prof. dr. rishi raj design and optimize a steam turbine

Prof. Dr. Rishi Raj Design And Optimize A Steam Turbine Power plant Hasan-1

The following project is to design and optimize a steam turbine power plant under the designed specification given as inlet pressure (1500Psi), inlet temperature (1000oF), power plant out put (150000KW) and maximum moisture level (13%). The main purpose of the designed project is to achieve the maximum efficiency without a large amount of heat being wasted. Ideally the efficiency is obtained as 35.3%, upon reheating the efficiency increases to 46.2%. after adding open feed water heaters the efficiency further increases to 48.2%, 49.2% and 50.4% respectively


Abbreviation Property Unit P Pressure psi T Temperature oF X Quality % h Specific Enthalpy Btu/lbm s Specific Entropy Btu/lbm.R v Specific Volume ft3/lbm m Mass Flow Rate lbm/s

Power kW qin Amount of Heat in Btu/lbm qout Amount of Heat out Btu/lbm f Saturated Liquid State - g Saturated vapor State - fg Difference between saturated liquid and saturated vapor state -

Table-1 (shows the abbreviation, property and units)


Power plant generates electrical power using fuels like coals, oil or a natural gas. A simple power plant consists of a pump, boiler, turbine and condenser. Fuel generally burned in the boiler and super heater heats the water to generates the steam. The steam is then further heated into superheated state once enters into the turbine, where the steam rotates the turbine to generate the power. In a steam, gas or hydraulic power plant the device that generates the electricity is turbine. A turbine is a machine in which the gradual change of momentum of a fluid generates a rotary motion. Turbines may be classified in several ways but the important one is that the turbines mainly classified on the basis of the fluid they run. The most common fluids on which the turbines run are water and steam, therefore the turbines run on the basis of water is called hydraulic turbines and similarly turbines run by steam called steam turbines. Flow of the fluid is other important criteria on which turbines can be classified further into three classes and the classes are as follows;

i. Radial-flow turbines- the class of the turbine in which the fluid flows from the center to the circumference of the wheel or vice versa. The following class of the turbine is further classified into outward flow and inward flow turbines.

ii. Parallel or Axial-flow turbines- the class in which the fluid flows parallel to the axis of the wheel or in a spiral co-axial with the wheel.

iii. Mixed-flow turbines- the class in which the fluid flows both as in radial and also as in parallel flow turbine.

WORKING PRINCIPLE

Figure-1 (Working Principle for a Turbine)1

The working principle is very simple once its mechanism is understood, in almost every turbine the same principle works. As the fluid passes through the turbine the work is done against the blades of the turbine, the blades are always attached to the shafts. Therefore


the shaft rotates and work is produced by the turbine as a consequence of the fluid passing through the turbine.


The concept of steam turbine is not new to the modern science technology because it has been postulated and given by some philosophers in early ages, the only difference is that the modern technology has made impression of steam turbine and power plants more easier than it looks.

• 2nd century B.C. the book, named “Pneumatics” written by the Egyptian philosopher Hero describes how the steam engine works.

• In 1577 A.D. the German mechanic has made a change in Hero’s steam engine, he has used broach instead of turnspit.

• In 1629 Italian Architect Branca introduced steam wheel or turbine in which jet of steam was projected against the series of vanes on a rotating wheel.

• In 1642 Kircher used Branca’s approach with two jets of vapor acting on its circumference instead of a one jet.

In modern era there are many modifications have been made from the previous steam turbine engines.


DESCRIPTION The most important and an efficient thermodynamic cycle that operates between the two specified temperature limits is Carnot Cycle. This is the first ever cycle that has been proposed for vapor power cycle, but there many of the impracticalities are associated with the Carnot cycle that can be easily eliminated by superheating the steam in the boiler and then further condensing it completely in the condenser. This further result in producing by another cycle named Rankine Cycle, which is the most ideal cycle for vapor power plants. The efficiency of the rankine cycle is usually limited vy the working fluid, the working fluid in this cycle follows a closed loop and constantly reused in the cycle. The water vapor is seen very often billowing from power station is generated by the cooling system and represents the waste heat that cannot be converted into useful work. In the following process steam is invisible until it comes in contact with cool, saturated air at which it condenses and then it leaves the cooling tower in the form of white billowy clouds.

Figure-2 (shows physical layout of Rankine Cycle)2


PROCESS OF RANKINE CYCLE The ideal rankine cycle does not involve any internal irreversiblities and consists of the following processes;

• Process 1-2- the process in an isentropic compression in which the working fluid is pumped from low to high pressure, and if the fluid is liquid then it requires very little input energy.

