Hydropower Plants - UniTrentorighetti/lezioni HPP/HPP/HPP_Gabl_Part1...Hydropower Plants Inlet...

48
Hydropower Plants Inlet Reservoir Surge tank Power station Penstock - Water hammer - Surge tank 1 Gallery / headrace tunnel Maurizio RIGHETTI, Roman GABL Environmental Fluid Mechanics / Hydropower Plants – 2015/16

Transcript of Hydropower Plants - UniTrentorighetti/lezioni HPP/HPP/HPP_Gabl_Part1...Hydropower Plants Inlet...

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Hydropower Plants

Inlet

Reservoir

Surge tank

Power

station

Penstock

- Water hammer- Surge tank

1

Gallery / headrace tunnel

Maurizio RIGHETTI, Roman GABL

Environmental Fluid Mechanics / Hydropower Plants – 2015/16

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What is … ???

2

= Pressure wave

every (!!!) change in the operation

Surge Tan

k

= Free surface -> Reflexion of the WH -> Limitation of the influence

= volume potential <-> kinetic energy

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3

Differences between water hammer and mass oscillation

m1

m2

manometer upper m1 long period (100-500 sec)lower m2 short period in the range of seconds

Pressure wave

Slow changes <- inertia

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Why is WH a problem?

4

!!! Pressure > resistance !!!

crack(Chaudhry 2014)

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5

Why is WH a problem?

!!! Pressure < resistance !!!

(Chaudhry 2014)

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What to do?

6(Chaudhry 2014)

- surge tank

- air chamber

- pressure-regulating valves

- slower changes->

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WH basic equations

7

Frictionless pipeWalls are rigid, constant cross section AInitial pressure ≈upstream head H0

Steady conditions v0

-> closing valve suddenly at t=0

before after

Travels as a wavefrontwith a wave velocity a

material parameter of the pipe

=

Change of the reference system ->observer travels with the wave

(Chaudhry 2014)

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8

Change of the reference system ->observer travels with the wave

(Chaudhry 2014)

WH basic equations

inflow outflow velocity

Resulting force F on the CV

Equal to the change of momentum

Range of 10 m/s 1000m/sFor wave front moving downstream

Pressure head increases for a reduction in velocity

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Wave Propagation

9(Chaudhry 2014)

Time t = e > 0

Time t = L/a

(Chaudhry 2014)

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10

Wave Propagation

(Chaudhry 2014)

Time t = L/a + e

Time t = 2*L/a

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11

Wave Propagation

(Chaudhry 2014)

Time t = 2*L/a + e

Time t = 3*L/a

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12

Wave Propagation

(Chaudhry 2014)

Time t = 3*L/a + e

Time t = 4*L/a

(Chaudhry 2014)

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13

Wave Propagation

with

(Giesecke and Heimerl 2014)

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14(Chaudhry 2014)

Wave Propagation

Refection at the reservoir

Refection at the valve

No friction!

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Wave Propagation … with friction

15

Refection at the reservoir

Refection at the valve

Lower !

(Chaudhry 2014)

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Reservoir <-> Dead end

16

Incident pressure wave F Reflected pressure wave f

with r = f/F

r = -1

r = 1 !!! Increasing !!!

(Chaudhry 2014)

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Change of the cross section / Junction

17

Incident pressure wave Reflected pressure wave

A1 > A2

v1 < v2

(Chaudhry 2014)

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Wave speed

18

water at 10°C:

= (Fluid Elastic Modulus/ Fluid Density)^0.5

… elastic pipe

Pipe Elastic Modulus and diameter / thickness

(Giesecke and Heimerl 2014)

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Wave speed

19

thick pipe

concrete and steal

with

with

only concrete only rock

(Giesecke and Heimerl 2014)

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How big is the effect of the WH?

• Joukowski

• Allievi

• Method of characteristics (MoC)

• Software

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Joukowski‘s theory

21

[m/s] * [m/s][m/s^2]

!!Max!!!

needed when closing time tS < TR

wave speed change of the velocity

gravitational acceleration ≈9.81 [m/s2]

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22

Joukowski‘s theory including closing time

(Giesecke and Heimerl 2014)

Reality:

-> Allievi

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How big is the effect of the WH?

