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Turbomachinery -40

Copyright © 2014,2015 by Jerry M. Seitzman. All rights reserved. AE4451 Propulsion

Turbomachinery

Turbines

Turbomachinery -41


Turbine Overview

• Configurations (axial, radial, mixed), analysis and other issues similar to compressors

• Compared to compressors

– higher loading ho/U2 (or specific work) and pressure

ratio per stage - why?

• favorable pressure gradient

– usually much higher temperature inlet

• higher temperature materials (strength) and/or blade cooling

Science, AAAS

2

Turbomachinery -42


Turbine Analysis

• Similar to compressor analysis

– Euler turbomachinery (conservation) equations

– and cascade flow to find c

• 12 nozzle

• 23 rotor

ii

rcrcm 1

ii

cucumW ii 11

Nozzle Rotor

1 2

3

Mechanics and Thermodynamics of Propulsion, Hill and Peterson

Turbomachinery -43


Turbine Cascade Analysis

uwc

z

c

Uw

3tan3

zcUc

2tan2

zcc

23 tantan13,2

U

c

U

cz


same form of blade loading eqn

for turbine as compressor

• Now blade moves upward (flip sign convention); again fixed r, ui=U

• Therefore for rotor, and constant cz

ii zz cw Ucwii

2

3,1Uho

2

3

Nozzle Rotor

3

Turbomachinery -44


Stage Pressure Ratio

• For adiabatic turbine with TPG/CPG

1

1

13

03

1 11

o

oo

st

ot

T

TT

p

pPr

1

2

1

23,111

1U

h

RT

U o

ost

1

1

2

3

1 3,211

U

c

RT

U

p

p

osto

oU

c

U

ho 3,23,1

2

• Stage pressure ratio still depends on

1. = f(U= r, c)

2. blade M=f(r, To1)

3. st

>1 as written

<0 for turbine

Turbomachinery -45


Axial Turbine Maps

• Larger stage pressure ratios and efficiencies then compressors

• Peak efficiency on-design

• For fixed RPM, larger pressure change (drop) at higher mass flowrate

– more work extracted per unit mass

• At high (corrected) mass flowrate, nozzle becomes choked


4

Turbomachinery -46


Blade Design: Degree of Reaction

• We have TWO blade

parameters to design

– rotor trailing edge (match 3)

– nozzle trailing edge (match 2)

• How to do this?

1.Degree of reaction, R

2.Stage exit condition

constraint (3)

23 tantan13,2

U

c

U

cz

Similar issue for

compressor; we just

“ignored” designing 1

2tan22

zccUw

3tan3

zcc

2tan2

zcc

3tan33

zccUw 3,23,2 wc


Turbomachinery -47


Degree of Reaction

• Recall

– allows us to distribute load (static pressure

change) between rotor and nozzle (or stator)

– how to relate static enthalpy change to

azimuthal velocity changes?

• KE !!

– for stationary blade, no work done

• e.g., nozzle blade

stagerotor hhR

KEhho 0

2v2 hho

22 2222

12 2221 cccchh zz 222

21 cc

if cz constant, and negligible cr

5

Turbomachinery -48


Degree of Reaction (Turbine)

• Rotor blades?? – are “stationary” in rotor’s

reference frame

• Reaction

123 zzz ccc

222

23 32 wwhh

13

23

hh

hhR

23

32

2

22

ccU

ww

2

11

2

33

23

chch

hh

oo

if c1 c3

13

23

oo hh

hh

23

22

tantan12

32

U

cU

wwR

zrelates design blade angles

to azimuthal KE change

23 tantan13,2

U

c

U

cz

Turbomachinery -49


Impulse Turbine


U

cU

U

w

U

cz 2tan23,23,2

13

23

h

hR

• R = 0

– all the pressure change occurs across the

nozzle, or the nozzle

creates high KE

022

32 ww

23

23 ww

2tan123,2

U

c

U

cz

23 tantan zz cc

23

22

tantan12

32

U

cU

ww

z

23,22 ww

6

Turbomachinery -50


Impulse Turbine


• So for impulse turbine, blade loading coeff.