• Process 2-3- the process is constant heat addition in boiler, the high pressure liquid enters in the boiler where it is treated at constant pressure externally using the heat source that converts high pressure liquid to a dry saturated vapor.

• Process 3-4- the process is an isentropic expansion in turbine, the power is generating when the dry saturated vapor expands through a turbine. Some of the condensation may be noticed in early stages, but mainly the temperature and the pressure of the dry vapor decreases.

• Process 3-4- the process is a constant pressure heat rejection in a condenser, at a constant temperature and pressure wet vapors become a saturated liquid as it enters in a condenser. At this stage of the process fluid undergoes a phase change due to the temperature and the pressure of the condenser is fixed by the temperature of the cooling coils.

Figure-3 (typical T-S diagram of Rankine Cycle between 0.06 bar and 50 bar)3

ACTUAL VAPOR CYCLE Vs IDEAL ONES Irreversibilities are the two common sources beside much other irreversibility in various components due to which the actual vapor cycle differs from the ideal Rankine cycle are fluid friction and heat loss. Fluid friction- It occurs when the pressure drops in the boiler, the condenser, and the piping between various components in resulting of this pressure drop steam leaves the boiler at very low pressure. The issue of the pressure drop can be resolved by pumping the excess amount of water at a high pressure than the ideal cycle, which requires larger pumps.


Heat loss- Another irreversibility that is loss of amount of heat into the surrounding from the steam as the steam flows through the various components. In maintaining the loss of heat so the net work output can be achieved at the same level, it requires more heat to be transferred to the steam in the boiler that will compensate for the heat losses, which resulting in the decrease of the cycle efficiency.

Figure-4 (typical T-S diagram of Actual Vs Ideal Cycle)4

EFFICIENCY INCREASE OF THE RANKINE CYCLCE The main and important ultimate goal is to achieve the high thermal efficiency of a power cycle so the cycle is considered to be the most efficient cycle. This can be done by either rejecting the heat at possibly very low pressure or by adding the heat at possibly high temperature. There are three ways that can help in achieving these two possible condition in improving the thermal efficiency of the cycle.

I. Lowering Condenser Pressure- When steam leaves in the condenser as a saturated mixture its temperature that is the saturation temperature corresponds to the pressure inside the condenser. It is important during this process the amount of heat to be rejected, therefore the overall thermal efficiency can be automatically increased by lowering the operating pressure of the condenser, which eventually lowers the steam temperature resulting in the heat being rejected.


Figure-5 (Shows Effect of Lowering Condenser Pressure on Cycle)5

II. Superheating Steam to High Temperature- Another important way of increasing the thermal efficiency of a cycle is to superheat the steam. This can be done when the amount of heat is being transferred to the steam then the average temperature can be increased without being increasing the boiler pressure by superheating the steam at a high temperature. Therefore due to the heat addition at this average temperature in increased then the overall thermal efficiency of the cycle will also increase simultaneously. Presently the highest steam temperature that has been allowed at the turbine inlet is about 620oC, which is equivalent to 1150oF.

Figure-6 (Shows Effect of Superheating Steam to Higher Temperature on

Cycle)6 III. Increasing Boiler Pressure- Another most efficient way to improve the

efficiency of the cycle is increase the pressure in the boiler. This is done when the amount of heat enters the boiler during this heat addition process the pressure at which the boiler is being operated must be increased, which automatically raises the temperature at which the boiling occurs. Due to this process the average temperature of the steam at which the amount of heat is transferred


increases that further resulting in increasing the overall thermal efficiency of the cycle.

Figure-7 (Shows Effect of Increasing Boiler Pressure on Cycle)7

ENERGY ANALYSIS OF RANKINE CYCLE In studying any cycle of the process the most important step is to analyze by mathematically or analytically. There are some important modifications or assumptions are needed to apply when energy analysis of the ideal rankine cycle is required to study. Such modification or assumptions are as follows;

• All the four cycles are associated with the rankine cycle such as pump, boiler, turbine and condenser.

• The flow of the process runs steadily because the devices that are used to run the process are steady-flow devices.

• The changes in kinetic and potential energy are relatively small during the steady flow therefore it can easily be neglected.

