Theory of turbo machinery / Turbomaskinernas teori
Dixon, chapter 9 Hydraulic Turbines
Hear ye not the hum of mighty workings? (KEATS, Sonnet No. 14). The power of water has changed more in this world than emperors or kings. (Leonardo da Vinci).
Hydraulic Turbines
Todays topics Introduction; Where and how much Types of turbines 9 Pelton 9 Francis 9 Kaplan
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Hydraulic Turbines, potential
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Hydraulic Turbines, production
Land
Produktion TWh
Per capita
Andel av elproduktion
Förändring mot 19952000
Kanada
344
-1%
Kina
321
+58%
Brasilien
304
+6%
USA
260
-20%
Ryssland
170
+7%
Norge
121
Syrien
119
+2%
Japan
96
+2%
Indien
78
+2%
Frankrike
67
-6%
Venezuela
66
+12%
Sverige
65
LTH / Kraftverksteknik / JK
99%
~ 50%
+1%
-6%
Wikipedia: Produktion 2000-2005
Hydraulic Turbines, large plants Name
Country
Year of completion
Total Capacity (MW)
Max annual electricity production (TW-hour)
Area flooded (km²)
Three Gorges Dam
China
2009
17,600 (August 2008); 22,500 (when complete)
>100
632
Itaipu
Brazil/Paragu ay
1984/1991/2003
14,000
90
1,350
Guri (Simón Bolívar)
Venezuela
1986
10,200
46
4,250
De största svenska kraftverken är: •
Harsprånget i Luleälven (945 MW)
•
Stornorrfors i Umeälven
•
Messaure i Luleälven
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Sverige Vattentillgången störst tidig sommar
Elbehovet i Sverige störst på vintern
Reglering-vattnet sparas Energi- eller effektbegränsningar? LTH / Kraftverksteknik / JK
Hydraulic Turbines Greenpeace (hemsida*): Fler stora vattenkraftsutbyggnader är inte försvarbara ur biologisk och ekologisk synpunkt: • •
Utbyggnaden av kraftverk i älvar och floder får stora konsekvenser för den biologiska mångfalden när stora områden sätts under vatten. Lekområden för fisk ödeläggs och vattenorganismer såväl som ett flertal andra växter, fåglar och djur påverkas negativt.
Småskalig vattenkraft å andra sidan fångar flodernas energi utan att ta bort för mycket vatten från deras naturliga flöde. Därför är den småskaliga vattenkraften en miljövänlig energikälla med stor tillväxtpotential.
* http://www.greenpeace.org/sweden/kampanjer/klimat/losningar/klimatvanlig-energi/vatten LTH / Kraftverksteknik / JK
Hydraulic Turbines Regeringen De fyra outbyggda huvudälvarna* ska bevaras
STEM Vattenkraft är en ren energikälla som ger stora mängder energi. Att anlägga nya vattenkraftverk orsakar dock stora skador i naturen. Därför byggs inga nya större kraftverk i vårt land.
*Nationalälvarna: Torne, Kalix, Pite och Vindelälven LTH / Kraftverksteknik / JK
Hydraulic Turbines 3 Gorges Dam The Three Gorges Project, including a 2,309-meter-long, 185-meter-high dam with 26 power generators, is being built on the middle reaches of the Yangtze, China's longest river. The project started 1993 and is assumed to be finished 2011, at what time the power output will be 22 500 MW. Water from upstream is flowing into the reservoir at a rate of 13,200 cubic meters per second. http://maps.google.com/maps?ll=30.83,111.01&spn=0.01,0.01&t=h&q=30.83,111.01 LTH / Kraftverksteknik / JK
Main types of Hydraulic Turbines
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Hydraulic Turbines Ωs =
Ω Q1/ 2
( gH )
3/ 4
Ω (P / ρ)
1/ 2
Ω sp =
( gH )
Ω sp = η Ωs
5/ 4
Ω sp = η Ωs
FIG. 9.1. Typical design point efficiencies of Pelton, Francis and Kaplan turbines.
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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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Hydraulic Turbines
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Hydraulic Turbines
Ohakuri Dam Bue Penstocks LTH / Kraftverksteknik / JK
Pelton Turbines
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Pelton Turbines
Lester Allan Pelton (no image) September 5, 1829 –March 14, 1908
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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 emergent velocity diagram is shown). LTH / Kraftverksteknik / JK
Pelton Turbines From Eulers turbine equation
ΔW = U1cθ 1 − U 2 cθ 2 For the Pelton turbine:
U1 = U 2 = U cθ 1 = c1 = U + w1 cθ 2 = U + w2 cos β 2 and thus Euler becomes
β2 cos β 2 < 0
ΔW = U ⎡⎣U + w1 − (U + w2 cos β 2 ) ⎤⎦ = U ( w1 − w2 cos β 2 ) LTH / Kraftverksteknik / JK
Pelton Turbines Friction looses are accounted for by relating relative velocities
w2 = kw1 where k is a loss factor less than 1.
