This document describes a project to optimize the performance of a steam turbine power plant using the Rankine cycle. Five different Rankine cycles are analyzed: 1) the ideal Rankine cycle, 2) the ideal Rankine cycle with reheat, 3) the ideal Rankine cycle with one open feedwater heater, 4) with two open feedwater heaters, and 5) with four open feedwater heaters. Sample calculations are shown for each cycle configuration to determine the thermal efficiency. The document also discusses cost effectiveness calculations to determine the most optimal design for the steam turbine.

1. James Li # 26
3/27/2015
Raj, Rishi
ME 43000
Spring 2015
Thermal Project # 1: Optimal Thermal Cycle for Steam Turbine Power Plant
1

2. Abstract: The purpose of this project is to make a design to optimize the performance a steam turbine using the
ideal Rankine cycle. For this project, five different Rankine cycles will be used to analyze the thermal
efficiency of the turbine with a minimum moisture content of 13 %. The given inlet power and inlet pressure of
150,000 KW and 2000 PSI respectively and the reheat temperature maximizes at 1000 degrees Fahrenheit. The
turbine will analyze first with the standard Ideal Rankine cycle, followed by adding another turbine using the
same cycle with a reheat, and finally by adding three water heaters in a process known as regeneration that
conserves fuel and improve work output. The efficiency and cost effectiveness will then be calculated to
determine the best design for the steam turbine.
Table of Contents
Nomenclature ……………………………………………………………………………………………. Pages 2-3
Introduction to the Steam Turbine …………………………………………………………………… Pages 3-4
Rankine Cycle Background……………………………………………………………………………… Pages 4-5
Sample Calculations …………………………………………………………………………………… Pages 5-22
• Step 1: Ideal Rankine Cycle [page 6]
• Step 2: Ideal Rankine Cycle with Reheat [page 7]
• Step 3: Ideal Rankine Cycle with Regeneration (cycle 3: one open feed water heater) [page 11]
• Step 3: Ideal Rankine Cycle with Regeneration (cycle 4: two open feed water heater) [page 14]
• Step 3: Ideal Rankine Cycle with Regeneration (cycle 5: four open feed water heater) [page 17]
• Cost Effectiveness [Page 22]
Excel Graphs and Spreadsheets…………………………………………………………………………… Pages 22 - 28
Discussions and conclusions………………………………………………………………………………Pages 28 - 30
Appendix ………………………………………………………………………………………………………
References …………………………………………………………………………………………………
Nomenclature
P = Pressure (Psi)
2

3. T = Temperature (degrees Fahrenheit)
h = enthalpy (btu/lbm)
s = entropy (btu/lbm.R)
v = specific volume (ft^3/lbm)
x = Steam Quality
m = mass flow rate (lbm/s)
q (in) = entering heat (btu/lbm)
q (out) = exiting heat (btu/lbm)
Wp = Work done by the pump (btu/lbm)
η = efficiency
f = saturated liquid state
g = saturated vapor state
fg = difference between saturated liquid state and saturated vapor state
Introduction to the Steam Turbine
The purpose of this design project is to find the most cost effective and efficient steam turbine cycle for
operating a steam turbine power plant. A steam turbine is a machine that provides mechanical energy by
converting pressurized steam into said energy when the steam hits the turbine’s rotating blades. When the
pressurized steam flows past and hits the rotating blade, it cools and transfers its thermal energy into the turbine
and rotates the shaft up to 3600 rpm to power the electricity generator. The thermal energy generated from the
cooling steam can extract the mechanical work needed to operate the electricity generators via power cycles
such as the Rankine, Rankine Reheat, and regenerative cycles. A steam turbine is also compact due to having
the steam flow spin the turbine blade continuously and expanding the steam in order to drive a machine. This
negates the need for a push-pull action or a piston operation, making the turbine useful for operating machines
where space is limited, such as on a ship.
The cycle used for this turbine would be the Ideal Rankine cycle and the cycle will be divided into
multiple stages in order to find the most efficient stage to operate the power plant. For this design, the turbine
3

4. will operate in three stages. The first stage of this turbine is based on the ideal Rankine cycle with a steam
quality minimum at around 87% or x = 0.87. The second stage of this turbine involves implementing a reheat to
the Rankine cycle of turbine. The cycle reheat involves separating the cycle into a high pressure stage and a low
pressure stage where the steam will be reheated during the high pressure stage prior to entering the low pressure
stage. The third stage consists of three different cycles. The first cycle of the third stage involves adding an
open feed water heater to the high pressure turbine. The second cycle involves adding another open feed water
heater to the intermediate pressure turbine and the third and final cycle involves adding two open feed water
heater to the low pressure turbine. For each stage, the efficiency will be calculated, analyzed and compared with
the other stages, after which the income of the stages will be calculated in order to determine whether or not the
improvements in efficiency would be worth designing a particular stage and cycle of the turbine.
Rankine Cycle Background
As mentioned earlier, the cycle used to operate the steam turbine power plant is the Rankine cycle. The
Rankine cycle is a variant of the Carnot cycle that is less efficient but more practical. The Carnot cycle is a
highly efficient vapor power cycle consisting of two isentropic (1  2 and 3  4) and two isothermal (2  3
and 4  1) processes as shown below:
Figure 1: Carnot Cycle
4

