Example Problem with Complete Solution

6F-3 : Heat, Work and Efficiency of a Water Vapor Power Cycle 8 pts
A piston-and-cylinder 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 433.2 kJ/kg 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 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 2 MPa P3 = P4 110 kPa
x1 0.14 Q23 = Q41 0 kJ/kg
P2 2 MPa W23 433.2 kJ/kg
x2 1 m 12 kg
Find: Part (a) PV Diagram
Part (b) Q12,Q23,Q34,Q41         ? kJ
W12,W23,W34,W41     ? kJ
Part (c) h ?
Diagram:
Part a.)
Assumptions: 1 -
 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.
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 1-2 : Eqn 2
Boundary work at for a constant pressure process, like step 1-2, 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 Q12 and W12.
Properties are determined from NIST WebBook:
Eqn 8
At P1 and x1: Vsat liq, 1 0.0011767 m3/kg
Vsat vap, 1 0.099585 m3/kg V1 0.014954 m3/kg
Eqn 9
Usat liq, 1 906.14 kJ/kg
Usat vap, 1 2599.1 kJ/kg U1 1143.2 kJ/kg
Hsat liq, 1 908.50 kJ/kg
Hsat vap, 1 2798.3 kJ/kg H1 1173.1 kJ/kg
Saturated vapor at P2: V2 0.099585 m3/kg
U2 2599.1 kJ/kg W12 2031.147 kJ
H2 2798.3 kJ/kg Q12 19502.68 kJ
Step 2 - 3
Apply the 1st Law, Eqn 1, to step 2-3 : Eqn 10
The specific heat transferred and specific work for step 2-3 are given in the problem statement.
Q23 0 kJ W23 5198.4 kJ
We can plug these values into Eqn 8 to determine DU23 : DU23 -5198.4 kJ
We already determined U2, so we can now determine U3 : Eqn 11
U3 2165.9 kJ/kg
We can use this value of U3 to determine the unknown quality, x3 , using : Eqn 12
Properties are determined from NIST WebBook:
At P3: Usat liq, 3 428.72 kJ/kg
Usat vap, 3 2508.7 kJ/kg x3 0.8352 kg vap/kg
Eqn 13
At P3 and x3: Vsat liq, 3 0.0010453 m3/kg
Vsat vap, 3 1.54946 m3/kg V3 1.2943 m3/kg
Eqn 14
Hsat liq, 3 428.84 kJ/kg
Hsat vap, 3 2679.2 kJ/kg H3 2308.3 kJ/kg
Step 3 - 4
Apply the 1st Law, Eqn 1, to step 3-4 : Eqn 15
Because step 3-4 is isobaric, just like step 1-2,
Eqn 7 is the simplified form of the 1st Law :
Eqn 16
To determine the properties at state 4, we make use of the relationship between the absolute Kelvin temperature scale and heat transferred in a Carnot Cycle.
Eqn 17
Solve Eqn 13 for Q34 : Eqn 18
TH = Tsat(P1) : TH 485.53 K Q34 -15080.8 kJ
TC = Tsat(P3) : TC 375.44 K Q34 -1256.7 kJ/kg
Now, we can use Q34 and Eqn 12 to determine H4 as follows:
Eqn 19 or : Eqn 20
H4 1051.56 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:
Eqn 21
At P4: Hsat liq, 4 428.84 kJ/kg
Hsat vap, 4 2679.2 kJ/kg x4 0.27672 kg vap/kg
Eqn 22
At P4 and x4: Vsat liq, 4 0.0010453 m3/kg
Vsat vap, 4 1.54946 m3/kg V4 0.42953 m3/kg
Eqn 23
Usat liq, 4 428.72 kJ/kg
Usat vap, 4 2508.7 kJ/kg U4 1004.31 kJ/kg
At last we have U4 and we can plug it into Eqn 11 to evaluate W34 :
Eqn 24
W34 -1141.45 kJ/kg
Step 4 - 1
The heat transferred for step 4-1 is given in the problem statement.
Apply the 1st Law, Eqn 1, to step 4-1 : Eqn 25
Solve Eqn 25 for W41 : Eqn 26
W41 -1666.19 kJ
Part c.) The efficiency of a Carnot Cycle is defined by: Eqn 27
Where : Eqn 28
And : Eqn 29
Qin 19502.7 kJ
Wcycle 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
1-2 19502.7 2031.1
2-3 0 5198.4
3-4 -15080.8 -1141.5
4-1 0.0 -1666.2
Cycle 4421.9 4421.9 The thermal efficiency of the process is : 22.7%