The reservoir:mechanistic model

Politecnico di Milano

NRMNRMLec08Lec08

Andrea Castelletti

2

Itaipu dam on the Parana river

Spillways in action

Dam

Penstocks and turbine hall

3

Three Gorges dam (China)

4

Typical Localization

Clan canyon damColorado river

5

penstock

Longitudinal section

barrier

water surface level

storage

intake tower

intakes

minimum intake level

surface spillway

maximum storagebottom outlet

6

Surface spillways

7

Bottom outlet

Loch Lagghan damScozia

8

surface spillway penstock

9

Dam-gate structures

Gates: a) rising sector b) vertical rising c) radial

10

Piave - S.Croce system

Hydropower reservoirs are often interconnected to form a network

Reservoir network

11

Features of reservoirs

• the active (or live) storage;

• the global stage-discharge curve of the spillways;

• the stage-discharge curve of the intake tower;

By the management point of view a reservoir is characterized by:

12

Stage-discharge curve (morning glory)

13

Causal network

1ts 1t

r tu

1th

ts

1ta

st = storage volume at time t

at+1 = inflow volume in [t ,t+1)

rt+1 = effective release volume in [t , t+1)

14

Causal network

tS

1tE1te

1ts 1t

a 1tr t

u

1th

ts

What is missed?

- evaporation

- r depends on a and e

15

Mechanistic model

tS

1tE1te

1ts 1t

a 1tr t

u

1th

ts

What is missed?

- evaporation

- r depends on a and e

tt sSS surface

16

Mechanistic model

tS

1tE1te

1ts 1t

a 1tr t

u

1th

ts

tt sSS surface

ttt sSeE 11 evaporation

17

Mechanistic model

tS

1tE1te

1ts 1t

a 1tr t

u

1th

ts

ttt sSeE 11 evaporation

tt sSS surface

1111 ttttt rEass storage

18

Mechanistic model

tS

1tE1te

1ts 1t

a 1tr t

u

1th

ts

ttt sSeE 11 evaporation

tt sSS surface

1111 ttttt rEass storage

tt shh level

19

Mechanistic model

tS

1tE1te

1ts 1t

a 1tr t

u

1th

ts

ttt sSeE 11 evaporation

tt sSS surface

1111 ttttt rEass storage

tt shh level

111 ,,, tttttt EausRrrelease

11111 ,,, tttttttttt eausRsSeass

20

Balance equation

Sean ttt 111 net inflow

11111 ,,, tttttttttt eausRsSeassbalance

111 ,, ttttttt nusRnss balance

111 ,, ttttttt nusRssnnet inflow estimator

Simplification: the storage is a cylinder S(st) = S

Pros

Cons

Using it when the storage is not actually a cylinder generates an error.

21

Storage-level relationship

By inverting h(.) one obtains the value of the storage by measuring the level, i.e. the only measurable quantity

There exist a biunivocal relationship between the level measured in a point and the storage.

tt shh Implicit assumption:

the water surface is always horizzontal.

Example: if the storage is a cylinder

arbitary constant

A negative storage represents the missing volume required to bring the water surface up to the level corresponding to the zero storage.

infh s s S

infshSs

tt shh

22

Storage-level relationship

By inverting h(.) one obtains the value of the storage by measuring the level, i.e. the only measurable quantity

There exist a biunivocal relationship between the level measured in a point and the storage.

tt shh Implicit assumption:

the water surface is always horizzontal.

tt shh

Non-cylindrical storage

batimetry of the reservoir (DEM)

The identification of h(.) can be performed in different ways, depending on which one of the following is known:

numerical computation point by point

interpolation tt sh ,historic time series

23

Examples of storage-level relationships

Campotosto reservoir (Italy)

60,000

80,000

100,000

120,000

140,000

160,000

180,000

200,000

220,000

1304 1306 1308 1310 1312 1314 1316 1318Level [m a.s.l.]

Sto

rag

e[1

03m

3]

Historic series

Regression line

Campotosto reservoir (Italy)

60,000

80,000

100,000

120,000

140,000

160,000

180,000

200,000

220,000

1304 1306 1308 1310 1312 1314 1316 1318Level [m a.s.l.]

Sto

rag

e[1

03m

3]

Historic series

Regression line

24

Examples of storage-level relationships

Piaganini reservoir (Italy)

0

100

200

300

400

500

600

700

800

387 388 389 390 391 392 393 394 395 396 397Level [m a.s.l.]

Sto

rag

e[1

03

m3]

Historic series

Piaganini reservoir (Italy)

0

100

200

300

400

500

600

700

800

387 388 389 390 391 392 393 394 395 396 397Level [m a.s.l.]