5. Despite this cycle’s efficiency, its process is highly idealized due to its pumping process being a mixture of
liquid and vapor. Modern technology cannot yet handle a liquid-vapor mixture process yet. The moisture
content generated from the steam quality is too high and corrodes the turbine’s blade rapidly. These issues make
the cycle not achievable in real life thus, this is where the Rankine cycle comes into play. The Rankine cycle
eliminates the liquid-vapor mixture and moisture content problems by vaporizing the steam at a constant
pressure instead of at a constant temperature like the Carnot cycle does. This means that the steam is
superheated in the boiler and condenses in the condenser so the liquid-vapor do not mix and the moisture
content is reduced. While this makes the Rankine Cycle more practical in real life, it needs some adjustments to
achieve its optimal efficiency and that is where its stages come in. The Parameters for this Particular design is
given below.
• Power Plant output: 150,000 KW
• Turbine inlet Pressure: 2000 Psi
• Reheat inlet temperature: 1000 degrees Fahrenheit
• Maximum moisture level (1-x): 13% or 0.13
The calculations will explain the steps necessary to achieve the highest efficiency for each stage with the given
parameters
Sample Calculations:
5

6. Step 1: Ideal Rankine Cycle
Figure 2: The Ideal Rankine Cycle
Cycle Steps
• 1  2: Isentropic Pump Compression (s1=s2)
• 2  3: Constant pressure heat addition inside of the boiler (P2=P3)
• 3  4: Isentropic turbine steam expansion (s3=s4)
• 4  1: Constant pressure heat addition inside of the condenser (P4=P1)
State 1 Test Pressure: P1 = 1 Psi (first value)
v1 = 0.01614 ft3/lbm (Appendix 2 Table A-5E)
h1 = 69.72 Btu/lbm (Appendix 2 Table A-5E)
State 2 P2 = 2000 Psi
wp (work) = v1*(P2-P1)*(144/778) = 5.971717 Btu/lbm
h2 = h1 + wp = 75.69172 Btu/lbm
State 3 P3 = P2 = 2000 Psi
T3 = 1000 degrees Fahrenheit
h3 = 1474.9 Btu/lbm (Appendix 2 Table A-6E)
s3 = 1.5606 Btu/(lbm*R)
State 4 P4 = P1 = 1 Psi
s4 = s3 = 1.5606 Btu/(lbm*R) (Appendix 2 Table A-6E)
6

7. sg4 = 1.9776 (Appendix 2 Table A-5E)
sfg4 = 1.84496 (Appendix 2 Table A-5E)
hg4 = 1105.4 Btu/lbm (Appendix 2 Table A-5E)
hfg4 = 1035.7 (Appendix 2 Table A-5E)
1-x4 (moisture content) = (sg4-s4)/(sfg4) = 0.2260
h4 = hg4 – (1-x4)*hfg4 = 871.3086 Btu/lbm
x4 (Steam quality) = 1 – (1-x4) = 0.7740
efficiency η = 1 – (h4-h1)/(h3-h2) = 0.4271 = 42.71 %
The process shown in the chart above is then repeated for test pressures 5, 10, 15, 20, 25, and 30 Psi and the
efficiency would be calculated accordingly (see table 1). Based on the trends shown (see Figures 6 and 7),
steam quality increases and efficiency decreases as pressure goes up The exit pressure used to proceed to the
next step and the efficiency of this cycle would be interpolated at steam quality: x4 = 0.87 (87%). The exit
pressure and efficiency interpolated at x4 = 0.87 was approximately 17 Psi and 34.20 % percent respectively.
With step 1 calculated, it is time to proceed to step 2 to improve upon this cycle.
Step 2: Ideal Rankine Cycle with Reheat
Figure 3: Ideal Rankine Cycle with Reheat
The Ideal Rankine Cycle with reheating involves adding another turbine to the previous cycle. For this stage,
the steam reenters the boiler and passes through the second turbine after passing through the first turbine (state
7

8. 4) with excess moisture eliminated. The second turbine has a pressure at around 10 – 40 percent ([0.1-0.4] *
Turbine inlet pressure) of the first turbine. The reheating also improves the steam quality due to the decrease in
excess moisture, therefore improving operational life and efficiency.
1  2: Isentropic pump compression (s1=s2)
2  3: Constant pressure heat addition inside of the boiler (P2=P3)
3  4: Isentropic expansion inside of the first high pressure turbine (s3=s4)
4  5: Constant pressure heat addition inside of the boiler (P4=P5)
5  6: Isentropic expansion inside of the second low pressure turbine (s5=s6)
6  1: Constant pressure heat rejection inside of the condenser (P6=P1)
Case 1: Constant turbine exit pressure at x = 0.87  17 Psi and variable reheat pressure ([0.1-0.4] * Turbine
inlet pressure)
State 1 P1 = 17 Psi
h1 = 187.21 Btm/lbm (Interpolated from table A-5E)
v1 = 0.016764 ft3/lbm (Interpolated from table A-5E)
sf = 0.322548 Btu/(lbm*R) (Interpolated from table A-5E)
sfg = 1.4154 Btu/(lbm*R) (Interpolated from table A-5E)
hfg = 965.654 Btu/lbm (Interpolated from table A-5E)
State 2 P2 = 2000 Psi
h2 = h1 + wp = h1 + v1*(P2-P1)*(144/778) = 193.363 Btu/lbm
State 3 P3 = P2 = 2000 Psi
T3 = 1000 degrees Fahrenheit
h3 = 1474.9 Btu/lbm (extracted from table A-6E)
s3 = 1.5606 Btu/(lbm*R) (extracted from table A-6E)
State 4 P4 = 200 Psi (first value)
s4 = s3 = 1.5606 Btu/(lbm*R)
h4 = 1211.253 Btu/lbm (interpolated from table A-6E)
State 5 P5 = P4 – 200 Psi
T5 = T3 = 1000 degrees Fahrenheit
8

