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Lunds universitet / Kraftverksteknik / JK
Theory of turbo machinery / Turbomaskinernas teori
Chapter 1
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
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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 =
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Performance characteristics
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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
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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 speed
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Specific speed
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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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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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Theory of turbo machinery / Turbomaskinernas teori
Chapter 2
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The first law
xWEnergy is transferred from fluid to the blades of the machine, positive work is at the rate
Heat transfer, , is positive from surrounding to machineQ
Mass flow, , enters at 1 and exits at 2m
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Euler’s pump and turbine equations
The work done on the fluid per unit mass (specific work) becomes:
01122 >−=Ω
==Δ θθτ cUcU
mmWW Ac
c(2.12a)
02211 >−==Δ θθ cUcUmWW t
t(2.12b)
FIG. 2.4. Control volume for a generalised turbomachine.
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Definitions of efficiencyConsider a turbine: The overall efficiency can be defined as
Mechanical energy available at coupling of output shaft in unit time
Maximum energy difference possible for the fluid in unit time η0 =
If mechanical losses in bearings etc. are not the aim of the analyses, the isentropic or hydraulic efficiency is suitable:
Mechanical energy supplied to the rotor in unit time
Maximum energy difference possible for the fluid in unit time ηt =
The Mechanical efficiency now becomes η0 / ηt
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Efficiency
Neglecting potential energy terms, the actual turbine rotor specific work becomes:
And, similarly, the ideal turbine rotor specific work becomes:
where the subscript s denotes an isentropic change from state 1 to state 2
( ) 222
21210201 cchhhhmWW xx −+−=−==Δ
( ) 222
21210201max,max, sssxx cchhhhmWW −+−=−==Δ
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EfficiencyIf the kinetic energy can be made useful, we define the total-to-total efficiency as
Which, if the difference between inlet and outlet kinetic energies is small, reduces to
( ) ( )sxxtt hhhhWW 02010201max, −−=ΔΔ=η
( ) ( )stt hhhh 2121 −−=η
(2.21)
(2.21a)
If the exhaust kinetic energy is wasted, it is useful to define the total-to-static efficiency as
( ) ( )sts hhhh 2010201 −−=η (2.22)
Since, here the ideal work is obtained between points 01 and 2s
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Efficiency
Efficiencies of compressors are obtained from similar considerations:
( ) ( )01020102 hhhh sc −−=η (2.28)
(2.28a)
Minimum adiabatic work input per unit time
Actual adiabatic work input to rotor per unit timeηc =
Which, if the difference between inlet and outlet kinetic energies is small, reduces to
( ) ( )1212 hhhh sc −−=η
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Small stage or polytropic efficiencyIf a compressor is considered to be composed of a large number of small stages, where the process goes from states 1 - x - y -…. - 2, we can define a small stage efficiency as
( ) ( ) ( ) ( ) ...11min =−−=−−== xyxysxxsp hhhhhhhhWW δδη
If all small stages have the same efficiency, then
However, since the constant pressure curves diverge:
WWp δδη ΣΣ= min
( ) ( ) ( )121 ... hhhhhhW xyx −=+−+−=Σδ
and thus
( ) ( )[ ] ( )121 ..... hhhhhh xysxsp −+−+−=η
( ) ( )1212 hhhh sc −−=η
( ) ( ) ( )121 ..... hhhhhh sxysxs −>+−+− and cp ηη >
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If T ds = dh - υ dp,
then for constant pressure:
(dh / ds)p = T
or
At equal values of T:
(dh / ds)p = constant
For a perfect gas, h = Cp T,
(dh / ds)p = constant
for equal h
Small stage or polytropic efficiency
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Small stage efficiency for a perfect gas
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Small stage efficiency for a perfect gas
For a turbine, similar analyses results in
( ) ( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−=
−− γγγγη
η1
1
2
1
1
2 11pp
pp p
t (2.38)
and
( ) γγη 1
1
2
1
2
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
p
pp
TT
(2.37)
Thus, for a turbine, the isentropic efficiency exceeds the polytropic (or small stage) efficiency.
