Example Problem with Complete Solution

6E-1 : Heat, Work and Efficiency of a Water Vapor Power Cycle 8 pts
One-half kilogram of water executes a Carnot power cycle. During the isothermal expansion, the water is heated until it is saturated vapor from an initial state where the pressure is 15 bar and the quality is 25%. The vapor then expands adiabatically to a pressure of 1 bar while doing 403.8 kJ/kg of work.
               
a.) Sketch the cycle on PV coordinates.      
b.) Evaluate the heat and work for each process, in kJ
c.) Evaluate the thermal efficiency.
               
Read : Apply the 1st Law (for a closed system) to get Q and W.
Use the first law applied to step 2-3 to determine U3 and x3.
The trick is to get Q34.
Use TC, TH, Q12 and the Carnot efficiency of this reversible cycle to determine Q34.
Given : P1 15 bar P3 1 bar
x1 0.25 Q23 = Q41 0 kJ/kg
P2 15 bar W23 403.8 kJ/kg
x2 1 m 0.5 kg
Find : Part (a) PV Diagram
Part (b) Q12,Q23,Q34,Q41         ? kJ
W12,W23,W34,W41     ? kJ
Part (c) h ?
Diagram :
Part a.)
Figure 1
Figure 2
Assumptions : - The system undergoes a Carnot cycle.
- Steps 1-2 and 3-4 are isothermal
- Steps 2-3 and 4-1 are adiabatic
- All steps are reversible
- The water inside the cylinder is the system and it is a closed system.
- Changes in kinetic and potential energies are negligible.
- Boundary work is the only form of work interaction wuring 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.
Equation 1
Eqn 1
Step 1 - 2
Apply the 1st Law, Eqn 1, to step 1-2 :
Equation 2
Eqn 2
Boundary work at for a constant pressure process, like step 1-2, can be determined from :
Equation 3
Eqn 3
Now, we can substitute Eqn 3 into Eqn 1 to get :
Equation 4
Eqn 4
The definition of enthalpy is:
Equation 5
Eqn 5
For isobaric processes, Eqn 5 becomes :
Equation 6
Eqn 6
Now, combine Eqns 4 and 6 to get :
Equation 7
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 Q12 and W12.
Properties are determined from NIST WebBook:
At P1 and x1:
Vsat liq, 1 0.001154 m3/kg Vsat vap, 0.13171 m3/kg
Equation 8
V1 0.033793 m3/kg
Usat liq, 1 842.83 kJ/kg
Usat vap, 2593.4 kJ/kg
Equation 9
U1 1280.5 kJ/kg
Hsat liq, 1 844.56 kJ/kg
Hsat vap, 2791 kJ/kg
Equation 10
H1 1331.2 kJ/kg
Saturated vapor at P2:
V2 0.13171 m3/kg
U2 2593.4 kJ/kg W12 73.438 kJ
H2 2791 kJ/kg Q12 729.92 kJ
Step 2 - 3
Apply the 1st Law, Eqn 1, to step 2-3 :
Equation 11 Eqn 8
The specific heat transferred and specific work for process 2-3 are given in the problem statement.
W23 201.9 kJ Q23 0 kJ
We can plug these values into Eqn 8 to determine DU23 :
DU23 -201.9 kJ
We already determined U2, so we can now determine U3 :
Equation 12
Eqn 9
U3 2189.6 kJ/kg
We can use this value of U3 to determine the unknown quality, x3 , using :
Equation 13
Eqn 10
Properties are determined from NIST WebBook:
At P3: Usat liq, 3 417.4 kJ/kg
Usat vap, 3 2505.6 kJ/kg
x3 0.849
At P3 and x3:
Vsat liq, 3 0.001043 m3/kg
Vsat vap, 3 1.6939 m3/kg
Equation 14
V3 1.4377 m3/kg
Hsat liq, 3 417.5 kJ/kg
Hsat vap, 3 2674.9 kJ/kg
Equation 15
H3 2333.3 kJ/kg
Step 3 - 4
Apply the 1st Law, Eqn 1, to step 2-3 :
Equation 16
Equation 17
Eqn 11
Because step 3-4 is isobaric, just like step 1-2, Eqn 7 is the simplified form of the 1st Law : :
Equation 18
Eqn 12
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.
Equation 19
Eqn 13
Solve Eqn 13 for Q34 : Equation 20
Eqn 14
TH = Tsat(P1) : TH 471.44 K
TC = Tsat(P3) : TC 372.76 K
Q34 -577.13 kJ
Q34 -1154.26 kJ/kg
Now, we can use Q34 and Eqn 12 to determine H4 as follows:
Equation 21
Eqn 15 or : Equation 22
Eqn 16
H4 1179.03 kJ/kg
Now that we know the values of two intensive properties at state 4, T4 and H4, we can evaluate all the other properties using the Saturation Tables in the Steam Tables.
Properties are determined from NIST WebBook:
At P4: Hsat liq, 4 417.5 kJ/kg
Hsat vap, 4 2674.9 kJ/kg
Equation 23
Eqn 17
x4 0.337
At P4 and x4: Vsat liq, 4 0.0010432 m3/kg
Vsat vap, 4 1.6939 m3/kg
Equation 24
V4 0.5721 m3/kg
Usat liq, 4 417.4 kJ/kg
Usat vap, 4 2505.6 kJ/kg
Equation 25
U4 1121.9 kJ/kg
At last we have U4 and we can plug it into Eqn 11 to evaluate W34 :
Equation 26
Eqn 18
W34 -43.26 kJ/kg
Step 4 - 1
The heat transferred for Process 4-1 is given in the problem statement.
Apply the 1st Law, Eqn 1, to step 4-1 :
Equation 27 Eqn 19
Solve Eqn 19 for W41 :
 
  Equation 28
 
  W41 -79.31 kJ
 
Part c.) The efficiency of a Carnot cycle is defined by:
Equation 29
Equation 30
Eqn 20
Where : Equation 31
Eqn 21
Equation 32
Eqn 22
Qin 729.9 kJ
Wcycle 152.8 kJ h 0.209
Or the efficiency can be determined in terms of reservoir temperatures:
Equation 33 Eqn 23
h 0.209
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : Process Q W
1-2 729.9 73.4
2-3 0 201.9
3-4 -577.1 -43.3
4-1 0 -79.3
Cycle 152.8 152.8
The thermal efficiency of the process is : 21%
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