1 energy conversion. 2 specific energy the specific energy of a hydro power plant is the quantity of...

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 m2/s2.

3

Gross Head

Hgr

Reference line

zres

ztw

twresgr zzH

4

Gross Specific Hydraulic Energy

In a hydro power plant, the difference between the level of the upper reservoir zres and the level of the tail water ztw is defined as the gross head:

Hgr = zres - ztw [m]

The corresponding gross specific hydraulic energy:

][ kgJHgE grgr

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Gross Power

grgrgr HgQEQP where:

Pgr is the gross power of the plant [W] is the density of the water [kg/m3] Q is the discharge [m3/s]

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Net Head

Reference line

z1

ztw

221

1 1

2 21

1 1

2 2

2

twabs atm tw n

twn abs atm tw

cch z h z H

g g

c cH h h z z

g

h1

c1

abs

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Net Head

g

chH pn

2

2

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

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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.

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Reaction turbines (Full turbines)

Francis Kaplan Bulb

12

Reaction forces in a curved channel

Newton’s 2.law in the x-direction:

xx

x

xx

dcQdt

dcdtcAdF

QAc

dtcdldt

dcdlAdF

c1y

c2y

c1x

c2x

dl

y

x

where:A is the area [m2] c is the velocity [m/s] is the density of the water [kg/m3] Q is the discharge [m3/s]

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c1y

c2y

c1x

c2x

dl

Newton’s 2.law in thex-direction:

xx dcQdF

Fx is the force that acts on the fluid particle from the wall. Rx is the reaction force that acts on the wall from the fluid: Rx = -Fx

y

x

FxRx

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Integrate the forces in the x-direction:

)()( 21

2

1

12 xxxxxx ccQccQdcQR

Integrate the forces in the y-direction: )( 21 yyy ccQR

Using vectors give: )( 21 ccQR

x

c1y

c2y

c1x

c2x

dl

y

RRy

Rx

15 Reaction forces in a curved channelForce-Momentum Equation

)( 21 ccQR

c1

c2

y

x

R

R1

R2

16Let the channel rotate around the point ”o”. What is the torque ?

Torque = force · arm:

)( 2211 rcrcQT uu

2

2u2222

1

1u1111

2211

cos

ccandcosra

cos

ccandcosra

)acac(QT

Let us define the u-direction as the normal of the radius (or tangent to the circle)

a1

a2

r1

r2

cu1

cu2

c1

c2

o

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Euler’s turbine equation

Power : P = T•[W]Angular velocity: [rad/s]Peripheral velocity: u = •r [m/s]

)( 2211 uu crcrQP

)( 2211 uu cucuQP

a1

a2

r1

r2

cu1

cu2

c1

c2

o

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)( 2211 uu cucuQP

Output power from the runner

nHQP

Available hydraulic power

19


n

uu

n

uu

Hg

cucu

HgQ

cucuQ

2211

2211 )(

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Velocity triangle

ru

v

c

cwu

cu

cm

26

cu1

u1

v1c1

1cm1

u2

v2

c2

2

D2

D1

Francis turbine

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c1

c2

v1

v2

u1

u2

28

c1v1

u1

c2

v2

u2

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Guidevanes

Runnerblades

Velocity triangles for an axial turbine

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c1v1

u1

c

c2v2

u2

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u1

v1

v2

c1

c2

u2

Inlet and outlet velocity diagram for reaction turbines

2

1 1

Runner vanes

Guidevanes

32

V1=C1- U

V2

33

Example1Francis turbine

D2 D1

B1

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Example1Francis turbine

D2 D1

Head: 150 mQ: 2 m3/sSpeed: 1000 rpmD1: 0,7 mD2: 0,3 mB1: 0,1 m: 0,96

Find all the information to draw inlet and outlet velocitytriangles

B1

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Example1Inlet velocity triangle

smD

ru 7,362

7,07,104

21

11

D1

u1

v1c1

sradn

7,10460

21000

60

2

36


cu1 u1

w1c1

1cm1 D

1

B1

sm

BD

Q

A

Qcm 1,9

1,07,0

2

1111

37


cu1 u1

w1c1

1cm1

sm

u

Hgc

Hg

cucu

nu

n

uu

4,337,36

13081,996,0

11

2211

We assume cu2 = 0

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cu1 u1

w1c1

1cm1

u1 = 36,7 m/scu1 = 33,4 m/scm1 = 9,1 m/s

0

11

11 70tan

u

m

cu

ca

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Example1Outlet velocity triangle

smD

ru 7,152

3,07,104

22

22

smcc mm 101,91,11,1 12

We assume: cu2 = 0and we choose: cm2 = 1,1· cm1

u2

v2

c2

2

0

2

22 5,32tan

u

ca m

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Exercise1Francis turbine

D2 D1

Head: 543 mQ: 71,5 m3/sSpeed: 333 rpmD1: 4,3 mD2: 2,35 mB1: 0,35 m: 0,96cm2 : 1,1· cm1

Find all the information to draw inlet and outlet velocity triangles

B1

41

Exercise 2Francis turbine

Speed: 666 rpmD1: 1,0 m: 0,96c1: 40 m/s1: 40o

Find: H1

1 energy conversion. 2 specific energy the specific energy of a hydro power plant is the quantity of...

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