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Small stage efficiency for a perfect gas
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Theory of turbo machinery / Turbomaskinernas teori
Chapter 3
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2D cascades
High hub-tip ratio (of radii)
• negligible radial velocities
• 2D cascades directly applicable
Low hub-tip ratio
• Blade speed varying
• Blades twisted from hub to tip
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2D cascades
FIG. 3.1. Compressor cascadewind tunnels. (a) Conventional low-speed, continuousrunning cascade tunnel (adapted from Carter et al. 1950).(b) Transonic/supersoniccascade tunnel (adapted from Sieverding 1985).
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2D cascades
FIG. 3.2. Compressor cascade and blade notation.
( )y x• Camber line
• Profile thickness ( )t x
a
x
t y
( )b y a=• Max camber
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2D cascades
FIG. 3.2. Compressor cascade and blade notation.
s• Spacing
• Stagger angle
• Camber angleChange in angle of the camber line
• Blade entry angle
• Blade exit angle
• Inlet flow angle
• Incidence
ξ
θ
1 'α
2 'α
1α
i
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2D cascades Efficiency of a compressor cascade
FIG. 3.6. Efficiency variation with average flow angle (adapted from Howell 1945).
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2D cascades Fluid deviation
FIG. 3.2. Compressor cascade and blade notation.
Incidence is chosen by designer
With limited number of blades:
2 2'α α≠So that the deviation may be defined as
2 2 'δ α α= −
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2D cascades Generalizing experimental results
2 2 'δ α α= −
Deviation by Howell: Nominal deviation a function of camber and space chord ratio:
( )* nm s lδ θ=
( ) *2
0.50.23 2 500
nm a l a=
= +
with the following constants for compressor cascades
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2D cascades Optimum space chord ratio of turbineblades (Zweifel)
FIG. 3.27. Pressure distribution around a turbine cascade blade (after Zweifel 1945).
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2D cascades Optimum space chord ratio of turbineblades (Zweifel)
22idY c bρ=
Maximum tangential load (force per unit span)
b is passage width, fig 3.27
Ratio of real to ideal load for minimum losses is around 0.8
( ) ( )22 1 22 cos tan tan 0.8T
id
Y s bY
Ψ α α α= = + ≈
For specified inlet and outlet angles s b or s l may be determined
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Theory of turbo machinery / Turbomaskinernas teori
Chapter 4
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Axial-flow Turbines: 2-D theory
FIG. 4.1. Turbine stage velocity diagrams.
Note direction of α2
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Axial-flow Turbines: 2-D theory
Assumptions:
• Hub to tip ratio high (close to 1)
• Negligible radial velocities
• No changes in circumferential direction (wakes and nonuniformoutlet velocity distribution neglected)
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Axial-flow Turbines: 2-D theory
( )01 03 2 3Δ y yW W m h h U c c= = − = +
01 02h h=
Please note: No work done in nozzle row:
With
And using above equations:
( )2 2 20 2 2x yh h c c c= + = +
Work done on rotor by unit mass of fluid
( ) ( )2 202 03 2 3 2 32x y y yh h h h c c U c c− = − + + = +
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Axial-flow Turbines: 2-D theory
FIG. 4.3. Soderberg’s correlation of turbine blade loss coefficient with fluid deflection (adapted from Horlock (1960).
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Axial-flow Turbines: 2-D theory
Corrections for
• Reynolds number
• Blade aspect ratio
Nozzles:
Rotors:
Tip clearance losses and disc friction not included
5Re 10≠
1 45* *10
Recorζ ζ⎛ ⎞
= ⎜ ⎟⎝ ⎠
( )( )* *1 1 0.993 0.021cor b Hζ ζ+ = + +
( )( )* *1 1 0.975 0.075cor b Hζ ζ+ = + +
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Axial-flow Turbines: 2-D theory
Design considerations
• Rotor angular velocity (stresses, grid phasing)
• Weight (aircraft)
• Outside diameter (aircraft)
• Efficiency (almost always)
• ………
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Axial-flow Turbines: 2-D theory
Consider a case with given
• Blade speed
• Specific work
• Axial velocity
U
( )2 3Δ y yW U c c= +
xc
The only remaining parameter to define is since
• Triangles may be constructed
• Loss coefficients determined from Soderberg
• Efficiencies computed from loss coefficients
2yc 3 2Δ
y yWc c
U= −
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Axial-flow Turbines: 2-D theory
FIG. 4.4. Variation of efficiency with cy2/U for several values of stage loadingfactor ΔW/U2 (adapted from Shapiro et al. 1957).