• The boiler and condenser do not involve any work. • The pump and turbine are assumed to be isentropic

The steady-flow energy equation can be written as; (Qin – Qout) + (Win – Wout) = (He – Hi) + KE + PE

Then after the modifications or assumptions the equation can further be written in terms of per unit mass that reduces as;

(qin – qout) + (win – wout) = (he – hi)

When pump (q = 0); wpump, in = h2 – h1 = v (p2 – p1)

When boiler (w = 0)


qin = h3 – h2

When turbine (q = 0) wturb, out = h3 – h4

When condenser (w = 0) qout = h4 – h1

Then thermal efficiency of the ideal rankine cycle can easily be calculated from the following equation;

ηth = wnet = 1- qout qin qin

Where wnet = qin – qout = wturb,out – wturb,in


It has been noticed that the thermal efficiency of the ideal rankine cycle increases as the boiler pressure increases, on the other hand it also increases the moisture content of the steam up to the level where it is unacceptable to the cycle. Ideally rankine cycle has solved the problem of efficiency but a problem arises as the excessive amount of moisture at the final stage of the turbine. Therefore reheating the ideal rankine cycle at the stages of the turbine before it enters the condenser can possibly resolve the issue. The two important possibilities that can help in resolving the issue of increase efficiency and an excess amount of impurities are as follows;

1. Before the steam enters the turbine it must be superheated to very high temperature.

2. The ideal rankine cycle is modified by reheating the steam as it leaves turbine to enter the condenser. In other words steam must be expanded in two stages and reheated in between the turbine and condenser stages.

Figure-8 (Shows the reheating ideal rankine cycle)8

T-s diagram shows the graphic presentation of the ideal rankine cycle after reheating between the two stages is given as follows;


Figure-9 (T-s Diagram of ideal reheating rankine cycle)9

The total heat input and the total turbine work can also be obtained for a reheat cycle as follows;

qin = qprimary + qreheat = (h3 – h2) + (h5 – h4)

And

wturb, out = wturb, I + wturb, II = (h3 – h4) + (h5 – h6)


Figure-10 (T-s Diagram heat addition process at low temperature) 10

The T-s diagram in figure-6 reveals that the another way to lower the average temperature and increase the cycle efficiency is that the heat must be transferred to the working fluid during the process between 2 and 2’ at a very low temperature. Therefore regeneration is the way to achieve the issue of lowering the temperature and improving the cycle efficiency. Regeneration not only improves the efficiency of the cycle on the same hand it prevents corrosion in the boiler by removing the air that leaks in at the condenser. The way an ideal regenerative rankine cycle achieve this improvement in cycle efficiency are as follows; OPEN FEED WATER HEATERS A mixing chamber in which steam is extracted from the turbine and further mixes with the feed water and the exits the pump, it is called open feed water heaters. It is important to know that ideally the mixture leaves the heater as a saturated liquid at the heater pressure.


Figure-11 (Shows Ideal Regenerative Cycle with an Open FWH)11

Figure-12 (Shows T-s Diagram for Ideal Regenerative Cycle with an Open FWH)12

ENERGY ANALYSIS OF OPEN FEEDWATER HEATER It is important when the work out put or input is required to obtain the quantities must be expressed in per unit mass of the steam through the boiler in the steam power plant calculation. Therefore the calculation for the steam in per unit mass is given as;

qin = h5 – h4

qout = (1- y)(h7 – h1)

wturb, out = (h5 – h6) + (1 – y)(h6 – h7)


wpump, in = (1 – y)wpmup I, in + wpump II, in

y = m6/m7

wpump I, in = v1(P2 – P1)

wpump II, in = v2(P4 – P3)

CLOSED FEED WATER HEATERS Beside an open FWH there is another important type of FWH that is used in steam power plant very often called closed feedwater heater. In this type of FWH the amount of heat is transferred from the extracted stream feed water without any mixing being done. There are many types of closed feedwater heater with various different pressures. The pressure for the streams are found to be different because the two streams do not mix at all at any stage. It is important to know that the extracted stream that is heated to the exit temperature leaves the heater as a saturated liquid at the extraction pressure ideally.

Figure-13 (Shows Ideal Regenerative Cycle with a Closed FWH)13


Figure-14 (Shows T-s Diagram of Ideal Regenerative Cycle with a Closed FWH)14

Figure-15 (Shows Ideal Regenerative Cycle with one Open & three Closed FWH)15


1) CYCLE 1-IDEAL RANKLINE BASIC CYCLE

Schematic diagram of cycle-1

T-s diagram of cycle-1

The cycle 1 was calculated simply to obtain the value of the exhaust pressure that satisfies the conditions of the highest efficiency of the cycle at the steam quality of 87%. The calculations were divided into two different methods. One method was to obtain the exhaust pressure with respect to steam quality and the other was with respect to the efficiency. However both the calculations required almost the same approach. The formulas that are required to obtain the exhaust pressure and the efficiency of the cycle are given as, State 1


P1=P4 (At this point on the curve it is a Saturated Liquid Region.) Where:

1

1

1

@1

@1

@1

Pf

Pf

Pf

ssvvhh

===

State 2 P2=P3 (At this point on the curve, it is in the Compressed Liquid Region.) s2=s1 The work of the pump must be calculated, and with that work the enthalpy value h2 must be calculated using:

12

12

121

40395.5)(

hWhhhW

PPvW

P

p

P

+=

−=

−=

State 3 Superheated Region

3

3

3

23

1000 sh

FTPP

°==

Note: Values will be obtained from Table A6E and interpolation if necessary. State 4

34

14

ssPP

==

To calculate h4 quality must first be calculated by the equation:

fg

f

sss

x−

= 44 ,

then if x4 <1, h4 can then be calculated using:

fgf hxhh 44 += , if not Table A6E must be used to acquire the value of h4. where the values for Sf, Sfg, hf, and hfg were obtained from Table A5E and interpolations. Now that all the enthalpy values have been found, the efficiency of the cycle must be calculated by using the following equations:

in

outth q

q−= 1η ,

where:

23

14

hhqhhq

in

out

−=−=


CALCULATING EXHAUST PRESSURE W.R.T.STEAM QUALITY

P1=P4 Psi

s4=s3 Btu/lbm.R

sf Btu/lbm.R

sg Btu/lbm.R

sfg Btu/lbm.R

x4

1 1.6007 0.13262 1.9776 1.84498 0.795716 10 1.6007 0.28362 1.7875 1.50388 0.875788 20 1.6007 0.33582 1.7319 1.39608 0.906023 30 1.6007 0.36821 1.6995 1.33129 0.925786

Table-1 (exhaust pressure w.r.t steam quality)

Graph-1


CALCULATING EXHAUST PRESSURE W.R.T.EFFICIENCY P1=P4

Psi x4

v1=vf ft3/lbm

h1=hf Btu/lbm

wpump,in Btu/lbm

h2 Btu/lbm

hg Btu/lbm

hfg Btu/lbm

h4 Btu/lbm

Qin Btu/lbm

Qout Btu/lbm

ηth %

1 0.795716 0.01614 69.72 4.47707 74.19707 1105.4 1035.68 893.8271 1416.603 824.1071 0.418251 10 0.875788 0.01659 161.25 4.574265 165.8243 1143.1 981.85 1021.142 1324.976 859.8924 0.351013 20 0.906023 0.01683 196.27 4.609295 200.8793 1156.2 959.93 1065.988 1289.921 869.7183 0.325758 30 0.925786 0.017 218.93 4.624395 223.5544 1164.1 945.17 1093.955 1267.246 875.0254 0.309506

Table-2 (exhaust pressure w.r.t. efficiency)

Graph-2


2) CYCLE 2-IDEAL RANKLINE CYCLE WITH REHEATING


T-s diagram of cycle-2 The following cycle was used to find the optimum value of reheat pressure in order to achieve the highest efficiency. The cycle is based on the two parts, first is to obtain the rehating pressure at the maximum efficiency and then in next step using the optimum reheating pressure is to obtain the exhaust pressure at the maximum moisture level or at the given steam quality as 87%.

1

2

3

4

5

6

T

s

P1 = P3 = 1500 psi

P1 = P6 = 9.34944 psi

Reheat P4 = 230 psi


CYCLE 2 (PART-1) State 1

16 PP = (At this point on the curve it is a Saturated Liquid) Where:

1

1

1

@1

@1

@1

Pf

Pf

Pf

ssvvhh

===

State 2

32 PP = (At this point on the curve, it is in the Compressed Liquid Region)

12 ss = The work of the pump must be calculated, and with that work the enthalpy value h2 must be calculated using:

12

12

121

40395.5)(

hWhhhW

PPvW

P

p

P

+=

−=

−=

State 3

3

3

3

23

1000 sh

FTPP

°==

Note: Values will be obtained from Table A6E and interpolation if needed. State 4 Superheated Region

434

34 )5.01.0(h

ssPP

=−=

Note: Value obtained by use of Table A6E and interpolation if needed. State 5 Inlet of Intermediate Turbine, Superheated Region.