9. h5 = 1529.6 Btu/lbm (extracted from table A-6E)
s5 = 1.843 Btu/(lbm*R) (extracted from table A-6E)
State 6 P6 = P1 = 17 Psi
s6 = s5 = 1.843 Btu/lbm
x6 = (s6-sf)/sfg = 1.07422
h6 = hf + x6*hfg = 1224.53553 Btu/lbm
Efficiency q(in) = (h3-h2) + (h5-h4) = 1599.884 Btu/lbm
q(out) = h6-h1 = 1037.32553 Btu/lbm
η = 1 – [q(out)/q(in)] = 0.351625
The process shown in the chart above is then repeated for P4 = 275, 350, 400, 450, 500, 600, 700, and 800 Psi
and the efficiency is calculated accordingly (see table 2). The optimal Efficiency for this process was 35.733 %
at exit pressure 500 Psi (see Figure 8). The pressure at optimal efficiency will then be used for step 2 case 2.
Case 2: Retaining exit pressure from Case 1 and changing the exit pressure to obtain efficiency at x = 0.87
State 1 P1 = 1 Psi (First value)
h1 = 69.72 Btu/lbm (extracted from table A-5E)
hfg = 1035.7 Btu/lbm (extracted from table A-5E)
v1 = 0.01614 ft3/lbm (extracted from table A-5E)
State 2 P2 = 2000 Psi
wp (work) = v1*(P2-P1)*(144/778) = 5.971717 Btu/lbm
h2 = h1 + wp = 75.69172 Btu/lbm
State 3 P3 = P2 = 2000 Psi
T3 = 1000 degrees Fahrenheit
h3 = 1474.9 Btu/lbm (extracted from table A-6E)
s3 = 1.5606 Btu/(lbm*R)
State 4 P4 = 500 psi
s4 = s3 = 1.5606 Btu/(lbm*R)
T4 = 602.9091 degrees Fahrenheit (interpolated from table A-6E)
h4 = 1300.334 Btu/lbm (interpolated from table A6-E)
9

10. State 5 P5 = P4 = 500 Psi
T5 = T3 = 1000 degrees Fahrenheit
s5 = 1.7376 Btu/(lbm*R) (extracted from Table A-6E)
h5 = 1521 Btu/lbm (extracted from Table A-6E)
sf = 0.13262 Btu/(lbm*R) (extracted from Table A-6E)
sfg = 1.84495 Btu/(lbm*R) (extracted from Table A-6E)
State 6 P6 = P1 = 1 Psi
x = (s6-sf)/sfg = 0.8669931
h6 = h1 + (x*hfg) = 970.708 Btu/lbm
Efficiency q(in) = (h3-h2)/(h5-h4) = 1619.874 Btu/lbm
q(out) = h6 – h1 = 900.988 Btu/lbm
η = 1 – [q(out)/q(in)] = 0.443791
The process from the case 2 chart was then repeated for P1 = 2  6 Psi and the efficiency is calculated
accordingly (see table 3). The optimal Efficiency and Exhaust pressure for this process was 44.37 % and 1 Psi
at x = 0.87 respectively (see Figures 9 and 10). After the efficiency and exhaust pressures are calculated, Step 3
will be initiated using the exhaust pressure as P1 in order to improve efficiency.
Step 3: Ideal Rankine Cycle with Regeneration (cycle 3: one open feed water heater)
10

11. Figure 4: Ideal Rankine Cycle with Regeneration (one water heater)
The first Regenerative Ideal Rankine Cycle involves adding an open feed water heater to the high
pressure turbine of the reheated Rankine cycle in order to further boost the cycle’s efficiency. By adding an
open feed water heater, the average temperature also increases. The temperature increase reduces the moisture
content, thus limiting potential moisture damage to the turbine. The process of the Regenerative Ideal Rankine
Cycle with one open feed water heater is shown on the next page.
1  2: Isentropic compression in the first pump
2  3: Constant pressure heat addition inside the open feed water heater
3  4: Isentropic compression in the second pump
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12. 4  5: Constant pressure heat addition inside the boiler
5  6: Isentropic steam expansion in high pressure turbine
6  3: Constant pressure heat regeneration at the open feedwater heater
6  7: Constant pressure heat addition inside the boiler
7  8: Isentropic steam expansion inside of the low pressure turbine
8 1: Constant pressure heat rejection inside of the condenser
State 1
P1 = 1 Psi (from step 2 case 2)
h1 = 69.72 Btu/lbm (extracted from table A-5E)
hfg = 1035.7 Btu/lbm (extracted from table A-5E)
v1 = 0.01614 ft3/lbm (extracted from table A-5E)
State 2
P2 = 500 Psi (from step 2 case 1)
wp = v1*(P2-P1)*(144/778) = 1.490689 Btu/lbm
h2 = h1 + wp = 71.2107 Btu/lbm
State 3
P3 = P2 = 500 Psi
h3 = 449.51 Btu/lbm (extracted from table A-5E)
v3 = 0.01975 ft3/lbm (extracted from table A-5E)
s3 = 0.649 Btu/(lbm*R) (extracted from table A-5E)
State 4
P4 = 2000 Psi
wp2 = v3*(P4-P3)*(144/778) = 5.48329 Btu/lbm
h4 = h3 + wp2 = 454.9932905 Btu/lbm
State 5
P5 = 2000 Psi
T5 = 1000 degrees Fahrenheit
s5 = 1.5606 (table A-6E)
State 6
P6 = 500 Psi (from cycle 2 case 2)
s6 = s5 = 1.5606 Btu/(lbm*R)
h6 = 1300.333818 Btu/lbm (interpolated from Table A-6E)
State 7
P7 = P6 = 500 Psi
T7 = 1000 degrees Fahrenheit
h7 = 1521 Btu/lbm (Table A-6E)
12