2
ΔStage loading factor: WU
flow coefficient: xcU
Aspect ratio: Hb
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Axial-flow Turbines: 2-D theory
Stage reaction, R
• Alternative description to
• Several definitions available
• Here:
2yc U
( ) ( )2 3 1 3R h h h h= − −
E.g: R = 0.5
( ) ( )2 3 1 3
2 3 1 2
0.5 h h h hh h h h
= − −
− = −
R = 0.5
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Axial-flow Turbines: 2-D theory
( ) ( )2 3 01 03R h h h h= − −
1 3c cFor a normal stage, =
Using eq. 4.4: and Euler( )2 22 3 2 3 2 0h h w w− + − =
( )2 23 2
2 32 y y
w wRU c c
−=
+
( )( )( )
3 2 3 2 3 2
2 3 22 y y
w w w w w wRUU c c
− + −= =
+
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Axial-flow Turbines: 2-D theory
FIG. 4.5. Velocity diagram and Mollier diagram for a zero reactionturbine stage.
( )3 2 3 2tan tan 0 if 2
xcRU
β β β β= − = =
Zero reaction stage
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Axial-flow Turbines: 2-D theory
FIG. 4.7. Velocity diagram and Mollier diagram for a 50% reactionturbine stage.
( )3 2 3 21 tan tan 0.5 if 2 2
xcRU
β α β α= + − = =
50% reaction stage
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Axial-flow Turbines: 2-D theory
Turbine blade cooling.
Why is the efficiency of the gas turbine comparable to that of a Rankine cycle?
(given that we do have to pay a considerable amount of energy to the compressor, whereas compression of water in the Rankine cycle is cheap)
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Theory of turbo machinery / Turbomaskinernas teori
Dixon, chapter 7
Centrifugal Pumps, Fans and Compressors
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Centrifugal Pumps, Fans and Compressors
( ) 4/3
2/1
4/31
2/11
gHNQN s ==
ψφ
Specific speed:
• Dimensionless flow to head ratio so that D is eliminated.
• Values of flow and head at max efficiency
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Centrifugal Pumps, Fans and Compressors
Three examples of characteristic curves for pumps of differing specificspeeds. a: radial impeller, nq ≈ 20; b: mixed flow impeller, nq ≈ 80; c: axial flow impeller, nq ≈ 200.
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Centrifugal Pumps, Fans and Compressors
Types:
• Axial, mixed and radial flow direction• Shrouded or unshrouded
Shrouded impellers
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• Impeller:• 2D, 3D, Backsweep?• Axial inlet, radial inlet
• Diffuser:• Vaneless, Vaned (Vane type)? • Diffuser ratio
Volute
Vanelessdiffuser
Impeller axial inlet
Main components
Pipe diffuserAirfoil diffuser
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Centrifugal Pumps, Fans and Compressors
FIG. 7.2. Radial-flow pump and velocity triangles.
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Centrifugal Pumps, Fans and Compressors
FIG. 7.1. Centrifugal compressor stage and velocitydiagrams at impeller entry and exit.
321
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Centrifugal Pumps, Fans and Compressors
[ ]2 1 21 1 22 0xW gH U c U c c U cθ θθ θ= = ⋅ ⋅ = ⋅− = =
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Centrifugal Pumps, Fans and Compressors
FIG. 7.3. Mollier diagram for the complete centrifugal compressorstage.