Introducing this into Eulers eq.:
ΔW = Uw1 (1 − k cos β 2 ) = U ( c1 − U )(1 − k cos β 2 ) Dividing by the available energy, c12 2 , yields a “runner” efficiency:
U⎛ U⎞ η R = 2ΔW c = 2 ⎜1 − ⎟ (1 − k cos β 2 ) c1 ⎝ c1 ⎠ 2 1
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Pelton Turbines η R ,max @ U = ν = 0.5 c1 Cos is a forgiving function:
β 2 = 165
cos (165 ) = = −0.966
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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 .
Pelton Turbines Surge tank reduces pressure spikes Gross head:
H G = zR − z N Effective head:
H E = H G − H F ( riction ) FIG. 9.7. Pelton turbine hydroelectric scheme.
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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
η N = c12 ( 2 gH E )
square of the blade speed:
KU 2
An overall efficiency of the machine (excluding penstock) may be defined: 2 ⎡ ⎛U ⎞ ⎤ ΔW − KU η0 = = ... = η N ⎢η R − 2 K ⎜ 2 ⎟ ⎥ gH E ⎢⎣ ⎝ c1 ⎠ ⎥⎦ 2
LTH / Kraftverksteknik / JK
Pelton Turbines The subtraction of energy by the U2 term displaces the optimum blade speed to jet speed ratio
FIG. 9.9. Variation of overall efficiency of a Pelton turbine with speed ratio for several values of windage coefficient, K .
LTH / Kraftverksteknik / JK
Pelton Turbines, controle Spear used for slow control
Deflector plate causes no “hammer”
FIG. 9.8. Methods of regulating the speed of a Pelton turbine: (a) with a spear (or needle) valve; (b) with a deflector plate. LTH / Kraftverksteknik / JK
Pelton Turbines, part load Controle by adjustment of needle valve: Hydraulic losses reduced at low load, but bearings and windage losses remain at constant speed FIG. 9.10. Pelton turbine overall efficiency variation with load under constant head and constant speed conditions.
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Francis Turbines
James Bicheno Francis May 18, 1815 – September 18, 1892 LTH / Kraftverksteknik / JK
Francis Turbines Reaction turbines
Pressure drop takes place in the turbine itself Water flow completely fills all part of the turbine Pivotable 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 Draft tube Shaped as a diffusor to minimize losses Turbine may be placed above tailwater surface FIG. 9.15. Location of draft tube in relation to vertical shaft Francis turbine.
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Cavitation may be an issue
Francis Turbines Flow is through the scroll into guide vanes and onto the runner
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Volute or scroll: Decreasing diameter => constant velocity
Francis Turbines Euler turbine equation
ΔW = U 2 cθ 2 − U 3cθ 3 If there is no swirl at exit (design point):
ΔW = U 2 cθ 2 Slip is present
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Francis Turbines, control Volume flow rate reduced by guide vanes Blade speed retained Rotor incidence high. Swirl at exit increases losses and risk for cavitation (why?) LTH / Kraftverksteknik / JK
FIG. 9.13. Comparison of velocity triangles for a Francis turbine for full load and at part load operation.
Kaplan Turbines
Viktor Kaplan November 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
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Yacyretà, Argentina
Kaplan Turbines
FIG. 9.17. Section of a Kaplan turbine and velocity diagrams at inlet to and exit from the runner. LTH / Kraftverksteknik / JK
Kaplan Turbines, free vortex Swirling flow at inlet, free vortex:
cθ 2 = K r cx = const. Flow angles become:
tan β 2 = U cx − tan α 2 = Ω r cx − K ( rcx ) tan β3 = U cx = Ω r cx
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No swirl at exit
Hydraulic Turbines, part load
FIG. 9.14. Variation of hydraulic efficiency for various types of turbine over a range of loading, at constant speed and constant head.
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Hydraulic Turbines, cavitation Two types: On the suction side of the runner near outlet On the centerline of the draft tube at off-design operation (Francis) The Thoma cavitation coefficient may be defined as
NPSH ( pa − pυ ) ( ρ g ) − z σ= = HE HE
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