Chalmers University of Technology - Gas Turbine Technology Lecture 3

5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3

1/44

Chalmers University of Technology

Lecture 3 Some more thermodynamics: Brief discussion of cycle efficiencies - continued

Ideal cycles II Heat exchanger cycle

Real cycles Stagnation properties, efficiencies,

pressure losses The Solar Mercury 50

Real cycles Mechanical efficiencies Specific heats (temperature variation) Fuel air ratio, combustion and cycle

efficiencies Bleeds

Jet engine nozzles Radial compressor I


2/44


Generalization of the Carnot efficiency

etemperaturhotaverageT

etemperaturcoldaverageT

T

T

T

T

H

L

H

L

H

Lcarnotth

11,

=

=

==

Is generalization of Carnot efficiency

to Brayton cycle possible?

Define average temp. to value that would

give the same heat transfer, i.e.:

sTTdsq

sTTdsq

Lout

Hin

==

==

1

4

3

2


3/44


Generalization of Carnot efficiency

But for the isobar we have,

2

3

0

2

3

2

3 lnlnlnT

Tc

P

PR

T

Tcs pp ==

=

Thus, the average temperature is obtained from (dp=0):

2

3

2323

3

2

3

22

3

ln

][ln

T

TTTTcdTcrelationscombineTds

TTcTsT HpppHH ======

Derive an expression for the lower average temperature in the same way.

PdvTdsdu =

Furthermore, we have Gibbs equation

(Cengel and Boles):

as well as: vdppdvdhpvhddudwdq === )(


4/44


Generalization of Carnot efficiency

( ) ( )

Carnotth

H

L

H

L

Braytonth

T

T

T

T

TT

TTTT

,

2

3

32

1

4

41

1

4

2

3

4

3

1

2

32

41

23

1243,

11

T

Tln

T

T

Tln

T

-1

T

T

T

T

T

T

T

T

T

T-1

===

=

===

=

=


5/44


When T4> T2a heat exchanger can be introduced.

Thi i true !hen" )1(21


6/44


#ere !e obtain the e$$icienc%" &ot inde'endento$ T3 (i'le c%cle

i inde'endent o$ t3)

*o!er out'ut i una$$ected b% heat exchanger ince the

turbine and co'reor !or+ are the ae a in the i'le c%cle.

t

r

T

TT

T

T

T

T

TT

T

TT

T

T

T

T

TT

TT

TT

TTTT

TTc

TTcTTc

p

pp

1

2

13

1

4

1

1

24

1

21

4

3

1

2

43

12

43

1243

3

1243

111

)1(

)1(

1)(

)(1

)(

)()(

)(

)()(

===

=

==

=

=

=

Theory 3.1 Ideal heat exchanger cycle


7/44


Heat exchange cycle

er% high e$$iciencie can be theoreticall% be obtained

#eat exchanger etallurgical liit !ill be releant.

T4 / 1000.0 /> 0>


8/44


Heat exchange cycle

o! 'reure ratio

/> high e$$icienc%

What ha''en

!ith the aerage

te'erature at

!hich heat i

addedre5ected

!hen the

'reure ratio

change in heat

exchange c%cle6

qinqin

qout qout

TH TH


9/44


Cycles with losses

a. Change in kinetic energy between inlet and outletmay not be negligible :

b. Fluid friction =>

- burners- combustion chambers- exhaust ducts

++

++= iinletsall

ii

ii

eexitsall

ee

ee gz

V

hmgz

V

hm!

2

2

22


10/44


Cycles with losses

c. Heat exchangers.Economic size =>terminal temperature

difference, i.e. T5 < T4.d. Friction losses in shaft, i.e.

the transmission of turbine power tocompressor. Auxiliary power requirementsuch as oil and fuel pumps.

e. and c pvary with temperature and gas

composition.


11/44


Cycles with losses

f. Efficiency is defined by SFC(specific fuel consumption = fuel

consumption per unit net work

output). Cycle efficiency

obtained using fuel heating value.

g. Cooling of blade roots and

turbine disks often require

approximately the same mass

flow of gas as fuel flow => air

flow is approximated as constantfor preliminary calculations. This

is done in this course.