• Joukowski

• Allievi

• Method of characteristics (MoC)

• Software

23

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Allievi

24

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Allievi‘s approach

25

Assumption

• No friction (IE = 0)

• Constant pipe (sin α = 0) with a fixed boundary condition (big reservoir)

• v << a

0D2

vvλsinαg

x

p

ρ

1

x

vv

t

v

= 0= 0= 0

= 0

0t

p

a²ρ

1

x

v

0x

hg

t

v

0t

h

g

x

v

and

0x

p

ρ

1

t

v

0x

pv

t

p

²aρ

1

x

v

Continuity equation

Momentum equation

Very good explanation in Jaeger (1977)

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x

a

26

Allievi‘s approach

Initial value

current valueDepending on time t and location x

(Chaudhry 2014)

a

xLtf

a

xLtfhtxh 210),(

a

xLtf

a

xLtf

a

gvtxv 210),(

Idea: two functions f1 and f2 are defined, which - are assumed as being complicated a priori … “and we don’t want it to know” - depend on the boundary conditions - independent of time t - f1+f2= pressure wave

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27(Chaudhry 2014)

A

Reservoir = constant -> Δh have to be 0 -> f1=-f2A

Allievi‘s approach

a

xLtf

a

xLtf 21

… the primary wave f1 is fully reflected at A total reflection with reversal of the sign

iii

i

tfa

Ltf

a

L

a

Ltf

a

Ltf

aLtt

21112

/

for

… the function f2(ti) is equal to –f1(ti-2L/a) one period TR earlier

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28

Allievi‘s approach

A

ALL r

2L t n T n n = 1, 2, 3, ... für x = L

a

B

B

Only at the main periods and at B

(Chaudhry 2014)

2 1 f (t ) f (t 2L / a)

n 0 1,n 1,n 1 h h f f

n 0 1,n 1,n 1

gv v f f

a

-> only f1

a

xLtf

a

xLtf

a

gvtxv 210),(

a

xLtf

a

xLtfhtxh 210),(

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29

Allievi‘s approach

n 0 1,n 1,n 1 h h f f

n 0 1,n 1,n 1

gv v f f

a

n n 1 0 1,n 1,n 2 h h 2h f f

n n 1 1,n 1,n 2

gv v f f

a

Initial value

current valueat time= n*2L/a

n 1 0 1,n 1 1,n 2 h h f f

n 1 0 1,n 1 1,n 2

gv v f f

a

previous valueat time= n*2L/a

n n 1 0 n 1 n

a h h 2h v v

g

equal

n n 1 0 n 1 n h f (h ,h ,v ,v )

hn is now a function depending on only known values

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30

0n n 1 0 n 1 n 1 n n

0

a v h h 2h h h

g h

Allievi‘s approach

rn

s

rn

s

T1 n

T

Tn

T

… closing

… opening

(Chaudhry 2014)

n n 1 0 n 1 n

a h h 2h v v

g

1st equation

A

A

nn 0 n

0

hv v

h

2nd equation neededfor examples:

B

B

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31

Allievi‘s approach 0n n 1 0 n 1 n 1 n n

0

a v h h 2h h h

g h

0

0r

hg

va2

0

n2

nh

h

0

0n2

nh

hh1

with

… pressure ration

… pipeline characteristic

… ratio of excess to steady pressure

nn1n1nr

2

1n

2

n 22

Allievi‘s equation (classical form)

nn1n1nr

2

1n

2

n 22 … n-th reflection tn=n ·Tr

2211r

2

1

2

2 22 … 2nd reflection t2=2 Tr

1100r

2

0

2

1 22 … 1st reflection t1= Tr

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Connecting Allievi with Joukowski

32

1h

h

0

02

0 r0

s

T1 0 1

T

1100r

2

0

2

1 22 … 1st reflection t1= Tr

Abrupt closure

11r

2

1 121

r

2

1 21

0

0

0

01

hg

va

h

hh

0v

g

ah

a/L2TT rS

11 1 0 by t1= 1

(Jaeger 1977)

… pipeline characteristic0

0r

hg

va2

0

0n2

nh

hh1

… ratio of excess to steady pressure

Remember:

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Slow closing and opening

33

rS TT

t

TR

(t)

2TR

3TR

4TR

5T =TR s

1

1

0

0

32

4 t

TR

(t)

2TR

3TR

4TR

5T =TR s

1

1

0

0

3

2

45

1T/T rS

rS TT

t

TR

0vg

ah

0vg

ah

hn

2TR 3TR 4TR 5TR 6TR

h0

t

TR

0vg

ah

0vg

ah

hn

2TR 3TR 4TR 5TR 6TR

h0

rS TT0

Fast closing

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Slow closing

34

TR 2TR 3TR 4TR 5TR

TS

12n

12

m

TR 2TR 3TR 4TR 5TR

TS

12n

12

m

1T/T rS rS TT

Asymptotic behaviour

S

rn

T

Tn1η

S

r1n

T

T1n1η

S

rn1n

T

Tηη

mnm1n

mnm1n hhhh

m1nn

m … asymptotes / final value

nn1n1nr

2

1n

2

n 22

n1nmr

2

m 222

S

rmr

2

mT

T222

1T

T

2T

T

2

2

S

rr

S

rrm

mnm1nr

2

m

2

m 22

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Exercise to Allievi

35

L, d, s, Q0

Pipe:Dout= 1,4 mThickness 20 mmE(steel)=2,1*1011N/m2

E(water)=2,06*109N/m2

L= 1400 mH=120 mQ0= 2,6 m³/s Valve:ζ = 380420·x-2,25 + 1,15 (x … position of the valve; 100 = open, 0 = closed)

???a

Joukowskihn(t)

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How big is the effect of the WH?

• Joukowski

• Allievi

• Method of characteristics (MoC)

• Software

36

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Calculation MoC

37

Continuity equation

Momentum equation

describing transient flow in closed conduits= hyperbolic partial differential equations

? solve ?

Method of characteristics (MoC)

L1 + unknown multiplier * L2 =

withf….friction factor

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Calculation MoC

38

total derivative

havewant

Eliminate the independent variable x ->Ordinary differential equation depending on t

if, this is satisfied then

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Calculation MoC

39

partial differential equation ordinary differential equation

for

valid everywhere in the x-t plane valid only on a straight line, if a is constant

= characteristic lines

(Chaudhry 2014)

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Example MoC

40

Region I:depending only on the initial conditions

back

furt

her

on

in t

ime

Region II:imposed by the downstream condition

excitations on both sides:

(Chaudhry 2014)

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Calculation MoC

41

for

known

wanted

friction losses (depending on QP)

first order approximation:(Q is constant from A to P for this term)

satisfactory results /short computational interval needed

(Chaudhry 2014)

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Calculation MoC

42

with

positive characteristic equation

negative characteristic equation

two equations = two unknowns: HP and QP

HP based on one characteristic equation

boundary conditionsfurther advanced methods

-> Chaudhry (2014)

(Chaudhry 2014)

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How big is the effect of the WH?

• Joukowski

• Allievi

• Method of characteristics (MoC)

• Software

43

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Software

44

Hydraulic System – EPFL

AFT ImpulseWater Hammer and Mass Oscillation(WHAMO) 3.0

… more information: Ghidaoui et al. (2005)

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Example WANDA

45

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WANDA

46

workflow

Most of the time!

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WANDA

47

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References

• Chaudhry, M. Hanif: Applied Hydraulic Transients. Springer, New York Heidelberg Dordrecht London, 2014. DOI: 10.1007/978-1-4614-8538-4

• Ghidaoui MS, Zhao M, McInnis DA, Axworthy DH. A Review of Water Hammer Theory and Practice. ASME. Appl. Mech. Rev. 2005;58(1):49-76. DOI:10.1115/1.1828050.

• Giesecke, J.; Heimerl S.: Wasserkraftanlagen – Planung, Bau, Betrieb. Springer, Berlin Heidelberg, 2014.DOI: 10.1007/978-3-642-53871-1

• Jaeger, Charles: Fluid Transients in Hydro-Electric Engineering Practice. Blackie, Glasgow London, 1977.

• WANDA User manual 4.3, Deltares, Delft Hydraulics, 2014.

48(Ghidaoui et al 2005)