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Centrifugal Pumps, Fans and Compressors
2cθ
2 2 22 2 2rc c cθ= −
( )22 22 2 2 2rc w U cθ= − −
( )2 2 2 2 22 2 2 2 2 2 22c c w U U c cθ θ θ− = − − +
Solving for 2 2U c
Setting them equal:
θ
2 2 22 2 2
2 2 2c U wU cθ+ −
= the same way1 1U cθ
Right triangle
Left triangle
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Centrifugal Pumps, Fans and Compressors
( ) ( ) ( )2 2 2 2 2 22 2 1 1 2 1 2 1 2 1
12xW gH U c U c c c U U w wθ θ⎡ ⎤= = ⋅ − ⋅ = − + − − −⎣ ⎦
Change in dynamic pressure Change in staticpressure
( ) ( )( ) ( ) ( )
2 2 2 22 1 2 1
2 2 2 2 2 22 1 2 1 2 1
change in static pressuretotal pressure change
U U w wR
c c U U w w
− − −= =
− + − − −
Reaction:
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Example
1 2r rc c=
[ ] 22 2 1 1 1 20 UgH U c U c ccθ θ θ θ= ⋅ − ⋅ ⋅= = =
Compare 2 pumps at
1. Same inlet velocity, radially directed:2. Constant radial velocity: 3. Same speed of rotation and same inner and outer diameter
1 1 10, rc c cθ = =
Consequences
1. Work:2. Change in dynamic pressure:
( ) ( )2
2 2 2 2 2 22 1 2 2 12 2 2d r
cP c c c c c θθ
ρρ ρΔ = − = + − =
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Example
2 2tP gH U cθρ ρΔ = = ⋅ ⋅The total pressure rise
22 2
2 2 22 2d
t
P c cP U c U
θ θ
θ
ρρ
Δ= =
Δ ⋅
The ratio of the dynamic and total pressure drop becomes:
2cθ2rcAt a fixed :
large β will decrease the dynamic part of the pressure rise
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Example
2 2U cθ=
2c2rc
2 2 2c Uθ =
2c2w
2 2rw c= 2β
Backward swept2 0β =
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SlipSlip
2 reduced decreasesc gHθ ⇒
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Centrifugal Pumps, Fans and Compressors
FIG. 7.7. Actual and hypothetical velocity diagrams at exit from an impeller with back swept vanes.
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Centrifugal Pumps, Fans and Compressors
FIG. 7.8. (a) Relative eddy without any throughflow. (b) Relative flow at impeller exit (throughflow added to relative eddy).
Stodola: A relative eddy with angular velocity 2 2U rΩ =
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Centrifugal Pumps, Fans and Compressors
FIG. 7.9. Flow model for Stodola slip factor.
• Slip velocity is the product of the relative eddy and the radius of a circle which can be inscribed within the channel
• With Z being the number of vanes:
'2 2 2sc c c dθ θ θ= − = Ω
( ) '2 22 cosd r Zπ β
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Example (same assumptions as before)
2 2rQ A c= ⋅
Head- Volume characteristics
where 2A is the exit area of the impeller 2cθ
2 2 2 2 2 2 2tan tanrc U c U Q Aθ β β= − = −2 2gH U cθ= ⋅
Combining these equations:
( )2 2 2 2tangH U U Q Aβ= ⋅ − Q
gH
22U
2 0β >
2 0β =
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Example
Losses:
a-b: Slip (finite number of vanes)
b-c: Friction
c-d: Other losses from 3d flows
d´ Instable region
2Q∝
Q
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Theory of turbo machinery / Turbomaskinernas teori
Dixon, chapter 9
Hydraulic Turbines 30° 49′ 15″ N, 111° 0′ 8″ E
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Main types of Hydraulic Turbines
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Hydraulic Turbines
sp
s
Ωη
Ω=
FIG. 9.1. Typical design point efficiencies of Pelton, Francis and Kaplan turbines.
sp
s
Ωη
Ω=
( )( )
1/ 2
5/ 4
/sp
P
gH
Ω ρΩ =
( )
1/ 2
3/ 4sQ
gHΩΩ =
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Hydraulic Turbines
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Hydraulic Turbines
Operating ranges of the main types of hydraulic turbines (Alvarez)
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Pelton Turbines
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Pelton Turbines
FIG. 9.5. The Pelton wheel showing the jet impinging onto a bucket and the relative and absolute velocities of the flow (only one-half of the emergentvelocity diagram is shown).
