Example Problem with Complete Solution

4F-4 : Heat and Work for a Cycle Executed in a Closed System Containing Ammonia 8 pts
Ammonia in a piston-and-cylinder device undergoes a 3-step thermodynamic cycle made up of the following three quasi-equilibrium processes.
Step 1-2: Isochoric heating from -20oC and 150 kPa up to 50oC
Step 2-3: Isothermal compression until the quality is 0.55 kg vap/kg, Q23 = -91.7 kJ
Step 3-1: Adiabatic expansion
a.) Sketch the process path for this cycle on a PV Diagram. Put a point on the diagram for each state and label it. Be sure to include and label all the important features for a complete PV Diagram for this system
b.) Calculate Qcycle and Wcycle in kJ/kg
c.) Determine whether this cycle is a power cycle or a refrigeration/heat-pump cycle?  Explain your reasoning.
Read : We are given T1 and P1, so we can determine any and all properties of the system using the Ammonia Tables.  In particular, we can evaluate the specific volume and we know that this does not change in step 1-2.  This gives us a 2nd intensive property for state 2 and allows us to evaluate all of the properties of state 2.  We expect T2 > T1.  Step 2-3 is an isothermal compression to a quality of x3 = 0.55.  Because T3 = T2, we will be able to evaluate all of the properties of state 3, again using the Ammonia Tables.  In each of the three steps, we know the value of either the heat or the work.  W12 = 0 because the process is isochoric.  Q23 is given and Q31 = 0 because the process is adiabatic.  So, when we apply the 1st Law to each step, there is just one unknown and we can evaluate it.  Once we know Q and W for each step, we can determine Qcycle and Wcycle because they are the sum of the Q's and W's for the steps that make up the cycle, respectively.
Given : T1 -20 oC Find : Q12 ??? kJ/kg
P1 150 kPa W23 ??? kJ/kg
T2 50 oC W31 ??? kJ/kg
T3 50 oC Qcycle ??? kJ/kg
x3 0.55 kg vap/kg Wcycle ??? kJ/kg
Q23 -91.7 kJ/kg Power or Refrigeration Cycle ?
Q31 0 kJ
Diagrams :
Part a.)
Assumptions :
1 - Changes in kinetic and potential energies are negligible.
2 - Boundary work is the only form of work that crosses the system boundary.
Equations / Data / Solve :
Part b.) Let's begin by writing the 1st Law for each of the three steps that make up the cycle, assuming that changes in potential and kinetic energies are negligible.
Step 1-2 :
Eqn 1
Step 2-3 :
Eqn 2
Step 3-1 :
Eqn 3
Step 1-2 is isochoric, so no boundary work occurs.  If we assume that boundary work is the only form of work interaction in this cycle, then W12 = 0.  Q31 = 0 because step 3-1 is adiabatic.
We can solve Eqns 1 - 3 to evaluate the unknowns Q12, W23 and W31.
Step 1-2 :
Eqn 4
Step 2-3 :
Eqn 5
Step 3-1 :
Eqn 6
Our next step must be to determine the value of the specific internal energy at states 1, 2 and 3 because, once we know these, we can use Eqns 4 - 6 to evaluate the unknowns Q12, W23 and W31.
Let's begin with state 1.  First, we must determine the phase or phases that exist in state 1.  We can accomplish this by comparing P1 to Psat(T1).
Psat(T1) 190.08 kPa
Since P1 < Psat(T1), we conclude that a superheated vapor exists in the cylinder at state 1.
We can determine U1 from the Superheated Ammonia Tables.  We can also determine V1 because we know V2 = V1 and the knowledge of this 2nd intensive variable for state 2 will allow us to evaluate U2.
V1 = V2 = 0.79779 m3/kg U1 1303.8 kJ/kg
Next, let's work on state 2.  We know the value of 2 intensive variables, T2 and V2, and we know that if a superheated vapor expands at constant volume, it must still be a superheated vapor.  Consequently, we can use the Superheated Ammonia Tables to determine U2 (and any other properties at state 2 that we want).
At T2 = 50oC, it turns out that V2 = 0.79779 m3/kg falls between 100 kPa and 200 kPa, so we must interpolate to determine U2.
At T2 = 50oC : V (m3/kg) U (kJ/kg) P (kPa)
1.56571 1425.2 100
0.79779 U2 P2 U2 1421.8 kJ/kg
0.77679 1421.7 200 P2 197.3 kPa
Now, let's work on state 3.  We know the temperature and the quality, so we can determine U3 using :
Eqn 7
We can use the Saturated Ammonia Tables to determine Usat vap and Usat liq at 50oC and then we can plug numbers into Eqn 7 to evaluate U3.
Usat liq 171.41 kJ/kg
Usat vap 263.69 kJ/kg U3 222.2 kJ/kg
Now, we can go back and plug the values of the specific internal energies into Eqns 4 - 6 to evaluate the unknowns Q12, W23, and W31.
Q12 118.04 kJ/kg W23 1107.94 kJ/kg
W31 -1081.60 kJ/kg
Next, we need to evaluate the specific work and specific heat transfer for the entire cycle.
The specific work for the cycle is the sum of the specific work for each step.
The specific heat transfer for the cycle is the sum of the specific heat transfer for each step.
Qcycle 26.34 kJ/kg Wcycle 26.34 kJ/kg
Verify: The assumptions cannot be verified from the information in the problem statement alone.
Answers : Part a.) See the diagram, above.
Part b.)   Q (kJ/kg) W (kJ/kg)
Step 1 - 2 118.0 0
Step 2 - 3 -91.7 1107.9
Step 3 - 1 0 -1081.6
Cycle 26.3 26.3
Part c.) Because Wcycle > 0, this is a power cycle !