6F3 :  Heat, Work and Efficiency of a Water Vapor Power Cycle  8 pts 

A pistonandcylinder device containing 12 kg of water carries out a Carnot power cycle. The maximum pressure is 2.0 MPa and the minimum pressure is 110 kPa.  
During the isothermal
expansion, the water is heated
from a quality of 14% until it is a saturated vapor. The cycle produces 500 kJkg of work during the adiabatic expansion. 

a.) Sketch the process path for the cycle on a PV Diagram b.) Calculate Q and W, in kJ, for each process in the cycle c.) Calculate the thermal efficiency of the cycle. 

Read :  Apply the 1st Law (for a closed system) to get Q and W.  
Use the 1st Law applied to step 23 to determine U_{3} and x_{3}.  
The trick is to get Q_{34}. Use T_{C}, T_{H}, Q_{12} and the Carnot Efficiency of this reversible cycle to determine Q_{34}.  
Given:  P_{1}  2  MPa  P_{3} = P_{4}  110  kPa  
x_{1}  0.14  ^{}  Q_{23 }= Q_{41}  0  kJ/kg  
P_{2}  2  MPa  W_{23}  500  kJ/kg  
x_{2}  1  ^{}  m  12  kg  
Find:  Part (a)  PV Diagram  ^{}  
Part (b)  Q_{12},Q_{23},Q_{34},Q_{41} ?  kJ  ^{}  
W_{12},W_{23},W_{34},W_{41} ?  kJ  ^{}  
Part (c)  h  ?  ^{}  
^{}  
Diagram:  
Part a.) 


Assumptions:  1  


  Steps 12 and 34 are isothermal.  
  Steps 23 and 41 are adiabatic.  
  All steps are reversible.  
2   The water inside the cylinder is the system and it is a closed system.  
3   Changes in kinetic and potential energies are negligible.  
4   Boundary work is the only form of work interaction during the cycle.  
Equations / Data / Solve:  
Part b.)  Begin by applying the 1st law for closed systems to each step in the Carnot Cycle. Assume that changes in kinetic and potential energies are negligible.  

Eqn 1  
Step 1  2  
Apply the 1st Law, Eqn 1, to step 12 : 

Eqn 2  
Boundary work at for a constant pressure process, like step 12, can be determined from :  

Eqn 3  
Now, we can substitute Eqn 3 into Eqn 1 to get : 

Eqn 4  
The definition of enthalpy is: 

Eqn 5  
For isobaric processes, Eqn 5 becomes : 

Eqn 6  
Now, combine Eqns 4 and 6 to get : 

Eqn 7  
We know the pressure and the quality of states 1 and 2, so we can use the Saturation Table in the Steam Tables to evaluate V and H for states 1 and 2 so we can use Eqns 3 and 7 to evaluate Q_{12} and W_{12}.  
Properties are determined from NIST WebBook:  

Eqn 8  
At P_{1} and x_{1}:  V_{sat liq, 1}  0.0011767  m^{3}/kg  
V_{sat vap, 1}  0.099585  m^{3}/kg  V_{1}  0.014954  m^{3}/kg  

Eqn 9  
U_{sat liq, 1}  906.14  kJ/kg  
U_{sat vap, 1}  2599.1  kJ/kg  U_{1}  1143.2  kJ/kg  


H_{sat liq, 1}  908.50  kJ/kg  
H_{sat vap, 1}  2798.3  kJ/kg  H_{1}  1173.1  kJ/kg  
Saturated vapor at P_{2}:  V_{2}  0.099585  m^{3}/kg  
U_{2}  2599.1  kJ/kg  W_{12}  2031.147  kJ  
H_{2}  2798.3  kJ/kg  Q_{12}  19502.68  kJ  
^{}  
Step 2  3  ^{}  
Apply the 1st Law, Eqn 1, to step 23 : 

