Theory of turbo machinery / Turbomaskinernas teori

Lunds universitet / Kraftverksteknik / JK

Introduction

Turbomachinery describes machines that transfer energy between a rotor and a fluid, including both turbines and compressors (source: Wiki).

Devices in which energy is transferred, either to, or from, a continuously flowing fluid by the dynamic action of one or more moving blade rows (Dixon)

Definition

The words rotor and continuous separate turbomachinesfrom reciprocating (piston) engines


Introduction

Close to all electric power is produced by turbomachines

They consume large parts of energy used in many industrial processes

They are integral parts of gas turbines used in e.g. aircraft engines and as (shaft-) power supply in oil and gas industry (for pumps and compressors) as well as propulsion of ships

Why a course on turbomachines?


Samples

WindpowerHydropowerTurbochargers of cars and trucks

Vacuum cleaners…

http://en.wikipedia.org/wiki/Image:Axial_compressor.gif


Introduction

Energy may flow to the fluid (increasing velocity and/or pressure) or from the fluid producing shaft power

Flowpath: Axial or radial machines (mixed flow)

Changes in density, compressible or incompressibleanalyses.

Impulse or reaction machines: Does the pressure change in the rotor, or in a set of nozzles before the rotor?

Classifications


Samples

FIG. 1.1. Diagrammatic form of various types of turbomachine.


Dixon, chapter 1

Diagrammatic form of various types of turbomachine.

http://upload.wikimedia.org/wikipedia/en/7/7b/Compressor_Types.png


If you have known one you have known all.

(TERENCE, Phormio.)


If a physical process satisfies physical dimension homogeneity and involves n dimensional variables, it can be reduced to a relation between only k variables

The reduction j = n - k equals the maximum number of variables which do not form a Π among themselves.

The reduction is always less or equal to the number of dimensions describing the variables.

The Π-theorem

Dimensional analyses and performance laws


1. List and count the n variables involved2. List the dimensions of the variables

3. Find j, guess j = number of dimension, if unsuccessful; j = j - 1

4. Select j scaling parameters which do not form a Π5. Add one additional variable and form a Π. Repeat

for the others.

6. Write the final dimensionless function.

The machinery (adopted from Frank M. White)

Π-Theorem


Moody chart


Dixon, chapter 1

Measures of pressure (head …

Shaft power

…


1 21

Δ , , , , , , ,...l lpgH f Q N DD D

ρ μρ

⎛ ⎞= = ⎜ ⎟⎝ ⎠

⎟⎠⎞

⎜⎝⎛= ,...,,,,,, 21

2 Dl

DlDNQf μρη

⎟⎠⎞

⎜⎝⎛= ,...,,,,,, 21

3 Dl

DlDNQfP μρ

Start of Dimension analyses

Incompressible fluid analyses


( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛== ,...,,, 21

2

342 Dl

DlND

NDQf

NDgH

μρψ

⎟⎟⎠

⎞⎜⎜⎝

⎛= ,...,,, 21

2

35 Dl

DlND

NDQf

μρη

⎟⎟⎠

⎞⎜⎜⎝

⎛== ,...,,,ˆ 21

2

3653 Dl

DlND

NDQf

DNPP

μρ

ρ

Find dimension-less groups



For geometrically similar machines, neglecting Reynolds-number dependence:

( ) ⎟⎠⎞

⎜⎝⎛== 342 ND

QfNDgHψ

⎟⎠⎞

⎜⎝⎛= 35 ND

Qfη

⎟⎠⎞

⎜⎝⎛== 3653

ˆNDQf

DNPP

ρ



For a pump:Net hydraulic power (transferred to the fluid): gHQPN ρ×=


PgHQ

PPN ρη ×

==

( )53

23

1 DNNDgH

NDQP ρ

η×××=

3 5ˆ /PP

N Dφψ η

ρ= =

for a turbine: N

PP

η =


Performance Characteristics


Performance characteristics


Variable geometry turbomachines


Variable geometry turbomachines

Let b represent the settings of the vanes

( )βφψ ,1f=

( )βφη ,2f=

Or alternatively:

( ) ( )ψφηφβ ,, 43 ff ==

And solve for β

( ) ⎟⎠⎞

⎜⎝⎛== 22355 ,,

DNgH

NDQff ψφη


Specific speed

An alternative representation can be obtained by eliminating the diameter

Define the dimensionless groups at maximum efficiency:

maxηη = 1φφ = 1ψψ = 1̂ˆ PP =

constant13 ==φNDQ

constant122 ==ψDN

gH

constant153 ==φρ DN

P


Specific speed

Eliminate D to obtain the following dimensionless parameters:

( ) 4/3

2/1

4/31

2/11

gHNQNs ==

ψφ

( )( ) 4/5

2/1

4/51

2/11 /ˆ

gHPNPNsp

ρψ

==

( ) 4/3

2/1

gHQ

sΩ

=Ω

( )( ) 4/5

2/1/gHP

spρΩ

=Ω

Dimensionless, directly proportional to N

Power specific speed, turbines

If speed of rotation is expressed in rad/s


Specific speedThe relation between Ns and Nsp

( )( )