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Pelton Turbines
1 1 2 2ΔW U c U cθ θ= −
( ) ( )1 2 2 1 2 2Δ cos cosW U U w U w U w wβ β⎡ ⎤= + − + = −⎣ ⎦
From Eulers turbine equation
For the Pelton turbine:
1 2U U U= =
1 1 1c c U wθ
and thus Euler becomes
= = +
2 2 2cosc U wθ β= +2β
2cos 0β <
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Pelton Turbines
2 1w kw=
where k is a loss factor less than 1.
Introducing this into Eulers eq.:
( )21 2
1 1
2Δ c 2 1 1 cosRU UW kc c
η β⎛ ⎞
= = − −⎜ ⎟⎝ ⎠
Dividing by the available energy, , yields a “runner” efficiency:
Friction looses are accounted for by relating relative velocities
21 2c
( ) ( )( )1 2 1 21 cos 1 cosW Uw k U c U kΔ β β= − = − −
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Pelton Turbines
FIG. 9.6. Theoretical variation of runner efficiency for a Pelton wheel with blade speed to jet speed ratio for several values of friction factor k .
2 165β =
,max
1
@
0.5
R
Uc
η
ν= =
Cos is a forgiving function:
( )cos 1650.966
=
= −
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Pelton Turbines
FIG. 9.7. Pelton turbine hydroelectric scheme.
Surge tank reduces pressure spikes
Gross head:
G R NH z z= −
Effective head:
( )E G F rictionH H H= −
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Pelton Turbines
More losses:
• Friction losses in penstock (pipe flow: moody chart)• Nozzle efficiency • Bearing friction and windage, assumed proportional to the
square of the blade speed:
( )21 2N Ec gHη =
2KU
An overall efficiency of the machine (excluding penstock) may bedefined:
22
0 21
... 2N RE
W KU UKgH c
Δη η η⎡ ⎤⎛ ⎞− ⎢ ⎥= = = − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
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Pelton Turbines
FIG. 9.9. Variation of overall efficiency of a Pelton turbine with speed ratio for severalvalues of windage coefficient, K .
The subtraction of energy by the U2
term displaces the optimum blade speed to jet speed ratio
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Pelton Turbines, controle
FIG. 9.8. Methods of regulating the speed of a Pelton turbine: (a) with a spear (or needle) valve; (b) with a deflector plate.
Spear used for slow control
Deflector plate causes no “hammer”
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Francis Turbines
James Bicheno FrancisMay 18, 1815 – September 18, 1892
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Francis Turbines
Reaction turbines
Pressure drop takes place in the turbine itselfWater flow completely fills all part of the turbinePivotable guide vanes are used for control (Francis)A draft tube is normally added on to the exit; it is considered an integral part of the turbine
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Francis Turbines
FIG. 9.15. Location of draft tube in relation to vertical shaft Francis turbine.
Draft tube
Shaped as a diffusor to minimize losses
Turbine may be placed above tailwater surface
Cavitation may be an issue
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Francis Turbines
Euler turbine equation
2 2 3 3ΔW U c U cθ θ= −
2 2ΔW U c
If there is no swirl at exit (design point):
θ=
Slip is present
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Francis Turbines, control
FIG. 9.13. Comparison of velocity triangles for a Francis turbine for full load and at part load operation.
Volume flow rate reduced by guide vanes
Blade speed retained
Rotor incidence high.
Swirl at exit increases losses and risk for cavitation(why?)
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Kaplan Turbines
Viktor KaplanNovember 27, 1876 – August 23, 1934
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Francis Turbines
FIG. 9.16. Part section of a Kaplan turbine in situ.
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Kaplan Turbines (Voith Siemens)
Cross section of a 9.5 m diameter Kaplan runner for the Yacyretá hydropower plant in Argentina
Yacyretà, Argentina
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Kaplan Turbines
FIG. 9.17. Section of a Kaplan turbine and velocity diagrams at inlet to and exit from the runner.
