  7.23  kg/min  b.)  W_{S}  6.34  kW  
m_{8}  4.15  kg/min  (The "" sign indicates that shaft work is done on the working fluid in the compressor.)  
c.)  Q_{23}  34.5  kW  
(The "" sign indicates that heat is transferred out of the working fluid in the condenser.)  