1 Energy Conversion. 2 Specific energy The specific energy of a hydro power plant is the quantity of potential and kinetic energy which 1 kilogram of.

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Presentation transcript:

1 Energy Conversion

2 Specific energy The specific energy of a hydro power plant is the quantity of potential and kinetic energy which 1 kilogram of the water delivers when passing through the plant from an upper to a lower reservoir. The expression of the specific energy is Nm/kg or J/kg and is designated as  m 2 /s 2 .

3 Gross Head H gr Reference line z res z tw

4 Gross Specific Hydraulic Energy In a hydro power plant, the difference between the level of the upper reservoir z res and the level of the tail water z tw is defined as the gross head: H gr = z res - z tw [m] The corresponding gross specific hydraulic energy:

5 Gross Power where: P gr is the gross power of the plant[W]  is the density of the water[kg/m 3 ] Q is the discharge[m 3 /s]

6 Net Head Reference line z1z1 z tw h1h1 c1c1 abs

7 Net Head

8 Impulse turbines (Partial turbines) The hydraulic energy of the impulse turbines are completely converted to kinetic energy before transformation in the runner

9 Impulse turbines (Partial turbines) Turgo Pelton

10 Reaction turbines (Full turbines) In the reaction turbines two effects cause the energy transfer from the flow to mechanical energy on the turbine shaft. Firstly it follows from a drop in pressure from inlet to outlet of the runner. This is denoted the reaction part of the energy conversion. Secondly changes in the directions of the velocity vectors of the flow through the canals between the runner blades transfer impulse forces. This is denoted the impulse part of the energy conversion.

11 Reaction turbines (Full turbines) FrancisKaplan Bulb

12 Reaction forces in a curved channel Newton’s 2.law in the x-direction: c 1y c 2y c 1x c 2x dl y x where: Ais the area[m 2 ] cis the velocity[m/s]  is the density of the water[kg/m 3 ] Qis the discharge[m 3 /s]

13 Reaction forces in a curved channel c 1y c 2y c 1x c 2x dl Newton’s 2.law in the x-direction: F x is the force that acts on the fluid particle from the wall. R x is the reaction force that acts on the wall from the fluid: R x = -F x y x FxFx RxRx

14 Reaction forces in a curved channel Integrate the forces in the x-direction: Integrate the forces in the y-direction: Using vectors give: x c 1y c 2y c 1x c 2x dl y R RyRy RxRx

15 Reaction forces in a curved channel Force-Momentum Equation c1c1 c2c2 y x R R1R1 R2R2

16 Let the channel rotate around the point ”o”. What is the torque ? Torque = force · arm: Let us define the u-direction as the normal of the radius (or tangent to the circle) a1a1 a2a2 r1r1 r2r2 c u1 c u2 c1c1 c2c2 o 

17 Euler’s turbine equation Power : P = T  [W] Angular velocity:  [rad/s] Peripheral velocity: u =  r [m/s] a1a1 a2a2 r1r1 r2r2 c u1 c u2 c1c1 c2c2 o

18 Euler’s turbine equation Output power from the runner Available hydraulic power

19 Euler’s turbine equation

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25 Velocity triangle v c cucu cmcm

26 c u1 u1u1 v1v1 c1c1 11 c m1 u2u2 v2v2 c2c2 22 D2D2 D1D1 Francis turbine

27 c1c1 c2c2 v1v1 v2v2 u1u1 u2u2

28 c1c1 v1v1 u1u1  c2c2 v2v2 u2u2

29 Guidevanes Runnerblades Velocity triangles for an axial turbine

30 c1c1 v1v1 u1u1 c  c2c2 v2v2 u2u2

31 u1u1 v1v1 v2v2 c1c1 c2c2 u2u2 Inlet and outlet velocity diagram for reaction turbines 22 11 11 Runner vanes Guidevanes

32 V 1 =C 1 - U V2V2

33 Example1 Francis turbine D2D2 D1D1 B1B1

34 Example1 Francis turbine D2D2 D1D1 Head:150 m Q:2 m 3 /s Speed:1000 rpm D 1 :0,7 m D 2 :0,3 m B 1 :0,1 m  :0,96 Find all the information to draw inlet and outlet velocity triangles B1B1

35 Example1 Inlet velocity triangle D1D1 u1u1 v1v1 c1c1

36 Example1 Inlet velocity triangle c u1 u1u1 w1w1 c1c1 11 c m1 D1D1 B1B1

37 Example1 Inlet velocity triangle c u1 u1u1 w1w1 c1c1 11 c m1 We assume c u2 = 0

38 Example1 Inlet velocity triangle c u1 u1u1 w1w1 c1c1 11 c m1 u 1 = 36,7 m/s c u1 = 33,4 m/s c m1 = 9,1 m/s

39 Example1 Outlet velocity triangle We assume: c u2 = 0 and we choose:c m2 = 1,1· c m1 u2u2 v2v2 c2c2 22

40 Exercise1 Francis turbine D2D2 D1D1 Head:543 m Q:71,5 m 3 /s Speed:333 rpm D 1 :4,3 m D 2 :2,35 m B 1 :0,35 m  :0,96 c m2 : 1,1· c m1 Find all the information to draw inlet and outlet velocity triangles B1B1

41 Exercise 2 Francis turbine Speed:666 rpm D 1 :1,0 m  :0,96 c 1 :40 m/s  1 :40 o Find: H  1