9C1 :  Ideal Rankine Cycle with Reheat  9 pts  
Water is the working fluid in an ideal Rankine cycle. The pressure and temperature at the turbine inlet are 1200 lbf/in2 and 1000°F, respectively, and the condenser pressure is 1 lbf/in2. The mass flow rate of steam entering the turbine is 1.4 X 106 lbm/h. The cooling water experiences a temperature increase from 60 to 80°F, with negligible pressure drop, as it passes through the condenser. The ideal Rankine cycle is modified to include reheat. In the modified cycle, steam expands through the firststage turbine to saturated vapor and then is reheated to 900°F. If the mass flow rate of steam in the modified cycle is the same as in ideal Rankine cycle, determine for the modified cycle:  
a.)  the net power developed, in Btu/h.  
b.)  the rate of heat transfer to the working fluid in the reheat process, in Btu/h.  
c.)  the thermal efficiency.  
Read:  Determine the specific enthalpy at each point in the process and then use the 1st Law to calculate (W_{S})_{cycle}, Q_{in}, and η.  
States 2 & 6 are completely determined from the given information. Use the fact that the pump and HP turbine are isentropic to fix states 1 & 3 respectively. Once state 3 is fixed, you know P_{3} and P_{4} = P_{3}. T_{4} is given, so state 4 is now fixed. Next, use the fact that the LP turbine is also isentropic to fix state 5.  
Once we know all the H values, we apply the 1st Law to the pump and turbines to determine W_{cycle}. Then, we apply the 1st Law to the boiler and the reheater to determine Q_{in}. Finally, we evaluate h from its definition.  
Given:  m_{dot}  1.4E+06  lb/h  T_{4}  900  ^{o}F  
P_{1}  1200  psia  P_{5}  1  psia  
P_{2}  1200  psia  P_{6}  1  psia  
T_{2}  1000  ^{o}F  
Find :  a.)  W_{cycle}  ?  Btu/h  
b.)  Q_{34}  ?  Btu/h  
c.)  η  ?  %  
Diagrams: 




Assumptions :  
 Each component in the cycle is analyzed as an open system operating at steadystate.  
 All of the processes are internally reversible.  
 The turbine and pump operate adiabatically and, therefore, isentropically.  
 Condensate exits the condenser as saturated liquid.  
 The effluent from the HP turbine is a saturated vapor.  
 No shaft work in the boiler or condenser.  
 Changes in kinetic and potential energies are negligible.  
Equations / Data / Solve :  
Stream  T (^{o}F) 
P (psia) 
H (Btu/lb_{m}) 
S (Btu/lb_{m}^{o}R) 
Phase  
1  102.1 
1200 
73.35 
0.1327 
Sub. Liq.  
2  1000 
1200 
1501.1 
1.6313 
Super. Vap.  
3  304.4 
71.59 
1182.0 
1.6313 
Sat'd Vap.  
4  900 
71.59 
1482.7 
1.9229 
Super. Vap.  
5  101.7 
1 
1074.8 
1.9229 
VLE  
6  101.7 
1 
69.77 
0.1327 
Sat'd Liq.  
Part a.)  The net shaft work for the reheat cycle is:  

Eqn 1  
Now, apply the 1st Law to the LP and HP turbine, as well as the pump.  
Assume each device is adiabatic, operating at steadystate and has negligible changes in kinetic and potential energies.  

Eqn 2  

Eqn 3  
So, we need to determine the enthalpy at every point in the cycle in order to determine W_{cycle}.  
States 2 & 6 are the only ones that are completely determined by the given information, so let's look up those properties in the steam tables or NIST Webbook first.  
T_{2}  1000  ^{o}F  T_{6}  101.69  ^{o}F  
H_{2}  1501.10  Btu/lb_{m}  H_{6}  69.769  Btu/lb_{m}  
S_{2}  1.6313  Btu/lb_{m}^{o}R  S_{6}  0.13271  Btu/lb_{m}^{o}R  
Now, because the pump and HP turbine are isentropic, S_{1} = S_{6} and S_{3} = S_{2}.  
Now, we know the values of two intensive properties at state 1 and we know both S_{3} and x_{3}, so we can fix these states and determine H_{1} and H_{3}.  
At 1200 psia:  
T (^{o}F) 
H Btu/lb_{m} 
S Btu/lb_{m} 