13. s7 = 1.7376 Btu/(lbm*R) (Table A-6E)
State 8
P8 = P1 = 1 Psi
s8 = s7 = 1.7376 Btu/lbm
sf8 = 0.13263 Btu/lbm (table A-5E)
sfg8 = 1.85595 (table A-5E)
x8 = (s8-sf8)/(sfg8) = 0.8699931
h8 = h1 + x8*hfg = 970.708 Btu/lbm
efficiency
y = (h3-h2)/(h6-h2) = 0.30778 (fraction of steam extracted)
q (in) = (h3-h4) + (1-y) * (h7-h6) = 1172.656 Btu/lbm
q (out) = (1-y) * (h8-h1) = 623.6821 Btu/lbm
η = 1 – [q(out)/q(in)] = 0.468146 = 46.8146 %
Mass Flow
Rate
Wnet = q (in) – q(out) = 548.974 Btu/lbm
m = Power*3412/Wnet = 932284.1883 lbm/hr
The efficiency showed an improvement over the previous cycle at x = 0.87, giving an efficiency of 46.8146 %.
The Mass Flow Rate was also calculated to be 932284.1883 lbm/hr (see table 4).
Step 3: Ideal Rankine Cycle with Regeneration (cycle 4: two open feed water heater)
13

14. Figure 5: Ideal Rankine Cycle with Regeneration (two water heaters)
The second Regenerative Ideal Rankine Cycle involves adding another open feed water heater to the
intermediate pressure turbine of the reheated Rankine cycle to give another enhancement to the previous cycle’s
efficiency. The process of this cycle is shown below:
1  2: Isentropic compression inside the first pump
2  3: Constant pressure heat addition inside of the second open feed water heater
3  4: Isentropic compression inside of the second pump
4  5: Constant pressure heat addition inside of the first open feed water heater
5  6: Isentropic compression inside of the third pump
6  7: Constant pressure heat addition inside of the boiler
7  8: Isentropic steam expansion inside of the low pressure turbine
8  9: Constant pressure heat addition inside of the boiler
8  5: Constant pressure heat regeneration inside of the first open feed water heater
9  11: Isentropic steam expansion inside of the intermediate pressure turbine
10  3: Constant pressure heat regeneration inside of the second open feed water heater
14

15. 11  1: Constant pressure heat rejection inside of the condenser
State 1
P1 = 1 Psi (from cycle 2 case 2)
h1 = 69.72 Btu/lbm(Table A-5E)
v1 = 0.01614 ft3/lbm (Table A-5E)
s1 = 0.13262 Btu/(lbm*R) (Table A-5E)
State 2
P2 = 250 Psi (assumed between 1 and 500 Psi)
s2 = s1= 0.13262 Btu/(lbm*R)
wp2 = v1*(P2-P1)*(144/778) = 0.74385 Btu/lbm
h2 = h1 + wp2 = 70.46385 Btu/lbm
State 3
P3 = P2 = 250 Psi
h3 = 376.09 Btu/lbm (Table A-5E)
v3 = 0.01865 ft3/lbm (Table A-5E)
s3 = 0.56784 Btu/(lbm*R) (Table A-5E)
State 4
P4 = 500 Psi (from step 2 case 1)
s4 = s3 = 0.56784 Btu/(lbm*R)
wp4 = v3*(P4-P3) * (144/778) = 0.86298 Btu/lbm
h4 = h3 + wp4 = 376.952982 Btu/lbm
State 5
P5 = P4 = 500 Psi
h5 = 449.51 Btu/lbm (Table A-5E)
v5 = 0.01975 ft3/lbm (Table A-5E)
s5 = 0.649 Btu/(lbm*R) (Table A-5E)
State 6
P6 = 2000 psi
s6 = s5 = 0.649 Btu/(lbm*R)
wp6 = v5*(P6-P5)*(144/778) = 5.48329 Btu/lbm
h6 = h5 + wp6 = 454.9933 Btu/lbm
State 7
P7 = P6 = 2000 Psi
T7 = 1000 degrees Fahrenheit
h7 = 1474.9 Btu/lbm (Table A-6E)
s7 = 1.5606 Btu/(lbm*R) (Table A-6E)
State 8 P8 = P4 = 500 Psi
s8 = s7 = 1.5606 Btu/(lbm*R)
15

16. h8 = 1300.333818 (Interpolated from table A-6E)
State 9
P9 = P8 = 500 Psi
T9 = 1000 degrees Fahrenheit
h9 = 1521 Btu/lbm (table A-6E)
s9 = 1.7376 Btu/(lbm*R) (table A-6E)
State 10
P10 = P3 = 250 Psi
s10 = s9 = 1.7376 Btu/(lbm*R)
h10 = 1419.881944 (Interpolated from table A-6E)
State 11
P11 = P1 = 1 Psi
s11 = s10 = 1.7376 Btu/(lbm*R)
sf = 0.13262 Btu/(lbm*R) (Table A-5E)
sfg = 1.84495 Btu/(lbm*R) (Table A-5E)
x = (s11-sf)/sfg = 0.869931434
hfg = 1035.7 (Table A-5E)
h11 = h1 + x*hfg = 970.7079867 Btu/lbm
Efficiency
y = (h5-h4)/(h8-h4) = 0.078577565
z = (1-y)*(h3-h2)/(h10-h2) = 0.20869054
q(in) = (1-y)*(h9-h8) + (h7 – h8) = 1223.23348 Btu/lbm
q(out) = (1-y-z)*(h11-h1) = 642.1628748 Btu/lbm
η = 1 – [q(out)/q(in)] = 0.475028 = 47.5028 %
Mass Flow
Rate
Wnet = q(in) – q(out) = 581.0706053 Btu/lbm
m = Power*3412/Wnet = 880787.9719 lbm/hr
The efficiency and mass flow rate for this cycle are 47.5028 % and 880787.9719 lbm/hr at x = 0.87
respectively (see table 5). The efficiency has improved and the mass flow rate has decreased from the previous
cycle
Step 3: Ideal Rankine Cycle with Regeneration (cycle 5: four open feed water heater)
16

