chapter 10 vapor and combined power · pdf filewe consider power cycles ... ideal and actual...

Chapter 10

Vapor and Combined Power Cycles

2

We consider power cycles where the working fluid undergoes a phase change. The best example of this cycle is the steam power cycle where water (steam) is the working fluid.

Carnot Vapor Cycle

3

The heat engine may be composed of the following components.

4

The working fluid, steam (water), undergoes a thermodynamic cycle as shown in one of the following T-s diagram.

Carnot Cycle

5

Carnot Cycle: The thermal efficiency of this cycle is given as:

η th Carnotnet

in

out

in

L

H

WQ

QQ

TT

, = = −

= −

1

1

Note the effect of TH and TL on ηth, Carnot.

1.) The larger the value of TH , the larger the ηth, Carnot

2.) The smaller the value of TL , the larger the ηth, Carnot

To increase the thermal efficiency in any power cycle, we try toincrease the maximum temperature at which heat is added.

6

Reasons why the Carnot cycle is not used

1.) Pumping process 1-2 requires the pumping of a mixture of saturated liquid and saturated vapor at state 1 and the delivery of a saturated vapor at state 2.

2.) To superheat the steam to take advantage of a higher temperature, elaborate controls are required to keep TH constant while the steam expands and does work.

To resolve the difficulties associated with the Carnot cycle, the Rankine cycle was devised.

7

Rankine Cycle

The simple Rankine cycle has the same component layout as the Carnot cycle shown above. The simple Rankine cycle continues the condensation process 4-1 until the saturated liquid line (state 1) is reached.

Ideal Rankine Cycle Processes

Process Description1-2 Isentropic compression in pump 2-3 Constant pressure heat addition in boiler 3-4 Isentropic expansion in turbine 4-1 Constant pressure heat rejection in condenser

8

1

2 3

4

P

s

Note that the Ideal Rankin Cycle on a P-s diagram is a rectangle.

Ideal Rankine Cycle Processes

Process Description1-2 Isentropic compression in pump 2-3 Constant pressure heat addition in boiler 3-4 Isentropic expansion in turbine 4-1 Constant pressure heat rejection in condenser

Heat In

Heat Out

TurbineWork Out

PumpWork In

9

Ideal Rankine Cycle ProcessesProcess Description

1-2 Isentropic compression in pump 2-3 Constant pressure heat addition in boiler 3-4 Isentropic expansion in turbine 4-1 Constant pressure heat rejection in condenser

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Ideal and Actual Rankine Cycle Processes

ηCs

a

Isentropic compressor workActual compressor work

ww

= = ηCs

a

h hh h

≅−−

2 1

2 1

11


ηTa

s

Actual turbine workIsentropic turbine work

ww

= =

12

These are used to determine h2a and h4a via the equations

and

13

Example 10-1a

Compute the thermal efficiency of an ideal Rankine cycle for which steam leaves the boiler as superheated vapor at 6 MPa, 350oC, and is condensed at 10 kPa.

We use the power system and T-s diagram shown below.P2 = P3 = 6 MPa = 6000 kPa; T3 = 350oC; P1 = P4 = 10 kPa

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Pump

The pump work is obtained from the conservation of mass and energy for steady-flow but neglecting potential and kinetic energy changes and assuming the pump is adiabatic and reversible.

& & &

& & &

& & ( )

m m mm h W m h

W m h hpump

pump

1 2

1 1 2 2

2 1

= =

+ =

= −

Since the pumping process involves an incompressible liquid, state 2 is in the compressed liquid region, we use a second method to find the pump work or the Δhacross the pump.Recall the property relation: dh = T ds + v dP.

Since the ideal pumping process 1-2 is isentropic, ds = 0.

Example 10-1a (Continued)

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The incompressible liquid assumption allows:v v consth h v P P≅ =− ≅ −

1

2 1 1 2 1

.( )

The pump work is calculated from

& & ( ) & ( )&

&( )

W m h h mv P P

wW

mv P P

pump

pumppump

= − ≅ −

= = −

2 1 1 2 1

1 2 1

Using the steam tables (A-5, next slide)

11

3

1

191.8110

.0.00101

f

f

kJh hkgP kPa

Sat liquid mv vkg

⎧ = =⎪= ⎫⎪⎬⎨⎭⎪ = =⎪⎩

w v P P

mkg

kPa kJm kPa

kJkg

pump = −

= −

=

1 2 1

3

30 00101 6000 10

6 05

( )

. ( )

.


