9B1 :  The Ideal Rankine Power Cycle  12 pts  
Water is the working fluid in a Rankine Cycle. Superheated vapor enters the turbine at 480°C. The net power output of the cycle is 100 MW. For this cycle :  
a.)  For a boiler pressure of 8 MPa, construct a plot of the mass flow rate of the working fluid, the rate of heat transfer into the working fluid in the boiler, the thermal efficiency of the cycle and the heat transfer rate out of the working fluid in the condenser as functions of condenser pressure over the range from 1 kPa and 100 kPa. Assume the cycle is an ideal Rankine Cycle.  
b.)  For a condenser pressure of 8 kPa, construct a plot of the rate of heat transfer into the working fluid in the boiler, the thermal efficiency of the cycle and the heat transfer rate out of the working fluid in the condenser as functions of boiler pressure over the range from 1 MPa and 25 MPa. Assume the cycle is an ideal Rankine Cycle.  
c.)  For a boiler pressure of 8 MPa and a condenser pressure of 8 kPa, determine the mass flow rate of the working fluid, the rate of heat transfer into the working fluid in the boiler, the thermal efficiency of the cycle and the heat transfer rate out of the working fluid in the condenser. The isentropic efficiencies of the pump and turbine are 70% and 85%, respectively.  
Read:  The key to parts (a) and (b) is that the cycle is an IDEAL Rankine Cycle. This means that the pump and turbine operate isentropically and that the condenser effluent is a saturated liquid.  
Constructing the plots in parts (a) and (b) requires looking up a lot of data. This can be done most efficiently using the Thermal/Fluids Toolbox plugin for Excel. This is the method I used for this solution. The next best method would be to use the NIST Webbook to generate data tables that can be copied and pasted into Excel.  
Given:  T_{2}  480  ^{o}C  W_{sh}  100  MW  
Part a.)  Part b.)  
P_{3} = P_{4}  8  kPa  P_{1} = P_{2}  8000  kPa  
Part c.)  
P_{1} = P_{2}  8  MPa  η_{s,pump}  70%  
P_{3} = P_{4}  8  kPa  η_{s,turb}  85%  
Find:  Part a.)  Part b.)  
Q_{H}  ???  MW  Q_{H}  ???  MW  
η_{th}  ???  η_{th}  ???  
Q_{C}  ???  MW  Q_{C}  ???  MW  
P_{1} = P_{2}  1 to 25  MPa  P_{3} = P_{4}  1 to 100  kPa  
Part c.)  
Q_{H}  ???  MW  
h_{th}  ???  
Q_{C}  ???  MW  
Diagrams : 




Assumptions:  
1   The pump and the turbine are adiabatic and reversible and, therefore, isentropic.  
2   Only flow work crosses the boundary of the boiler and condenser.  
Equations / Data / Solve :  
Part a.)  Let's begin with a very detailed analysis of the problem for a single condenser pressure of 8 kPa.  
Then, we can present a table of results for all of the other condenser pressures so we can construct the plot that is required.  
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.  
State  T (^{o}C) 
P (kPa) 
X 
H (kJ/kg) 
S (kJ/kgK) 

1  Sub Liq  41.76 
8000 
N/A 
181.46 
0.5915 

2  Super Vap  480 
8000 
N/A 
3347.9 
6.6576 

3  VLE  41.51 
8 
0.7944 
2082.4 
6.6576 

4  Sat Liquid  41.51 
8 
0 
173.65 
0.5915 

Additional data that may be useful.  
Sat Vap  41.51 
8 
1 
2576.5 
8.2278 

Sat Liquid  41.51 
8 
0 
173.6 
0.5915 

Sat Vap  295.06 
8000 
1 
2757.5 
5.7423 

Sat Liquid  295.06 
8000 
0 
1317.0 
3.2072 

The values in the table that are shown in bold with a shaded background are the values we will determine.  
One approach to solving cycle problems of this nature is to work your way around the cycle until you have evaluated all the properties to complete the table shown above. Then, you can go back and apply the 1st Law to each process in the cycle to evaluate Q and W_{s} as need. That is the approach I will take.  
In this problem it makes the most sense to begin at either state 2 or state 4 because these states are completely fixed and we can look up all of the relevant properties. I will begin at state 2.  
Because the turbine is isentropic, I know that S_{3} = S_{2} :  
S_{3}  6.6576  kJ/kgK  
Now, we know the values of two intensive properties at state 3, so this state is completely fixed and we can look up all of its properties. Because S_{sat liq} < S_{3} < S_{sat vap} , state 3 is a twophase mixture, T_{3} = T_{sat} at 8 kPa and we must determine the quality so that we can determine the enthalpy.  

Eqn 1  
T_{3}  41.51  ^{o}C  
x_{3}  0.7944  kg vap/kg  

Eqn 2  
H_{3}  2082.4  kJ/kg  
We can look up all the properties at state 4 because the state is completely fixed by the fact that the fluid is a saturated liquid leaving the condenser in an Ideal Rankine Cycle and because we know the pressure is 8 kPa.  
Next, we proceed to state 1 using the fact that the pump is also isentropic: S_{1} = S_{4}.  
S_{1}  0.5915  kJ/kg  
Now, we know the values of two intensive properties at state 4, so this state is completely fixed and we can look up all of its properties. Because S_{4} < S_{sat liq} , state 4 is a subcooled liquid. We can look up its properties in the subcooled liquid property table for water, the NIST Webbook or we can use the Thermal/Fluids Toolbox (TFT). The data tables and NIST Webbook require interpolation, but the TFT does not !  
At 8 MPa :  
T (^{o}C)  H (kJ/kg)  S(kJ/kgK)  
40  174.14  0.56846  
T_{1}  H_{1}  0.59150  T_{1}  41.76  ^{o}C  
50  215.71  0.69915  H_{1}  181.46  kJ/kg  
Now, we have all the information we need to apply the 1st Law to each device in the cycle to determine Q and W for each device. This process is made easier by our assumptions that no shaft work crosses the boundary of the boiler and condenser and that both the pump and the turbine are adiabatic. The four relevant forms of the 1st law are :  

Eqn 4  
Boiler : 

Eqn 5  
Turbine : 

Eqn 6  
Condenser : 

Eqn 7  
Pump : 

Eqn 8  
Plugging values into Eqns 5  8 yields :  
Q_{boil}  3166.4  kJ/kg  W_{turb}  1265.4  kJ/kg  
Q_{cond}  1908.8  kJ/kg  W_{pump}  7.818  kJ/kg  
W_{cycle}  1257.6  kJ/kg  
Now, we can determine the mass flow rate from the given value of the power output of the cycle and use the mass flow rate to evaluate the heat transfer rates in the condenser and boiler.  

Eqn 9  Eqn 10  
m_{dot}  79.52  kg/s  Q_{boil}  251.8  MW  
Q_{cond}  151.8  MW  
Finally, we can calculate the thermal efficiency of this cycle.  

Eqn 11  
η_{th}  39.72% 