17. Figure 6: Ideal Rankine Cycle with Regeneration (four water heaters)
The third and final Regenerative Ideal Rankine Cycle involves adding two open feed water heater to the lower
pressure turbine of the reheated Rankine cycle to give a third enhancement to the previous cycle’s efficiency.
The process of this cycle is shown on the below:
1  2: Isentropic compression at the first pump
2  3: Constant pressure heat addition at the fourth open feed water heater
3  4: Isentropic compression at the second pump
17

18. 4  5: Constant pressure heat addition at the third open feed water heater
5  6: Isentropic compression at the third pump
6  7: Constant pressure heat addition at the second open feed water heater
7  8: Isentropic compression at the fourth pump
8  9: Constant pressure heat addition at the first open feed water heater
9 10: Isentropic compression at the fifth pump
10  11: Constant pressure heat addition inside of the boiler
11  12: Isentropic steam expansion inside of the high pressure turbine
12  9: Constant pressure heat regeneration at the first open feed water heater
12  13: Constant pressure heat addition at the boiler
14  7: Constant pressure heat regeneration at the second open feed water heater
13  15: Isentropic steam expansion inside of the intermediate pressure turbine
16  5: Constant pressure heat regeneration at the third open feed water heater
17  3: Constant pressure heat regeneration at the fourth open feed water heater
15  18: Isentropic steam expansion inside of the low pressure turbine
State 1
P1 = 1 Psi (from cycle 2 case 2)
h1 = 69.72 Btu/lbm(Table A-5E)
v1 = 0.01614 ft3/lbm (Table A-5E)
s1 = 0.13262 Btu/(lbm*R) (Table A-5E)
State 2
P2 = 50 Psi (Assumed for low pressure turbine)
s2 = s1 = 0.13262 Btu/(lbm*R)
Wp1 = v1*(P2-P1)*(144/778) = 0.14638 Btu/lbm
h2 = h1 + Wp1 = 69.8663 Btu/lbm
State 3 P3 = P2 = 50 Psi
h3 = 250.21 Btu/lbm(Table A-5E)
v3 = 0.01727 ft3/lbm (Table A-5E)
18

19. s3 = 0.41125 Btu/(lbm*R) (Table A-5E)
State 4
P4 = 100 Psi (Assumed for low pressure turbine)
s4 = s3 = 0.41125 Btu/(lbm*R)
Wp2 = v3*(P4-P3)*(144/778) = 0.159825 Btu/lbm
h4 = h3 + Wp2 = 69.8663 Btu/lbm
State 5
P5 = P4 = 100 Psi
h5 = 298.51 Btu/lbm (Table A-5E)
v5 = 0.01774 ft3/lbm (Table A-5E)
s5 = 0.47427 Btu/(lbm*R) (Table A-5E)
State 6
P6 = 250 Psi (Assumed for intermediate pressure turbine)
s6 = s5 = 0.47427 Btu/(lbm*R)
Wp3 = v6*(P6-P5)*(144/778) = 0.49252 Btu/lbm
h6 = h5 + Wp3 = 299.0025244 Btu/lbm
State 7
P7 = P6 = 250 Psi
h7 = 376.09 Btu/lbm (Table A-5E)
v7 = 0.01865 ft3/lbm (Table A-5E)
s7 = 0.56784 Btu/(lbm*R) (Table A-5E)
State 8
P8 = 500 Psi (from step 2 cycle 1)
s8 = s7 = 0.56784 Btu/(lbm*R)
Wp4 = v8*(P8-P7)*(144/778) = 0.862982 Btu/lbm
h8 = h7 + Wp4 = 376.952982 Btu/lbm
State 9
P9 = P8 = 500 Psi
h9 = 449.51 Btu/lbm (Table A-5E)
v9 = 0.01975 ft3/lbm (Table A-5E)
s9 = 0.649 Btu/(lbm*R) (Table A-5E)
State 10 P10 = 2000 Psi
s10 = s9 = 0.649 Btu/(lbm*R)
Wp5 = v10*(P10-P9)*(144/778) = 5.483290488 Btu/lbm
19

