9C1 :  Ideal Rankine Cycle with Reheat  9 pts 

Water is the working fluid in an ideal Rankine cycle with reheat. The steam at the highpressure turbine inlet is at 1500 psia and 800^{o}F and the effluent is saturated vapor.  
The steam is reheated to 750^{o}F before it enters the low pressure turbine where the steam is let down to 20 psia. If the mass flow rate of steam is 126
lb_{m}/s, determine... 

a.) The net power output in million Btu per hour (mmBtu/h) b.) The heat transfer rate in the reheat process in mmBtu/h c.) The thermal efficiency of the cycle 

Read :  Determine the specific enthalpy of each stream in the process and then use the 1st Law to calculate (W_{S})_{cycle}, Q_{in}, and h.  
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}  126  lb_{m}/s  T_{2}  800  ^{o}F  
4.54E+05  lb_{m}/h  T_{4}  750  ^{o}F  
P_{1}  1500  psia  P_{5}  20  psia  
P_{2}  1500  psia  P_{6}  20  psia  
Find:  a.)  W_{cycle}  ???  Btu/h  
b.)  Q_{34}  ???  Btu/h  c.)  h  ???  %  
Diagram:  



Assumptions:  1   Each component in the cycle is analyzed as an open system operating at steadystate.  
2   All of the processes are internally reversible.  
3   The turbine and pump operate adiabatically and are internally reversible, so they are also isentropic.  
4   Condensate exits the condenser as saturated liquid.  
5   The effluent from the HP turbine is a saturated vapor.  
6   No shaft work crosses the system boundary of the boiler or condenser.  
7   Changes in kinetic and potential energies are negligible.  
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  T (^{o}F) 
P (psia) 
X (lb_{m} vap/lb_{m}) 
H (Btu/lb_{m}) 
S (Btu/lb_{m}^{o}R) 
Phase  
1  229.30  1500  N/A  201.00  0.33605  Sub. Liq.  
2  800  1500  N/A  1364.0  1.5075  Super. Vap.  
3  422.18  316.11  1  1204.6  1.5075  Sat'd Vap.  
4  750  316.11  N/A  1395.1  1.6928  Super. Vap.  
5  227.92  20  0.9712  1129.3  1.6928  VLE  
6  227.92  20  0  196.40  0.33605  Sat'd Liq.  
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  0  1500  1  1169.8  1.3372  
Sat Liquid  0  1500  0  612.08  0.80900  
Sat Vap  0  20  1  1157.0  1.7331  
Sat Liquid  0  20  0  196.40  0.33605  
Part a.)  The net shaft work for the reheat cycle is: 

Eqn 1  
Now, apply the 1st Law to the LP and HP turbines, 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 in every stream in the cycle in order to determine W_{cycle}.  
States 2 & 6 are the only streams that are completely determined by the given information, so let's look up the properties of those streams in the Steam Tables or NIST Webbook first.  
T_{2}  800  ^{o}F  T_{6}  227.92  ^{o}F  
H_{2}  1364.0  Btu/lb_{m}  H_{6}  196.40  Btu/lb_{m}  
S_{2}  1.5075  Btu/lb_{m}^{o}R  S_{6}  0.33605  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}. Because stream 1 is a subcooled liquid, it is easiest and most accurate to use the NIST Webbook instead of the Steam Tables.  
At 1500 psia:  T (^{o}F)  H Btu/lb_{m}  S Btu/lb_{m}  
220  191.67  0.32241  Interpolation yields:  
T_{1}  H_{1}  0.33605  T_{1}  229.30  ^{o}F  
230  201.71  0.33708  H_{1}  201.01  Btu/lb_{m}  
Sat. Vap. :  T (^{o}F)  P (psia)  H Btu/lb_{m}  S Btu/lb_{m}  Interpolation yields:  
417.35  300  1204.1  1.5121  P_{3}  316.18  psia  
T_{3}  P_{3}  H_{3}  1.5075  T_{3}  422.18  ^{o}F  
423.31  320  1204.7  1.5064  H_{3}  1204.6  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}  1395.1  Btu/lb_{m}  S_{4}  1.6928  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}  1.6928  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 20 psia  
T_{sat}  227.92  ^{o}F  Since S_{sat liq} < S_{5} < S_{sat vap}, state 5 is a saturated mixture.  
S_{sat liq}  0.33605  Btu/lb_{m}^{o}R  
S_{sat vap}  1.7331  Btu/lb_{m}^{o}R  T_{5}  227.92  ^{o}F  
Determine x_{3S} from the specific entropy, using: 

Eqn 4  
x_{5}  0.9712  lb_{m} vap/lb_{m}  
Then, we can use the quality to determine H_{3S}, using: 

Eqn 5  
At 20 psia  H_{sat liq}  196.400  Btu/lb_{m}  
H_{sat vap}  1106.2  Btu/lb_{m}  H_{5}  1079.96  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 work for each device as well as the entire cycle.  
W_{S,pump}  2.090E+06  Btu/h  
W_{S,LPturb}  1.430E+08  Btu/h  
W_{S,HPturb}  7.233E+07  Btu/h  W_{cycle}  2.132E+08  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 6  
Since we already know m_{dot} and all the H values, we can immediately plug values into Eqn 6 :  
Q_{34}  8.645E+07  Btu/h  
Part c.)  The thermal efficiency of a power cycle is defined by: 

Eqn 7  
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 7 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 8  
Since we already know m_{dot} and all the H values, we can immediately plug values into Eqn 8 :  
Q_{12}  5.275E+08  Btu/h  
Now, we can plug values into Eqn 7 :  Q_{in}  6.140E+08  Btu/h  
h  34.72%  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  a.)  W_{cycle}  2.13E+08  Btu/h  
b.)  Q_{34}  8.65E+07  Btu/h  c.)  h  34.7% 