# Example Problem with Complete Solution

10C-1 : Analysis of a Dual Evaporator V-C Refrigeration System 10 pts
The special R-134a 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 0oC 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 -15oC and 250 kPa, respectively.
The R-134a leaves each evaporator as a saturated vapor and it leaves the condenser as a saturated liquid.
Calculate...
a.) m6 and m7 in kg/min
b.) WS,comp in kW
c.) Qcond 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 steady-state 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: Qin,1 5 tons P3 800 kPa
Qin,2 3 tons x3 0 kg vap/kg
T6 -15 oC x6 1 kg vap/kg
P2 800 kPa P7 250 kPa
x7 1 kg vap/kg
Find: Part (a) m6 ??? kg/min Part (b) Wcomp ??? kW
m8 ??? kg/min Part (c) Qout ??? kW
Diagram: The process flow diagram was provided in the problem statement. Assumptions: 1 - Each component is an open system operating at steady-state.
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/kg-K)
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 m6 and m8 are the given values for Qin,1 and Qin,2.  We can use these values when we apply the 1st Law to each evaporator to determine m6 and m8.
Each evaporator operates at steady-state, 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 R-134a 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 !
H3 243.65 kJ/kg H6 389.63 kJ/kg
H4 243.65 kJ/kg H7 396.08 kJ/kg
H5 243.65 kJ/kg H8 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 m6 7.227 kg/min
m8 4.152 kg/min
Part b.) We need to determine WS for the compressor. We can accomplish this by applying the 1st Law to the compressor. The compressor operates at steady-state, 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 H1 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 steady-state, 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: H1. Eqn 10
Now, we can plug values into Eqns 9 & 10 to evaluate m1 and H1 :
m1 11.38 kg/min H1 391.98 kJ/kg
Now, we need to work on evaluating H2.  The key to determining H2 is the fact that the compressor is both adiabatic and internally reversible, so it is isentropic.  S2 = S1.  We can lookup S1 because we know H1 and we know that: Eqn 11
The Saturated R-134a Tables or the NIST Webbook tells us: Psat(-18oC) = 163.94 kPa
P1 163.94 kPa
In evaluating S1, we must first determine whether stream 1 is a superheated vapor or a saturated mixture.
This is easier using the NIST Webbook than the R-134a Tables because no interpolation is required.
At P = 163.94 kPa : Hsat liq 180.14 kJ/kg
Hsat vap 389.63 kJ/kg
Since H1 > Hsat vap, state 1 is a superheated vapor.
T (oC) H (kJ/kg) S (kJ/kg-K)
-13 391.30 1.7435 Interpolation yields : T1 -12.17 oC
T1 391.98 S1 S1 1.7462 kJ/kg-K
-10 393.80 1.7531 S2 1.7462 kJ/kg-K
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 (P2 & S2), so we can go back to the R-134a Tables or NIST Webbook and determine H2 by interpolation.
At P = 800 kPa :
T (oC) H (kJ/kg) S (kJ/kg-K)
40 424.59 1.7436
T2 H2 1.7462 Interpolation yields : T2 40.78 oC
45 429.74 1.7599 H2 425.40 kJ/kg-K
Finally, we can plug values back into Eqn 7 :
WS -380.20 kJ/min WS -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 steady-state, 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 m1, H2 and H3, so we cannow plug values into Eqn 12:
Q23 -2068.2 kJ/min Q23 -34.47 kW
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : a.) m6 7.23 kg/min b.) WS -6.34 kW
m8 4.15 kg/min   (The "-" sign indicates that shaft work is done on the working fluid in the compressor.)
c.) Q23 -34.5 kW
(The "-" sign indicates that heat is transferred out of the working fluid in the condenser.) 