# Example Problem with Complete Solution

9C-1 : Ideal Rankine Cycle with Reheat 9 pts
Water is the working fluid in an ideal Rankine cycle with reheat. The steam at the high-pressure turbine inlet is at 1500 psia and 800oF and the effluent is saturated vapor.
The steam is reheated to 750oF 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 lbm/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 (WS)cycle, Qin, 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 P3 and P4 = P3.  T4 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 Wcycle.  Then, we apply the 1st Law to the boiler and the reheater to determine Qin.  Finally, we evaluate h from its definition.
Given: mdot 126 lbm/s T2 800 oF
4.54E+05 lbm/h T4 750 oF
P1 1500 psia P5 20 psia
P2 1500 psia P6 20 psia
Find: a.) Wcycle ??? Btu/h
b.) Q34 ??? Btu/h c.) h ??? %
Diagram:  Assumptions: 1 - Each component in the cycle is analyzed as an open system operating at steady-state.
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
(oF)
P
(psia)
X
(lbm vap/lbm)
H
(Btu/lbm)
S
(Btu/lbm-oR)
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
(oF)
P
(psia)
X
(lbm vap/lbm)
H
(Btu/lbm)
S
(Btu/lbm-oR)
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 steady-state 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 Wcycle.
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.
T2 800 oF T6 227.92 oF
H2 1364.0 Btu/lbm H6 196.40 Btu/lbm
S2 1.5075 Btu/lbm-oR S6 0.33605 Btu/lbm-oR
Now, because the pump and HP turbine are isentropic, S1 = S6 and S3 = S2.
Now, we know the values of two intensive properties at state 1 and we know both S3 and x3, so we can fix these states and determine H1 and H3. 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 (oF) H Btu/lbm S Btu/lbm-
220 191.67 0.32241 Interpolation yields:
T1 H1 0.33605 T1 229.30 oF
230 201.71 0.33708 H1 201.01 Btu/lbm
Sat. Vap. : T (oF) P (psia) H Btu/lbm S Btu/lbm- Interpolation yields:
417.35 300 1204.1 1.5121 P3 316.18 psia
T3 P3 H3 1.5075 T3 422.18 oF
423.31 320 1204.7 1.5064 H3 1204.6 Btu/lbm
Now that we know P4, we can use it with T4 to fix state 4 and use the Steam Tables or NIST Webbook to evaluate H4 and S4.
H4 1395.1 Btu/lbm S4 1.6928 Btu/lbm-oR
We now know S4 and we know that the LP turbine is also isentropic, so S5 = S4.
S5 1.6928 Btu/lbm-oR
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
Tsat 227.92 oF Since Ssat liq < S5 < Ssat vap, state 5 is a saturated mixture.
Ssat liq 0.33605 Btu/lbm-oR
Ssat vap 1.7331 Btu/lbm-oR T5 227.92 oF
Determine x3S from the specific entropy, using: Eqn 4
x5 0.9712 lbm vap/lbm
Then, we can use the quality
to determine
H3S, using: Eqn 5
At 20 psia Hsat liq 196.400 Btu/lbm
Hsat vap 1106.2 Btu/lbm H5 1079.96 Btu/lbm
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.
WS,pump -2.090E+06 Btu/h
WS,LP-turb 1.430E+08 Btu/h
WS,HP-turb 7.233E+07 Btu/h Wcycle 2.132E+08 Btu/h
Part b.) The amount of heat absorbed in the reheat step is Q34.  We can determine it by applying the 1st Law to the reheater, Step 3-4.  The reheater operates at steady-state, 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 mdot and all the H values, we can immediately plug values into Eqn 6 :
Q34 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 Wcycle and Q34, so here we need to evaluate Q12 so we can use Eqn 7 to evaluate h.
We can determine Q12 by applying the 1st Law to the boiler, Step 1-2.  The boiler operates at steady-state, 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 mdot and all the H values, we can immediately plug values into Eqn 8 :
Q12 5.275E+08 Btu/h
Now, we can plug values into Eqn 7 : Qin 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.) Wcycle 2.13E+08 Btu/h
b.) Q34 8.65E+07 Btu/h c.) h 34.7% 