• Relates blade loading to nozzle exit angle

• From equation, rotor blade angles given by

22

tan1223

U

c

U

c

U

hzstageo

222

1tanU

h

c

U stageo

z

zc

U 223 tantantan

<0

Turbomachinery -51


Impulse Turbine


• To let largest power per unit mass flow rate large 2

– tends to produce high velocities and po losses

– practical limit, ~70-75

• Further possible constraint

– no exit swirl (c3=0)

233,2 ccc 23,2 cc

Uc2

U

c

U

c2323 12

22

U

hstageo

zcU2tan 2

2tan1223

U

c

U

cz

zcU3tan,

7

Turbomachinery -52


50% Reaction Turbine

• R 0.5

– balanced p drop across stage

– if no exit swirl

3232

323,2

22

wwww

wwcU

22

tan213,2

U

c

U

c

U

hzstageo

12

U

hstageo

3,2

32

2

22

cU

wwR

23 tantan

23 tantan13,2

U

c

U

cz

23,2 cc

zcU1

22 tan

half loading of impulse: less power/stage Mechanics and Thermodynamics of Propulsion, Hill and Peterson

3,232 tantan cccU zz

Turbomachinery -53


Compressor-Turbine Matching


• Another part of design/operational requirement

• Need to “match” compressor and turbine stages on same spool

• Steady oper- ation match 1. N (RPM) 2. 3.

• Iterative procedure

mW

8

Turbomachinery -54


Turbine Stresses/Operational Limits

• Turbine blades experience large stresses: bending, thermal and centrifugal (rotor: 104-105 g)

• Materials exhibit significant loss of strength and enhanced creep at high T

– low strength at modern engine To4 (high ST, th) To4 >1400C (2500F)


Turbomachinery -55


Turbine Inlet Temperature Evolution

www.virginia.edu/ms/research/wadley/high-temp.html

• Solutions

– high temperaturematerials

– blade cooling

– TBC (thermal barrier coatings) Ni superalloys

single crystal super alloys

9

Turbomachinery -56


Turbine Blade

Cooling • Usually use

compressor (bleed) air

• Configurations

– internal passages

– external • film cooling

• tip cooling

• Heat transfer designed to

– focus on “hot” spots and initial stages

– minimize stress concentration

Gas Turbine Theory, Cohen, Rogers and Saravanamuttoo

Turbomachinery -57


Turbine Blade

Cooling

• Rotor and

nozzle cooling

configurations

Gas Turbine Theory, Cohen, Rogers and Saravanamuttoo

10

Turbomachinery -58


Introduction to Heat Transfer

vhot (=c or w)

AQ

zTouter

zTinner

z

x

x

• Consider a simplified version of a (half) turbine blade

• Inner cooling only

– neglect film and tip cooling for now

– hot gas (combustor products) flows over outer surface

– “cold” gas (bleed air) flowing over inner surface

– turbine blade “wall” in between

• How to analyze this “heat transfer” problem?

Turbomachinery -59


Conduction Heat Transfer

• Start with description of (conduction) heat transfer through the wall

– assume one-dimensional

– top side of wall uniform temp. (Touter)

– bottom side of wall uniform temp. (Tinner)

• Look at energy equation

– differential CV

– steady

• Need model for

– Fourier’s Law (1d)

