9B3 :  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, T_{H} = 82^{o}F, and cool water deep beneath the surface as the heat sink, T_{C} = 48^{o}F. Ammonia is the working fluid.  


The boiler produces saturated ammonia vapor at 80^{o}F and the condenser effluent is saturated liquid ammonia at 50^{o}F. 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 steadystate 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 82^{o}F in order to absorb heat from the warmer seawater. This explains why the ammonia boils at 80^{o}F. A similar argument regarding the vaporliquid equilibrium in the condenser leads us to the choice of 50^{o}F for the saturation temperature of the ammonia in the condenser.  
Given:  T_{2}  80  ^{o}F  T_{C}  48  ^{o}F  
x_{2}  1  lb_{m} vap/lb_{m}  T_{H}  82  ^{o}F  
T_{3}  50  ^{o}F  (T_{sw, in})_{boil}  80  ^{o}F  
x_{4}  0  lb_{m} vap/lb_{m}  (T_{sw, in})_{cond}  48  ^{o}F  
^{}  
h_{S,pump}  75%  h_{S,turb}  85%  ^{}  
Find:  a.)  h_{th}  ???  %  b.)  h_{max}  ???  %  
Diagram:  A good flow diagram was provided in the problem statement.  
TS Diagram : 


Assumptions:  1   Each process in the cycle operates at steadystate.  
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 (^{o}F) 
P (psia) 
X (lb_{m} vap/lb_{m}) 
H (Btu/lb_{m}) 
S (Btu/lb_{m}^{o}R) 

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 (^{o}F) 
P (psia) 
X (lb_{m} vap/lb_{m}) 
H (Btu/lb_{m}) 
S (Btu/lb_{m}^{o}R) 

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 Q_{12} and Q_{34} 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 steadystate, 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 H_{1} from the Steam Tables or NIST Webbook because we know that the boiler effluent in a Rankine Cycle is a saturated vapor at 80^{o}F.  
P_{2}  153.13  psia  H_{2}  630.36  Btu/lb_{m}  
In order to fix state 1 and evaluate H_{1}, we must use the isentropic efficiency of the pump. The feed to the pump is a saturated liquid at 80^{o}F. So, we can lookup S_{4} in the Steam Tables or NIST Webbook.  
P_{4}  89.205  psia  S_{4}  0.21024  Btu/lb_{m}^{o}R  
H_{4}  97.828  Btu/lb_{m}  
The definition of isentropic efficiency for a pump is : 

Eqn 3  
We can solve Eqn 2 for H_{1} : 

Eqn 4  
Next, we need to determine H_{1S}. For an isentropic pump :  S_{1S}  0.21024  Btu/lb_{m}^{o}R  
Now, we know the value of two intensive properties at state 1S: S_{1S} and P_{1} (because the boiler is isobaric in a Rankine Cycle, P_{1} = P_{2}.  
At P = 153.13 psia :  T (^{o}F)  H Btu/lb_{m}  S Btu/lb_{m}^{o}R  
50  97.935  0.20985  
T_{1S}  H_{1S}  0.21024  T_{1S}  50.18  ^{o}F  
55  103.528  0.22077  H_{1S}  98.13  Btu/lb_{m}  
Now ,we can plug values into Eqn 4 to determine H_{1} :  H_{1}  98.23  Btu/lb_{m}  
Next, we can plug values into Eqn 2 to evaluate Q_{12} :  Q_{12}  532.12  Btu/lb_{m}  
We can determine Q_{34} by applying the 1st Law, with all the same assumptions made about the boiler.  

Eqn 5  
We already know H_{4}, so we need to evaluate H_{3}. To do this, we use the isentropic efficiency of the turbine.  
In order to fix state 3 and evaluate H_{3}, we must use the isentropic efficiency of the turbine. The feed to the turbine is a saturated vapor at 80^{o}F. So, we can lookup S_{2} in the Steam Tables or NIST Webbook.  
P_{4}  89.205  psia  S_{2}  1.1982  Btu/lb_{m}^{o}R  
H_{2}  630.36  Btu/lb_{m}  
The definition of isentropic efficiency for a turnine is : 

Eqn 6  
We can solve Eqn 6 for H_{3} : 

Eqn 7  
Next, we need to determine H_{3S}. For an isentropic turbine :  S_{3S}  1.1982  Btu/lb_{m}^{o}R  
Now, we know the value of two intensive properties at state 3S: S_{3S} and P_{3} (because the condenser is isobaric in a Rankine Cycle, P_{3} = P_{4}.  
At P = 89.205 psia :  S_{sat liq}  0.21024  Btu/lb_{m}^{o}R  Since S_{sat liq} < S_{3S} < S_{sat vap}, state 3S is a saturated mixture. 

S_{sat vap}  1.2447  Btu/lb_{m}^{o}R  
Determine x_{3S} from the specific entropy, using: 

Eqn 8  
x_{3S}  0.9551  lb_{m} vap/lb_{m}  
Then, we can use the quality to determine H_{3S}, using: 

Eqn 9  
At P = 89.205 psia :  H_{sat liq}  97.828  Btu/lb_{m}  
H_{sat vap}  625.07  Btu/lb_{m}  
H_{3S}  601.38  Btu/lb_{m}  
Now ,we can plug values into Eqn 7 to determine H_{1} :  H_{3}  605.73  Btu/lb_{m}  
Next, we can plug values into Eqn 5 to evaluate Q_{34} :  Q_{34}  507.90  Btu/lb_{m}  
We can now plug values into Eqn 1 to evaluate the thermal efficiency of the Rankine Cycle.  
h_{th}  4.55%  
The maximum thermal efficiency for a power cycle operating between two thermal reservoirs at T_{H} and T_{C} is the Carnot Efficiency :  

Eqn 10  
T_{C}  507.67  ^{o}R  
T_{H}  541.67  ^{o}R  h_{max}  6.28%  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  a.)  h_{th}  4.55%  h_{max}  6.28% 