chalmers university of technology lecture 3 some more thermodynamics : brief discussion of cycle...

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


Generalization of the Carnot efficiency

etemperaturhotaverageT

etemperaturcoldaverageT

T

T

T

T

H

L

H

L

H

Lcarnotth

11,

Is generalization of Carnot efficiencyto 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


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

TT

TTTcdTcrelationscombineTds

T

TcTsT 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 )(


Generalization of Carnot efficiency

CarnotthH

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

TT

ln

T

-1

T

T

T

T

T

T

T

T

T

T-1


When T4 > T2 a heat exchanger can be introduced. This is true when: )1(21

tr

Heat exchange cycle


Here we obtain the efficiency: Not independent of T3!!! (simple cycleis independent of t3)

Power output is unaffected by heat exchangers since the turbine and compressor work are the same as in the simple cycle.

t

r

TT

T

T

T

T

TT

T

TT

T

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

53

1243

111)1(

)1(1

)(

)(1

)(

)()(

)(

)()(

Theory 3.1 – Ideal heat exchanger cycle


Heat exchange cycle

Very high efficiencies can be theoretically be obtained! Heat exchanger metallurgical limits will be relevant.

T4 = 1000.0 K => 70%


Heat exchange cycle

Low pressure ratio=> high efficiency

What happens with the average temperature at which heat is added/rejected when the pressure ratio changes in heatexchange cycle?

qinqin

qout qout

TH TH


Cycles with lossesa. Change in kinetic energy between inlet and outlet

may not be negligible :

b. Fluid friction =>- burners- combustion chambers- exhaust ducts

iinletsalli

iii

eexitsalle

eee gz

Vhmgz

VhmWQ

2

2

22


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 to compressor. Auxiliary power requirement such as oil and fuel pumps.

e. γ and cp vary with temperature and gas composition.


Cycles with lossesf. 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. Thisis done in this course.


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

be neglected but the kinetic energy can not!

0

1

21

11

0

2

22

22

2

2

0102

22]output single -input single[

22

gzV

hmgzV

hm

gzV

hmgzV

hmWQ

hh

iinletsalli

iii

eexitsalle

eee

• 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


Stagnation properties

0102

21

1

22

2

0102

22hh

Vh

Vhwq

hh

For a perfect gas we get the stagnation temperature T0, according to:

ppp c

VTT

VTcTc

2

2

2

0

2

0


Stagnation pressure• Defined in same manner as stagnation temperature (no

heat or work transfer) with added restriction– retardation is thought to occur reversibly

• Thus we define the stagnation pressure p0 by:

• Note that for an isentropicprocess between 02 and 01 we get

100

T

T

P

P

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


Compressor and turbine efficienciesIsentropic efficiency (compressors and turbines are approximately

adiabatic => if expansion is reversible it is isentropic). The isentropic efficiency is for the compressor is:

0

'0

'

0

'0

Tc

Tc

h

h

p

pc

pp cc ,'Where are the averaged specific heats of the temperature

intervals 01-02´ and 01-02 respectively.


Compressor and turbine efficiencies

• Ideal and mean temperature differences are not very different. Thus it is a good approximation to assume:

• We therefore define:

pp cc '

0102

0102

TT

TTc

• Similarly for the turbine:

0403

0403

TT

TTt


Compressor and turbine efficiencies

Using produces the

frequently used expressions:

1

04

03

04

03

1

01

02

01

02 and

P

P

T

T

P

P

T

T

(2.11) 1

1

01

02010102

P

PTTT

c

(2.12) 1

1

1

04

03

030403

PP

TTT t


Turbine efficiency options• If the turbine exhausts directly

to atmosphere the kinetic energy is lost and a more proper definition of efficiency would be:

1

1

1

03

030403

a

t

PP

TTT

• In practice some of the kinetic energy is recovered in an exhaust diffuser => turbine pressure ratio increases.

• Here we put p04=pa for gas turbines exhausting into atmosphere and think of ηt as taking both turbine and exhaust duct losses into account


Turbine diffusers

Recovered energy


Heat-exchanger efficiency

020525,060446, TTcTTc pp

Conservation of energy (neglecting energy transfer to surrounding):

In a real heat-exchanger T05 will no longer equal T04 (T05 <T04). We introduce heat exchanger effectiveness as:

0204

0205

TT

TTessEffectiven

• Modern heat exchangers are designed to

for effectiveness values above 90%. Use of stainless steel requires T04 around 900 K (or less). More advanced steal alloys can be used up to 1025 K.


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 ΔPha

– Gas passage pressure loss ΔPhg

– Losses depend on heat exchanger effectiveness. A 4% pressure loss is a reasonable starting point for design.

engine)(aircraft %63

turbine)gas l(industria %32

1

02

02

02020203

p

p

p

p

P

p

p

pPP

b

b

hab


The Solar Mercury 50• 4.3 MW output• η = 40.5 %• System was designed from scratch to allow high performance integration of heat-exchanger


Mechanical losses

Turbine power is transmitted directly from the turbine without intermediate gearing => (only bearing and windage losses). We define the transmission efficiency ηm:

010212,

1TTcW p

mturbine

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

shaft. We will assume ηm=0.99 for calculations.