5

5

35

45

sh

TTPP

==

State 6

56

16

ssPP

==

To calculate h6, quality must first be calculated by the equation:

fg

f

sss

x−

= 66 ,


then if x6 <1 h6 can then be calculated using:

fgf hxhh 66 += , if not Table A6E must be used to acquire the value of h6. where the values for Sf, Sfg, hf, and hfg were obtained from Table A5E and interpolations. Now that all the enthalpy values have been found, the efficiency of the cycle must be calculated by using the following equations:

in

outth q

q−= 1η ,

where:

)()( 4523

16

hhhhqhhq

in

out

−+−=−=

p4=p5 Psi

h2 Btu/lbm

h3 Btu/lbm

h4 Btu/lbm

h5 Btu/lbm

h6 Btu/lbm

Qin Btu/lbm

Qout Btu/lbm

ηth %

180 162.4403 1490.8 1237.064 1530.1 1183.606 1621.396 1025.736 0.367375 225 162.4403 1490.8 1258.473 1528.9 1165.784 1598.787 1007.914 0.369576 300 162.4403 1490.8 1287.622 1526.7 1147.314 1567.438 989.4433 0.368751 450 162.4403 1490.8 1338.15 1522.4 1114.682 1512.61 956.812 0.367443

Table-3 (data for obtaining reheat pressure w.r.t. efficiency)

Graph-3


CYCLE 2 (PART-2) State 1

16 PP = (At this point on the curve it is a Saturated Liquid) Where:

1

1

1

@1

@1

@1

Pf

Pf

Pf

ssvvhh

===

State 2

32 PP = (At this point on the curve, it is in the Compressed Liquid Region)


12

12

121

40395.5)(

hWhhhW

PPvW

P

p

P

+=

−=

−=

State 3

3

3

3

23

1000 sh

FTPP

°==

Note: Values will be obtained from Table A6E and interpolation if needed. State 4 Superheated Region

434

4 230h

sspsiP

==

Note: Value obtained by use of Table A6E and interpolation if needed. State 5 Inlet of Intermediate Turbine, Superheated Region.

5

5

35

45

sh

TTPP

==

State 6

56

16

ssPP

==


fg

f

sss

x−

= 66 ,


then if x6 <1 h6 can then be calculated using:

fgf hxhh 66 += , if not Table A6E must be used to acquire the value of h6. Where the values for Sf, Sfg, hf, and hfg were obtained from Table A5E and interpolations. Now that all the enthalpy values have been found, the efficiency of the cycle must be calculated by using the following equations:

in

outth q

q−= 1η ,

where:

)()( 4523

16

hhhhqhhq

in

out

−+−=−=

P6 Psi

x6

h6 Btu/lbm

Qout Btu/lbm

Qin Btu/lbm

ηth %

10 1.026391 1168.981 1007.731 1590.95 0.366585 9.34944 1.022325 1163.714 1005.844 1594.334 0.369113

8 1.014077 1152.92 1002.06 1601.353 0.374242 6 0.998912 1132.817 994.7969 1614.206 0.383724 5 0.989683 1120.357 990.1775 1622.054 0.389553 4 0.978761 1105.533 984.6334 1631.345 0.396428 3 0.965238 1086.993 977.5931 1642.859 0.404944 2 0.947141 1061.714 967.6936 1658.255 0.416439 1 0.918507 1021.018 951.298 1682.577 0.434618

0.95052 0.916512 1018.178 950.1484 1684.27 0.435869 0.81643 0.910617 1009.627 946.5869 1689.262 0.439645 0.69904 0.904714 1001.124 943.0743 1694.257 0.44337 0.59659 0.898808 992.4938 939.4338 1699.25 0.447148 0.50745 0.892881 983.8097 935.7397 1704.245 0.450936 0.43016 0.88696 975.1762 932.1062 1709.247 0.454669 0.36334 0.881024 966.4151 928.3351 1714.24 0.458457 0.30578 0.875078 957.6001 924.5201 1719.243 0.462251 0.25638 0.869122 948.8283 920.7483 1724.242 0.465998 0.21413 0.863158 939.9166 916.8466 1729.255 0.469803

Table、、、、、

-、4 (data for obtaining exit pressure & eff、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、

iciency at maximum moisture level)、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、


Graph-4


3) REGENERATIVE IDEAL RANKIN CYCLE In the following cycle open feed water heaters are introduced continuously at the end of same reheat Rankin cycle. The following cycle has been divided into three cycles,

CYCLE 3 (CYCLE-2 + 1 OPEN FWH)

Design Criteria: Power Output of Power Plant: 150,000 kW Turbine Inlet Pressure: 1500 psi Steam Quality: 0.87 Reheating Regeneration with one open feed water heater

The Open Feed Water Heater can be classified as a regenerative tool. Regeneration improves cycle efficiency and also provides a solution to prevent corrosion in the boiler. This process also helps control the large volumetric flow rate of the steam at its final stages of the turbine. Figure 5 below shows an Ideal Reheat Rankine Cycle with Regeneration.




