Lecture note steam power cycle

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Vapor Power Cycles

Vapor Power Plant 1

Introduction

Chapter Objective

The objective of this chapter is to carry out first law and

second law analysis on a vapor power plant in which the

working fluid is alternatively vaporized and condensed as it

completes a thermodynamics cycle.

What is a Vapor Power Plant?

A vapor power plant is a thermodynamics heat engine used to

produce mechanical power output from energy sources such

as fossil fuel, nuclear (uranium) or solar energy.

Vapor Power Plant 2

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Introduction Cont’d

The power output is typically used to drive electrical

generator, to produce electricity for our everyday use.

The plant uses water as a working fluid, which will be

alternately vaporized and condensed as it undergoes a

complete thermodynamics cycle.

Note: The processes taking place in actual power generating

system are complicated. To carry out thermodynamics study

on the system, we will develop a simplified model of the

system.

Vapor Power Plant 3

Vapor Power Plant4

Simplified Model for Analysis

Figure 1 A simplified model for a fossil-fuel vapor

power plant

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Simplified Model for Analysis

• Subsystem A: This is where the energy conversion process occurs. Heat

energy, obtained from thermal energy sources such as fossil-fuel and

nuclear, is converted into mechanical work, in a form of a rotating shaft.

• Subsystem B: This subsystem supplies the energy required to vaporize the

liquid water. In fossil-fuel plants, this is accomplished by heat transfer from

the hot gases produced from the combustion of the fossil-fuel, to the liquid

water passing through the tubes and drums in the boiler.

• Subsystem C: It comprises of a cooling water circuit. Cooling water is used

to cool off the wet vapor exiting the turbine, thus condensing it back into the

liquid water. The hot cooling water is sent to a cooling tower, where the heat

energy taken up in the condenser is rejected to the atmosphere.

• Subsystem D: It comprises of an electric generator, which is connected to

the turbine via a shaft. The shaft rotates as the steam expands to lower

pressure through the turbine. It drives the generator, which produces

electrical power output.

Vapor Power Plant 5

Focus of the Analysis

Vapor Power Plant 6

We will focus our analysis on Subsystem A, shown in Figure 2.

Figure 2 Subsystem A of the plant

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Basic Components

Vapor Power Plant 7

The functions of the basic components of Subsystem A are:

a) Boiler: to transform liquid water into vapor (steam) of high

pressure and temperature.

b) Turbine-Generator: to transform the kinetic energy of the vapor

into mechanical power (rotating shaft). The mechanical power is

used to drive an electric generator, to produce electricity.

c) Condenser: to cool off the wet vapor exiting the turbine and

transform it back into the liquid water.

d) Feed-water Pump: to deliver the water exiting the condenser back

into the boiler, thus completing one thermodynamics cycle.

Revision

• The 1st Law of Thermodynamics

Net heat received by any cyclic device is the same with

the net work produced.

• The 2nd Law of Thermodynamics

It is impossible for any device that operates on a cycle to

receive heat from a single reservoir and produce a net

amount of work. Certain amount of heat must be rejected

to the surrounding.

Vapor Power Plant 8

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The 1st Law of Thermodynamics

Vapor Power Plant 9

∑ ∑= WQ

netnet WQ =

netWQQ =− ⋯21

The 2nd Law of Thermodynamics

Vapor Power Plant 10

Cyclic

device

Q1

Q2

Wt

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Revision

Isentropic Process

A process during which the entropy remains constant. It also

can be recognized as internally reversible, adiabatic process.


)./(0 12 KkgkJssors ==∆

Energy Balance for Steady-flow Systems

The first law or energy balance relation for a general steady-flow system

is,

For single-stream (one-inlet-one exit) systems, the inlet and exit states

are denoted by subscripts 1 and 2 for simplicity. The mass flow rate

through the entire control volume remains constant

( ) and is denoted by . Then the energy balance for single-

stream steady-flow systems becomes

21

••

= mm

∑∑

+

∇+−

+

∇+=−

••••

��

inleteachfor

ii

ii

exiteachfor

ee

ee gzhmgzhmWQ22

22


•

m

( )

−+

∇−∇+−=−

•••

12

2

1

2

212

2zzghhmWQ

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Energy Balance for Steady-flow Systems

If the fluid experiences a negligible change in its kinetic and

potential energies as it flows through the control volume

(that is, ∆KE = 0, ∆PE = 0), then the energy equation for a

single-stream steady-flow system reduces further to

(1)


( )12 hhmWQ −=−•••

Performance of Steam Plant

1. Specific steam consumption (s.s.c.)

Defined as the steam flow rate in kg/hr required to develop 1 kWatt of power output.