16


17

Now, h2 is found from

2 1

6.05 191.81

197.86

pumph w h

kJ kJkg kg

kJkg

= +

= +

=Boiler

To find the heat supplied in the boiler, we apply the steady-flow conservation of mass and energy to the boiler. If we neglect the potential and kinetic energies, and note that no work is done on the steam in the boiler, then

& & &

& & &

& & ( )

m m mm h Q m h

Q m h hin

in

2 3

2 2 3 3

3 2

= =

+ =

= −


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We find the properties at state 3 from the superheated tables (A-6, next slide) as

33

33

3043.96000

350 6.3357o

kJhP kPa kg

kJT C skg K

⎧ =⎪= ⎫⎪⎬⎨

= ⎭⎪ =⎪ ⋅⎩

The heat transfer per unit mass is

3 2

(3043.9 197.86)

2845.1

inin

Qq h hm

kJkg

kJkg

= = −

= −

=

&

&


19


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Example 10-1a (Continued): The Turbine

The turbine work is obtained from the application of the conservation of mass and energy for steady flow. We assume the process is adiabatic and reversible and neglect changes in kinetic and potential energies.

& & &

& & &

& & ( )

m m mm h W m hW m h h

turb

turb

3 4

3 3 4 4

3 4

= =

= +

= −We find the properties at state 4 from the steam tables by noting s4 = s3 = 6.3357 kJ/kg-K and asking three questions.

4

4

4

4

10 : 0.6492 ; 8.1488

??

?

f g

f

f g

g

kJ kJat P kPa s skg K kg K

is s sis s s s

is s s

= = =⋅ ⋅

<

< <

<

Yes

21


22

4 4

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6.3357 0.6492 0.7587.4996

f fg

f

fg

s s x s

s sx

s

= +

− −= = =

4 4

191.81 0.758(2392.1)

2005.0

f fgh h x h

kJ kJkg kgkJkg

= +

= +

=

The turbine work per unit mass is

3 4

(3043.9 2005.0)

1038.9

turbw h hkJkg

kJkg

= −

= −

=


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The net work done by the cycle is

(1038.9 6.05)

1032.8

net turb pumpw w wkJkg

kJkg

= −

= −

=

The thermal efficiency is 1032.8

2845.1

0.363 36.3%

netth

in

kJw kg

kJqkg

or

η = =

=


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ηCs

a

Isentropic compressor workActual compressor work

ww

= = ηCs

a

h hh h

≅−−

2 1

2 1

25


ηTa

s

Actual turbine workIsentropic turbine work

ww

= =

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Deviation from Actual Cycles

1.) Piping losses--frictional effects reduce the available energy content of the steam.2.) Turbine losses--turbine isentropic (or adiabatic) efficiency.

4a4s

3

s

T P3

P4η turb

actual

isentropic

a

s

ww

h hh h

= =−−

3 4

3 4

The actual enthalpy at the turbine exit (needed for the energy analysis of the next component) is

h h h ha turb s4 3 3 4= − −η ( )

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1.) Pump losses--pump isentropic (or adiabatic) efficiency.

2a

2s

1

s

T P2

P1

η pumpisentropic

actual

s

a

ww

h hh h

= =−−

2 1

2 1

The actual enthalpy at the pump exit (needed for the energy analysis of the next component) is

h h h hapump

s2 1 2 11

= + −η

( )

2.) Condenser losses--relatively small losses that result from cooling the condensate below the saturation temperature in the condenser.

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Example 10-1b

Redo Example 10-1a assuming that the isentropic efficiency of the pump (compressor) is ηC = 0.8 and the isentropic efficiency of the turbine is ηT = 0.9.

We use the power system and T-s diagram shown below.P2 = P3 = 6 MPa = 6000 kPa; T3 = 350oC; P1 = P4 = 10 kPa

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Example 10-1b (Continued)

From the results in Example 10-1a, we have

h2s = h2 = 197.86 kJ/kgand

h4s = h4 = 2005.0 kJ/kg. Then

and

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The value of h4a is still between hf@10kPa and hg@10kPa and we find that

which is smaller than 1038.9 kJ/kg (Example 10.1a) by a factor of 0.9 (ηT), as it must be, and,

which is larger than 6.05 kJ/kg (Example 10.1a) by a factor of 0.8 (ηC), as it must be.

which is larger than 0.758 (Example 10.1a) We also have

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and

which is smaller than 0.363, obtained in Example 10-1a,

Then

and

which is smaller than 1032.8 kJ/kg, obtained in Example 10-1a,

which is smaller than 2845.1 kJ/kg, obtained in Example 10-1a,

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Ways to improve the simple Rankine cycle efficiency

1.) Superheat the vapor: Average temperature is higher during heat addition.Moisture is reduced at turbine exit since x4’ > x4.

Original Cycle is1-2-3-4-1

Modified Cycle is1-2-3’-4’-1

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2.) Lower condenser pressure: Less energy is lost to surroundings.Moisture is increased at turbine exit since x4’ < x4.


Modified Cycle is1’-2’-3-4’-1’

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3.) Increase boiler pressure (for fixed maximum temperature): Availability of steam is higher at higher pressures. Moisture is increased at turbine exit since x4’ < x4.


Modified Cycle is1-2’-3’-4’-1

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Reheat Rankine Cycle

As the boiler pressure is increased in the simple Rankine cycle, not only does the thermal efficiency increase, but also the turbine exit moisture increases. The reheat cycle allows the use of higher boiler pressures and provides a means to keep the turbine exit moisture (x6 > x4’) at an acceptable level. The cycle 1-2-3-4’-1 is the no-reheat cycle

T-s diagram for the reheat cycle 1-2-3-4-5-6-1.Note that T3 = T5.

