grup konversi energi - htgr

8/8/2019 Grup Konversi Energi - HTGR

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Gas Turbine Power Plants

Gas Turbine Power Plants are lighter and more compactthan vapor power plants. The favorable power-output-to-weight ratio for gas turbines make them suitable fortransportation.

Air-standard Brayton Cycle

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Q&W&

CV CV

For steady-state: 0 =-+

(h -h )

in out

m&m&

W&in

1 .. 2 Adiabatic compression=

(h -h )

21

m&

Q&in

2 .. 3 Heat addition =(h -h )

32

m&

W&out

3 .. 4 Adiabatic expansion =(h -h )

34

m&

Q&out

4 .. 1 Heat removal =


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(h -h )

41

m&

Cycle Thermal Efficiency:

m&h -hQ&out

41

.=1-

=1-

Brayton

m&h -h

cycleQ&in

32

Back work ratio:

&

W

m&h -hin

21

=

bwr =

m&h -


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hW&out

34

2


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Ideal Air-standard Brayton Cycle (processes arereversible)

1 .. 2 Isentropic compression2 .. 3 Constant pressure heat addition3 .. 4 Isentropic expansion

4 .. 1 Constant pressure heat removal

QinQout.....

P

2

.

.

..

.

For the isentropic process 1.. 2

P

r

P

r

=

2

1

P

1

.

.

..

.


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P

4

.

.

..

.

For the isentropic process 3 .. 4

P

r

P

r

=

4

3

P

3

3


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Ideal Cold Air-standard Brayton Cycle

For isentropic processes 1 .. 2 and 3.. 4

k -1 k -1T .

P .kT .P .k

22 44

=. .....and =. ...

..

TP TP

1 .1 .3 .3 .

kkPP .T .k -1 .T .k -1 TT

232 323

Since = thus ......=.......=

PPTT TT

14 .1 ..4 .14

Thermal Efficiency

h -hc (T -T )T (T / T -1)

41 P 4 1 141


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.Brayton =1-=1-=1-

h -hc (T -T )

T (T / T -1)

constk 32 P 3 2 232

TT TT

23 43

recall =.=

TT TT

14 12

T 1

.=1-1 =1-

Brayton

k -1

constk2

T (P2

P1 )k4


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Efficiency increases with increased pressure ratio across

the compressorBack work ratio

&

&

W

m&cP (T2 -T1) T2 -T1m&Wcomp

in

bwr =

=

==

&

&

m&Wm&c (T -T )T -TWout

turb

P 34 34

Typical BWR for the Brayton cycle is 40 - 80% comparedto < 5% for the Rankine cycle.

Recall, reversible compressor work is given by .12 vdPSince gas has a much larger specific volume than liquidmuch more power is required to compress the gas from P1to P2 in the Brayton cycle compared to the Rankine cyclefor which liquid is compressed.

The turbine inlet temperature is limited by metallurgical

factors, e.g., Tmax = 1700K

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Gas Turbine Irreversibilities

In the ideal Brayton cycle all 4 processes are assumedreversible, thus processes 2-3 and 4-1 are constantpressure and processes1-2 and 3-4 are isentropic.

The constant pressure assumption does not normally incurany great errors but the compressor and turbine processesare far from isentropic

Ideal (reversible) processes:1 - 2s and 3 - 4s

Actual (irreversible) processes:1 - 2 and 3 - 4

These irreversiblities are taken into account by:

.W&t ..W&c ..

.

..

.m&h -h .m&

.h -h

.34 s 2s 1

.turb =&=.=&=

comp

.W .h -h .W .h -h

t 34sc 21

.

..


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.

mm

.&

.s .&.

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Efficiency versus Power

Consider two Brayton cycles A and B with a similarturbine inlet temperatures T3

Since

.

.

..

.

P

2

P

1

.

...

.

.

...

A

.>

P

2

P

1

.

.

..

.

B


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.

A

>

.

B

Since (enclosed area 1-2-3-4)B > (enclosed area 1-2-3-4)A

&

&

&

..

.

...

..

.

..

.

m&

W

cycle

W

cycle

W

cycle, A

A

B


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=

>

&

W

m&

m&

m&

.

.

.

.

B

A

cycle,B

In order for cycle A to produce the same amount of net

power as cycle B, i.e.,

&

W

cycle, A

=


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&

W

cycle,B

, need m&

A

>

m&

B

.

Higher mass flow rate requires larger (heavier) equipmentwhich is a concern in transportation applications

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Increasing Cycle Power

The net cycle power is: W&cycle =W&t -W&

c

The cycle power can be increased by either increasing theturbine output power or decreasing the compressor inputpower.