(2)

The lower the s.c.c the more compact the steam plant.

2. Work ratio (wr)

Defined as the ratio of the net work produced by the plant to the work produced by the turbine, i.e.,

(3)

where Wt = turbine work

Wp = pump or compressor work

( )3600 3600. . . / .

net net t p

ms s c kg kW hr

W W W W= = =

−


t

pt

t

net

W

WW

W

Wwr

−==

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3. Thermal efficiency (ηth)

Defined as the ratio of net work produced by the plant to

the amount of heat added to the working fluid in the

boiler i.e.,

(4)


%100%100 xQ

WWx

Q

W

in

pt

in

netth

−==η


4. Isentropic efficiency (ηis)

The actual expansion and pumping processes

are adiabatic but not reversible. Thus, they

are not isentropic.

For the expansion process in the turbine,

(5)

For the pumping process in the feed-pump,

(6)( )( )3'

4

34,

hh

hh

workActual

workIsentropicpis −

−==η


( )( )21

'

21,

hh

hh

workIsentropic

workActualtis −

−==η

T

s

1

2’23

4

4’

Figure 3

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5. Back work ratio

Defined as the ratio of the work supplied to the feed-water pump to the work

produced by the turbine, i.e.,

(7)

6. Efficiency ratio

Defined as,

(8)efficiencycycleRankineIdeal

cycle actual of efficiency thermal=ratioη


t

p

W

Wbwr =

• The most efficient cycle is the Carnot cycle

for given temperatures of source and sink

(T1 and T2).

• It also can be called as the ideal heat engine.

• The cycle for a wet vapor is shown in Figure 4

and a brief summary of the essential features

is as follows,

4 to 1: heat is supplied at constant temperature

and pressure.

1 to 2: the vapor expands isentropically from

the high pressure and temperature to the low pressure.

2 to 3: the vapor, which is wet at 2, has to be cooled to state

point 3.

3 to 4: isentropic compression from 3 to 4. From 4 the cycle is

repeated.

The Carnot Cycle


ss3 = s4 s1 = s2

Figure 4

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• Although the Carnot cycle is the most efficient cycle,

there are several problems such as,

1. the work ratio is low

2. it is difficult to stop the condensation

at point 3 and then compressed it just

to state 4

3. the steam at outlet from the turbine is wet,

so it can damage the turbine blades

4. the 3 to 4 process is bad for the compressor because

two phases compression is not practical.

• The efficiency of the cycle can be calculated by,

(9)

The Carnot Cycle

1

2

1

2 11T

Tor

Q

Qthth −=−= ηη


Figure 4 Repeated

ss3 = s4 s1 = s2


The Rankine Cycle

Basic(Ideal/actual)

WithSuperheat

(Ideal/actual)

Reheat Cycle(Ideal/actual)

Regenerative Cycle

With Open-type

Feedwater Heater

(Ideal/actual)

Regenerative Cycle

With Closed-type

Feedwater Heater

(Ideal/actual)

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Basic Rankine Cycle

The working fluid undergoes a thermodynamicscycle as it flows through each component of theplant. The cycle is called an ideal Rankine cycle,and is shown on a temperature-entropy (T-s)diagram, in Figure 5.


Basic Rankine Cycle


Figure 2 Repeated

Figure (2-3) Repeated

Figure 5 An ideal

Rankine Cycle

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Basic Rankine Cycle

By comparing with the Carnot Cycle, this basicRankine Cycle is more practical since, it is moreconvenient to allow condensation process toproceed to completion.

In addition, the working fluid is water, and this canbe conveniently pumped to boiler pressure asshown at point 4.


Basic Rankine Cycle

The Cycle Analysis

The steam flows round the cycle and each process isanalyzed using the steady flow energy equation(sfee).

