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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

    ⎟⎠⎞

    ⎜⎝⎛== 3653ˆ ND

    QfDN

    PPρ

    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

  • 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 ==ψDNgH

    constant153 ==φρ DNP

  • Lunds universitet / Kraftverksteknik / JK

    Specific speed

    Eliminate D to obtain the following dimensionless parameters:

    ( ) 4/32/1

    4/31

    2/11

    gHNQNs ==ψ

    φ

    ( )( ) 4/5

    2/1

    4/51

    2/11 /ˆ

    gHPNPNsp

    ρψ

    ==

    ( ) 4/32/1

    gHQ

    ( )( ) 4/5

    2/1/gHP

    spρΩ

    Dimensionless, directly proportional to N

    Power specific speed, turbines

    If speed of rotation is expressed in rad/s

  • Lunds universitet / Kraftverksteknik / JK

    Specific speed

  • Lunds universitet / Kraftverksteknik / JK

    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 =

  • 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

  • Lunds universitet / Kraftverksteknik / JK

    Compressor and expander maps

  • Lunds universitet / Kraftverksteknik / JK

    Theory of turbo machinery / Turbomaskinernas teori

    Chapter 2

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    Euler’s pump and turbine equations

    The work done on the fluid per unit mass (specific work) becomes:

    01122 >−=Ω

    ==Δ θθτ cUcU

    mmWW Acc (2.12a)

    02211 >−==Δ θθ cUcUmWW tt (2.12b)

    FIG. 2.4. Control volume for a generalised turbomachine.

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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

    ( ) 22221210201 cchhhhmWW xx −+−=−==Δ

    ( ) 22221210201max,max, sssxx cchhhhmWW −+−=−==Δ

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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 >

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    Small stage efficiency for a perfect gas

  • Lunds universitet / Kraftverksteknik / JK

    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.

  • Lunds universitet / Kraftverksteknik / JK

    Small stage efficiency for a perfect gas

  • Lunds universitet / Kraftverksteknik / JK

    Theory of turbo machinery / Turbomaskinernas teori

    Chapter 3

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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).

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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

    i

  • Lunds universitet / Kraftverksteknik / JK

    2D cascades Efficiency of a compressor cascade

    FIG. 3.6. Efficiency variation with average flow angle (adapted from Howell 1945).

  • Lunds universitet / Kraftverksteknik / JK

    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= −

  • Lunds universitet / Kraftverksteknik / JK

    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=

    ( ) *20.50.23 2 500

    nm a l a=

    = +

    with the following constants for compressor cascades

  • Lunds universitet / Kraftverksteknik / JK

    2D cascades Optimum space chord ratio of turbineblades (Zweifel)

    FIG. 3.27. Pressure distribution around a turbine cascade blade (after Zweifel 1945).

  • Lunds universitet / Kraftverksteknik / JK

    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

    ( ) ( )2 2 1 22 cos tan tan 0.8Tid

    Y s bY

    Ψ α α α= = + ≈

    For specified inlet and outlet angles s b or s l may be determined

  • Lunds universitet / Kraftverksteknik / JK

    Theory of turbo machinery / Turbomaskinernas teori

    Chapter 4

  • Lunds universitet / Kraftverksteknik / JK

    Axial-flow Turbines: 2-D theory

    FIG. 4.1. Turbine stage velocity diagrams.

    Note direction of α2

  • Lunds universitet / Kraftverksteknik / JK

    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)

  • Lunds universitet / Kraftverksteknik / JK

    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− = − + + = +

  • Lunds universitet / Kraftverksteknik / JK

    Axial-flow Turbines: 2-D theory

    FIG. 4.3. Soderberg’s correlation of turbine blade loss coefficient with fluid deflection (adapted from Horlock (1960).

  • Lunds universitet / Kraftverksteknik / JK

    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+ = + +

  • Lunds universitet / Kraftverksteknik / JK

    Axial-flow Turbines: 2-D theory

    Design considerations

    • Rotor angular velocity (stresses, grid phasing)

    • Weight (aircraft)

    • Outside diameter (aircraft)

    • Efficiency (almost always)

    • ………

  • Lunds universitet / Kraftverksteknik / JK

    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= −

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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 32 3 1 2

    0.5 h h h hh h h h

    = − −

    − = −

    R = 0.5

  • Lunds universitet / Kraftverksteknik / JK

    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

    − + −= =

    +

  • Lunds universitet / Kraftverksteknik / JK

    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 2xcRU

    β β β β= − = =

    Zero reaction stage

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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)

  • Lunds universitet / Kraftverksteknik / JK

    Theory of turbo machinery / Turbomaskinernas teori

    Dixon, chapter 7

    Centrifugal Pumps, Fans and Compressors

  • Lunds universitet / Kraftverksteknik / JK

    Centrifugal Pumps, Fans and Compressors

    ( ) 4/32/1

    4/31

    2/11

    gHNQN s ==ψ

    φ

    Specific speed:

    • Dimensionless flow to head ratio so that D is eliminated.

