thermal project 1

James Li # 263/27/2015Raj, RishiME 43000Spring 2015Thermal Project # 1: Optimal Thermal Cycle for Steam Turbine Power Plant

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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 [2]

Excel Graphs and Spreadsheets…………………………………………………………………………… Pages 22 - 28

Discussions and conclusions………………………………………………………………………………Pages 28 - 30

Appendix ………………………………………………………………………………………………………

References …………………………………………………………………………………………………

Nomenclature

P = Pressure (Psi)

T = Temperature (degrees Fahrenheit)2

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

will operate in three stages. The first stage of this turbine is based on the ideal Rankine cycle with a steam 3

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

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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:

Step 1: Ideal Rankine Cycle5

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)

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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

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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

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)


3 4: Isentropic expansion inside of the first high pressure turbine (s3=s4)


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)


h2 = h1 + wp = h1 + v1*(P2-P1)*(144/778) = 193.363 Btu/lbm

State 3 P3 = P2 = 2000 Psi


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)

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h4 = 1211.253 Btu/lbm (interpolated from table A-6E)

State 5 P5 = P4 – 200 Psi

T5 = T3 = 1000 degrees Fahrenheit




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)


hfg = 1035.7 Btu/lbm (extracted from table A-5E)

v1 = 0.01614 ft3/lbm (extracted from table A-5E)


wp (work) = v1*(P2-P1)*(144/778) = 5.971717 Btu/lbm

h2 = h1 + wp = 75.69172 Btu/lbm

State 3 P3 = P2 = 2000 Psi



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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)


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)


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.

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Step 3: Ideal Rankine Cycle with Regeneration (cycle 3: one open feed water heater)

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

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3 4: Isentropic compression in the second pump

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)


hfg = 1035.7 Btu/lbm (extracted from table A-5E)


State 2


wp = v1*(P2-P1)*(144/778) = 1.490689 Btu/lbm

h2 = h1 + wp = 71.2107 Btu/lbm

State 3

P3 = P2 = 500 Psi




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


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)

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State 7

P7 = P6 = 500 Psi


h7 = 1521 Btu/lbm (Table A-6E)

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).

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Step 3: Ideal Rankine Cycle with Regeneration (cycle 4: two open feed water heater)

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



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

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10 3: Constant pressure heat regeneration inside of the second open feed water heater

11 1: Constant pressure heat rejection inside of the condenser

State 1


h1 = 69.72 Btu/lbm(Table A-5E)

v1 = 0.01614 ft3/lbm (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)


State 4


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


v5 = 0.01975 ft3/lbm (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



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State 8

P8 = P4 = 500 Psi

s8 = s7 = 1.5606 Btu/(lbm*R)

h8 = 1300.333818 (Interpolated from table A-6E)

State 9

P9 = P8 = 500 Psi


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

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Step 3: Ideal Rankine Cycle with Regeneration (cycle 5: four open feed water heater)

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

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2 3: Constant pressure heat addition at the fourth open feed water heater

3 4: Isentropic compression at the second pump

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


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


State 1



v1 = 0.01614 ft3/lbm (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


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v3 = 0.01727 ft3/lbm (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


v5 = 0.01774 ft3/lbm (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


v7 = 0.01865 ft3/lbm (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



v9 = 0.01975 ft3/lbm (Table A-5E)

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State 10

P10 = 2000 Psi

s10 = s9 = 0.649 Btu/(lbm*R)

Wp5 = v10*(P10-P9)*(144/778) = 5.483290488 Btu/lbm

h10 = h9 + Wp5 = 454.9932905 Btu/lbm

State 11

P11 = P10 = 2000 Psi


v11 = 0.39479 ft3/lbm (Table A-6E)



State 12

P12 = P9 = 500 Psi

s12 = s11 = 1.5606 Btu/(lbm*R)


m12 = (h9-h8)/(h12-h8) = 0.078577

State 13

P13 = P12 = 500 Psi


v13 = 1.70094 ft3/lbm (Table A-6E)

h13 = 1521 Btu/lbm (Table A-6E)


m13 = 1- m12 = 0.9214

State 14

P14 = P7 = 250 Psi

s14 = s13 = 1.7376 Btu/(lbm*R)


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)


m16 = (1-m13-m14)*(h5-h4)/(h16-h4) = 0.039046183

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State 17

P17 = P3 = 50 Psi

s17 = s16 = 1.7376 Btu/(lbm*R)


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

h18 = h1 + x18*hfg = h1 + x18*1035.7 = 970.7079867 Btu/lbm

Efficiency

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]

