4F4 :  Heat and Work for a Cycle Executed in a Closed System Containing Ammonia  8 pts 

Ammonia
in a pistonandcylinder
device undergoes a 3step thermodynamic cycle made up of the following three quasiequilibrium processes. 

Step 12: Isochoric heating from 20^{o}C and 150
kPa up to 50^{o}C Step 23: Isothermal compression until the quality is 0.55 kg vap/kg, Q_{23} = 91.7 kJ Step 31: 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 Q_{cycle} and W_{cycle} in kJ/kg c.) Determine whether this cycle is a power cycle or a refrigeration/heatpump cycle? Explain your reasoning. 

Read :  We are given T_{1} and P_{1}, 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 12. This gives us a 2nd intensive property for state 2 and allows us to evaluate all of the properties of state 2. We expect T_{2} > T_{1}. Step 23 is an isothermal compression to a quality of x_{3} = 0.55. Because T_{3} = T_{2}, 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. W_{12} = 0 because the process is isochoric. Q_{23} is given and Q_{31} = 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 Q_{cycle} and W_{cycle} because they are the sum of the Q's and W's for the steps that make up the cycle, respectively.  
Given :  T_{1}  20  ^{o}C  Find :  Q_{12}  ???  kJ/kg  
P_{1}  150  kPa  W_{23}  ???  kJ/kg  
T_{2}  50  ^{o}C  W_{31}  ???  kJ/kg  
T_{3}  50  ^{o}C  Q_{cycle}  ???  kJ/kg  
x_{3}  0.55  kg vap/kg  W_{cycle}  ???  kJ/kg  
Q_{23}  91.7  kJ/kg  Power or Refrigeration Cycle ?  
Q_{31}  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 12 : 

Eqn 1  
Step 23 : 

Eqn 2  
Step 31 : 

Eqn 3  
Step 12 is isochoric, so no boundary work occurs. If we assume that boundary work is the only form of work interaction in this cycle, then W_{12} = 0. Q_{31} = 0 because step 31 is adiabatic.  
We can solve Eqns 1  3 to evaluate the unknowns Q_{12}, W_{23} and W_{31}.  
Step 12 : 

Eqn 4  
Step 23 : 

Eqn 5  
Step 31 : 

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 Q_{12}, W_{23} and W_{31}.  
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 P_{1} to P_{sat}(T_{1}).  
P_{sat}(T_{1})  190.08  kPa  
Since P_{1} < P_{sat}(T_{1}), we conclude that a superheated vapor exists in the cylinder at state 1.  
We can determine U_{1} from the Superheated Ammonia Tables. We can also determine V_{1} because we know V_{2} = V_{1} and the knowledge of this 2nd intensive variable for state 2 will allow us to evaluate U_{2}.  
V_{1} = V_{2} =  0.79779  m^{3}/kg  U_{1}  1303.8  kJ/kg  
Next, let's work on state 2. We know the value of 2 intensive variables, T_{2} and V_{2}, 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 U_{2} (and any other properties at state 2 that we want).  
At T_{2} = 50^{o}C, it turns out that V_{2 }= 0.79779 m^{3}/kg falls between 100 kPa and 200 kPa, so we must interpolate to determine U_{2}.  
At T_{2} = 50^{o}C :  V (m^{3}/kg)  U (kJ/kg)  P (kPa)  
1.56571  1425.2  100  
0.79779  U_{2}  P_{2}  U_{2}  1421.8  kJ/kg  
0.77679  1421.7  200  P_{2}  197.3  kPa  
Now, let's work on state 3. We know the temperature and the quality, so we can determine U_{3} using :  

Eqn 7  
We can use the Saturated Ammonia Tables to determine U_{sat vap} and U_{sat liq} at 50^{o}C and then we can plug numbers into Eqn 7 to evaluate U_{3}.  
U_{sat liq}  171.41  kJ/kg  
U_{sat vap}  263.69  kJ/kg  U_{3}  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 Q_{12}, W_{23}, and W_{31}.  
Q_{12}  118.04  kJ/kg  W_{23}  1107.94  kJ/kg  
W_{31}  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.  
Q_{cycle}  26.34  kJ/kg  W_{cycle}  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 W_{cycle} > 0, this is a power cycle ! 