# Example Problem with Complete Solution

9B-3 : Vapor Power Cycle Based on Temperature Gradients in the Ocean 9 pts
A Rankine Power Cycle uses water at the surface of a tropical ocean as the heat source, TH = 82oF, and cool water deep beneath the surface as the heat sink, TC = 48oF. Ammonia is the working fluid. The boiler produces saturated ammonia vapor at 80oF and the condenser effluent is saturated liquid ammonia at 50oF. The isentropic efficiencies of the pump and turbine are 75% and 85%, repectively.
Calculate the...
a.) Thermal efficiency of this Rankine Cycle
b.) Thermal efficiency of a Carnot Cycle operating between the same two thermal reservoirs

Read : Assume the process operates at steady-state and that changes in potential and kinetic energies are negligible.
We want the boiler pressure to be as high as possible and the condenser pressure to be as low as possible in order to maximize the thermal efficiency of the cycle. But the saturation temperature of the ammonia at the boiler pressure must be less than 82oF in order to absorb heat from the warmer seawater. This explains why the ammonia boils at 80oF. A similar argument regarding the vapor-liquid equilibrium in the condenser leads us to the choice of 50oF for the saturation temperature of the ammonia in the condenser.
Given: T2 80 oF TC 48 oF
x2 1 lbm vap/lbm TH 82 oF
T3 50 oF (Tsw, in)boil 80 oF
x4 0 lbm vap/lbm (Tsw, in)cond 48 oF
hS,pump 75% hS,turb 85%
Find: a.) hth ??? % b.) hmax ??? %
Diagram: A good flow diagram was provided in the problem statement.
TS Diagram : Assumptions: 1 - Each process in the cycle operates at steady-state.
2 - The power cycle operates on the Rankine Cycle.
3 - Changes in kinetic and potential energies are negligible.
4 - The pump and the turbine are both adiabatic.
Equations / Data / Solve:
Let's organize the data that we need to collect into a table.  This will make it easier to keep track of the values we have looked up and the values we have calculated.
Stream State T
(oF)
P
(psia)
X
(lbm vap/lbm)
H
(Btu/lbm)
S
(Btu/lbm-oR)
1S Sub Liq 50.18 153.13 N/A 98.132 0.21024
1 Sub Liq 50.27 153.13 N/A 98.233 0.21044
2 Sat Vap 80 153.13 1 630.36 1.1982
3S Sat Mix 50 89.205 0.9551 601.38 1.1982
3 Sat Mix 50 89.205 0.9633 605.73 1.2068
4 Sat Liq 50 89.205 0 97.828 0.21024
Additional data that may be useful.
State T
(oF)
P
(psia)
X
(lbm vap/lbm)
H
(Btu/lbm)
S
(Btu/lbm-oR)
Sat Vap 80 153.13 1 630.36 1.1982
Sat Liquid 80 153.13 0 131.86 0.27452
Sat Vap 50 89.205 1 625.07 1.2447
Sat Liquid 50 89.205 0 97.828 0.21024
Part a.) The thermal efficiency of a
power cycle can be determined using : Eqn 1
We need to evaluate Q12 and Q34 so we can use Eqn 1 to evaluate the thermal efficiency of the cycle.
Apply the 1st Law to the boiler, assuming it operates at steady-state, changes in kinetic and potential energies are negligible and no shaft work crosses the boundary of the boiler. Eqn 2
We can get a value for H1 from the Steam Tables or NIST Webbook because we know that the boiler effluent in a Rankine Cycle is a saturated vapor at 80oF.
P2 153.13 psia H2 630.36 Btu/lbm
In order to fix state 1 and evaluate H1, we must use the isentropic efficiency of the pump.  The feed to the pump is a saturated liquid at 80oF.  So, we can look-up S4 in the Steam Tables or NIST Webbook.
P4 89.205 psia S4 0.21024 Btu/lbm-oR
H4 97.828 Btu/lbm
The definition of
isentropic efficiency for a pump is : Eqn 3
We can solve Eqn 2 for H1 : Eqn 4
Next, we need to determine H1S.  For an isentropic pump : S1S 0.21024 Btu/lbm-oR
Now, we know the value of two intensive properties at state 1S: S1S and P1 (because the boiler is isobaric in a Rankine Cycle,  P1 = P2.
At P = 153.13 psia : T (oF) H Btu/lbm S Btu/lbm-oR
50 97.935 0.20985
T1S H1S 0.21024 T1S 50.18 oF
55 103.528 0.22077 H1S 98.13 Btu/lbm
Now ,we can plug values into Eqn 4 to determine H1 : H1 98.23 Btu/lbm
Next, we can plug values into Eqn 2 to evaluate Q12 : Q12 532.12 Btu/lbm
We can determine Q34 by applying the 1st Law, with all the same assumptions made about the boiler. Eqn 5
We already know H4, so we need to evaluate H3.  To do this, we use the isentropic efficiency of the turbine.
In order to fix state 3 and evaluate H3, we must use the isentropic efficiency of the turbine.  The feed to the turbine is a saturated vapor at 80oF.  So, we can look-up S2 in the Steam Tables or NIST Webbook.
P4 89.205 psia S2 1.1982 Btu/lbm-oR
H2 630.36 Btu/lbm
The definition of
isentropic efficiency for a turnine is : Eqn 6
We can solve Eqn 6 for H3 : Eqn 7
Next, we need to determine H3S.  For an isentropic turbine : S3S 1.1982 Btu/lbm-oR
Now, we know the value of two intensive properties at state 3S: S3S and P3 (because the condenser is isobaric in a Rankine Cycle,  P3 = P4.
At P = 89.205 psia : Ssat liq 0.21024 Btu/lbm-oR Since Ssat liq < S3S < Ssat vap,
state 3S is a saturated mixture.
Ssat vap 1.2447 Btu/lbm-oR
Determine x3S from the
specific entropy, using: Eqn 8
x3S 0.9551 lbm vap/lbm
Then, we can use the quality
to determine
H3S, using: Eqn 9
At P = 89.205 psia : Hsat liq 97.828 Btu/lbm
Hsat vap 625.07 Btu/lbm
H3S 601.38 Btu/lbm
Now ,we can plug values into Eqn 7 to determine H1 : H3 605.73 Btu/lbm
Next, we can plug values into Eqn 5 to evaluate Q34 : Q34 -507.90 Btu/lbm
We can now plug values into Eqn 1 to evaluate the thermal efficiency of the Rankine Cycle.
hth 4.55%
The maximum thermal efficiency for a power cycle operating between two thermal reservoirs at TH and TC is the Carnot Efficiency : Eqn 10
TC 507.67 oR
TH 541.67 oR hmax 6.28%
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : a.) hth 4.55% hmax 6.28% 