    • Values of flow and head at max efficiency

  • Lunds universitet / Kraftverksteknik / JK

    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.

  • Lunds universitet / Kraftverksteknik / JK

    Centrifugal Pumps, Fans and Compressors

    Types:

    • Axial, mixed and radial flow direction• Shrouded or unshrouded

    Shrouded impellers

  • Lunds universitet / Kraftverksteknik / JK

    • 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

  • Lunds universitet / Kraftverksteknik / JK

    Centrifugal Pumps, Fans and Compressors

    FIG. 7.2. Radial-flow pump and velocity triangles.

  • Lunds universitet / Kraftverksteknik / JK

    Centrifugal Pumps, Fans and Compressors

    FIG. 7.1. Centrifugal compressor stage and velocitydiagrams at impeller entry and exit.

    321

  • Lunds universitet / Kraftverksteknik / JK

    Centrifugal Pumps, Fans and Compressors

    [ ]2 1 21 1 22 0xW gH U c U c c U cθ θθ θ= = ⋅ ⋅ = ⋅− = =

  • Lunds universitet / Kraftverksteknik / JK

    Centrifugal Pumps, Fans and Compressors

    FIG. 7.3. Mollier diagram for the complete centrifugal compressorstage.

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    Centrifugal Pumps, Fans and Compressors

    ( ) ( ) ( )2 2 2 2 2 22 2 1 1 2 1 2 1 2 112xW 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:

  • Lunds universitet / Kraftverksteknik / JK

    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 θθρρ ρ

    Δ = − = + − =

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    Example

    2 2U cθ=

    2c2rc

    2 2 2c Uθ =

    2c2w2 2rw c= 2β

    Backward swept2 0β =

  • Lunds universitet / Kraftverksteknik / JK

    SlipSlip

    2 reduced decreasesc gHθ ⇒

  • Lunds universitet / Kraftverksteknik / JK

    Centrifugal Pumps, Fans and Compressors

    FIG. 7.7. Actual and hypothetical velocity diagrams at exit from an impeller with back swept vanes.

  • Lunds universitet / Kraftverksteknik / JK

    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 =

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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 =

  • Lunds universitet / Kraftverksteknik / JK

    Example

    Losses:

    a-b: Slip (finite number of vanes)

    b-c: Friction

    c-d: Other losses from 3d flows

    d´ Instable region

    2Q∝

    Q

  • Lunds universitet / Kraftverksteknik / JK

    Theory of turbo machinery / Turbomaskinernas teori

    Dixon, chapter 9

    Hydraulic Turbines 30° 49′ 15″ N, 111° 0′ 8″ E

  • Lunds universitet / Kraftverksteknik / JK

    Main types of Hydraulic Turbines

  • Lunds universitet / Kraftverksteknik / JK

    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 =

  • Lunds universitet / Kraftverksteknik / JK

    Hydraulic Turbines

  • Lunds universitet / Kraftverksteknik / JK

    Hydraulic Turbines

    Operating ranges of the main types of hydraulic turbines (Alvarez)

  • Lunds universitet / Kraftverksteknik / JK

    Pelton Turbines

  • Lunds universitet / Kraftverksteknik / JK

    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).

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    Pelton Turbines

    2 1w kw=

    where k is a loss factor less than 1.

    Introducing this into Eulers eq.:

    ( )21 21 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= − = − −

  • Lunds universitet / Kraftverksteknik / JK

    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

    =

    = −

  • Lunds universitet / Kraftverksteknik / JK

    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= −

  • Lunds universitet / Kraftverksteknik / JK

    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⎡ ⎤⎛ ⎞− ⎢ ⎥= = = − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

  • Lunds universitet / Kraftverksteknik / JK

    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 U2term displaces the optimum blade speed to jet speed ratio

  • Lunds universitet / Kraftverksteknik / JK

    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”

  • Lunds universitet / Kraftverksteknik / JK

    Francis Turbines

    James Bicheno FrancisMay 18, 1815 – September 18, 1892

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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?)