The changes in kinetic energy and potential energyare neglected.


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Basic Rankine Cycle

The Cycle Analysis

a) Boiler

Since there is no work interaction

between the working fluid and the

surrounding, W = 0. Hence, the

amount of heat added to the working

fluid in the boiler is

(10)

b) Turbine

Since the expansion process is assumed to be isentropic (reversible

adiabatic), then Q = 0. Thus the amount of mechanical work produced by the

turbine is

(11)


( ) kgkJhhqq in /4114 −==−


( ) kgkJhhww t /2121 −==−

Basic Rankine Cycle

The Cycle Analysis

c) Condenser

There is no work interaction,

so that W = 0. Hence, the amount

of heat rejected from the working

fluid to the cooling water is

(12)

d) Feed-water pump

Since the pumping process is assumed to be isentropic (reversible

adiabatic), then Q = 0. Thus the amount of work supplied to the feed-water

pump is

(13)

where v is the specific volume (m3/kg) of water at pressure p3.


( ) kgkJhhqq out /3232 −==−


( ) ( ) kgkJppvhhww p /3433443 −=−==−

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Example 1

A steam power plant operates between a boiler pressure of

50 bar and a condenser pressure of 0.035 bar. Calculate for

these limits the thermal efficiency, the work ratio, and the

specific steam consumption:

a) for a Rankine cycle with dry saturated steam at entry

to the turbine

b) for a Rankine cycle with the turbine isentropic

efficiency of 85 %.

Sketch the cycle on a T-s diagram.


The Solutions

( ) ( )( )

( ) ( )( )

441.0

11552794

2.87011559.17852794

41

3421

=

−−−−

=

−−−−

=−

==hh

hhhh

q

ww

q

w

in

ct

in

netthη


a) h1 = hg@50 bar = 2794 kJ/kg

s1 = sg@50 bar = s2 = 5.973 kJ/kgK

= sf + [email protected] bar

= 0.391 + x2 (8.130)

x2 = 0.6866

h2 = hf + [email protected] bar

= 112 + 0.6866 (2438)

= 1785.9 kJ/kg

s4 = sf@50 bar = s3 = 2.921 kJ/kg

= sf + [email protected] bar

= 0.391 + x3 (8.130)

x3 = 0.311


= 112 + 0.311 (2438)

= 870.2 kJ/kg

h4 = hf@50 bar = 1155 kJ/kg

T

s

1

23

4

50 bar

0.035 bar

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The Solutions

( ) ( )( )

( ) ( )( )

717.0

9.17852794

2.87011559.17852794

21

3421

=

−−−−

=

−−−−

=−

=hh

hhhh

w

wwwr

t

ct

( ) ( )

( ) ( )hrkWkg

hhhhwwcss

ct

./98.4

2.87011559.17852794

3600

36003600...

3421

=

−−−=

−−−=

−=


T

s

1

23

4

50 bar

0.035 bar

The Solutions

( )( )

( ) ( ) ( ) kgkJxhhxhhwhh

hh

w

wtisactual

Isentropic

actualtis /9.8569.1785279485.021,

'

21

21

'

21, =−=−=−=→

−−

== ηη

( )( ) ( ) ( ) ( )

kgkJhh

hhwhh

hh

w

w

cis

actual

actual

Isentropic

cis /33585.0

2.8701155

,

343

'

4

3

'

4

34, =

−=

−=−=→

−−

==η

η

kgkJhhq

kgkJhkgkJhh

in /8.15882.12052794

/2.12052.870335/335

'

41

'

3

'

44

=−=−=

=+=→=−

328.08.1588

3359.856

14

=−

=−

==−q

ww

q

w ct

in

netthη


b)

hrkWkgw

css

w

wwr

net

t

net

./9.63359.856

36003600...

609.09.856

3359.856

=−

==

=−

==

T

s

1

23

4

50 bar

0.035 bar

2’

4’

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The Solutions

( ) ( )( )

375.041

3421

14

=−

−−−=

−==

− hh

hhhh

q

ww

q

w pt

in

netthη

( ) ( ) kgkJxppvhhwp /9.410035.050001.0 2

34334 =−=−≈−=

kgkJWhh p /9.1169.411234 =+=+=


c)

( ) ( )( )

( ) ( )hrkWkg

hhhhwcss

hh

hhhh

w

wwr

net

t

net

./59.336003600

...

