1

Turbomachinery Lecture 3

- Compressibility- Isentropic- Area, Mass Flow Functions- C-D Nozzle

2

Turbomachinery• Definition:

– A turbomachine transfers energy to or from a fluid flowing continuously through a casing by the dynamic action of a rotor and by the flow conditioning of a stator.

• Works on a fluid to produce power or flow (and pressure rise)

• Adds energy to fluid................Pump or Compressor– Fan: pressure rise up to 1 lbf/in2

– Blower: pressure between 1 - 40 lbf/in2

– Compressor: pressure rise > 40 lbf/in2

• Extracts energy from fluid............Turbine

– Pressure changes due to motion of parts or displacement of boundaries

3

Compressible Flow• Density varies making continuity & momentum more

difficult to solve.

because varies with velocity.

• Also, can't integrate Bernoulli directly

• Compressible flow problems can be solved iteratively using continuity, state et. al.

cosm AC

.2

2

constVdP

4

• Example:m = 50 lb/sec A = 200 sq.in.

P0 = 14.7 psia = 30

T0 = 519 R

• GuessC = 646.8 ft/sec

2

0

22

2

2 2

2

646.8 / sec519

. .2 32.174 778.16 .24

.sec

6008.8 / sec

484.19deg

p

CT T

C

ftT

ft lbm ft lbf BTUlbf BTU lbm R

ft R

T R

Compressible Flow

5

Compressible Flow

• Pressure

/ 1

00

TP P

T

3.5484.19

14.7 11.529519

P psi

6

Compressible Flow• Density can now be found from state:

11.529 144

53.349 484.19

0.06427 / .lbm cu ft

RT

P

7

Compressible Flow• Mass Flow:

• note:

• 19%>

cos

0.06427 646.8 cos30 200 /144

50.00 / sec

m AC

m

m lb

0 0.0765 / .lbm cu ft

8

Compressible Flow

• Mach Number Functions:

– Easily calculated & clarify physics

• Mach number & acoustic speed are critical concepts!

9

Compressible Flow

0

1

p

dPIsentropically TdS dh

dPc dT RT

PdT dP

T P

State P RT

dP RdT RTd

T

dTd

P

dP

10

Compressible Flow• Using isentropic relation between pressure &

temperature derivatives:– Use adiabatic state law

P

dPd

P

dP

1

d

P

dP

1

2dP Pa RT

d

1P CT

11

Compressible Flow

• Using equation of state, acoustic speed in an ideal gas is [from kinetic theory]:

• By definition Mach Number is:2

2

2

V dynamic pressure

p static pressureV VM

a a V kinetic energy

RT thermal energy

1716

287

Ta gRT

T

12

Compressible Flow

• Static & Total properties as functions of Mach number: 2

0 2

Vh h

g

20

20

12

12

p

p

T V

T gc T

T R V

T c gRT

0 211

2

TM

T

13

Compressible Flow – Critical Velocity

• What does subscript * mean? It means value of variable when M=1 [sonic]

• Vcr is only function of gas [] and stagnation props.

2

0

2 2 22 220 0

0

2

1

1 1 2 1 2 2 1 1

2

1

cr crcr

cr

Vh h

a V V RTa VV

V RT

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Compressible Flow

• The relation between static & stagnation properties is isentropic. Then:

/ 120 1

12

PM

P

1/ 10 21

12

M

15

Compressible Flow• The relation between compressible and Bernoulli [B-p.55]

12

0

2 2

2 40

2 2 22

2

1/ 1

2

1(1 ) 1 ( 1) / 2 ... &

1 2

2/ 1 ...

2 2

2 2 2 / 2

n

p p M

Binomial expansion for small x is x nx n n x n x M

For small M one gets p p M M

V V VBut since pM p p

a p

The

2 24

0

2

0

21 ...

2 4 2

0.3, 2.3% ( ).2

V Mexpanded isentropic equationbecomes p p M

Vfor M p is in error from Bernoulli p

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Compressible Flow Relationships

• Mass Flow parameter [=0]

0

0

m VA

dm VdA AdV VAd

dm dA dV d

m A V

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Compressible Flow Relationships

• Area-Mach number differential relation

• Area-Mach number integral relation

2

22

11

MdA dV dpM

A V M p

1

2 121 2 1

11 2

AM

A M

More on next chart

18

Compressible Flow Relationships

• What does subscript * mean?– For all flow variables it means value of

variable when M=1 [sonic]

– For area A* this is reference area for choking flow [M=1]

• Note this area is a minimum or throat

1

2 121 2 1

11 2

AM

A M

More on next chart

19

Compressible Flow Relationships

Flow textbooks

-www.engr.uconn.edu/barbertj- Compressible

- Aero Calculator- calcbody2

20

Compressible Flow Relations

21

2

22

11

MdA dV dpM

A V M p

Of interesthere

Of interesthere

22

2

22

11

MdA dV dpM

A V M p

Over-expanded

23

24

25

26

Compressible Flow Examples

0 00 0

: 450 1890 1.5

3.671 6938 1.45 652.5

1.4 1716 450 1040

1.5 1040 1560

s s

s s

s s

Given T R p psf M

p Tp psf T R

p T

a a RT fps

V fps

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Compressible Flow Examples01 01 *

*

0 0

*

0

10 300 / 6

/ 6 0.097

1.006 9.94 1.002 299.4

1.4 287 299.4 346.8

33.6

/ 6 3.368

63.13 0.