20. h10 = h9 + Wp5 = 454.9932905 Btu/lbm
State 11
P11 = P10 = 2000 Psi
T11 = 1000 degrees Fahrenheit
v11 = 0.39479 ft3/lbm (Table A-6E)
h11 = 1474.9 Btu/lbm (Table A-6E)
s11 = 1.5606 Btu/(lbm*R) (Table A-6E)
State 12
P12 = P9 = 500 Psi
s12 = s11 = 1.5606 Btu/(lbm*R)
h12 = 1300.333818 Btu/lbm (interpolated from table A-6E)
m12 = (h9-h8)/(h12-h8) = 0.078577
State 13
P13 = P12 = 500 Psi
T13 = 1000 degrees Fahrenheit
v13 = 1.70094 ft3/lbm (Table A-6E)
h13 = 1521 Btu/lbm (Table A-6E)
s13 = 1.7376 Btu/(lbm*R) (Table A-6E)
m13 = 1- m12 = 0.9214
State 14
P14 = P7 = 250 Psi
s14 = s13 = 1.7376 Btu/(lbm*R)
h14 = 1419.994431 Btu/lbm (interpolated from table A-6E)
m14 = (1 - m13)*(h7-h6)(/(h14-h6) = 0.063363642
State 15 (Irrelevant for this particular calculation)
State 16
P16 = P5 = 100 Psi
s16 = s14 = 1.7376 Btu/(lbm*R)
h16 = 1308.273494 Btu/lbm (interpolated from table A-6E)
m16 = (1-m13-m14)*(h5-h4)/(h16-h4) = 0.039046183
State 17
P17 = P3 = 50 Psi
s17 = s16 = 1.7376 Btu/(lbm*R)
h17 = 1245.05 Btu/lbm (interpolated from table A-6E)
m17 = (1-m13-m14-m16)*(h4-h3)/(h17-h3) = 0.125685634
State 18 P18 = P1 = 1 Psi
s18 = s17 = 1.7376 Btu/(lbm*R)
x18 = (s18-s1)/sfg = (s18-s1)/ 1.84495 = 0.869931434
20

21. h18 = h1 + x18*hfg = h1 + x18*1035.7 = 970.7079867 Btu/lbm
Efficienc
y
q(in) = (h11-h10) + (1-m12)*(h13-h12) = 1223.23348 Btu/lbm
q(out) = (1-m12-m14-m16-m17)*(h18- h1) = 624.6792759 Btu/lbm
η = 1 – [q(out)/q(in)] = 0.489321306 = 48.93213061 %
Mass
Flow Rate
Wnet = q(in) – q(out) = 598.5542041 Btu/lbm
m = power*3412/Wnet = 855060.4047 (lbm/hr)
The efficiency and mass flow rate for this cycle are 48.9321% and 855060.4047 lbm/hr at x = 0.87
respectively (see table 6). The efficiency has improved and the mass flow rate has decreased again from the
previous cycle. [Note: Mass flow rates for step 4 has already been calculated in step 3]
Cost Effectiveness
Formula: annual income = Power (Kw) * 8765.81 (hours/year) * (Efficiency/100) * $0.16 (cost of
electricity in kwh)
Cycle Type Efficiency (%) Annual income
($ in Millions)
Profits gained
($ in Millions)
Ideal Rankine Cycle 34.2011 71.95 N/A
Ideal Rankine Cycle with
reheating 44.3745 93.35 21.4
21

22. Ideal Rankine Cycle with
1 Feed water Heater 46.8146 98.49 26.54
Ideal Rankine Cycle with
2 Feed water Heater 47.5208 99.97 28.02
Ideal Rankine Cycle with
4 Feed water Heater 48.3213 101.66 29.71
Based on the cost effectiveness calculation in the chart above, it is safe to conclude that reheating and adding
feed water heaters to the turbine will improve its efficiency
Excel Graphs and Spreadsheets
Test Pressure P1 (Psi) P2 = P3 T3 v1
1 2000 1000 0.01614
5 2000 1000 0.01641
10 2000 1000 0.01659
15 2000 1000 0.01672
20 2000 1000 0.01683
25 2000 1000 0.01692
30 2000 1000 0.017
h1 wp h2 h3 s3 = s4 hg4 hfg4 sg4 sfg4
69.72 5.971717 75.69172 1474.9 1.5606 1105.4 1035.7 1.9776 1.84495
130.18 6.059466 136.2395 1474.9 1.5606 1130.7 1000.5 1.8438 1.60894
161.25 6.110579 167.3606 1474.9 1.5606 1143.1 981.82 1.7875 1.50391
181.21 6.142988 187.353 1474.9 1.5606 1150.7 969.47 1.7549 1.44441
196.27 6.167827 202.4378 1474.9 1.5606 1156.2 959.93 1.7319 1.39606
208.52 6.185152 214.7052 1474.9 1.5606 1160.6 952.03 1.7141 1.3606
218.93 6.198663 225.1287 1474.9 1.5606 1164.1 945.21 1.6995 1.33132
1- x4 (moisture
content) h4
steam quality
(x4) efficiency
0.226022385 871.308615 0.773977615 0.4271127
0.176016508 954.595484 0.823983492 0.384149
0.15087339 994.969488 0.84912661 0.3623753
0.134518592 1020.28826 0.865481408 0.3483125
0.122702463 1038.41423 0.877297537 0.3381774
0.112817874 1053.194 0.887182126 0.3297275
0.104332542 1065.48384 0.895667458 0.322633
Steam quality At 87% (0.87) Exit pressure = 16.91 --> 17 Psi
Efficiency = 0.3420114291= 34.2011 %
Table 1: Step 1
22

23. Steam quality versus exit pressure
y = 0.0357Ln(x) + 0.7705
R
2
= 0.9944
0.76
0.78
0.8
0.82
0.84
0.86
0.88
0.9
0.92
0 5 10 15 20 25 30 35
Exit Pressure (Psi)
SteamQuality
Figure 6: Steam Quality vs. Exit Pressure
Efficiency versus exit pressurey = -0.0307Ln(x) + 0.4301
R
2
= 0.9946
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0 5 10 15 20 25 30 35
Exit Pressure (Psi)
Efficiency
Figure 7: Efficiency vs. Exit Pressure
23