995.0

3421

21

3421

=−−−

==

=−

−−−==

T

s

1

23

4

50 bar

0.035 bar

The Solutions

85.021

'

21, =

−−

==hh

hh

w

w

isentropic

actualtisη ( ) kgkJhhhh /9.85685.0 21

'

21 =−=−→

( ) ( )( )

32.041

34

'

21

14

=−

−−−=

−==

− hh

hhhh

q

ww

q

w pt

in

netthη

( ) ( )( )

( ) ( )hrkWkg

hhhhwcss

hh

hhhh

w

wwr

net

t

net

./23.436003600

...

845.0

34

'

21

'

21

34

'

21

=−−−

==

=−

−−−==


d)

T

s

1

23

4

50 bar

0.035 bar

2’

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Rankine Cycle with Superheating

• Improvement in the basic Rankine cycle.

• Steam temperature at inlet to the turbine is increased at

boiler pressure, thus increasing the average temperature

of heat addition.

• Increase the cycle efficiency.

• Steam exits the turbine is more dry, i.e., the dryness

fraction, x2 of the steam increases.

• Specific steam consumption drops.



Technique of Superheating

The saturated steam exiting the boiler is passed

through a second bank of smaller tubes located

within the boiler, which is heated by the hot gases

from the furnace.


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Hot well

Receiver

Collecting steam from

others boiler

Provide storage

for the condensate

Degree ofsuperheat, ∆T

Superheated temperature, Tsh

Saturation temperatureTs at boiler pressure

1

Figure 6 Ideal Rankine cycle

with superheating

Example 2

Reconsider the vapor power cycle of Example 1.

Calculate it’s thermal efficiency and s.s.c. if the steam

exiting the boiler is heated to 500oC before entering

the turbine. Assume the pump work is small and can

be neglected.

Sketch the cycle on a T-s diagram.


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The Solutions


s

0.035 bar

h1 = 3433 kJ/kg

s2 = s1 = 6.975 kJ/kg = sf + [email protected] bar

= 0.391 + x2 (8.310)

x2 = 0.81


= 112 + 0.81 (2438) = 2086.8 kJ/kg

h3 = [email protected] bar = 112 kJ/kg

h4 = h3 pump work neglected

( )( )

( )( )405.0

1123433

8.20863433

41

21

=

−−

=

−−

=−

==hh

hh

q

ww

q

w

in

pt

in

netthη

( )

hrkWkg

hhwwcss

pt

./67.2

2.1346

3600

36003600...

21

=

=

−=

−=

Rankine Cycle with Reheating

• Improvement in the Superheat Rankine cycle.

• The average heat addition is increased in another way.

• As the steam is expanded in the turbine it is withdrawn at

the point where it just about becomes wet and then reheated

to a high temperature.

• The dryness fraction of the steam exiting the turbine stages

is further increased, which is the desired effect.

• The steam is reheated at constant pressure in process 2 to 3.

• The specific steam consumption is improved (decrease).

• Usually, the steam is reheated to the inlet temperature of the

high-pressure turbine.


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3

4

5

6

Figure 7 A steam power plant

with reheating process

Low pressure

turbine

1

2

3

45

6

T


with superheat and reheat


The Cycle Analysis

a) Heat supplied

The heat added to the working fluid is given by

(14)

b) Work output

The work developed by the turbine is given by

(15)


( ) ( ) kgkJhhhhq

qqq

totalin

totalin

/2361,

3216,

−+−=

+= −−

( ) ( ) kgkJhhhhw

www

totalt

totalt

/3421,

4321,

−+−=

+= −−

Low-pressure

turbine

1

2

3

45

6

T

Figure 8 Repeat

High-pressure

turbine

Reheating

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The Cycle Analysis

c) Work input

The work supplied to the feed-water pump is given by

(16)


( ) ( ) kgkJppvhhwwp /5655665 −=−== −

Low-pressure

turbine

1

2

3

45

6

T

Figure 8 Repeated

High-pressure

turbine

Reheating

Example 3

In a reheat Rankine cycle, steam enters the high-pressureturbine at 15 MPa and 600°C and is condensed in acondenser at a pressure of 10 kPa. If the moisture content ofthe steam at the exit of the low-pressure turbine is not toexceed 10.4 percent, determine:

a) the pressure at which the steam should be reheated

b) thermal efficiency of the cycle

Assume that the steam is reheated to the inlet temperatureof the high-pressure turbine.