s ss s

s s

ss

Consider isentropic flow in C D nozzle

p atm T K A A

Subsonic A A M

p Tp atm T K

p T

a a RT mps

V mps

Supersonic A A M

pp

p

0

1 /2 1

0 *

0

1584 3.269 91.77

192 646.7

2

1

ss

s

Tatm T K

T

a a mps V mps

p Am VA if choked

RT

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Compressible Flow

• Mass Flow Parameters:

VRT

P

A

m

AVm

cos

cos

1/ 2

0

0cos

Tm V g

PA RT TgRT

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Compressible Flow Relationships

• Mass flow parameters

0 0

0 0

00 1

0 2 12

1/ 2

0 2

( , )

11

2

11

2ss

m VA

m p V pV RT M

A RT a RT

m T TpM f M

p A p T R

m T R MFP

p AM

m T RFP M M

p A

Note: FPo, FPs are similar, but different f[M] powers

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Compressible Flow Relationships• Mass flow parameters

00 1

0 2 12

01

0 2 12

11

2

11

2

m T R MFP

p AM

p A Mm

RTM

How to get more mass flow, i.e. greater thrust, more power?

1

2 10

0

, 1

2

1

if choked at throat M

p Am

RT

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Compressible Flow Relationships

• Mass flow parameters units– m in lbm/sec

– p0A in lbf [spatial dimensions cancel]

– T0 in degs. Rankine

– A is sometimes frontal area Acos

00

0

0

0

0

0

0

0

1

1716 /1.4

32.2

1.0888

RTmFP

p A g

m T gR

p A g

m T

p A

m T

p A

For air

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Compressible Flow Examples

01 01 *

*

0 0

1 /2 1

0 *

0

10 300 / 6

/ 6 3.368

63.13 0.1584 3.269 91.77

192 646.7

2

1

s ss s

s

Consider isentropic flow in C D nozzle

p atm T K A A

Supersonic A A M

p Tp atm T K

p T

a a mps V mps

p Am VA if choked

RT

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Example2

0 0

2*

*

00

0

2 ,

0.5 1 300

1.4, 0.5 1.340 1.49

( , ) 353.6 / sec

Air in duct of A m has flow such that

M p atm T K

AFor M A m

A

area to choke

p Am FP M kg

T R

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Static Pressure Mass Flow Parameter

• Defining: FP = Flow parameter=f(M)

• For Air

• Can be inverted

1/ 220 1

1cos 2s

RTmFP M M

PA g

01.0883coss

m TFP

PA

2/1

2

1

1211

sFPM

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Total Pressure Mass Flow Parameter• Introducing P0:

• No explicit solution for M • FPs is single valued, FPo is not• FPo max = 0.5787 for =1.4• FPo max always at M=1

1/ 2

00

0 0cos

Tm P gP M

A P RT T

1/ 2

0 00

0 0cos

RT Tm PFP M

P A g P T

1

2 12

0

11

2FP M M

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Calculate FPo• From Previous Example:

m = 50 lb/sec A = 200 sq.in.

P0 = 14.7 psia = 30

T0 = 519 R

• Rearrange FPo

00

0

0.4869cos

RTmFP

P A g

1 /2/ 12

0

11 0.5997

2calc guessM FP M M

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Mass Flow Parameters

Be careful: FPs single valued, FPo double values

38

Total Pressure Mass Flow Parameter

• Consider FPt:

• For fixed , a fixed value of

produces the same Mach number - regardless of the level of pressure, temperature or molecular weight (R).

1

2 120

00

11

cos 2

RTmFP M M

P A g

0

0 cos

m RT

P A

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Total Pressure Mass Flow Parameter

• Defines common flow parameters.

• Valid for flow with one gas.

• Corrected flow.

0

0 cos

m RT

P A

0

0

m T

P

0

0

/ 519

/14.7

m T m

P

40

Other Parameters[Covered in Lecture 4]

• Ideal gas equation for Mach number leads to speed parameters, also for a single gas.

• Speed parameter

• Corrected speed

0

N

T

N

0 0 14.696 518.7

in inP T

41

Significance of Flow & Speed Parameters

• A device operating at the same speed parameter and flow parameter has the same Mach numbers, velocity diagrams, flow angles etc, regardless of the level of physical speed, pressure & temperature.

42

Flow and Speed Parameters• Conditions: same gas, high Reynolds number, same

clearances, and same • Speed and Flow parameters are used for turbine

maps

0

5

10

15

20

25

30

35

1.0 1.5 2.0 2.5 3.0

Exp Ratio

Mrt

T/P

N/rtT

43

Corrected Flow & Speed Parameters

• Corrected Flow and Corrected speed used for compressor maps

3

4

5

6

7

8

9

50 60 70 80 90 100 110

Corrected Flow lb/sec

Pre

ssu

re R

atio

Corrected Speed

44

Flow Parameter

• Again, Consider FP0:

• Unlike P, T & R; cannot be "corrected".

• Changing , changes relation between FP0 and Mach number!

1

2 120

00

11

cos 2

RTmFP M M

P A g

45

Flow ParameterGamma Effect On Continuity

0.60

0.65

0.70

0.75

0.80

0.85

0.90

0.95

1.00

1.0 1.2 1.4 1.6 1.8

Gamma

Ma

ch N

um

eb

er

Air

Helium

Butane

FPT = .560

Message: More complex gasses choke at a lower Mach number

46

Example Solution to Mass Flow Parameter

0 1.0719 484.18degT

T RT

646.87 / secV M gRT ft

/ 1

0 0 11.53P T

P psiP T

47

Area Ratio

• A/A* is flow area / flow area at M = 1.0

0 1

0

1 / 2 / 12

1 / 2 / 1

*

11

2* 1

12

MFPA

A FP

MA

AM