12/44


Stagnation properties For high-speed flows, the potential energy of the fluid can still

be neglected but the kinetic energy can not!

++

++=

=

++

++=

0

1

2

111

0

2

2

222

2

2

0102

22]out'utingle-in'utingle[

22

gzV

hmgzV

hm

gzV

hmgzV

hm!

hh

iinletsall

ii

ii

eexitsall

ee

ee

It is convenient to combine the static temperature and the kinetic

energy into a single term called the stagnation (or total) enthalpy,

h0=h+V2/2, i.e. the energy obtained when a gas is brought to rest

without heat or work transfer


13/44


Stagnation properties

0102

21

1

22

2

0102

22hh

Vh

Vhwq

hh

=++=

For a perfect gas we get the stagnation temperature T0,

according to:

p

ppc

VTT

VTcTc 22

2

0

2

0 +=+=


14/44


Stagnation pressure

Defined in same manner as stagnation temperature (noheat or work transfer) with added restriction

retardation is thought to occur reversibly

Thus we define the

stagnation pressure p0by:

Note that for an isentropic

process between 02 and 01

we get

100

=

TT

PP

1

01

021

01

1

1

2

2

021

01

11

1

21

2

02

01

1

1

2

2

02

01

02

=

=

==

T

T

T

T

T

T

T

T

T

T

T

T

T

T

P

P

P

P

P

P

P

P


15/44


Compressor and turbine efficiencies

Isentropic efficiency (compressors and turbines are approximatelyadiabatic => if expansion is reversible it is isentropic).The isentropicefficiency is for the compressor is:

0

7

0

7

0

7

0

TcTc

hh

p

pc ==

pp cc ,7

Where are the averaged specific heats of the temperature

intervals 01-02and 01-02 respectively.


16/44



Ideal and mean temperature

differences are not very

different. Thus it is a good

approximation to assume:

We therefore define:

pp cc 7

0102

0102

TT

TTc

=

Similarly for the turbine:

0403

0403

TT

TTt

=


17/44



Using produces the

frequently used expressions:

1

04

03

04

03

1

01

02

01

02 and

== PP

TT

PP

TT

(2.11)1

1

01

02010102

=

P

PTTT

c

(2.12)1

1

1

04

03

030403

=

P

PTTT

t


18/44


Turbine efficiency options

If the turbine exhausts directlyto atmosphere the kinetic energyis lost and a more properdefinition of efficiency would

be:

1

1

1

03

030403

=

a

t

P

PTTT

In practice some of the kinetic energy is recovered in an

exhaust diffuser => turbine pressure ratio increases.

Here we put p04=pafor gas turbines exhausting intoatmosphere and think oftas taking both turbine and

exhaust duct losses into account


19/44


Turbine diffusers

Recovered

energy


20/44


Heat-exchanger efficiency

( ) ( )0202,080448, TTcTTc pp =

Conservation of energy (neglecting energy transfer to surrounding):

In a real heat-exchanger T05will no longer equal T04(T05


21/44


Pressure losses burners & heat-exchangers

Burner pressure losses

Flame stabilizing & mixing

Fundamental loss (Chapter 7 + Rayleigh-line

appendix A.4)

Heat exchanger pressure loss

Air passage pressure loss

Pha Gas passage pressure loss Phg

Losses depend on heat exchanger effectiveness. A

4% pressure loss is a reasonable starting point for

design.

engine)(aircra$t83

turbine)gal(indutria32

1

02

02

0202

0203

=

p

p

p

p

P

p

p

pPP

b

b

hab


22/44


The Solar Mercury 504.3 MW output= 40.5 %System was designed from

scratch to allow high performance

integration of heat-exchanger

C f


23/44


Mechanical losses

Turbine power is transmitted directly from the turbine

without intermediate gearing => (only bearing and windage

losses). We define the transmission efficiency m:

( )010212,1

TTc pm

turbine =

Usually power to drive fuel and oil pumps are transmitted from the

shaft. We will assumem=0.99 for calculations.