24. Exit pressure at x = 0.87 (Pe) = 17 Psi
Pa = 15 Psi
Pb = 20 Psi
hf1 = 181.21 btu/lbm
hf2 = 196.21 btu/lbm
hf = h1 = (Pe -Pa)*(hf2-hf1)/(Pb-Pa) + hf1 = 187.21 btu/lbm
vf2 = 0.01683 ft^3/lbm
vf1 = 0.01672 ft^3/lbm
vf = (Pe -Pa)*(vf2-vf1)/(Pb-Pa) + vf1 = 0.016764
P2 = P3 = 2000 Psi
h2 = h1 + wp = h1 + [vf*(P2-Pe)]*(144/778) = 193.363 btu/lbm
T3 = 1000 degrees Fahrenheit
h3 = 1474.9 Btu/lbm
s3 = 1.5606
sf1 = 0.31370 Btu/lbm-R
sf2 = 0.33582 Btu/lbm-R
sfg1 =1.44441 Btu/lbm-R
sfg2 = 1.39606 Btu/lbm-R
hfg1 = 969.47 Btu/lbm
hfg2 = 959.93 Btu/lbm
sf = (Pe-Pa)*(sf2-sf1)/(Pb-Pa) + sf1 = 0.322548 Btu/lbm-R
sfg = (Pe-Pa)*(sfg2-sfg1)/(Pb-Pa) + sfg1 = 1.4154 Btu/lbm-R
hfg = (Pe-Pa)*(hfg2-hfg1)/(Pb-Pa) + hfg1 = 965.654 Btu/lbm-
P4 = P5 (0.1-0.4)*P(given) h4 h5 s5 = s6 x6
200 1211.253 1529.6 1.843 1.074220715
275 1240.311 1527.4 1.8068 1.048644906
350 1263.607 1525.3 1.7791 1.029074467
400 1276.906 1523.9 1.7636 1.018123499
450 1289.302 1522.4 1.7499 1.008444256
500 1300.334 1521 1.7376 0.999754133
600 1320.431 1518.1 1.716 0.984493429
700 1337.987 1515.2 1.6974 0.971352268
800 1353.809 1512.2 1.6812 0.95990674
h6 = hf + (x6)*hfg q(in) q(out) efficiency
1224.53553 1599.884 1037.32553 0.351625
1199.838148 1568.626 1012.628148 0.354449
1180.939875 1543.23 993.729875 0.356071
1170.365029 1528.531 983.155029 0.356797
1161.01823 1514.635 973.8082296 0.357067
1152.626578 1502.203 965.4165777 0.357333
1137.890018 1479.206 950.6800181 0.357304
1125.200203 1458.75 937.9902029 0.35699
1114.147783 1439.928 926.9377832 0.356261
Optimal Efficiency
35.7333% at 500 Psi (P4)
Table 2: Step 2 Case 1
24

25. Pressure versus Efficiency
0.351
0.352
0.353
0.354
0.355
0.356
0.357
0.358
0 100 200 300 400 500 600 700 800 900
Pressure (Psi)
Efficiency
Series1
Figure 8: Pressure versus Efficiency
P1 = P6 P2 = P3 P4 = P5 h1 = hf hfg
1 2000 500 69.72 1035.7
2 2000 500 94.02 1021.7
3 2000 500 109.4 1012.8
4 2000 500 120.9 1006
5 2000 500 130.18 1000.5
6 2000 500 138.02 995.88
v1 = vf wp (work) h2 T3 = T5 h3 s3 = s4
0.01614 5.971717 75.69172 1000 1474.9 1.5606
0.01623 6.002013 100.022 1000 1474.9 1.5606
0.0163 6.024882 115.4249 1000 1474.9 1.5606
0.01636 6.044032 126.944 1000 1474.9 1.5606
0.01641 6.059466 136.2395 1000 1474.9 1.5606
0.01645 6.071192 144.0912 1000 1474.9 1.5606
T4 h4 s5 = s6 h5 sf sfg x
602.9090909 1300.334 1.7376 1521 0.13262 1.84495 0.869931
602.9090909 1300.334 1.7376 1521 0.17499 1.74444 0.895766
602.9090909 1300.334 1.7376 1521 0.2009 1.68489 0.912048
602.9090909 1300.334 1.7376 1521 0.21985 1.64225 0.924189
602.9090909 1300.334 1.7376 1521 0.23488 1.60894 0.933981
602.9090909 1300.334 1.7376 1521 0.24739 1.58155 0.942247
h6 q(in) q(out) efficiency
970.708 1619.874 900.988 0.443791
1009.224 1595.544 915.2041 0.4264
1033.122 1580.141 923.7219 0.415418
1050.635 1568.622 929.7345 0.407292
1064.628 1559.327 934.4484 0.400736
1076.384 1551.475 938.3645 0.395179
Efficiency at x = 0.87 Exhaust pressure at x = 0.87
0.443745318 1 Psi
44.37%
Table 3: Step 2 Case 2
25

26. Pressure versus steam quality
0.86
0.87
0.88
0.89
0.9
0.91
0.92
0.93
0.94
0.95
0 1 2 3 4 5 6 7
Pressure (Psi)
SteamQuality
Figure 9: Pressure vs. Steam Quality
Pressure vs efficiency
0.39
0.4
0.41
0.42
0.43
0.44
0.45
0 1 2 3 4 5 6 7
Pressure (Psi)
Efficiency
Figure 10: Pressure vs. Efficiency
26