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Example 3


Example 4 (Test 1 – 2004/05-1)

A steam power plant operates on the actual Rankine cycle between

boiler pressure of 60 bar and condenser pressure of 0.1 bar. Steam

enters the high-pressure turbine at 500oC and is expanded to a

pressure of 16 bar. Steam is then reheated to 400oC before it

expands in the low-pressure turbine to the condenser pressure.

The isentropic efficiencies for each turbine are 90% and the pump

is 100%. Show a complete schematic diagram of the plant and the

cycle on a T-s diagram. Considering the pump work determine,

(i) the net work output of the plant (kJ/kg);

(ii) the heat supplied to the plant (kJ/kg);

(iii) the thermal efficiency of the plant (%);

(iv) the specific steam consumption of the plant (kg/kWhr);

(v) the thermal efficiency of the plant, if it operates on the ideal

Rankine cycle (%).Vapor Power Plant 44

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Example 4


160 bar

P2 = P3 = 16 bar

2

3

4

56

60 bar5 4

6

1

2

3

60 bar

0.1 bar

2’

4’

15 bar

ToC

s

1

2

3

6

0.1 bar

60 bar

16 bar

2’

4

4’

5

500

400

Example 5 (Test 1 – 2005/06-1)

A steam power plant operates on the actual Rankine cycle.Steam enters the high-pressure turbine at 40 bar and 350oCand leaves at 10 bar. Steam is then reheated at constantpressure to a temperature of 350oC before it expands to 0.03bar in the low-pressure turbine. If the isentropic efficienciesof both turbines are 84% and 80%, respectively, calculate

(i) the work output per kg of steam;

(ii) the heat supplied per kg of steam;

(iii)the thermal efficiency;

(iv)the specific steam consumption.

Show the cycle on a T-s diagram. For this case, neglect the feed pump term.


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Example 5


140 bar

P2 = P3 = 10 bar

2

3

4

56

40 bar5 4

6

1

2

3

40 bar

0.03 bar

2’

4’

10 bar

0.03 bar

0.03 bar

350

The Regenerative Cycle


What is Regeneration Process?

• In a regenerative cycle, the feed-water is preheated in a

feed-water heater (FWH), using some amount of steam

bled off the turbine, before it is delivered back into the

boiler. This is shown in Figure 9.

• The preheating process occurs in the FWH at a

constant

pressure. The steam required for heating the feed-

water is bled off the turbine at certain bleeding

pressure, Pbleed.

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Figure 9 Ideal regenerative

Rankine cycle power plant



Purpose of Regeneration Process

• The main purpose of regeneration process is to

increase the thermal efficiency of the cycle.

• If the feed-water is preheated before entering the

boiler, then less heat will be required to transform

the feed-water into steam, in the boiler.

• As a result, thermal efficiency of the plant increases.

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Types of Feed-water Heater (FWH)

There are two types of feed-water heater or heat

exchanger: an open-type and a closed-type.

a) Open-type Feed-water Heater

• An open-type FWH is basically a “mixing chamber”.

• The feed-water is preheated by direct mixing with the

steam extracted from the turbine.

• The plant can use more than one open feed-water

heater.

• Each open-type FWH requires one extra pump.

Open-type Feedwater Heater


Figure 10 Ideal regenerative

cycle using open-type FWH3

1

4

2

Condenser

Pump 1

Pump 2

steam exits open FWH =

saturated liquid

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Analysis of open-type FWH

The mass of the steam extracted from the boiler, y,

is determined by doing an energy balance on the

feed-water heater.

i.e.

Solve to give,

(17)

The choice of Bled-Off Pressure

(18)


( ) ( )outin

hmhm ∑∑ = ..

