basics in the thermodynamic analyses of the gas turbine power plant

Basics in the Thermodynamic Analyses

of theGas Turbine Power Plant

Prof. R. ShanthiniDept. of C&P EngineeringUniversity of PeradeniyaSri Lanka

Comp-ressor

atmosphericair

Combustionchamber

fuel

Gasturbine

gasesto the stack

Gen

compressed air

hot gases

Gen stands for Electricity Generator

Compressorshaft

Turbineshaft

atmospheric air

(WGT)out

fuel

Gen

compressed air

hot gases

Comp-ressor

Combustionchamber

Gasturbine

gasesto the stack

W stands for work flow rate and GT stands for Gas Turbine

atmospheric air

(WGT)out

Comp-ressor

(QCC)in

Gen

compressed air

hot gasesCombustion

chamber

Gasturbine

gasesto the stack

Q stands for heat flow rate and CC stands for Combustion Chamber

atmospheric air

(WGT)out

Comp-ressor

(WC)in

(QCC)in

1

2 3

4Gen

compressed air

hot gasesCombustion

chamber

Gasturbine

gasesto the stack

W stands for work flow rate and C stands for Compressor

atmospheric air

(WGT)out

Comp-ressor

(QCC)in

1

2

4Gen

hot gases

compressed air 3

Combustionchamber

Gasturbine

gasesto the stack

(WC)in

(WGT)out

3

4

+ (QGT)out

= m ( h – h )3 4g

+ m ( C – C ) / 23 42 2

g

hot gases

gasesto the stack

Gasturbine

m stands for mass flow rate of gas,

h stands for enthalpy,

C stands for speed of gas flow

g stands for gravitational acceleration, &

Z stands for height above reference level

Steady flow energy equationapplied to the flow across turbine:

g

+ m g ( Z – Z ) 3 4g r

r

+ (QGT)out=

+ m ( C – C ) / 23 42 2

g

Assumptions: - Adiabatic condition prevails across the gas turbine- Kinetic energy changes are negligible compared to enthalpy changes

= m ( h – h )3 4g+

- (QGT)out 3

4

hot gases

gasesto the stack

Gasturbine

+ m g ( Z – Z ) 3 4g r

- Potential energy changes are ignored

(WGT)out

(WGT)out

= 3

4

hot gases

gasesto the stack

Gasturbine

Assumptions: - Adiabatic condition prevails across the gas turbine- Kinetic energy changes are negligible compared to enthalpy changes- Potential energy changes are ignored

= m ( h – h )3 4g+

(WGT)out

= m ( h – h )3 4g 3

4

hot gases

gasesto the stack

Gasturbine

Assumptions: - Adiabatic condition prevails across the gas turbine- Kinetic energy changes are negligible compared to enthalpy changes- Potential energy changes are ignored

(WGT)out

= m ( h – h )3 4g

Assumption: - Gases flowing through the turbine behave as ideal gases

(WGT)out

= m C ( T – T )3 4g pg

3

4

hot gases

gasesto the stack

Gasturbine

(WGT)out

3

= m C ( T – T )3 4g pg

T 4

T 3

Specific heat of gas at constant pressure

Mass flow rate of gas

4

Temperature at the outlet

Temperature at the inlet

Gasturbine

m ( h – h )3 4g=

3

T 4

T 3 =

4

fixed

changes

fixed

free to choose, but we fix it at some value

= ?

(WGT)out

= m C ( T – T )3 4g pg

Gasturbine

3

T 4

T 3 =

4

= ?

P 4 =

P 3 =

should be as small as possible

T 4

How small should T4 be ?

(WGT)out

= m C ( T – T )3 4g pg

Gasturbine

To get maximum work output from the turbine,

at the given P 3 P 4and

(P stands for pressure)

3

T 4

T 3

4

P 4

P 3

(WGT)out

= m C ( T – T )3 4g pg

T

Specific Entropy (s)

3

4s

P3

P4

4

real flow

ideal flow

Gasturbine

3

T 4s

P 3

4

T 4s 3

T = P ( 4

P 3 )

(-1)/

(WGT)out,ideal

= m C ( T – T ) 3 4sg pg

T 3

P 4

Gasturbine

(WGT)out

= m C ( T – T )3 4g pg

For the ideal flow (ideal gas at constant specific entropy):

Therefore,

where is the isentropic constant

3

4

T =

(WGT)out

(WGT)out,ideal

m C ( T – T )3 4g pg

m C ( T – T )3 4sg pg

=

Turbine Efficiency

-=

T 3

T 3

T 4

T 4s -

Gasturbine

3

4

TT4 = T3 ( T3 T4s– – )

T 4s 3 T = P ( 4 P 3 )

(-1)/

(WGT)out

(WGT)out,ideal

= T

= T m C ( T – T ) 3 4sg pg

Gasturbine

Governing equations:

Let‘s do some Excel sheet calculations across the turbine

3

4

Gasturbine= 350 kg/s m g

C pg = 1.1 kJ/kg.s T 4

Data:

T 3

= 1200 K

P 4= 1 bar

T = 88%

γ = 1.3

(WGT)out

Determine

for P3 varying in the range of 2 to 15 bar

300

400

500

600

700

800

900

1000

1100

1200

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Turbine Inlet Pressure P3 (in bar)

Tu

rbin

e O

utl

et T

emp

erat

ure

T4

(in

K)

at the specified efficiency

under ideal conditions

Turbine outlet temperature increaseswith decreasing turbine efficiency

= 88%T

0

50

100

150

200

250

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Turbine Inlet Pressure P3 (in bar)

Tu

rbin

e W

ork

Ou

tpu

t

(WG

T )o

ut (

in M

W)

under ideal conditions

at the specified efficiency

= 88%T

Turbine work output decreaseswith decreasing turbine efficiency

atmospheric air

gasesto the stack

(WGT)out

fuel

Gen

compressed air

hot gases

Comp-ressor

Combustionchamber

Gasturbine

atmospheric air

gasesto the stack

(WGT)out

Comp-ressor

(WC)in

(QCC)in

1

23

4Gen

compressed air

hot gasesCombustion

chamber

Gasturbine

atmospheric air

gasesto the stack

(WGT)out

Comp-ressor

(WC)in

(QCC)in

1

2

4Gen

3compressed air

hot gasesCombustion

chamber

Gasturbine

atmosphericair

Comp-ressor

(WC)in

1

= (QC)out

+ m ( h – h )2 1a

+ m ( C – C ) / 22 12 2

a

2

compressed air

Subscript a stands for air

+ m g ( Z – Z ) 3 4a r

Steady flow energy equationapplied to the flow across compressor:

atmosphericair

Comp-ressor

(WC)in

1

= (QC)out

+ m ( h – h )2 1a

+ m ( C – C ) / 22 12 2

a

2

compressed air

+ m g ( Z – Z ) 3 4a r

Assumptions: - Adiabatic condition prevails across the compressor- Kinetic energy changes are negligible compared to enthalpy changes- Potential energy changes are ignored

atmosphericair

Comp-ressor

(WC)in

1

= + m ( h – h )2 1a

2

compressed air

Assumptions: - Adiabatic condition prevails across the compressor- Kinetic energy changes are negligible compared to enthalpy changes- Potential energy changes are ignored

atmosphericair

Comp-ressor

(WC)in

1

= m ( h – h )2 1a

= m C ( T – T )2 1a pa

2

compressed air

Assumption: - Air flowing through the compressor behaves as an ideal gas

1

2

T 1

T 2

(WC)in

= m C ( T – T )2 1a pa

T at the inlet

T at the outlet

Specific heat of air at constant pressure

Mass flow rate of air

Comp-ressor

1

2

T 1

T 2

(WC)in

= m C ( T – T )2 1a pa

fixed

changes

fixed

free to choose, but we fix it at some value

=

= ?

Comp-ressor

1

T 1

=

P 1

=

T 2 = ?

(WC)in

= m C ( T – T )2 1a pa

; P 2 = 2

should be as small as possible

T 2

How small should T2 be ?

at the given P 1 P 2and

Comp-ressor

To give minimum work input to the compressor,

T

3

4s

P3=P2

4real flow

idealflow

(WC)in

= m C ( T – T )2 1a pa

1

T 1

P 1

T 2

P 2

2

1

2s2

P4=P1

Comp-ressor

Specific Entropy (s)

1

T 2s 1

T = P ( 2

P 1 )

(-1)/

T 1

=

P 1

=

T 2s = ?

(WC)in,ideal

= m C ( T – T )2s 1a pa

P 2 = 2

(WC)in

= m C ( T – T )2 1a pa

Comp-ressor

For the ideal flow (ideal gas at constant specific entropy):

Therefore,

1

C =

(WC)in,ideal

(WC)in

= T

2s

T 2

T 1

T 1

--

2 m C ( T – T )

2s 1a pa

m C ( T – T )2 1a pa

=

Compressorefficiency

Comp-ressor

T 2s 1 T = P ( 2 P 1 )

(-1)/

1

2

CT2 = T1 ( T2s T1+ – )/

= C m C ( T – T ) 2s 1a pa /

Comp-ressor

in(WC) (WC)

in,ideal= C/

Governing equations:

Let‘s do some Excel sheet calculations across the compressor

= 350 kg/s m a

T 2

Data:

T 1

= 300 K

P 1= 1 bar

C = 85%

γ = 1.4

Determine

for P2 varying in the range of 2 to 15 bar

1

2

Comp-ressor

(WC)in

C pa = 1.005 kJ/kg.s

300

400

500

600

700

800

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Compressor Outlet Pressure P2 (in bar)