AQ

outerT

innerT

z

x

x

inQ

T dx

outQ

outin QmcdTdt

dQ

outin QQ

Q

dx

dTk

A

Q

Thermal Conductivity

11

Turbomachinery -60


Conduction and Thermal Conductivity

• For steady, uniform material

– T gradient is a constant

– so T varies linearly through wall

• Thermal conductivity

– insulators like ceramics have much lower conductivities than metals

• so TBC will produce much lower heat flux for same temperature gradient

kA

Q

dx

dT

x

TouterTinnerTMaterial k (W / mK) at 1000C

Nickel Super Alloys 20-30

Ceramic TBC’s 1-2

AQ

outerT

innerT

z

x

x

Turbomachinery -61


Effect of Adding TBC Coating

• Example Ni alloy with

• Now add 500m TBC

dx

dTk

A

Q

x

TouterTinnerT

AQouterT

innerT

xalloy

AQ

outerT

innerT

xalloy

xTBC

midT

25.3

5

70025

m

MW

mm

K

mK

W

TBCalloyA

Q

A

Q

alloy

innermidalloy

TBC

midouterTBC

x

TTk

x

TTk

TBCTBCalloyalloy

TBCTBCinneralloyalloyouter

midkxkx

kxTkxTT

xalloy5mm

kalloy=25W/mK

Touter=1400K

Tinner700K

KTmid 9622

3.15

26225

m

MW

mm

K

mK

W

A

Q

midTMost of the temperature drop occurs across TBC, much lower metal T and lower heat transfer

xTBC0.5mm

kalloy=1.5W/mK

12

Turbomachinery -62


Convective Heat Transfer

• Examine heat transfer between gas flow and blade wall

• Convective heat transfer

– due to fluid moving over surface

– thermal boundary layer develops, like momentum boundary layer

• Model

– so Twall varies downstream

– e.g., for laminar flow over flat plate

v

AQ

zTwall

z

x ,gasT

wallgas TTh

A

Q ,

Convective Heat Transfer Coeff.

PrRehh z ,Reynolds number

zshearshear Re

Prandtl number

PrThermal

diffusivity

pck

2132332.0v

x

pg

RePrc

h

g

2132664.0v

L

pg

RePrc

h

g

Stanton Number averaged over full length

Turbomachinery -63


Convective Heat Transfer - External

• Example

hot air

• Analysis

L

L4cm p=10atm

Tgas=1850K

Twall1400K

v=250m/s

Pr=0.7 v=310-4m2/s

v

AQ

zTwall

z

x ,gasT

21

1

32

1 v332.0

mmpgmm RePrch

g

1

T

pc

gpg

zRez

v

Km

kW

Km

kW2

5.0667.0

298337.0626332.0

2

2

1 4140018009 mMWKKm

kW

A

Q mm

5.0

24

667.0

103

001.02507.0250

128.1

28.1

1850

013.1332.0

sm

msmsm

K

MPa

Km

kWhh LxL 2

9.22

2,3.1 mMW

A

Q totalL

much higher heat load around leading edge

13

Turbomachinery -64


Turbine Blade Analysis

• In our two examples

– conduction through TBC-

coated alloy

– convective heat transfer into

blade

• So together they represent a single problem

• Next step is to investigate bleed air cooling

requirement

23.1 mMW

K1400

K700

alloy

K962TBC

K1850sm250

23.1~ mMW

23.1~ mMW

conductionconvectionAQAQ

Turbomachinery -65


Cooling – Convection Internal Flow

• In pipe/channel flow can’t assume infinite flow

– boundary layers meet and central flow changes with axial distance

• Now

• New expressions for h, e.g., for round tubes

– turbulent flow, profile still developing

– averaged over channel length

v

Q

innerT

coolT

Q

coolantbulkinner TThperimeterL

Q

A

Q,

Bulk avg. temp.

055.0

318.0036.0

L

dPrRe

d

kh d

14

Turbomachinery -66


Turbine Blade Analysis • Assuming same information

in previous examples AND 2mm height channels with span = 80% of blade chord, with 500K, 30 m/s inlet bleed air, negligible spacing betweenchannels

• Much less than the cooling requirement from previous parts of the analysis – need to enhance the coolingfilm cooling

23.1 mMW

K1400

K700

alloy

K962TBC

K1850sm250

bleedinnercool TThA

Q

2

1

2191 mWh

vhot (=c or w)

AQ

zTouter

zTinner

z

x

x

vhot (=c or w)

zTouter

zTinner

220 mkWA

Qcool

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