( ) 2/1

2/1

4/3

4/5

2/1/⎟⎟⎠

⎞⎜⎜⎝

⎛==

gQHP

NQgH

gHPN

NN

s

sp

ρρ

η1

=Ω

Ω=

s

sp

s

sp

NN

η=Ω

Ω=

s

sp

s

sp

NN

For a pump and a turbine respectively

From the definition of the hydraulic efficiency, we obtain


Specific speed (from Japixe-Baines)


Specific speed


Pumps in pipe systems

The pump head (uppfordringshöjden):

2 22 1 2 1

2stat fp p c cH H h

g gρ− −

= + + + Δ

where

HeadH =

2 1 static pressure differencep p− =

hight differencestatH =

2 22 1 squared velocity differenciesc c− =

friction lossesfhΔ =


Deplacement pumps• A piston moves for and back in a cylinder• Valves ensure the right direction of the flow


Cavitation

Formation of bubbles in liquid, if pressure decreases so that temperature is above boiling pointIn practice, cavitation onset is influenced by dissolved gases and presence of boiling nuclei.

As pressure increases, the bubbles collapse and generate strong pressure waves.

If the collapse occurs near walls, the pressure waves will cause cavitation erosion


Cavitation

Fairly low pressures are required to cause cavitation in water

Increased temperatures will of cause trigger cavitation at higher pressures

Boiling point of water


Compressible gas flow relations

For compressible fluids with large pressure changes, large velocity differences occur over the stages.

It is convenient to combine the enthalpy, h, with the kinetic energy of the fluid to the stagnation enthalpy:

220 chh +=

If the fluid is brought to rest through a reversible process with no heat transfer (adiabatic) the state change is called isentropic



For a perfect gas:

TCh p=

( )1−= γγRCp

And the stagnation temperature can be defined by:

pCcTT

2

2

0 +=

( ) ( )2

112

1122

0 MRTc

TT

−+=−+= γγ

γ

RTcacM γ==

Where M is the Mach Number:



Gibbs’s relation

Isentropic (ds = 0) retardation to zero velocity

so that

1d d dT s h pρ

= −

1 dd d dpph C T p RTpρ

= = =

TT

TT

RC

pp p d

1dd

−==γγ

pRT

ρ⎛ ⎞=⎜ ⎟⎝ ⎠


Compressible gas flow relationsIntegration yields

or

Using the gas law:

Tp ln1

constantlnln−

+=γγ

12100

211

−−⎟⎠⎞

⎜⎝⎛ −+=⎟

⎠⎞

⎜⎝⎛=

γγ

γγ

γ MTT

pp

( )RTp=ρ ( )( )000 TTpp=ρρor

11

211

00

211

−−⎟⎠⎞

⎜⎝⎛ −+=⎟

⎠⎞

⎜⎝⎛=

γγ γρρ M

TT

We obtain:


Compressible fluid analysesPerformance parameters for compressible flow (3 relations among 8 variables)

Dimensional analyses(3 relations among 5 variables)

( )γρμη ,,,,,,,, 01010 amDNfPh s =Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Δ γμ

ρρρ

η ,,,,,01

201

301

5301

220

aNDND

NDmf

DNP

DNh s

Note: ND is proportional to blade velocity, thus:

μρ 2

01ND Reynolds Number

01aND

Blade Mach Number


Compressible fluid analyses



Consider temperature ratio for isentropic pressure change from p01 to p02

For a perfect gas, the enthalpy change becomes

γγ 1

01

02

01

02

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

pp

TT s

( )( )[ ]110102010 −=Δ − γγppTCh ps

Substitute ( )1−= γγRCpand divide by

01201 RTa γ=

( )1

0 02 02012

01 01 01 01

1 11

sh p pR T fa RT p p

γ γΔ γ

γ γ

−⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − =⎜ ⎟ ⎜ ⎟− ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦



Similarly for the flow coefficient:

201

012

0101

012

0101 DpRTm

DRTpRTm

Dam

γγρ==

And for the power coefficient, using ( ) 201 DNDm ρ∝

( )( ) ( ) 01

020

2201

053

01

ˆTT

NDTC

NDNDDTCm

DNPP pp Δ

≡Δ

=Δ

==ρρ



Substituting these relations into the result from dimensional analyses:

For a specific machine, handling one gas, γ, R and D can be omitted. If further the Reynolds number dependence is neglected,the following simplification results:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Δ ,,,,0101

01

01

0

01

02

TN

PTm

fTT

PP η

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Δ γμ

ργγ

η ,,,,,2

01

01012

01

01

0

01

02 NDRT

NDPDRTm

fTT

PP

However, this relation is not dimensionless


Compressor and expander maps


Instationarity of flow

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