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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/AIdeal Rankine Cycle with

reheating 44.3745 93.35 21.4Ideal Rankine Cycle with

1 Feed water Heater 46.8146 98.49 26.54Ideal Rankine Cycle with

2 Feed water Heater 47.5208 99.97 28.02Ideal 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 v11 2000 1000 0.016145 2000 1000 0.01641

10 2000 1000 0.0165915 2000 1000 0.0167220 2000 1000 0.0168325 2000 1000 0.0169230 2000 1000 0.017

h1 wp h2 h3 s3 = s4 hg4 hfg4 sg4 sfg469.72 5.971717 75.69172 1474.9 1.5606 1105.4 1035.7 1.9776 1.84495130.18 6.059466 136.2395 1474.9 1.5606 1130.7 1000.5 1.8438 1.60894161.25 6.110579 167.3606 1474.9 1.5606 1143.1 981.82 1.7875 1.50391181.21 6.142988 187.353 1474.9 1.5606 1150.7 969.47 1.7549 1.44441196.27 6.167827 202.4378 1474.9 1.5606 1156.2 959.93 1.7319 1.39606208.52 6.185152 214.7052 1474.9 1.5606 1160.6 952.03 1.7141 1.3606218.93 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 0.42711270.176016508 954.595484 0.823983492 0.3841490.15087339 994.969488 0.84912661 0.36237530.134518592 1020.28826 0.865481408 0.34831250.122702463 1038.41423 0.877297537 0.33817740.112817874 1053.194 0.887182126 0.3297275

22

0.104332542 1065.48384 0.895667458 0.322633Steam quality At 87% (0.87) Exit pressure = 16.91 --> 17 Psi

Efficiency = 0.3420114291= 34.2011 %Table 1: Step 1

Steam quality versus exit pressure

y = 0.0357Ln(x) + 0.7705R2 = 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)

Stea

m Q

ualit

y

Figure 6: Steam Quality vs. Exit Pressure

23

Efficiency versus exit pressurey = -0.0307Ln(x) + 0.4301R2 = 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)

Efficie

ncy

Figure 7: Efficiency vs. Exit Pressure

Exit pressure at x = 0.87 (Pe) = 17 PsiPa = 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 Psih2 = 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-Rsf2 = 0.33582 Btu/lbm-Rsfg1 =1.44441 Btu/lbm-Rsfg2 = 1.39606 Btu/lbm-R

hfg1 = 969.47 Btu/lbmhfg2 = 959.93 Btu/lbm

sf = (Pe-Pa)*(sf2-sf1)/(Pb-Pa) + sf1 = 0.322548 Btu/lbm-Rsfg = (Pe-Pa)*(sfg2-sfg1)/(Pb-Pa) + sfg1 = 1.4154 Btu/lbm-Rhfg = (Pe-Pa)*(hfg2-hfg1)/(Pb-Pa) + hfg1 = 965.654 Btu/lbm-

P4 = P5 (0.1-0.4)*P(given) h4 h5 s5 = s6 x6200 1211.253 1529.6 1.843 1.074220715

24

275 1240.311 1527.4 1.8068 1.048644906350 1263.607 1525.3 1.7791 1.029074467400 1276.906 1523.9 1.7636 1.018123499450 1289.302 1522.4 1.7499 1.008444256500 1300.334 1521 1.7376 0.999754133600 1320.431 1518.1 1.716 0.984493429700 1337.987 1515.2 1.6974 0.971352268800 1353.809 1512.2 1.6812 0.95990674

h6 = hf + (x6)*hfg q(in) q(out) efficiency 1224.53553 1599.884 1037.32553 0.3516251199.838148 1568.626 1012.628148 0.3544491180.939875 1543.23 993.729875 0.3560711170.365029 1528.531 983.155029 0.3567971161.01823 1514.635 973.8082296 0.3570671152.626578 1502.203 965.4165777 0.3573331137.890018 1479.206 950.6800181 0.3573041125.200203 1458.75 937.9902029 0.356991114.147783 1439.928 926.9377832 0.356261

Optimal Efficiency 35.7333% at 500 Psi (P4)

Table 2: Step 2 Case 1

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)

Effic

ienc

y

Series1

Figure 8: Pressure versus Efficiency

P1 = P6 P2 = P3 P4 = P5 h1 = hf hfg1 2000 500 69.72 1035.72 2000 500 94.02 1021.73 2000 500 109.4 1012.84 2000 500 120.9 10065 2000 500 130.18 1000.56 2000 500 138.02 995.88

v1 = vf wp (work) h2 T3 = T5 h3 s3 = s40.01614 5.971717 75.69172 1000 1474.9 1.56060.01623 6.002013 100.022 1000 1474.9 1.56060.0163 6.024882 115.4249 1000 1474.9 1.5606