27. P1 = P8 (Psi) P2 = P3 = P6 = P7 h1 = hf hfg v1 = vf wp = work h2 = h1 + wp
1 500 69.72 1035.7 0.01614 1.490689 71.21068874
h3 = hf3 v3 = vf3 s3 = s4 = sf P4 = P5 (Psi) wp2 h4 = h3 + wp2
449.51 0.01975 0.649 2000 5.48329 454.9932905
T5 h5 s5 = s6 h6 T7 h7 s7 = s8
1000 1474.9 1.5606 1300.333818 1000 1521 1.7376
sf8 sfg8 x8 h8 y
0.13262 1.84495 0.869931 970.708 0.30778
q(in) q (out) efficiency
1172.656 623.6821 0.468146
efficiency at .87
(%) Wnet (btu/lbm) Mass Flow Rate (lbm/hr)
46.81459012 548.974236 932284.1883
Table 4: Step 3 Cycle 3
P1 h1 = hf v1 = vf s1 = sf
1 69.72 0.01614 0.13262
P2 s2 wp2 (work) h2
250 0.13262 0.743850694 70.46385069
P3 h3 v3 s3
250 376.09 0.01865 0.56784
P4 s4 wp4 h4
500 0.56784 0.862982005 376.952982
P5 h5 v5 s5
500 449.51 0.01975 0.649
P6 s6 wp6 h6
2000 0.649 5.483290488 454.9932905
P7 T7 h7 s7
2000 1000 1474.9 1.5606
P8 s8 h8
500 1.5606 1300.333818
P9 T9 h9 s9
500 1000 1521 1.7376
P10 s10 h10
250 1.7376 1419.881944
P11 s11 sf sfg
1 1.7376 0.13262 1.84495
x hfg h11
0.869931434 1035.7 970.7079867
y z q (out) q (in)
0.078577565 0.20869054 642.1628748 1223.23348
27

28. efficiency at 0.87 Wnet (btu/lbm)
Mass flow rate
(lbm/hr)
0.475028369 581.0706053 880787.9719
47.50283693
Table 5: Step 3 Cycle 4
P1 h1 v1 s1
1 69.72 0.01614 0.13262
P2 s2 Wp1 h2
50 0.13262 0.146380257 69.86638026
P3 h3 v3 s3
50 250.21 0.01727 0.41125
P4 s4 Wp2 h4
100 0.41125 0.159825193 250.3698252
P5 v5 h5 s5
100 0.01774 298.51 0.47427
P6 s6 Wp3 h6
250 0.47427 0.492524422 299.0025244
P7 v7 h7 s7
250 0.01865 376.09 0.56784
P8 s8 Wp4 h8
500 0.56784 0.862982005 376.952982
P9 v9 h9 s9
500 0.01975 449.51 0.649
P10 s10 Wp5 h10
2000 0.649 5.483290488 454.9932905
P11 T11 v11 h11
2000 1000 0.39479 1474.9
s11
1.5606
P12 s12 h12 m12
500 1.5606 1300.333818 0.078577565
P13 T13 v13 h13
500 1000 1.70094 1521
s13 m13
1.7376 0.921422435
P14 s14 h14 m14
250 1.7376 1419.994432 0.063363642
28

29. P16 s16 h16 m16
100 1.7376 1308.273494 0.039046183
P17 s17 h17 m17
50 1.7376 1245.05 0.125685634
P18 s18 x18 h18
1 1.7376 0.869931434 970.7079867
efficiency (%) q(in) q(out) Wnet
0.489321306 1223.23348 624.6792759 598.5542041
48.93213061
Mass flow rate (lbm/hr)
855060.4047
Table 6: step 3 cycle 5
Discussions and Conclusion
It is shown through the project that one can improve the efficiency of standard Rankine cycle via
reheating (34.2 to 44.37 percent) and adding feed water heaters (34.2 to 46.8146, 47.5208, and 48.3213 with
one, two and four water heaters respectively) to the turbine. The design proved that adding more open
feed water heaters to the turbine cycle will improve the efficiency of the steam turbine and lower the
turbine’s mass flow rate. The improved efficiency resulted in increased annual incomes and the
diminished mass flow rate resulted in less steam required to obtain the same amount of work. However,
it should be noted that efficiency gains starts to decrease as more feed water heaters are added until it
tapers off at a certain point. This will eventually leaves potential gains not worth the cost effectiveness
at a certain amount of feed water heaters being added.
Based on the calculations in step 3, it is shown in the cost effectiveness section that the 5th
cycle, the
Ideal Rankine Cycle with 4 Feed water Heater, produced that most gains out of all the other cycles. However in
terms actual cost effectiveness, the 4th
cycle, the Ideal Rankine Cycle with 2 Feed water Heater, is the most
effective because water heater cost extra money to maintain and the gains from adding four more water heaters
basic Rankine cycle turbine is barely anymore than the gains from merely adding two water heaters to said
Rankine cycle turbine (29.71 million dollars versus 28.02 million dollars respectively), meaning the cost needed
to maintain essentially double the amount of water heaters, outweighs the minimal potential gains produced.
29

30. In conclusion, this project was very helpful in demonstrating how a steam turbine power plant works.
The process in calculating the Rankine cycle’s efficiency proved that the Power Plant’s Performance can be
improved via reheating and regeneration, showing that the theory behind optimizing the power plant’s
performance has indeed been confirmed.
Appendix
30

31. 31

32. 32

33. 33

34. 34

35. 35

36. 36

37. References
[1]: Thermodynamics, an Engineering Approach. Yunus A. Cengel & Michael A. Boles. Seventh edition.
McGraw Hill Publications
[2]: Lecture notes: ME: 43000 Thermo-Fluid Systems Analysis and Design, Professor Rishi Raj.
[3]: www.google.com , Google images of the Rankine cycle
[4]: Woodford, Chris (19th July 2014), Steam Turbine, Explain that Stuff!, retrieved from
http://www.explainthatstuff.com/steam-turbines.html
37