Example Problem with Complete Solution

10C-1 Analysis of a Dual Evaporator V-C Refrigeration System 10 pts
The figure shows the schematic diagram of a vapor-compression 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 low-temperature evaporator operates at -18°C with saturated vapor at its exit and has a refrigerating capacity of 3 tons. The higher-temperature 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 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 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: Qin,1 3 tons P2 10 bar
Qin,2 2 tons P3 10 bar
T6 -18 oC P7 3.2 bar
Find : Part (a) m6 ??? kg/min
m8 ??? kg/min
Part (b) Wcomp ??? kW
Part (c) Qout ??? kW
Diagram: Process flow diagram provided in the problem statement.
Assumptions :
- Each component is open system operating at steady-state.
- 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
(oC)
P
(kPa)
H
(kJ/kg)
S
(kJ/kg-K)
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 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 and determine m6 and m8.
Each evaporator operates at steady-state, 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 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 255.50 kJ/kg H6 387.79 kJ/kg
H4 255.50 kJ/kg H7 400.04 kJ/kg
H5 255.50 kJ/kg H8 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 m6 4.7849 kg/min
m8 2.9196 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 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 H1 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 steady-state, 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: H1.
Eqn 10
Now, we can plug values into Eqns 9 & 10 to evaluate m1 and H1 :
m1 7.7045 kg/min H1 392.43 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 tell us:
Psat(-18oC) = 1.446 bar
P1 1.446 bar
In evaluating S1, we must first determine whether it 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 = 1.446 bar :
Hsat liq 176.23 kJ/kg Since H1 > Hsat vap, state 1 is a superheated vapor.
Hsat vap 387.79 kJ/kg
T (oC) H (kJ/kg) S (kJ/kg-K)
-18 387.79 1.7396 T1 -12.36 oC
T1 392.43 S1 S1 1.7575 kJ/kg-K
-10 394.37 1.7650 S2 1.7575 kJ/kg-K
Now, we know the values of two intensive variables at state 2 (P2 & S2), so we can go back to the steam tables or NIST Webbook and determine H2 by interpolation.
At P = 10 bar :
T (oC) H (kJ/kg) S (kJ/kg-K)
50 430.88 1.7482
T2 H2 1.7575 T2 52.88 oC
60 441.53 1.7806 H2 433.94 kJ/kg-K
Finally, we can plug values back into Eqn 7 :
WS -319.83 kJ/min WS -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 steady-state, 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 m1, H2 and H3, so we cannow plug values into Eqn 12:
Q23 -1374.8 kJ/min Q23 -22.914 kW
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.