According to the cycle pressures at each location are distributed as, P1 = P8 = 0.306271Psi P2 = P3 = P6 = P7 = 230Psi P4 = P5 = 1500Psi State 1

psiPP 306271.081 == (At this point on the curve it is a Saturated Liquid) Where:

RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/06519.0

/01604.0

/1165.33

1

1

1

@1

3@1

@1

State 2

psiPP 23032 == (At this point on the curve, it is in the Compressed Liquid Region)


lbmBtuhWhhhW

PPvW

P

p

P

/7983.33

40395.5)(

12

12

121

=+=

−=

−=


State 3 PsiPPPP 2307623 ==== (At this point on the curve it is a Saturated Liquid)

Where:

RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/5582.0

/01855.0

/836.367

3

3

3

@3

3@3

@3


lbmBtuhPPv

hssPP

/1955.37240395.5

)(3

3434

34

54 =+−

=

==

Note: Value obtained by use of Table A6E and interpolation if needed. State 5

RlbmBtuslbmBtuh

FTpsiP

⋅==

°==

/6007.1/8.1490

10001500

5

5

5

5

Note: Values will be obtained from Table A6E and interpolation if needed. Inlet of High Pressure Turbine, Superheated Region. State 6 Reheat Pressure

lbmBtuhsspsiP

/599.1260230

656

6=

==

Note: Values will be obtained using A6E and interpolation if needed. State 7

RlbmBtuslbmBtuh

TTPsiP

⋅==

==

/82722.1/76.1528230

7

7

57

7

Note: Values will be obtained from Table A6E and interpolation if needed. Inlet of Intermediate Turbine, Superheated Region. State 8

78

18

ssPP

==


875108.088 =

−=

fg

f

sss

x ,



lbmBtuhxhh fgf /6466.95788 =+= , if not Table A6E must be used to acquire the value of h8. where the values for Sf, Sfg, hf, and hfg were obtained from Table A5E and interpolations. Now that all the enthalpy values have been found, the efficiency of the cycle must be calculated by using the following equations:

( ) %8.481001 =×−=in

outth q

qη ,

where:

lbmBtuhhyhhqlbmBtuhhyq

in

out

/776.1313))(1()(/7876.672))(1(

6745

18

=−−+−==−−=

y represents the fraction of steam extracted from the turbine and it is derived as follows:

26

23

hhhh

y−−

=

Substituting all enthalpy values and y into efficiency formula yields thη = 48.8% Using the design criteria of power plant output of 150,000 KWT and rearranging the

power formula to calculate for mass flow rate, the equation becomes:

netWPowerm 3412×

=

lbmBtuqqW outinnet /9885.640=−=

hrlbmm /7984542.798454 ≈=


CYCLE 4 (CYCLE-3 + 1 OPEN FWH) Design Criteria: Power Output of Power Plant: 150,000 KW Reheating and Regeneration with two Open feed Water Heaters Beginning of Low Pressure Turbine is 40 -50 psi.




According to the cycle the pressure distribution at each location are as follows; P1 = P11 = 0.306271Psi P2 = P3 = P10 = 115.1531Psi P4 = P5 = P8 = P9 = 230Psi P6 = P7 = 1500Psi State 1


RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/065191.0

/01604.0

/165.33

1

1

1

@1

3@1

@1

State 2

psiPPP 1531.1151032 === (At this point on the curve, it is in the Compressed Liquid Region)



lbmBtuhWhhhW

PPvW

P

p

P

/5059.33

40395.5)(

12

12

121

=+=

−=

−=

State 3

1023 PPP == (At this point on the curve it is a Saturated Liquid) Where:

RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/4878.0

/01785.0

/1531.305

3

3

3

@3

3@3

@3

State 4

psiPPP 230854 ===

34 ss =

lbmBtuhPPvh /648.30540395.5

)(3

3434 =+

−=

State 5

845 PPP == Where:

RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/5582.0

/01855.0

/836.367

5

5

5

@5

3@5

@5


lbmBtuhPPvhssPP

/1955.37240395.5

)(5

5656

56

76 =+−

=

==

Note: Value obtained by use of Table A6E and interpolation if needed. State 7

RlbmBtuslbmBtuh

FTpsiP

⋅==

°==

/6007.1/8.1490

10001500

7

7

7

7

Note: Values will be obtained from Table A6E and interpolation if needed. Inlet of High Pressure Turbine, Superheated Region.