100 
71.243 
0.12896 

T_{1} 
H_{1} 
0.13271 
T_{1}  102.12  ^{o}F  
105 
76.213 
0.13780 
H_{1}  73.351  Btu/lb_{m}  
Sat Vap:  T (^{o}F) 
P (psia) 
H Btu/lb_{m} 
S Btu/lb_{m}^{o}R 

302.91 
70 
1181.6 
1.6330 

T_{3} 
P_{3} 
H_{3} 
1.6313 

312.02 
80 
1184.1 
1.6223 

P_{3}  71.589  psia  
T_{3}  304.4  ^{o}F  H_{3}  1182.00  Btu/lb_{m}  
Now that we know P_{4}, we can use it with T_{4} to fix state 4 and use the steam tables or NIST Webbook to evaluate H_{4} and S_{4}.  
H_{4}  1482.70  Btu/lb_{m}  S_{4}  1.9229  Btu/lb_{m}^{o}R  
We now know S_{4} and we know that the LP turbine is also isentropic.  
So, S_{5} = S_{4}.  
S_{5}  1.9229  Btu/lb_{m}^{o}R  
We now know the values of two intensive properties at state 5, so the state is fixed and we can evaluate its properties. We begin by determining which phases are present in state 5.  
At 1 psia :  
T_{sat}  101.69  ^{o}F  Since S_{sat liq} < S_{5} < S_{sat vap}, state 5 is a saturated mixture.  
S_{sat liq}  0.13271  Btu/lb_{m}^{o}R  
S_{sat vap}  1.9789  Btu/lb_{m}^{o}R  T_{5}  101.69  ^{o}F  
Determine x_{3S} from the specific entropy, using:  

Eqn 4  
x_{5}  0.9697  
Then, we can use the quality to determine H_{3S}, using:  

Eqn 9  
At 1 psia :  
H_{sat liq}  69.769  Btu/lb_{m}  
H_{sat vap}  1106.2  Btu/lb_{m}  H_{5}  1074.76  Btu/lb_{m}  
At last we know all the properties at all of the states in the reheat cycle and we can use Eqns 1  3 to evalaute the shaft wor for each device as well as the entire cycle.  
W_{S,pump}  5.015E+06  Btu/h  
W_{S,LPtur}  5.711E+08  Btu/h  
W_{S,HPtur}  4.467E+08  Btu/h  W_{cycle}  1.013E+09  Btu/h  
Part b.)  The amount of heat absorbed in the reheat step is Q_{34}. We can determine it by applying the 1st Law to the reheater, Step 34. The reheater operates at steadystate, has no shaft work interaction and has negligible changes in kinetic and potential energies. The appropriate form of the 1st Law is:  

Eqn 10  
Since we already know m_{dot} and all the H values, we can immediately plug values into Eqn 10 :  
Q_{34}  4.210E+08  Btu/h  
Part c.)  The thermal efficiency of a power cycle is defined by:  

Eqn 11  
In parts (a) and (b) we determined W_{cycle} and Q_{34}, so here we need to evaluate Q_{12} so we can use Eqn 11 to evaluate h.  
We can determine Q_{12} by applying the 1st Law to the boiler, Step 12. The boiler operates at steadystate, has no shaft work interaction and has negligible changes in kinetic and potential energies. The appropriate form of the 1st Law is:  

Eqn 12  
Since we already know m_{dot} and all the H values, we can immediately plug values into Eqn 12 :  
Q_{12}  1.999E+09  Btu/h  
Now, we can plug values into Eqn 11 :  Q_{in}  2.420E+09  Btu/h  
η  41.86%  
Verify :  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  a.)  W_{cycle}  1.01E+09  Btu/h  
b.)  Q_{34}  4.21E+08  Btu/h  
c.)  η  41.9% 