Sto

rag

e[1

03

m3]

Historic series

25

Examples of storage-level relationships

Piaganini reservoir (Italy)

0

100

200

300

400

500

600

700

800

387 388 389 390 391 392 393 394 395 396 397Level [m a.s.l.]

Sro

rag

e[1

03

m3]

Data from 1988 to 1992

Data from 1993 to 2001

Data of February 19933rd February 1993

Piaganini reservoir (Italy)

0

100

200

300

400

500

600

700

800

387 388 389 390 391 392 393 394 395 396 397Level [m a.s.l.]

Sro

rag

e[1

03

m3]

Data from 1988 to 1992

Data from 1993 to 2001

Data of February 19933rd February 1993

26

Surface-storage relationship t tS S s

Can be determined with similar techniques.

27

Continuous-time model of reservoir

( ) , , ,ds t

a t i t s t e t S s t r t s t p tdt

s(t) = storage volume at time t [m3]

a(t) = inflow rate at time t [m3/s]

e(t) = evaporation per surface area unit at time t [m/s]

i(t,s(t)) = infiltration [m3/s]

S(s(t)) = surface area [m2] r(t,s(t),p(t)) = release when the dam gate are open of p [m3/s]

28

It depends on the storage-discharge functions and the

position p of the intake sluice gates

It depends on the storage-discharge functions and the

position p of the intake sluice gates

( ) , , ,ds t

a t i t s t e t S s t r t s t p tdt

Simplifications

Cylindric reservoir n(t) = a(t)-e(t)S net inflow

i = 0 almost always true, at least in artificial reservoirs

n(t)

29

Instantaneous storage-discharge relationships

• s min , s max : bounds of the regulation range• s* : storage at wich spillways are activated

s min s max s*

spillway

open gates

maximum release

minimum release

s(t)

limited

min,norN

maxN

max,norN

minN

max, maximum allowed release norN min, minimum allowed release norN

( )r t

30

Example of storage-discharge relationship

flow

ra

te [

m3/s

]

storage [Mm3]

N max (•)

N min (•)

~

~

31

Modelling for managing

The continuos-time model can not be used for managing:

• decision are taken in discrete time instants• data are not always collected continuously

discretize

32

st+1 = st +nt+1 -rt+1

nt+1 = net inflow volume in [t ,t+1)

+nt+1

Discrete model of a reservoir

t t +1

nt+1= net inflow volume in [t , t+1)

nt+1

we assume it uniformly distributed

st = storage volume at time t

st

rt+1 = volume actually released in [t , t+1)

-rt+1

( )n t

33

The release function

rt+1 = Rt (st ,nt+1 ,ut)

1

1max,, ,t t t

tnor

t

V s n N s d

Maximum dischargeable

volume in [t , t+1)

1

1min,, ,t t t

tnor

t

v s n N s d

Minimum dischargeable

volume in [t , t+1)

rt+1

nt+1

ut

ut

rt+1given st & ut

given st, & nt+1

Vt

vt

45°

release decision

with

min,( ) ( ) ( , ( ))

( ) [ , 1)

n r

t

os n s

s t s t t

N

      

max,( ) ( ) ( , ( ))

( ) [ , 1)

nor

t

s n s

s t s t t

N

with

34

An example of minimum and maximum release (Campotosto, Italy)

n = 50 m3/s

n = 0 m3/s

flow

ra

te [

m3/s

]

storage [Mm3]

35

Set of feasible controls U(st)

feasible control

U(st)flo

w r

ate

[m

3/s

]

storage [Mm3]

depends on the inflow!

1 1min( ) : ( , ) ( ma, x )t t tt tt t t t tU n ns u v s u V s 1 1min( ) : ( , ) ( ma, x )t t tt tt t t t tU n ns u v s u V s

36

ut

st given

Vt (st,min{nt+1})

vt(st,min{nt+1})

rt+1

45°

Vt (st,max{nt+1})

vt(st,max{nt+1})

U(st)

Set of feasible controls U(st)

37

Remarks

If nt+1 is known

Alternative ways of formulating the mass balance equation

rt+1 = ut

1 1

1

t t

t t

n rh h

S S

ht = level at time t rt+1 = Rt(st ,nt+1 ,ut) = Rt (ht ,nt+1 ,ut) actual release in [t , t +1)

1 1 1t t t th h n r nt+1= net inflow in terms of level

rt+1= actual release in terms of level

vt(st ,nt+1) rt+1 Vt(st ,nt+1)

vt(st ,nt+1) ut Vt(st ,n t+1)

constraint already included in Rt(•)