State 8 Reheat Pressure

lbmBtuhssPP

/559.1260878

48 =

==


RBtu/lbm82722.1s/76.15289

979

89

⋅==

== lbmBtuh

TTPP


lbmBtuhss

PPP/46.142610

910

2310 =

===

Note: Values will be obtained from Table A6E and interpolation if needed. Exit of Intermediate Turbine, Superheated Region. State 11

1011

111

ssPP

==


875108.01111 =

−=

fg

f

sss

x ,



( ) %19.491001 =×−=in

outth q

qη ,

where:

lbmBtuhhhhyqlbmBtuhhzyq

in

out

/339.1369)())(1(/6933.695))(1(

6789

111

=−+−−==−−−=


where y and z are steam extractions from the turbine and are calculated by:

48

45

hhhh

y−−

= and )(

))(1(

210

23

hhhhy

z−

−−=

Substituting all enthalpy values and y into efficiency formula yields thη = 49.2% Using the design criteria of power plant output of 150,000 KWT and rearranging the


netWPowerm 3412×

=


hrlbmm /7597479.759746 ≈=


CYCLE 5 (CYCLE-4+ 2 OPEN FWH) Design Criteria: Power Output of Power Plant: 150,000 KW Reheating and Regeneration with Four Open Feed Water Heaters Beginning of Low Pressure Turbine is 40 -50 psi.



3

4

8

T

13

17

15

10

12

16

5

6 7

2

9

1

11

14


According to the cycle the pressure distribution at each location are given as; P1 = P17 = 0.306271Psi P2 = P3 = P16 = 28.75Psi P4 = P5 = P15 = 57.5Psi P6 = P7 = P14 = 115Psi P8 = P9 = P12 = P13 = 230Psi P10 = P11 = 1500Psi State 1


RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/065191.0

/01604.0

/165.33

1

1

1

@1

3@1

@1

State 2

psiPPP 75.281632 === (At this point on the curve, it is in the Compressed Liquid Region)


RlbmBtuhWhhhW

PPvW

P

p

P

⋅=+=

−=

−=

/221.33

40395.5)(

12

12

121

State 3

1623 PPP == (At this point on the curve it is a Saturated Liquid) Where:

RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/3645.0

/01698.0

/3275.216

3

3

3

@3

3@3

@3

State 4

psiPPP 5.571554 ===

34 ss =

lbmBtuhPPvh /4178.21640395.5

)(3

3434 =+

−=


State 5 1545 PPP ==

Where:

RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/4234.0

/01735.0

/31.259

5

5

5

@5

3@5

@5

State 6

psiPPP 1151476 ===

56 ss =

lbmBtuhPPvh /4946.25940395.5

)(5

5656 =+

−=

State 7

1467 PPP == Where:

RlbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

⋅==

==

==

/4876.0

/01785.0

/785.308

7

7

7

@7

3@7

@7

State 8

psiPPPP 230131298 ====

78 ss =

lbmBtuhPPvh /1649.30940395.5

)(7

7878 =+

−=

State 9

131289 PPPP === Where:

lbmBtuss

lbmftvv

lbmBtuhh

Pf

Pf

Pf

/5582.0

/018546.0

/838.367

9

9

9

@9

3@9

@9

==

==

==

State 10

PsiPP 15001110 ==

910 ss =

lbmBtuhPPvh /1966.37240395.5

)(9

910910 =+

−=


State 11

RlbmBtuslbmBtuh

FTpsiP

⋅==

°==

/6007.1/8.1490

10001500

11

11

11

11

Note: Values will be obtained from Table A6E and interpolation if needed. Inlet of High Pressure Turbine, Superheated Region. State 12 Superheated Region

lbmBtuhss

PsiPPPP/559.1260

230112

1112

981312 =

=====

Note: Value obtained by use of Table A6E and interpolation if needed. State 13 Reheat Pressure

RlbmBtuslbmBtuh

FTPsiPPPP

⋅==

°=====

/82722.1/76.1528

1000230

13

13

13

981213


lbmBtuhss

PsiPPP/046.1387

11514

1314

6714 =

====


lbmBtuhss

PsiPPP/047.1338

5.5715

1514

4515 =

====

Note: Values will be obtained from Table A6E and interpolation if needed. State 16

lbmBtuhss

PsiPPP/03.1259

75.2816

1615

2316 =

====

Note: Values will be obtained from Table A6E and interpolation if needed. State 17

1716

117 306271.0ss

PsiPP=

==


875108.01717 =

−=

fg

f

sss

x ,




( ) %47.501001 =×−=in

outth q

qη ,

where:

lbmBtuhhhhyqlbmBtuhhuxzyq

in

out

/264.1370)())(1(/6843.678))(1(

10111213

117

=−+−−==−−−−−=

where y, z, x, and u are steam extractions from the turbine and are calculated by:

812

89

hhhh

y−−

= , )(

))(1(

614

67

hhhhy

z−

−−= ,

)())(1(

416

45

hhhhzy

x−

−−−= , and

)())(1(

217

23

hhhhxzy

u−

−−−−=

Substituting all enthalpy values and u, x, y, and z into efficiency formula yields an efficiency of thη = 50.5% Using the design criteria of power plant output of 150,000 KWT and rearranging the


netWPowerm 3412×

=


hrlbmm /7400459.740044 ≈=


• The exhaust pressure was found as 9.3944Psi from the graph-1 (steam quality Vs exhaust pressure, similarly the maximum efficiency was found 35.26% (approximately 35.3%) from the grph-2 (efficiency Vs Exhaust pressure). This precise estimation of the pressure and efficiency satisfies the condition of the maximum moisture level 13% or in other words at 87% steam quality the exit pressure and efficiency was obtained as 9.34944Psi and 35.3% respectively.