The no re-heat cycle is 1-2-3-4’-1

4’

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Reheat Rankine Cycle: Energy-Flow Equations

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Rankine Cycle with ReheatComponent Process First Law Result

Boiler Isobaric qin = (h3 - h2) + (h5 - h4) Turbine Isentropic wout = (h3 - h4) + (h5 - h6) Condenser Isobaric qout = (h6 - h1) Pump Isentropic win = (h2 - h1) = v1(P2 - P1)

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The thermal efficiency is given by

or

which reduces to

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1

2 3

4

P

s

The Ideal Reheat Rankin Cycle on a P-s diagram is an L shape.

Ideal Reheat Rankine Cycle (1-2-3-4-5-6-1) P-s Diagram

Heat Out 6-1

HP-TurbineWork Out

5-6

PumpWork In

1-2 5

6

LP-Turbine Work Out 3-4

Heat In 4-5

Heat In 2-3

4’

The Ideal No Reheat Rankine Cycle is

1-2-3-4’-1

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Example 10-2

Compare the thermal efficiency and turbine-exit quality at the condenser pressure for a simple Rankine cycle and the reheat cycle when the boiler pressure is 4 MPa, the boiler exit temperature is 400oC, and the condenser pressure is 10 kPa. The reheat takes place at 0.8 MPa and the steam leaves the reheater at 400oC.

P1 = P6 = 10 kPa; P2 = P3 = 4 MPa; P4 = P5 = 0.8 MPa; T3 = T5 = 400oC

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Note that:

s1 = s2

s3 = s4

s5 = s6

We first have from Table A-5:

v1 = vf@10 kPa = 0.001010 m3/kg and

h1 = hf@10 kPa = 191.81 kJ/kg. Then

wpump = v1(P2 – P1) = 4.03 kJ/kg = h2 – h1

so that h2 = 195.84 kJ/kg.

Example 10-2 (Continued)

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We also have from Table A-6:

h3 = h@4 MPa,400oC = 3214.5 kJ/kg and s3 = s@4 MPa,400oC = 6.7714 kJ/(kg.K)

so that qin,23 = h3 – h2 = 3018.66 kJ/kg. We also have s4 = s3 = 6.7714 kJ/(kg.K)

and P4 = 0.8 MPa. Since s4 > [email protected] MPa = 6.6616 kJ/(kg.K), we have a

superheated vapor and hence (Table A-6 and interpolation) h4 = 2818.59 kJ/kg.


43

Now we have:

wHP-turb = h3 – h4 = 395.91 kJ/kg and s4 = s3 = 6.7714 kJ/(kg.K)

We also have P5 = 0.8 MPa and T5 = 400oC, so that h5 = 3267.7 kJ/kg, and

s5 = 7.5735 kJ/(kg.K) = s6. Then qin,45 = h5 – h4 = 449.11 kJ/kg.


44

Since s6 = 7.5735 kJ/(kg.K)

0.6492 kJ/(kg.K) = sf@10 kPa < s6 < sg@10 kPa = 8.1488 kJ/(kg.K),

we have a saturated mixture with quality

x6 = (s6 - sf@10 kPa )/(sg@10 kPa - sf@10 kPa ) = 0.923and hence

h6 = hf@10 kPa + x6(hg@10 kPa - hf@10 kPa) = 2400.4 kJ/kg.


45

Now we have:

wLP-turb = h5 – h6 = 867.3 kJ/kg and then

wturb = wHP-turb + wLP-turb = 1263.21 kJ/kg, so that

wout = wturb - wpump = 1259.18 kJ/kg and

qin = qin,23 + qin,45 = 3467.77 kJ/kg.


The efficiency is then

wout/qin = 1259.18/3467.77

or, 0.363.

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Example 10-2 (Continued), if there were no reheating: 1-2-3-4’-1

With no reheat we get: s4’ = s3 = 6.7714 kJ/(kg.K), and x4’ = (s4’ - sf@10 kPa )/(sg@10 kPa - sf@10 kPa ) = 0.816 < x6 = 0.923, andh4’ = hf@10 kPa + x4’(hg@10 kPa - hf@10 kPa) = 2144.56 kJ/kg, andwturb = h3 – h4’ = 1069.94 kJ/kg andwout = wturb - wpump = 1065.91 kJ/kg, and qin = qin,23 = 3018.66 kJ/kg.

4’

The efficiency is then wout/qin = 1065.91/3018.66 or 0.356 < 0.363

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Multiple Reheat Rankine Cycle

The multiple reheat cycle allows the use of higher boiler pressures and provides a means to keep the turbine exit moisture (x > 0.85 to 0.90) at an acceptable level.

Question: What would this look like on a P-s diagram? Answer: A Staircase

chapter 10 vapor and combined power · pdf filewe consider power cycles ... ideal and actual...

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