NO releases of interest)( ttt sUu

38

CONCLUSIONS

1 1 1 1 1

1 1 1

1 1

( , , , )

( , , , )

( )

t t t t t t t t t t

t t t t t t

t t

t t

t t t

t t

s s a e S s R s u a e

r R s u a e

h h s

S S s

E e S s

u U s

Model of a reservoir in operation

outputs

feasible controls

state transition

39

CONCLUSIONS

1 1 1 11

1 1 1

1 1

( , , , , ) if

0 if

( ,

>0

=0

, , , )

( , )

t t t t t t t t tt

t t t t t t

p p

p

p

p

p

t t

t t

t t t

t t

p

s a e S s R s u a es

r R s u a e

h h

u u

u

u

u

s

S S

u

s

E e S s

u U s

U

Model of a reservoir to be constructed

1 1( , , , , )t t t tp

tR u a es u

40

Natural lakeh

e

s

a r

hmin

n t a t e t

r

s

(s - smin)

0 if ssmin

N(s) =

s t n t r t

N(s(t))

net or effective inflow

Stage-discharge function( ) ( ( ))r t N s t

smin

41

min min

1 +

s s s s e n

t

e d

t

s t n t N s t (s - smin)

Remark: s(t+1) depends on s(t) only if = .

T is the so-called time constant of the reservor.

Linearization and time constant

r

ssmin

t t+1

Linear continuos system

( )s t As t Bn t

0

(0) t

A tAts t e s n e d

Lagrange formula

Remark: s(t+1) depends on s(t) only if = .

T is the so-called time constant of the reservor.

42

min min

1 +

s s s s e n

t

e d

t

s t n t N s t (s - smin)

Linearization and time constant

r

ssmin

t t+1

Meaning of T

By assuming =T=1/ one gets

T is the time required for the storage to reach 1/3 of its initial value.

1min min1s t s s t s e

min min

1 +

s s s s e n

t

e d

t

43

min min

1 +

s s s s e n

t

e d

t

s t n t N s t (s - smin)

Linearization and time constant

r

ssmin

t t+1

An accurate modelling does require Shannon’s or Sampling theorem

An accurate modelling does require Shannon’s or Sampling theorem

Remark: s(t+1) depends on s(t) only if = .

T is the so-called time constant of the reservor.

44

LAKE S T=

[km2] [daysi]

Maggiore 212.0 7.4Lugano 48.9 8.7Varese 15.0 34.7Alserio 1.5 8.0Pusiano 5.2 15.0Como 146.0 7.7Iseo 61.0 7.8Garda 370.0 86.6

The modelling time-step for lakes with T = 8 is about 1 day.

For ll these lakes the catchment areas is relatively small compared to the lake surface.

The outlet mouth has not yet reached an equilibrium condition.

For most of the lakes T is nearly 8 days.

Time constant and the Lombardy lakes

45

dB1/T

• Incoming waves whose frequency is smaller than 1/T are not smoothed.

E.g.: flood waves from snow melt.

• Waves with frequency greater than 1/T are smoothed.E.g.: flood waves from storms.

Bode diagram

Buffering effecti.e. low-pass filtering

46

The modelling time-step depends upon the state

Non-linear model: T is not defined

It would be useful to have models with a time-step that changes with s, however this is not possible with the algorithms nowadays available.

Solution: use models with different at different time steps.

changes with s

T changes with the value s around which the system is linearized

Linearization of the system

To be sure that the system is well

represented by the model: 0,1* T

47

r

h

Comparison between two lakes

min

nh h

2 >1 12h h

12h t h t

Average level

2

2

1

10 0h h

.

min n t

h h hS S

1) ( )

2) 0

n t n.h

hmax

h

t

48

Lake shore inhabitants

*

minmax 0t

n Ph t h h

S

Downstream users *

t

TPr t n e

S

happy with lake 2 ( T small)

Happy with lake 1 ( T big )

Comparison between two lakes:impulsive flood

CONFLICTCONFLICTh

t

r

t

*

min 0t

Tn Ph t h e t

S

Average levelAverage level*

min

nh

*

min

nh

Impulse response Impulse response

t

TPe

S

t

TPe

S

n

t

n*

P

49

Lake shore population big

Downstream users small

Which compromise ?

Natural lake Regulated lake

Different stage-discharge

functions in different time

instants

Natural regime curve

Natural curve

Curves for different gate positions

r

h

Lake regulation

50

Downstream users

Months

t

t

Lake popul.

h(t)

r

Lake regulation

t

t

Readings

IPWRM.Theory Ch.5