• At the maximum efficiency the optimum value of reheat pressure was found as 230Psi from graph-3 (efficiency Vs pressure), and from that reheating pressure of 230Psi at the maximum steam quality as 87% the exit pressure and efficiency were found as 0.306271Psi and 46.2% respectively.

• At the given design specification the efficiency and mass flow rate was calculated and given as;

Cycle Condition Efficiency %

Mass flow rate

lbm/hr

Mass flow rate

lbm/min

Mass flow rate

lbm/sec Cycle-3 Cycle 2 + 1

open FWH 48.8 798454 13308 222

Cycle-4 Cycle 3 + 1 open FWH

49.2 759747 12662 211


50.5 740045 12334 206

Table-5 (cycle with efficiency and mass flow rate)


The main objective of the designing and optimizing the following steam turbine cycle is to achieve the highest efficiency keeping the certain design specifications such as inlet pressure (1500Psi), inlet temperature (1000oF), power plant output power (150000 KW) and maximum moisture level (13%). Therefore it has been noticed clearly that from the ideal rankin cycle to the regenerative rankin cycle efficiency is increased as it was expected to be increase, however at one point in the cycles the increase of the efficiency is relatively slower than the efficiency that was obtained ideally or by reheating the steam. Also it may effect on the quality of the steam as well by adding few impurities in the steam as well. The increment in increasing the efficiency was calculated and given as;

Cycle Condition Efficiency %

Increase in Efficiency

% Cycle-1 Ideal Rankin cycle 35.3 N/A Cycle-2 Reheat Rankin cycle 46.2 10.9 Cycle-3 Regenerative cycle-

2 + 1 open FWH 48.2 2


49.2 1


50.4 1.2

Table-6 (All cycles with the calculated efficiency)


References:

1. http://www.frings.com/wEnglisch/upload/bilder/beluefter_prinzip_fett.gif 2. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach,

The Ideal Rankine Cycle, Vapor and combined Power Cycle, pg#553, 5th edition. 3. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach,

T-s diagram of Ideal Rankine Cycle, Vapor and combined Power Cycle, pg#553, 5th edition.

4. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, Deviation of Actual Vapor Cycles From Idealized Ones , Vapor and combined Power Cycle, pg#558, 5th edition.

5. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, T-s Diagram (Effect of Pressure on Cycle), Vapor and combined Power Cycle, pg#560, 5th edition.

6. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, T-s Diagram (Effect of Superheating on Cycle) , Vapor and combined Power Cycle, pg#561, 5th edition.

7. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, T-s Diagram (Effect of Increasing Boiler Pressure on Cycle) , Vapor and combined Power Cycle, pg#561, 5th edition.

8. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, Ideal Reheat Rankine Cycle , Vapor and combined Power Cycle, pg#565, 5th edition.

9. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, T-s diagram of Ideal Reheat Rankine Cycle, Vapor and combined Power Cycle, pg#565, 5th edition.

10. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, T-s diagram of heat addition at low temperature, Vapor and combined Power Cycle, pg#568, 5th edition.

11. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, Ideal Regenrative Cycle with an Open FWH, Vapor and combined Power Cycle, pg#569, 5th edition.

12. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, T-s Diagram of Ideal Regenrative Cycle with an Open FWH, Vapor and combined Power Cycle, pg#569, 5th edition.

13. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, Ideal Regenrative Cycle with a Closed FWH, Vapor and combined Power Cycle, pg#570, 5th edition.

14. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, T-s Diagram of Ideal Regenrative Cycle with a Closed FWH, Vapor and combined Power Cycle, pg#570, 5th edition.

http://www.frings.com/wEnglisch/upload/bilder/beluefter_prinzip_fett.gif


15. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, Ideal Regenrative Cycle with one Open and three Closed FWH, Vapor and combined Power Cycle, pg#571, 5th edition.

Acknowledgement

1. Cengel A. Yunus, Boles A. Michael, Thermodynamics An Engineering Approach, Vapor and combined Power Cycle, pg#551-576, 5th edition.

prof. dr. rishi raj design and optimize a steam turbine

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