Co

mp

ress

or

Ou

tlet

Tem

per

atu

re T

2 (i

n K

)

at the specified efficiency

under ideal conditions

= 85%C

Compressor outlet temperature increaseswith decreasing compressor efficiency

0

20

40

60

80

100

120

140

160

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Compressor Outlet Pressure P2 (in bar)

Co

mp

ress

or

Wo

rk I

np

ut

(WC)in

(in

MW

)

at the specified efficiency

under ideal conditions

= 85%C

Work input to the compressor increaseswith decreasing compressor efficiency

atmospheric air

gasesto the stack

(WGT)out

fuel

Gen

compressed air

Comp-ressor

hot gases

(WC)in

Wnet

Combustionchamber

Gasturbine

= (WC)in

(WGT)out

Wnet -

(WC)in,ideal

= (WGT)out,ideal-

Wnet,ideal

Net work output from the turbine is the power available for electricity generation

Net work output under ideal conditions is the maximum power available for electricity generation

0

50

100

150

200

250

300

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Compressor outlet Pressure P2 (in bar) =

Turbine Inlet Pressure P3 (in bar)

Turbine work output (in MW)

Compressor work input (in MW)

Net work output from the Gas Turbine System (in MW)

= 85%C

T= 88%and

0

10

20

30

40

50

60

70

80

90

100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Compressor outlet Pressure P2 (in bar) =

Turbine Inlet Pressure P3 (in bar)

Net

Wo

rk O

utp

ut

fro

m t

he

Gas

Tu

rbin

e S

yste

m (

in M

W)

under ideal conditions

at the specified efficiencies

= 85%C

T= 88%and

atmospheric air

gasesto the stack

(QCC)in

1

23

4Gen

compressed air

hot gases

WnetComp-ressor

Combustionchamber

Gasturbine

atmospheric air

gasesto the stack

(QCC)in

1

23

4Gen

compressed air

hot gases

WnetComp-ressor

Combustionchamber

Gasturbine

(QCC)in,ideal

= m ( h – h ) 3 2a

Assumptions: - Kinetic energy changes are negligible

- Potential energy changes are ignored

- Fuel flow rate is negligible compared to the air flow rate

2 3compressed air

hot gasesCombustionchamber

fuel

2 3compressed air

hot gasesCombustionchamber

fuel

= m C ( T – T )3 2a pa

Assumption: - Air flowing through the compressor behaves as an ideal gas

(QCC)in,ideal

= m ( h – h ) 3 2a

(QCC)in

2 3compressed air

hot gasesCombustionchamber

fuel

= m C ( T – T )3 2a pa

is the compressor efficiency

(QCC)in,ideal

= CC/

CC

CC/

Let‘s do some Excel sheet calculations across the combustion chamber

2 3compressed air

hot gasesCombustionchamber

fuel

= 350 kg/s m a

CC = 80%

C pa = 1.005 kJ/kg.s

P 2

P 3

=

Determine

(QCC)in

= 80%CC

0

50

100

150

200

250

300

350

400

450

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Compressor outlet Pressure P2 (in bar) = Turbine Inlet

Pressure P3 (in bar)

Heat input at the specified efficiency (in MW)

Heat input inder ideal conditions (in MW)

= 80%CC

atmospheric air

(QCC)in

1

2

compressed air

Comp-ressor

Combustionchamber

gasesto the stack

3

4Gen

hot gases

WnetGas

turbine

= th

Wnet

(QCC) in

Thermal efficiency

= 80%CC

0

50

100150

200

250

300

350400

450

500

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Compressor outlet Pressure P2 (in bar) = Turbine Inlet

Pressure P3 (in bar)

Heat input at the specifiedefficiency (in MW)

Net work output from the GasTurbine System (in MW)

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Compressor outlet Pressure P2 (in bar) = Turbine Inlet

Pressure P3 (in bar)

Overall Thermal Efficiency underideal conditions

Overall Thermal Efficiency at thespecified unit efficiencies

C

CC = 80%

= 85%

T= 88%

atmospheric air

gasesto the stack

(QCC)in

1

23

4Gen

compressed air

hot gases

Wnet

Heat Loss?

Comp-ressor

Combustionchamber

Gasturbine

= -(QCC)in Wnet

0

50

100

150

200

250

300

350

400

450

500

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Compressor outlet Pressure P2 (in bar) = Turbine Inlet Pressure P3 (in bar)

Heat input at the specified efficiency (in MW)

Heat loss at the specified unit efficiencies (in MW)

Net work output from the Gas Turbine System (in MW)

atmospheric air

gasesto the atmospherethrough the stack

(QCC)in

1

23

4Gen

compressed air

hot gases

Wnet

Heat Loss

Comp-ressor

Combustionchamber

Gasturbine

Heat is lost with the turbine exhaust gases to the atmosphere through the stack

Should not we make good use of all that heat

that is not only getting wasted

but also pollute the environment in many ways?