( ) ( ) ( ) 326 .1.1. hhyhy =−+

( )( )26

23

hh

hhy

−−

=

bleedbleed

condsboilers

bleed

Tatpressuresaturatedp

TTT

=→

+=

2

,,


Open

FWH

(1 – y)

23

(1)

6

(y)



Open

FWH

(1 – y)

2

3(1)

6

(y)

Note: y is chosen so that the condition

of point 3 is saturated liquid.

3

2

1

4

s

Figure 10 Repeated

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Analysis of the cycle

a) Heat supplied

The amount of heat added to the working

fluid,

qin = q4-5 = (h5 – h4) (19)

b) Turbine work

The total amount of work produced by

the turbine,

wt = (h5 – h6)+(1-y)(h6 – h7) (20)




Analysis of the cycle

c) Heat rejected

The heat rejected to the cooling water in

the condenser,

qout = q7-1 = (1-y)(h7 – h1) (21)

d) Feed-water pump work

The total amount of work supplied to

the feed-water pumps,

wp = (1-y)(h2 – h1)+(h4 - h3) (22)

Note: The heat added to the working fluid

in the FWH = (h3 – h2)


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Example 5

A steam power plant operates on the ideal

regenerative cycle with one feed-water heater. Steam

enters the turbine at 15 MPa and 600°C and is

condensed to a condenser pressure of 10 kPa. Some

steam is extracted from the turbine at a pressure of 1.2

MPa and enters the open-type feed-water heater.

Determine:

(a) the amount of steam extracted from the turbine,

and,

(b) thermal efficiency of the cycle.


Example 5


Pump 11 Pump 1

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30



b) Closed-type Feed-water Heater

• Each open-type FWH requires one extra pump, thus, the plant cost

increases.

• This can be improved by using closed-type FWH.

• The feed-water does not mix freely with the bled off steam.

• There is only a heat transfer from the steam to the feed-water in the

closed-type FWH.

• Since there is no mixing, the bled off steam can be at different pressure

than the feed-water.

• The condensate exiting the closed-type FWH (state 7) is throttled back

into the condenser and mix with the feed-water in the condenser.

• The mixture of condensate and feed-water is then pumped back into the

boiler to repeat the cycle.

Closed-type Feedwater Heater


Throttle valve


With closed-type FWHFigure 11

s

Ideally, T9 = T7

Pbleed = p6

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Analysis of closed-type FWH

Mass of Bled-Off Steam

By performing energy balance on the

closed-type FWH,

i.e.,

Simplify gives the fraction of the steam

bled off,

(23)


( ) ( )∑∑ = outin hmhm ..

( ) ( ) ( ) ( ) 9746 .1..1. hhyhhy +=+

( )( )76

49

hh

hhy

−−

=


s

Figure 12 Repeated

T9 = T7

Pbleed = P6


Analysis of the Cycle

For 1 kg of steam flowing through the cycle,

a) Heat supplied

The heat supplied to the working

q9-1 = (1)(h1–h9) (24)

b) Turbine work

The work produced by the turbine

w1-2 = (1)(h1–h6)+(1-y)(h6-h2) (25)

c) Heat rejected

The heat rejected to the cooling water,

q2-3 = (1-y)(h2-h8)+(1)(h8-h3) (26)


s

Figure 12 Repeated

T9 = T7

Pbleed = P6

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Analysis of the Cycle

d) Feed-water pump work

The feed-water pump work input,

w3-4 = v3(p4-p3) (27)


s

Figure 12 Repeated

T9 = T7

Pbleed = P6


Alternate Scheme

Sometimes, two closed-type FWH are used as shown inFigure 13.


y1 kg (y1 + y2) kg

Figure 13 Regenerative plant using two closed-type FWH

(1 – y1 – y2) kg

Ideally, T6 = T11

T5 = T9

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Mass of Extracted Steam

The mass of steam extracted for feed-water heater 1,

y1.h7 + h5 = y1.h11 + h6 (28)

h6 = h11 = h12

and that for feed-water heater 2,

y2.h8 + y1.h12 + h4 = (y1 + y2)h9 + h5 (29)

h5 = h9 = h10

Note: in liquid region, h = hf@T


Figure 13 Repeated

(1 – y1 – y2) kg

Ideally, T6 = T11

T5 = T9


Cycle Performance

a) Heat supplied

The heat supplied to the working fluid,

qin = q6-1 = (1).(h1+h6) (30)

b) Work output

The work output of the turbine

wt = (h1- h7) + (1-y1)(h7- h8)+(1-y1-y2) (h8- h2) (31)