Lunds universitet / Kraftverksteknik / JK
Hydraulic Turbines, cavitation
Two types:
On the suction side of the runner near outletOn the centerline of the draft tube at off-design operation (Francis)
The Thoma cavitation coefficient may be defined as
( ) ( )a
E E
p p g zNPSHH H
υ ρσ
− −= =
Lunds universitet / Kraftverksteknik / JK
Theory of turbo machinery / Turbomaskinernas teori
Dixon, chapter 10
Wind Turbines
Lunds universitet / Kraftverksteknik / JK
Statistisk beskrivning
Wizelius
Frekvensfördelningar:
• Weibull
• Rayleigh
Dvs Weibull m. k = 2
( )( )2
2
2
exp 2xx
f x σσ
−=
( ) ( )1
expk kk x xf x σσ σ
−⎛ ⎞ ⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
Lunds universitet / Kraftverksteknik / JK
Single stream tube analyses
1xc2xc
3xc
Assumptions:
Steady uniform flow, upstream and at the disc.No flow rotation produced by discFlow contained by stream tubeIncompressible flow
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Mass flow 2 2xm c Aρ=
Single stream tube analyses
Axial force ( )1 3x xX m c c= −
Energy loss of the wind ( )2 21 3 2W x xP m c c= −
Disk power ( )2 1 3 2x x x xP Xc m c c c= = −
Setting :WP P=
( ) ( ) ( )2 21 3 2 1 3 2 1 32 2x x x x x x x xm c c c m c c c c c− = − ⇒ = +
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Axial flow induction factor
Rewriting power ( ) ( )21 3 2 2 2 1 3x x x x x xP m c c c A c c cρ= − = −
Using 3 2 12x x xc c c= − the power becomes
( ) ( )2 22 2 1 2 1 2 2 1 22 2x x x x x x xP A c c c c A c c cρ ρ= − + = −
It is convenient to introduce an axial flow induction factor
( )1 2 1x x xa c c c= −
( )232 12 1xP a A c aρ= −
What does this mean?
Lunds universitet / Kraftverksteknik / JK
The power coefficient
max 1 2xQ c A=
The total available power, P0, in the upstream vind may be defined from maximum possible volume flow and maximum obtainable pressure drop
and maximum obtainable pressure drop 2max 1 2xp cρΔ =
3max max 0 2 1 2xQ p P A cρΔ = =
A power coefficient may now be defined:
( ) ( )23
22 13
0 2 1
2 14 1
2x
Px
a A c aPC a aP A c
ρρ
−= = = −
Lunds universitet / Kraftverksteknik / JK
Wind Turbines
( )22 1
4 12X
x
XC a aA cρ
= = −
Axial Force Coefficient:
( )24 1PC a a= −
Power Coefficient:
,max 16 27 @ 1 3PC a= =
Lunds universitet / Kraftverksteknik / JK
Optimum power coefficient
1x
RJCΩ
=
Tip speed ratio
• Typical values in GT:< 1 (.5)
• Typical values WT:5-10
• Tip speeds still far from transonic
1x
RJCΩ
=
Lunds universitet / Kraftverksteknik / JK
Wind Turbines
Power output range
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Types of Wind Turbines
Horisontal Axis Wind Turbine (HAWT)
totalhöjd 62-72mrotordiameter 44mnavhöjd 40-50mProduktionskostnad: ~ 6 Mkr
Vid 10 m/s:3 2 3
0 3 2 1.2 4 10 2 912kWP Ac Dρ π= = × ≈
0 16 27 912 503 kWPP C P= = × ≈
Till detta kommer turbinens ”normala” verkningsgrad, mekaniska förluster, generatorns vekningsgrad mm.