Eqn 10  
^{}  
The specific heat transferred and specific work for step 23 are given in the problem statement.  
Q_{23}  0  kJ  W_{23}  6000  kJ  
We can plug these values into Eqn 8 to determine DU_{23} :  DU_{23}  6000  kJ  
We already determined U_{2}, so we can now determine U_{3} : 

Eqn 11  
U_{3}  2099.1  kJ/kg  
We can use this value of U_{3} to determine the unknown quality, x_{3} , using : 

Eqn 12  
Properties are determined from NIST WebBook:  
At P_{3}:  U_{sat liq, 3}  428.72  kJ/kg  
U_{sat vap, 3}  2508.7  kJ/kg  x_{3}  0.8031  kg vap/kg  

Eqn 13  
At P_{3} and x_{3}:  V_{sat liq, 3}  0.0010453  m^{3}/kg  
V_{sat vap, 3}  1.54946  m^{3}/kg  V_{3}  1.2445  m^{3}/kg  

Eqn 14  
H_{sat liq, 3}  428.84  kJ/kg  
H_{sat vap, 3}  2679.2  kJ/kg  H_{3}  2236.0  kJ/kg  
Step 3  4  
Apply the 1st Law, Eqn 1, to step 34 : 

Eqn 15  
Because step 34 is isobaric, just like step 12, Eqn 7 is the simplified form of the 1st Law : 

Eqn 16  
To determine the properties at state 4, we make us of the relationship between the absolute Kelvin temperature scale and heat transferred in a Carnot Cycle.  

Eqn 17  
Solve Eqn 13 for Q_{34} : 

Eqn 18  
T_{H} = T_{sat}(P_{1}) :  T_{H}  485.53  K  Q_{34}  15080.8  kJ  
T_{C} = T_{sat}(P_{3}) :  T_{C}  375.44  K  Q_{34}  1256.7  kJ/kg  
Now, we can use Q_{34} and Eqn 12 to determine H_{4} as follows:  

Eqn 19  or : 

Eqn 20  
H_{4}  979.29  kJ/kg  
Now that we know the values of two intensive properties at state 4, T_{4} and H_{4}, we can evaluate all the other properties using the Saturation Tables in the Steam Tables.  
Properties are determined from NIST WebBook:  

Eqn 21  
At P_{4}:  H_{sat liq, 4}  428.84  kJ/kg  
H_{sat vap, 4}  2679.2  kJ/kg  x_{4}  0.24461  kg vap/kg  

Eqn 22  
At P_{4} and x_{4}:  V_{sat liq, 4}  0.0010453  m^{3}/kg  
V_{sat vap, 4}  1.54946  m^{3}/kg  V_{4}  0.37980  m^{3}/kg  

Eqn 23  
U_{sat liq, 4}  428.72  kJ/kg  
U_{sat vap, 4}  2508.7  kJ/kg  U_{4}  937.51  kJ/kg  
At last we have U_{4} and we can plug it into Eqn 11 to evaluate W_{34} :  

Eqn 24  
W_{34}  1141.45  kJ/kg  
Step 4  1  
The heat transferred for step 41 is given in the problem statement.  
Apply the 1st Law, Eqn 1, to step 41 : 

Eqn 25  
Solve Eqn 25 for W_{41} : 

Eqn 26  
W_{41}  2467.79  kJ  
Part c.)  The efficiency of a Carnot Cycle is defined by: 

Eqn 27  
Where : 

Eqn 28  
And : 

Eqn 29  
Q_{in}  19502.7  kJ  
W_{cycle}  4421.9  kJ  h  0.2267  
Or the efficiency can be determined
in terms of reservoir temperatures: 

Eqn 30  
h  0.2267  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  Process  Q  W  
12  19502.7  2031.1  
23  0  6000.0  
34  15080.8  1141.5  
41  0.0  2467.8  
Cycle  4421.9  4421.9  The thermal efficiency of the process is :  22.7% 