0.01636 6.044032 126.944 1000 1474.9 1.5606

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0.01641 6.059466 136.2395 1000 1474.9 1.56060.01645 6.071192 144.0912 1000 1474.9 1.5606

T4 h4 s5 = s6 h5 sf sfg x602.9090909 1300.334 1.7376 1521 0.13262 1.84495 0.869931602.9090909 1300.334 1.7376 1521 0.17499 1.74444 0.895766602.9090909 1300.334 1.7376 1521 0.2009 1.68489 0.912048602.9090909 1300.334 1.7376 1521 0.21985 1.64225 0.924189602.9090909 1300.334 1.7376 1521 0.23488 1.60894 0.933981602.9090909 1300.334 1.7376 1521 0.24739 1.58155 0.942247

h6 q(in) q(out) efficiency970.708 1619.874 900.988 0.4437911009.224 1595.544 915.2041 0.42641033.122 1580.141 923.7219 0.4154181050.635 1568.622 929.7345 0.4072921064.628 1559.327 934.4484 0.4007361076.384 1551.475 938.3645 0.395179

Efficiency at x = 0.87 Exhaust pressure at x = 0.870.443745318 1 Psi

44.37%Table 3: Step 2 Case 2

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)

Stea

m Q

ualit

y

Figure 9: Pressure vs. Steam Quality

26

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)

Effic

ienc

y

Figure 10: Pressure vs. Efficiency

P1 = P8 (Psi) P2 = P3 = P6 = P7 h1 = hf hfg v1 = vf wp = work h2 = h1 + wp1 500 69.72 1035.7 0.01614 1.490689 71.21068874

h3 = hf3 v3 = vf3 s3 = s4 = sf P4 = P5 (Psi) wp2 h4 = h3 + wp2449.51 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 1.7376sf8 sfg8 x8 h8 y

0.13262 1.84495 0.869931 970.708 0.30778q(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 932284.1883Table 4: Step 3 Cycle 3

P1 h1 = hf v1 = vf s1 = sf1 69.72 0.01614 0.13262

P2 s2 wp2 (work) h2250 0.13262 0.743850694 70.46385069

P3 h3 v3 s3250 376.09 0.01865 0.56784

P4 s4 wp4 h4500 0.56784 0.862982005 376.952982

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P5 h5 v5 s5500 449.51 0.01975 0.649

P6 s6 wp6 h62000 0.649 5.483290488 454.9932905

P7 T7 h7 s72000 1000 1474.9 1.5606

P8 s8 h8500 1.5606 1300.333818

P9 T9 h9 s9500 1000 1521 1.7376

P10 s10 h10250 1.7376 1419.881944

P11 s11 sf sfg1 1.7376 0.13262 1.84495

x hfg h110.869931434 1035.7 970.7079867

y z q (out) q (in)0.078577565 0.20869054 642.1628748 1223.23348

efficiency at 0.87 Wnet (btu/lbm)Mass flow rate

(lbm/hr)0.475028369 581.0706053 880787.971947.50283693

Table 5: Step 3 Cycle 4

P1 h1 v1 s11 69.72 0.01614 0.13262

P2 s2 Wp1 h250 0.13262 0.146380257 69.86638026

P3 h3 v3 s350 250.21 0.01727 0.41125

P4 s4 Wp2 h4100 0.41125 0.159825193 250.3698252

P5 v5 h5 s5100 0.01774 298.51 0.47427

P6 s6 Wp3 h6250 0.47427 0.492524422 299.0025244

P7 v7 h7 s7250 0.01865 376.09 0.56784

P8 s8 Wp4 h8

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500 0.56784 0.862982005 376.952982

P9 v9 h9 s9500 0.01975 449.51 0.649

P10 s10 Wp5 h102000 0.649 5.483290488 454.9932905

P11 T11 v11 h112000 1000 0.39479 1474.9

s111.5606

P12 s12 h12 m12500 1.5606 1300.333818 0.078577565

P13 T13 v13 h13500 1000 1.70094 1521

s13 m131.7376 0.921422435

P14 s14 h14 m14250 1.7376 1419.994432 0.063363642

P16 s16 h16 m16100 1.7376 1308.273494 0.039046183

P17 s17 h17 m1750 1.7376 1245.05 0.125685634

P18 s18 x18 h181 1.7376 0.869931434 970.7079867

efficiency (%) q(in) q(out) Wnet0.489321306 1223.23348 624.6792759 598.554204148.93213061

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

29

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.

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

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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

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http://www.google.com/

thermal project 1

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