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

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

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

Why is WH a problem?

4

!!! Pressure > resistance !!!

crack(Chaudhry 2014)

5

Why is WH a problem?

!!! Pressure < resistance !!!

(Chaudhry 2014)

What to do?

6(Chaudhry 2014)

- surge tank

- air chamber

- pressure-regulating valves

- slower changes->

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)

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

Wave Propagation

9(Chaudhry 2014)

Time t = e > 0

Time t = L/a

(Chaudhry 2014)

10

Wave Propagation

(Chaudhry 2014)

Time t = L/a + e

Time t = 2*L/a

11

Wave Propagation

(Chaudhry 2014)

Time t = 2*L/a + e

Time t = 3*L/a

12

Wave Propagation

(Chaudhry 2014)

Time t = 3*L/a + e

Time t = 4*L/a

(Chaudhry 2014)

13

Wave Propagation

with

(Giesecke and Heimerl 2014)

14(Chaudhry 2014)

Wave Propagation

Refection at the reservoir

Refection at the valve

No friction!

Wave Propagation … with friction

15

Refection at the reservoir

Refection at the valve

Lower !

(Chaudhry 2014)

Reservoir <-> Dead end

16

Incident pressure wave F Reflected pressure wave f

with r = f/F

r = -1

r = 1 !!! Increasing !!!

(Chaudhry 2014)

Change of the cross section / Junction

17

Incident pressure wave Reflected pressure wave

A1 > A2

v1 < v2

(Chaudhry 2014)

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)

Wave speed

19

thick pipe

concrete and steal

with

with

only concrete only rock

(Giesecke and Heimerl 2014)

How big is the effect of the WH?

• Joukowski

• Allievi

• Method of characteristics (MoC)

• Software

20

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]

22

Joukowski‘s theory including closing time

(Giesecke and Heimerl 2014)

Reality:

-> Allievi

How big is the effect of the WH?

• Joukowski

• Allievi

• Method of characteristics (MoC)

• Software

23

Allievi

24

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

a²

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)

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

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

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),(

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

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

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

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:

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

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

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)

How big is the effect of the WH?

• Joukowski

• Allievi

• Method of characteristics (MoC)

• Software

36

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

Calculation MoC

38

total derivative

havewant

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

if, this is satisfied then

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)

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)

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)

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)

How big is the effect of the WH?

• Joukowski

• Allievi

• Method of characteristics (MoC)

• Software

43

Software

44

Hydraulic System – EPFL

AFT ImpulseWater Hammer and Mass Oscillation(WHAMO) 3.0

… more information: Ghidaoui et al. (2005)

Example WANDA

45

WANDA

46

workflow

Most of the time!

WANDA

47

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)