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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
Lunds universitet / Kraftverksteknik / JK
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?
Lunds universitet / Kraftverksteknik / JK
Samples
WindpowerHydropowerTurbochargers of cars and trucks
Vacuum cleaners…
Lunds universitet / Kraftverksteknik / JK
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
Lunds universitet / Kraftverksteknik / JK
Samples
FIG. 1.1. Diagrammatic form of various types of turbomachine.
Lunds universitet / Kraftverksteknik / JK
Dixon, chapter 1
Diagrammatic form of various types of turbomachine.
Lunds universitet / Kraftverksteknik / JK
If you have known one you have known all.
(TERENCE, Phormio.)
Lunds universitet / Kraftverksteknik / JK
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
Lunds universitet / Kraftverksteknik / JK
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
Lunds universitet / Kraftverksteknik / JK
Moody chart
Lunds universitet / Kraftverksteknik / JK
Dixon, chapter 1
Measures of pressure (head …
Shaft power
…
Lunds universitet / Kraftverksteknik / JK
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
Lunds universitet / Kraftverksteknik / JK
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛== ,...,,, 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
Incompressible fluid analyses
Lunds universitet / Kraftverksteknik / JK
For geometrically similar machines, neglecting Reynolds-number dependence:
( ) ⎟⎠⎞
⎜⎝⎛== 342 ND
QfNDgHψ
⎟⎠⎞
⎜⎝⎛= 35 ND
Qfη
⎟⎠⎞
⎜⎝⎛== 3653
ˆNDQf
DNPP
ρ
Incompressible fluid analyses
Lunds universitet / Kraftverksteknik / JK
For a pump:Net hydraulic power (transferred to the fluid): gHQPN ρ×=
Incompressible fluid analyses
PgHQ
PPN ρη ×
==
( )53
23
1 DNNDgH
NDQP ρ
η×××=
3 5ˆ /PP
N Dφψ η
ρ= =
for a turbine: N
PP
η =
Lunds universitet / Kraftverksteknik / JK
Performance Characteristics
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Performance characteristics
Lunds universitet / Kraftverksteknik / JK
Variable geometry turbomachines
Lunds universitet / Kraftverksteknik / JK
Variable geometry turbomachines
Lunds universitet / Kraftverksteknik / JK
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
Lunds universitet / Kraftverksteknik / JK
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
Lunds universitet / Kraftverksteknik / JK
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
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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
Lunds universitet / Kraftverksteknik / JK
Specific speed (from Japixe-Baines)
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Specific speed
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Specific speed
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Specific speed
Lunds universitet / Kraftverksteknik / JK
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 =
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Deplacement pumps• A piston moves for and back in a cylinder• Valves ensure the right direction of the flow
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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
Lunds universitet / Kraftverksteknik / JK
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
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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
Lunds universitet / Kraftverksteknik / JK
Compressible gas flow relations
Lunds universitet / Kraftverksteknik / JK
Compressible gas flow relations
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:
Lunds universitet / Kraftverksteknik / JK
Compressible gas flow relations
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⎛ ⎞=⎜ ⎟⎝ ⎠
Lunds universitet / Kraftverksteknik / JK
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:
Lunds universitet / Kraftverksteknik / JK
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
Lunds universitet / Kraftverksteknik / JK
Compressible fluid analyses
Lunds universitet / Kraftverksteknik / JK
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
−⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − =⎜ ⎟ ⎜ ⎟− ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
Lunds universitet / Kraftverksteknik / JK
Compressible fluid analyses
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
Lunds universitet / Kraftverksteknik / JK
Compressible fluid analyses
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
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Compressor and expander maps
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Instationarity of flow