Ch l U i it f T h l


24/44


Temperature variation of specific heat

We have already established:

cp=f1(T)

cv=f2(T)

Since=cp/cvwe

have =f3(T)

The combustion product thermodynamic

properties will depend on T and f (fuel air

ratio)

Ch l U i it f T h l


25/44


Pressure dependency?

$H$H 22221 +

=

P

P

P

P

P

P

%

H$

$H

22

2

2

1

At 1500 K dissociation begins to have animpact on cp and.

22

2

1C$$C$ +

Detailed gas tables for afterburnersmay include pressure effects. We exclude them in thiscourse.

Ch l U i it f T h l


26/44


Temperature variation of specific heat

333.1,9+g114:

400.1,9+g100

==

==

gpg

apa

c

c

In this course we use:

Since gamma and cpvary in opposing senses some of the

error introduced by this approximation is cancelled.

Ch l U i it f T h l


27/44


Calculate fthat gives T03for given T02? Use first law forcontrol volumes (q=w=0) and that enthalpy is a pointfunction(any path will produce the same result)

Determining the fuel air ratio

( ) ( ) ( ) ( ) ( ) 0

02203 2;:2;:2;:1++=+ #p#papg T#cTcH#Tc#

fis small (typically around 0.02) and cpfis also small => last term

is negligible. The equation determines f.



28/44


Combustion temperature rise

Hypothetic fuel:

86.08% carbon

13.92% hydrogen

H25= - 43100 kj/kg

Curves ok for kerosene

burned in dry air. Not ok

in afterburner (fin0).



29/44


Shaft cycle performance parameters

&w

#nconsumptio#uelspeci#ic'(C ==

pnet

net

#!

we##iciencycycle

,

==



30/44


Bleeds

Combustor and turbineregions require most of thecooling air.

Anti-ice

Rule of thumb: take air asearly as possible (less workput in)

Accessory unit cooling (oilsystem, aircraft powersupply (generator), fuel

pumps) Air entering before rotor

contributes to work!



31/44


dragoentuinta+ethrutoentugro

a) mCmC

Thrust&et

momentumo#changeo#Rate

=

=

Aircraft propulsion thrust generation



32/44


( )

thrustpressure

a))

velocity*ircra#t

a) pp*CCm

Thrust&et

+

=

Jet engine principles of thrust generation



33/44


Jet engine principles of thrust generation

0

22040

2

44

2

00

040

==++===

hhV

hV

hwq

hh

00 Tch p=

040 TT =

No heat or work transfer in the jet engine nozzle

Stagnation temperature

is constant



34/44


Mach number relations for stagnation properties

We have already introduced the

stagnation temperature as:pc

VTT

2

2

0 +=

and shown that (revision task): Rccvp

+=

The Mach number is defined as:

RT

V

soundo#speed

V

a

V+

===

additional loss =>

2 -doneor. =



42/44

Overall pressure rise:

1

01

2

01

03 1

+=

Tc

-

p

p

p

c

P03is here used to denote the pressure at

compressor exit. P02is reserved for the stagnation

pressure between the impeller and the diffuser

vanes



43/44

Example 4.1a

=1.04,= 0.90

N = 290.0 rev/s,

D = 0.5 m

Deye,tip = 0.3, Deye,root= 0.15 m = 9.0 kg/s

T01= 295 K

P01= 1.1 bar

c= 0.78

Compute pressure ratio and

power required



44/44

Learning goals

Understand why the Carnot cycle can be used

for qualitative arguments also for the

Joule/Brayton cycle

Be able to state reasonable loss levels for gasturbine components turbine and compressor

performance are given in Lecture !" and include

them in cycle analysis

#now how to compute cycle efficiencies for the

heat e$changer cycle

Chalmers University of Technology - Gas Turbine Technology Lecture 3

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