Lunds universitet / Kraftverksteknik / JK
Vestas V90-3.0 MW
Vestas V90-3.0 MW
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Theory of turbo machinery / Turbomaskinernas teori
Gas turbines
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Simple Cycle
Compressor
h
Expander
s
p1
p2
Air
Combustor
Fuel
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Ideal Cycle
a) Compression and expansion are reversible and adiabatic, i.e. isentropic
b) The change of kinetic energy of the working fluid between the inlet and outlet of each component is negligable.
c) Pressure losses are neglected.d) The working fluid is the same in the entire cycle and it is a
perfect gas with constant specific heats.e) The mass flow of gas is the same throughout the cyclef) The heat-exhanger is a counterflow type with “complete”
heat transfer
Lunds universitet / Kraftverksteknik / JK
Ideal Cycle
0ΔQ h W= +
( ) ( ) ( )12 02 01 2 1 2 1pW h h h h c T T= − − = − − = − −
( ) ( )23 3 2 3 2pQ h h c T T= − = −
( ) ( )34 3 4 3 4pW h h c T T= − = −
Compressor
Combustor
Expander (turbine)
steady flow energy equation
Lunds universitet / Kraftverksteknik / JK
Ideal Cycle
( )1 32
1 4
TT rT T
γ γ−= =
32
1 4
pprp p
= =where r is pressure ratio
The efficiency equals the ratio of net work output and supplied heat
( ) ( )( )
3 4 2 134 12
23 3 2
p p
p
c T T c T TW WQ c T T
η− − −−
= =−
Cycle efficiancy
Lunds universitet / Kraftverksteknik / JK
Ideal Cycle
( ) ( )4
14 1 1 1
24 4 22
1 1 1
111 1 1 1
1 1
TT T T T
T rT T TTT T T
γ γ
η−
⎛ ⎞−⎜ ⎟− ⎛ ⎞⎝ ⎠= − = − = − = − ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎝ ⎠− −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
i.e. a function of r and γ only!
Lunds universitet / Kraftverksteknik / JK
Ideal Cycle
( ) ( )( )
( )( )13 4 2 1 31
1 1 1
11 1P
T T T T TW rc T T T r
γ γγ γ
−−
− − − ⎛ ⎞= = − − −⎜ ⎟⎝ ⎠
( ) ( )34 12 3 4 2 1p pW W W c T T c T T= − = − − −
Normalising with 1PC T
Net work
3
1
T tT
=
Lunds universitet / Kraftverksteknik / JK2009-10-19 Magnus Genrup 110
CRS
Ideal Cycle
Lunds universitet / Kraftverksteknik / JK
Typical industrial engine
2009-10-19 Magnus Genrup 111
Opt pressure ratio’sCOT=1643K
• S/C η: 25.06
• C/C η: 18:08
• Spec. work: 14.36
2 stage compressor turbine, rotor blade temp <900°C, COT-SOT = 70°C
.3
.32
.34
.36
.38
.4
Ther
mal
Effi
cien
cy
300 350 400 450 500
Specific Power [kW/(kg/s)]
Pressure Ratio = 10 ... 26 Burner Exit Temperature = 1473 ... 1773 [K]
147
3
152
3
157
3
162
3
167
3
172
3
177
3
0.51 0.52
0.53
0.54
0.55
0.56
Dotted Lines = (PWSD*0.985*0.985+cp_val7)/(WF* [g/(kN*s)]
10
12
14
16
18
20 22
24 26 WCLTq2 iterated for T_m_T=1173ZW2Rstd iterated for W2=100WCLNq2 iterated for cp_val1=70
Example!ηcc
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Recuperated Cycle
Compressor Expander
Heat Exchanger
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Evaporative Cycle
Compressor Expander
Liquid water
Evaporator
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O2 fired cycle Cycle
ASU
Air
O2
N2
Compressor Expander
EGR
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H2 fired Cycle
Reformer
Air
CmHnH2O(Air)
Compressor Expander
H2, CO2, (N2)
CO2Separation
Combustor
H2 (N2)
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Combined cycles
Magnus Genrup 116
Steam Turbine (condensing)
100 % fuel
15 °CGas Turbine
2-pressure HRSG
520…540 °C
27 °C
31 °C
Courtesy to Alstom
Lunds universitet / Kraftverksteknik / JK
Physics of combustion differs
• Spray formation• Evaporation• Mixing• Ignition• Combustion• Emission
formation• Temperature
distribution• ……..
Air blast atomizer, diffusion combustion