Example Problem with Complete Solution

9C-1 : 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 first-stage 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 (WS)cycle, Qin, 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 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 1.4E+06 lb/h T4 900 oF
P1 1200 psia P5 1 psia
P2 1200 psia P6 1 psia
T2 1000 oF
Find : a.) Wcycle ? Btu/h
b.) Q34 ? Btu/h
c.) η ? %
Diagrams:
Assumptions :
- Each component in the cycle is analyzed as an open system operating at steady-state.
- 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
(oF)
P
(psia)
H
(Btu/lbm)
S
(Btu/lbm-oR)
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 steady-state 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 Wcycle.
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.
T2 1000 oF T6 101.69 oF
H2 1501.10 Btu/lbm H6 69.769 Btu/lbm
S2 1.6313 Btu/lbm-oR S6 0.13271 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.
At 1200 psia:
T (oF)
H Btu/lbm
S Btu/lbm-
100
71.243
0.12896
T1
H1
0.13271
T1 102.12 oF
105
76.213
0.13780
H1 73.351 Btu/lbm
Sat Vap:
T (oF)
P (psia)
H Btu/lbm
S Btu/lbm-oR
302.91
70
1181.6
1.6330
T3
P3
H3
1.6313
312.02
80
1184.1
1.6223
P3 71.589 psia
T3 304.4 oF H3 1182.00 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 1482.70 Btu/lbm S4 1.9229 Btu/lbm-oR
We now know S4 and we know that the LP turbine is also isentropic.
So, S5 = S4.
S5 1.9229 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 1 psia :
Tsat 101.69 oF Since Ssat liq < S5 < Ssat vap, state 5 is a saturated mixture.
Ssat liq 0.13271 Btu/lbm-oR
Ssat vap 1.9789 Btu/lbm-oR T5 101.69 oF
Determine x3S from the specific entropy, using:
Eqn 4
x5 0.9697
Then, we can use the quality to determine H3S, using:
Eqn 9
At 1 psia :
Hsat liq 69.769 Btu/lbm
Hsat vap 1106.2 Btu/lbm H5 1074.76 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 wor for each device as well as the entire cycle.
WS,pump -5.015E+06 Btu/h
WS,LP-tur 5.711E+08 Btu/h
WS,HP-tur 4.467E+08 Btu/h Wcycle 1.013E+09 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 10
Since we already know mdot and all the H values, we can immediately plug values into Eqn 10 :
Q34 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 Wcycle and Q34, so here we need to evaluate Q12 so we can use Eqn 11 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 12
Since we already know mdot and all the H values, we can immediately plug values into Eqn 12 :
Q12 1.999E+09 Btu/h
Now, we can plug values into Eqn 11 : Qin 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.) Wcycle 1.01E+09 Btu/h
b.) Q34 4.21E+08 Btu/h
c.) η 41.9%