Temperature variation of specific heat

We have already established:

cp=f1(T)

cv=f2(T)

Since γ =cp/cv we have γ=f3(T)

The combustion product thermodynamic properties will depend on T and f (fuel air ratio)


Pressure dependency?

OHOH 222 2

1

P

P

P

P

P

P

K

HO

OH

22

2

2

1

At 1500 K dissociation begins to have an impact on cp and γ.

222

1COOCO

Detailed gas tables for afterburners may include pressure effects. We exclude them in this course.


Temperature variation of specific heat

333.1 , J/kg 1148

400.1 , J/kg 1005

gpg

apa

c

c

In this course we use:

Since gamma and cp vary in opposing senses some of the error introduced by this approximation is cancelled.


Calculate f that gives T03 for given T02? Use first law for control volumes (q=w=0) and that enthalpy is a point function (any path will produce the same result)

Determining the fuel air ratio

0

022503 2982982981

fpfpapg TfcTcHfTcf

f is small (typically around 0.02) and cpf is also small => last term is negligible. The equation determines f.


Combustion temperature rise

Hypothetic fuel:

86.08% carbon

13.92% hydrogen

ΔH25 = - 43100 kj/kg

Curves ok for kerosene

burned in dry air. Not ok

in afterburner (fin≠0).


Shaft cycle performance parameters

Nw

fnconsumptiofuelspecificSFC

pnet

net

fQ

wefficiencycycle

,


Bleeds• Combustor and turbine

regions require most of the cooling air.

• Anti-ice• Rule of thumb: take air as

early as possible (less work put in)

• Accessory unit cooling (oil system, aircraft power supply (generator), fuel pumps)

• Air entering before rotor contributes to work!


drag momentum intake thrustmomentum gross

aj mCmC

ThrustNet

momentumofchangeofRate

Aircraft propulsion – thrust generation


thrustpressure

ajj

velocityAircraft

aj ppACCm

ThrustNet

Jet engine – principles of thrust generation


Jet engine – principles of thrust generation

022 0405

24

4

25

500

0405

hhV

hV

hwq

hh

00 Tch p

0405 TT

No heat or work transfer in the jet engine nozzle

Stagnation temperature is constant


Mach number relations for stagnation properties

We have already introduced the stagnation temperature as:

pc

VTT

2

2

0

and shown that (revision task): Rcc vp

The Mach number is defined as:

RT

V

soundofspeed

V

a

VM

*

The specific heat ratio γ is defined:

v

p

c

c


Mach number relations for stagnation properties

Thus:

120

2

11

M

P

P

2

11

2

1

2

1

12

202

22

0

MT

TTMT

TMT

R

c

c

RTMTT p

p

100

T

T

P

Pbut we defined: which directly gives:


Nozzle efficienciesNozzle may operate choked or unchoked:

54

504

TT

TTj

pc

Cnte equivaleTemperatur

2

25


Nozzle efficiencies

2

1

2

11 2

5

05

5

04

M

T

T

T

T

Critical pressure for irreversible nozzle is obtained from:

c

cj TT

TT

04

04

10404

cc T

T

P

P

which gives: 1

04

111

1

1

j

cP

P


• Impeller - work is transferred to accelerate flow and increase pressure

• Diffuser - recover high speed generated in impeller as pressure

Basic operation of radial compressor


2222

1122

termprewhirl

1122

prewhirl without casesfor

UCrC

rCrCworklTheoretica

rCrC

momentumangularofchangeofRate

torquelTheoretica

ww

ww

ww

Radial compressoroperation

• Typical design takes 50 % of increase in static pressure in diffuser

• Conservation of angular momentum governs performance:


Slip factor

vanesimpellerofnumbernn

63.01

Due to inertia of flow Cw2 < U:

22 UworklTheoretica

Stanitz formula for estimating σ


Power input factor -

• Power is put into overcoming additional friction not related to the flow in the impeller channels

• Converts energy to heat => additional loss =>2 UdoneWork


Overall pressure rise:

1

01

2

01

03 1

Tc

U

p

p

p

c

• P03 is here used to denote the pressure at compressor exit. P02 is reserved for the stagnation pressure between the impeller and the diffuser vanes


Example 4.1a• ψ =1.04, σ = 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

• ηc = 0.78

• Compute pressure ratio and power required


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 gas turbine components (turbine and compressor performance are given in Lecture 4) and include them in cycle analysis

• Know how to compute cycle efficiencies for the heat exchanger cycle

chalmers university of technology lecture 3 some more thermodynamics : brief discussion of cycle...

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