Example Problem with Complete Solution

9B-3 : Vapor Power Cycle Based on Temperature Gradients in the Ocean
9 pts
At a particular location in the ocean, the temperature near the surface is 80°F, and the temperature at a depth of 1500 ft is 46°F. A power plant based on the Rankine cycle, with ammonia as the working fluid, has been proposed to utilize this naturally occurring temperature gradient to produce electrical power. The power to be developed by the turbine is 8.2 X 108 Btu/h. A schematic diagram is shown. For simplicity, the properties of seawater can be taken as those of pure water.Assume the isentropic efficiency of the turbine is 80% and the isentropic efficiency of the pump is 70%. Assume that the ammonia leaving the boiler is saturated vapor and that the ammonia leaving the condenser is saturated liquid.
     
a.) Estimate the net power output of the plant, in Btu/h if the pumps used to circulate seawater through the evaporator and condenser heat exchangers require a total power input of 2.55 X 108 Btu/h.
   
b.) Determine the seawater flow rates through the boiler and condenser, in lbm/h.
   
c.) Estimate the thermal efficiency of the proposed cycle and compare it to the thermal efficiency of a reversible power cycle operating between thermal reservoirs at 80°F and 46°F.
           
     
Read : Assume the process is at steady-state and that changes in potential and kinetic energy are negligible.
Assume that seawater has the same properties as pure water and therefore CP = 1.0 Btu/(lbm oR).
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 efficiency of the cycle. But the saturation temperature of the ammonia at the boiler pressure must be less than 77.5oF in order to absorb heat from the warmer seawater. So, let's assume that ammonia boils at 75oF. 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 75 oF (Tsw, in)boil 80 oF
T3 50 oF (Tsw, out)boil 77.5 oF
TC 46 oF (Tsw, in)cond 46 oF
TH 80 oF (Tsw, out)cond 47.7 oF
Wturb 8.2E+08 Btu/h ηS,turb 80%
Wsw, pumps -2.55E+08 Btu/h ηS,pump 70%
CP,sw 1 Btu / lbm-oR
Find : a.) ηth ? % ηmax ? %
b.) Wcycle ? Btu/h
c.) msw, cond ? lbm/h msw, boiler ? lbm/h
Diagram:   Diagram already provided in the problem statement.
Assumptions: - Each system operates at steady-state.
- The power cycle operates on the Rankine Cycle.
- Changes in kinetic and potential energies are negligible.
- The pump and the turbine are both adiabatic.
- The seawater has properties of pure water, assumed to be an incompressible liquid: CP = 1.0 Btu/(lbm oR).
Equations / Data / Solve :  
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 oreates at steady-state, changes in kinetic and potential energy are negligible and no shaft work crosses the boundary of the boiler.
 
Eqn 2
We can get a value for H2 from the steam tables or NIST Webbook because we know that the boiler effluent in a Rankine Cycle is a saturated vapor at 75oF.
P2 140.59 psia H2 699.48 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 50oF.  So, we can lookup S4 in the steam tables or NIST Webbook.
P4 89.205 psia S4 0.39148 Btu/lbm-oR
H4 167.66 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 determin H1S.
For an isentropic pump :
S1S 0.39148 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 CycleP1 = P2.
At P = 140.59 psia : T (oF) H Btu/lbm S
Btu/lbm-oR
50 167.75 0.39117
T1S H1S 0.39148 T1S 50.14 oF
55 173.34 0.40209 H1S 167.91 Btu/lbm
Now ,we can plug values into
Eqn 4 to determine H1 :
H1 168.02 Btu/lbm
Next, we can plug values into
Eqn 2 to evaluate Q12 :
Q12 531.46 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 75oF.  So, we can lookup S2 in the steam tables or NIST Webbook.
P4 89.205 psia S2 1.3869 Btu/lbm-oR
H2 699.48 Btu/lbm
 
The definition of isentropic
efficiency for a turbine is :
Eqn 6
We can solve Eqn 6 for H3 : Eqn 7
Next, we need to determin H3S.  For an isentropic turbine : S3S 1.3869 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 CycleP3 = P4.
At P = 89.205 psia : Ssat liq 0.39148 Btu/lbm-oR Since Ssat liq < S3S < Ssat vap, state 3S is a saturated mixture.
Ssat vap 1.4260 Btu/lbm-oR
 
Determine x3S from the specific entropy, using: Eqn 8
x2S 0.9622

lbm vap/lbm

Then, we can use the quality to
determine H3S, using:
  Eqn 9
At P = 89.205 psia : Hsat liq 167.66 Btu/lbm
Hsat vap 694.91 Btu/lbm H3S 674.98 Btu/lbm
Now ,we can plug values
into Eqn 7 to determine H1 :
H3 679.88 Btu/lbm
Next, we can plug values
into Eqn 5 to evaluate Q34 :
Q34 -512.22 Btu/lbm
We can now plug values into Eqn 1 to evaluate the thermal efficiency of the Rankine Cycle.
ηth 3.62%
The maximum thermal efficiency for a power cycle operating between two thermal reservoirs at TH and TC is the Carnot Efficiency :
                 
Eqn 10
TC 505.67 oR
TH 539.67 oR ηmax 6.30%
Part b.) The power output of the cycle is the net shaft work rate for all of the processes that make up the cycle.  Shat work from the turbine is used to drive the seawater pumps and the ammonia pump, but no shaft work is consumed or produced in the boiler or condenser.  Therefore :
 
Eqn 11
We can use the given value of WS,turb with the 1st Law to detemine the mass flow rate of ammonia through the cycle, as follows.  The 1st Law, applied to the adiabatic turbine  with negligible changes in kinetic and potential energies, is :
   
Eqn 12 Solving for mdot : Eqn 13
Fortunately, we already know H2 and H3 from
part (a), so we can now plug values into Eqn 13 :
mNH4 4.18E+07 lbm / h
Now, we can apply the 1st Law to the adiabatic pump (with negligible changes in kinetic and potential energies).  The appropriate form of the 1st Law is:
Eqn 14
Plugging in values yields: Ws,pump -1.486E+07 Btu/h
Now, we have all the values we need
to evaluate Wcycle using Eqn 11 :
Wcycle 5.501E+08 Btu/h
Part c.) We can determine the flow rates of the sea water through the boiler and condenser by applying the 1st Law to just the seawater as it passes through each of these devices.
1st Law, steady-state, no shaft work, negligible changes in kinetic and potential energies…
Condenser seawater : Eqn 15
Boiler seawater : Eqn 16
We can now solve Eqns 15 & 16 for the mass flow rate of the seawater through the condenser and the boiler, respectively.
Eqn 17
 
Eqn 18
msw, con 1.261E+10 lbm/h   msw, boil 8.895E+09 lbm/h
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answer: a.) ηth 3.62% ηmax 6.30%
b.) Wcycle 5.501E+08 Btu/h
c.) msw, cond 1.261E+10 lbm/h msw, boil 8.895E+09 lbm/h
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