Figure 13 Repeated

(1 – y1 – y2) kg

Ideally, T6 = T11

T5 = T9

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Comparison

Comparisons between open-type and closed-type feed-water

heaters are summarized as follows:

Aspects Open-type Closed-type

Design Simpler More complex

Cost Cheaper More expensive

Heat Transfer Good Less effective

Feed Pump Many Single/Less


Example 6

An ideal regenerative steam power plant operates between

a pressure limit of 40 bar and 0.2 bar. The steam enters the

turbine at 450°C and exits the turbine with a dryness

fraction of 0.86. Some amount of steam is extracted from

the turbine at a pressure of 4 bar and enters a closed-type

feed-water heater. Neglecting feed pump work, determine:

(a) the mass of steam extracted from the turbine,

(b) thermal efficiency of the cycle,

(c) specific steam consumption, and

(d) condenser heat load.


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Example 6


Throttle valves

Ideally,

T9 = T7

4 bar

450oC40 bar

0.2 bar

Example 7

Consider a steam power plant that operates on an ideal reheat-

regenerative Rankine cycle with one open feed-water heater, one

closed feed-water heater, and one reheater. Steam enters the

turbine at 140 bar and 520oC and is condensed in the condenser

at a pressure of 0.1 bar. Some steam is extracted from the high

pressure (H.P) turbine at 40 bar for the closed feed-water heater,

and the remaining steam is reheated at the same pressure to

520oC. The condensate exiting the closed feed-water heater and is

throttled to the open feed-water heater at 7 bar. Some steam of

the low-pressure (L.P) turbine is extracted from the turbine at 7

bar for the open feed-water heater. Saturated liquid from the open

feed-water heater is pumped to 140 bar before entering the closed

feed-water heater and passes to the boiler. Determine (a) the

fraction of steam extracted from each turbine as well as (b) the

thermal efficiency of the cycle.Vapor Power Plant 70

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36

Example 8

Stim meninggalkan dandang pada 140 bar dan 520°C. Stim

kemudiannya memasuki turbin peringkat pertama dan keluar pada

tekanan 40 bar. Sebahagian stim yang keluar dari turbin ini dibekalkan

kepada sebuah pemanas air suapan tertutup, dan selebihnya

dipanaskan semula kepada suhu 520°C. Stim yang dipanaskan semula

memasuki turbin peringkat kedua dan dikembangkan kepada tekanan

pemeluwap 0.1 bar. Sebahagian stim dijujuh dari turbin peringkat

kedua pada tekanan 7 bar untuk digunakan oleh pemanas air suapan

terbuka. Stim jujuhan daripada turbin peringkat pertama memeluwap

apabila ia melalui pemanas tertutup dan didikitkan ke pemanas

terbuka pada 7 bar. Cecair tepu air yang meninggalkan pemanas

terbuka melalui sebuah pam yang meningkatkan tekanan kepada 140

bar sebelum memasuki pemanas tertutup dan dialirkan kepada

dandang. Dengan anggapan kitar adalah unggul dan mengabaikan kerja

pam, tentukan (a) peratus stim memasuki turbin peringkat pertama

yang memasuki pemanas tertutup, (b) peratus stim memasuki turbin

peringkat kedua yang dialirkan kepada pemanas terbuka, dan (c)

kecekapan kitar. Vapor Power Plant 71

Example 8


520oC

12

(y1)

10 11

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(1)

7

T3 = 420oC

4

(y1)

11 12

P2 = P3 = 4 Mpa

1 Hig

h-

Pre

ssur

e

turb

ine

10

86

Heater

P

2 HeaterP

1

3

5

9

Lo

w-

Pre

ssure

turb

ine

P4 = 800 kPa

(y2)

P5 = 10 kPa

(1 – y1 – y2)

2Boiler

Condense

r

P1 = 13 Mpa

T1 = 600oC

Date post:	18-Jan-2015
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Lecture note steam power cycle

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