Gas Turbine with Reheat

The turbine work can be increased by using reheat, as was

shown in the Rankine cycle32

Compressora b41The turbine is split into two stages and a secondcombustor is added where additional heat can be added

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TT

Recall: 2 =3 so, isobars on T-s diagram diverge

TT

1 4'

Q&in,2

3aQ&1,in4Note:

T

b

hb - h4 > ha - h4

24

1

s

The total turbine work output without reheat is:

&

W =[(h -h )+(h -h )]m&

basic 3 aa 4'

The total turbine work output with reheat is:

&&&

Wturbine =Wt,1 +Wt,2 =[(h3 -ha )+(hb -h4 )]m&w/ reheat

Since hb - h4 > ha - h4 W&turbine >W&

basicw/ reheat


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Compression with Intercooling

The compressor power can be reduced by compressing instages with cooling between stages.

TT

Recall: 2 =3 so, isobars on T-s diagram diverge

TT

1 4'

22dh2 hc > h2 hd

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The compressor power input without intercooling is:

&

W =[(h -

h )+(h -h )]m&

basic 2' cc 1

The total compressor power input with intercooling is:

&&&

Wcomp =Wc,1 +Wc,2 =

[(hc -h1 )+(h2 -hd )]m&w/ reheat

Since h2 hc > h2 hd W&W

cycle cyclew/ reheat basic

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Different approach: The reversible work per unit mass fora steady flow device is .vdP , so

2&

2 c 2'

.W .

c

..=.vdP =.vdP +.vdP

Without intercooling : m&

..basic 11 c

=area b-1-c-2'-a

&2 c 2

.W .

c

..=.vdP =.vdP +.vdP

With intercooling : m&

..w/ int1 1 d

=area b-1-c-d-2-aSince area(b-1-c-2-a) > area(b-1-c-d-2-a)

&&

.W ..W .

cc

..>..

m&m&


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..basic ..w/int

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Aircraft Gas Turbines

Gas turbine engines are widely used to power aircraftbecause of their high power-to-weight ratio

Turbojet engines used on most large commercial and

military aircraftIdeal air-standard jet propulsion cycle:

Nozzle Diffusera21 4 5313


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Normally compression through the diffuser (a-1), andexpansion through the nozzle (4-5) are taken as isentropic

inQ&outQ&In the ideal jet propulsion engine the gas is not expanded

to ambient pressure Pa.

Instead the gas expands to an intermediate pressure P4such that the power produced is just sufficient to drive thecompressor, no net cycle power produced (W&cycle =0),

thus

&&

Wc Wt

=

m&m&

(h -h )=(h -h4 )

21 3

After the turbine the gas expands to ambient pressure P5which is the same as Pa.

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Apply the steady-state conservation of energy equation tothe Diffuser and Nozzle

22

CVCV in out

Q&W&.V ..V .

0 =

-

+..hin +..-..hout +..m&m&22

...

.

Diffuser slows the flow to a zero velocity relative to theengine:

22

1 a

VV

h +=h +

1 a

22

2

V

a

Diffuser (a .. 1) h =h +


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1 a

2

2

V

a

T =T +for constant k

1 a

2c

P

Nozzle accelerates the gas leaving the turbine (turbineexit velocity negligible compared to nozzle exit velocity):

22

45

VV

h +

=h +

45

22

2(h4 -h5 )Nozzle (4 .. 5) V5 =

V =

2c (T -T )for constant k

5

P 45

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The gas velocity leaving the nozzle is much higher thanthe velocity of the gas entering the diffuser, this change inmomentum produces a propulsive force, or thrust Ft

F =m&

(V -Va )

t 5

Where V is flow velocity relative to engine

For aircraft under cruise conditions the thrust justovercomes the drag force on the aircraft .. fly at highaltitude where the air is thinner and thus less drag

To accelerate the aircraft increase thrust by increasing V5

In military aircraft afterburners are used to get verylarge thrust for short take-offs on aircraft carriers

An afterburner is simply a reheat device!

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Other Propulsion Systems

Turboprop TurbofanSubsonic ramjet

In turbofan bypass flow produces additional thrust fortake-off. During cruise thrust comes from turbojet

In a ramjet engine there is no compressor or turbine,compression is achieved gasdynamically.

Ramjet engines produce no thrust when stationary thusmust be coupled with a turbojet engine to get off theground

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Supersonic Ramjet Engine

The flow is decelerated to subsonic velocity before theburner via a series of shock waves.

Combustion occurs at constant pressure

chokedSupersonicfree streamflowSupersonicexhaust flowflow

Turbojet-ramjet combination:

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Supersonic Combustion Ramjet (SCRAMJET) Engine

At very high Mach numbers the air temperature getsextremely hot after deceleration through the diffuser

Va

2

T1 =Ta +2cP

For Mach 6 flight speed, the air temperature just beforethe burner reaches about 1550K. At this temperature theair dissociates resulting in a drop in enthalpy

At flight speeds greater than Mach 6 (hypersonic) betterto burn fuel- in supersonic air stream

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