  • Lunds universitet / Kraftverksteknik / JK

    Kaplan Turbines

    Viktor KaplanNovember 27, 1876 – August 23, 1934

  • Lunds universitet / Kraftverksteknik / JK

    Francis Turbines

    FIG. 9.16. Part section of a Kaplan turbine in situ.

  • Lunds universitet / Kraftverksteknik / JK

    Kaplan Turbines (Voith Siemens)

    Cross section of a 9.5 m diameter Kaplan runner for the Yacyretá hydropower plant in Argentina

    Yacyretà, Argentina

  • Lunds universitet / Kraftverksteknik / JK

    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

    ( ) ( )aE 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

  • Lunds universitet / Kraftverksteknik / JK

    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− = − ⇒ = +

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    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

  • Lunds universitet / Kraftverksteknik / JK

    Theory of turbo machinery / Turbomaskinernas teori

    Gas turbines

  • Lunds universitet / Kraftverksteknik / JK

    Simple Cycle

    Compressor

    h

    Expander

    s

    p1

    p2

    Air

    Combustor

    Fuel

  • Lunds universitet / Kraftverksteknik / JK

    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 3 11 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

  • Lunds universitet / Kraftverksteknik / JK

    Recuperated Cycle

    Compressor Expander

    Heat Exchanger

  • Lunds universitet / Kraftverksteknik / JK

    Evaporative Cycle

    Compressor Expander

    Liquid water

    Evaporator

  • Lunds universitet / Kraftverksteknik / JK

    O2 fired cycle Cycle

    ASU

    Air

    O2

    N2

    Compressor Expander

    EGR

  • Lunds universitet / Kraftverksteknik / JK

    H2 fired Cycle

    Reformer

    Air

    CmHnH2O(Air)

    Compressor Expander

    H2, CO2, (N2)

    CO2Separation

    Combustor

    H2 (N2)

  • Lunds universitet / Kraftverksteknik / JK

    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

    Theory of turbo machinery / Turbomaskinernas teori �Incompressible fluid analysesIncompressible fluid analysesPerformance characteristicsSpecific speedSpecific speedSpecific speedSpecific speedPumps in pipe systemsCompressible fluid analysesCompressor and expander mapsTheory of turbo machinery / Turbomaskinernas teori �The first lawEuler’s pump and turbine equationsDefinitions of efficiencyEfficiencyEfficiencyEfficiencySmall stage or polytropic efficiencySmall stage or polytropic efficiencySmall stage efficiency for a perfect gasSmall stage efficiency for a perfect gasSmall stage efficiency for a perfect gasTheory of turbo machinery / Turbomaskinernas teori �2D cascades 2D cascades2D cascades2D cascades2D cascades Efficiency of a compressor cascade2D cascades Fluid deviation2D cascades Generalizing experimental results2D cascades Optimum space chord ratio of turbine blades (Zweifel)2D cascades Optimum space chord ratio of turbine blades (Zweifel)Theory of turbo machinery / Turbomaskinernas teori �Axial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryAxial-flow Turbines: 2-D theoryTheory of turbo machinery / Turbomaskinernas teori �Centrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsExampleExampleExampleCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsCentrifugal Pumps, Fans and CompressorsExample (same assumptions as before)ExampleTheory of turbo machinery / Turbomaskinernas teori �Hydraulic TurbinesHydraulic TurbinesHydraulic TurbinesPelton TurbinesPelton TurbinesPelton TurbinesPelton TurbinesPelton TurbinesPelton TurbinesPelton TurbinesPelton TurbinesPelton Turbines, controleFrancis TurbinesFrancis TurbinesFrancis TurbinesFrancis TurbinesFrancis Turbines, controlKaplan TurbinesFrancis TurbinesKaplan Turbines (Voith Siemens)Kaplan TurbinesHydraulic Turbines, cavitationTheory of turbo machinery / Turbomaskinernas teori �Wind TurbinesWind TurbinesTypes of Wind TurbinesVestas V90-3.0 MWTheory of turbo machinery / Turbomaskinernas teori �Simple CycleIdeal CycleIdeal CycleIdeal CycleIdeal CycleIdeal CycleTypical industrial engineRecuperated CycleEvaporative CycleO2 fired cycle CycleH2 fired CycleCombined cyclesPhysics of combustion differs