10 Viscous Flow Open Channel

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Monroe L. Weber-Shirk School of Civil andEnvironmental Engineering

Open Channel Flow

June 14, 2011



Hydraulic radius

Steady-Uniform Flow:

Force Balance

U

W

U

W sin U

(x

a

b

c

d

Shear force

Energy grade line

Hydraulic grade line

Shear force =________

sin !(( x P x Ao

X UK

UK X sin

P

A

o!

hR =

P

AU

U

Usi

cos

si$S

W cos U

g

V

2

2

Wetted perimeter = __

Gravitational force = ________

XoP (x

P

K%(x sin U

Dimensional analysisRelationship between shear and velocity? ___________________



Geometric parameters

___________________

___________________

___________________

Write the functional relationship

P

A R

h!Hydraulic radius ( R

h)

Channel length (l )

Roughness (I)

Open Conduits:

Dimensional Analysis

V

gl

!

ReVl V

Q

Uniform flow

Kinetic Energy

Potential Energy

R e r p

h h

l C f

R R

I ¨ ¸! © ¹

ª º,M,W



Pressure Coefficient for Open

Channel Flow?

2

2C

V

p p

V

(

2

2C

V

ghl hl !

2

f 2

Cf

V

l S S !

l S hl f !

l h p K (Pressure Coefficient

Mechanical Energy Loss

Friction slope coefficient

Friction slope

The friction slope is the slope of the EGL. The friction slope is

the same as the bottom slope (S o) for steady, uniform flow.

Kinetic Energy



Dimensional Analysis

R eS

h h

l C f

R R

I ¨ ¸! © ¹

ª º

P!l

RC h

S f

P!

l

R

V

l S h

2

f 2

P

h R gS V f

! h RS V f

2

P!

f ,Re

S

h h

l f

R R

I ! Head loss w length of channel

f ,R e

h

S h

R

f l R

I ! !

2

f 2

f

V

l gS S !

(like f in Darcy-Weisbach)

2

f 2

l

h

l

h S l RP! !



Chezy formula

Open Channel Flow Formulas

VAQ !

2/13/21oh S A R

nQ !

1/2

o

2/3

h SR 1

nV !

hS R g

V P

2!

Dimensions of n?

Isn

only a function of roughness?

hSRC !

Manning formula (MKS units!)

NO!

T /L1/3



Manning Formula

The Manning n is a function of the boundar y

roughness as well as other geometric parameters

in some unknown way... ____________________

_______________________________

Hydraulic radius for wide channels

1/2

o

2/3

hR 1

nV !

P h

A bh

P b h 2

Rbh

b h

h

2

Channel curvature (bends)Cross section geometr y

1 2

P 1

< P 2

Rh1

> Rh2

h

b



Why Use the Manning Formula?

Tradition

Natural channels are geometrically complex and

the errors associated with using an equation thatisn¶t dimensionally correct are small comparedwith our inability to characterize stream geometr y

Measurement errors for Q and h are large

We only ever deal with water in channels, so wedon¶t need to know how other fluids wouldrespond



Values of Manning n

Lined Canals n

Cement plaster 0.011

Untrea ted gunite 0.016

Wood, planed 0.01

Wood, unplaned 0.013Concrete, trow led 0.01

Concrete, wood orms, un inished 0.015

Rubble in cement 0.0 0

Aspha lt, smooth 0.013

Aspha lt, rough 0.016

N atural Channels

G ra vel beds, stra ight 0.0 5

G ra vel beds plus la rge boulders 0.040

Earth, stra ight, w ith some gra ss 0.0 6

Earth, w inding, no vegeta tion 0.030

Earth , w inding w ith vegeta tion 0.050

The worst

channel

has«Roughness at

many scales!



Example: Manning Formula

What is the flow capacity of a finished

concrete channel that drops 1.2 m in 3 km?

1

2

3 m

1.5 m

solution



Depth as f(Q)

Find the depth in the channel when the flow

is 5 m3/s

Hydraulic radius is function of depth

Area is a function of depth

Can¶t solve explicitly

Use trial and error or solver

/ 3 /

h oQ A S

n!



Open Channel Energy

Relationships

L

p V p V z z h

g g E E

K K

2 2

1 2

1 2

2 2o f

V V y S x y S x

g g ( (

Turbulent flow (E $ 1)

z - measured from

horizontal datum

y - _____________

Pipe flow

Energy Equation for Open Channel Flow

11 o f y x y x

g g

( (

From diagram

2g

V2

1

1E

2g

V2

2

2E

xS (

2 y

1 y

x(

L f h S x= D

depth of flow



Specific Energy

The sum of the depth of flow and the

velocity head is the specific energy:

g

V y E

2

2

If channel bottom is horizontal and no head loss E E !

y - _______ energy

g

V - _______ energy

For a change in bottom elevation

E h E !

xS E xS E o(!(

f 1

potential

kinetic

h

EGL

HGL

1 E

E



Specific Energy

In a channel with uniform discharge, Q

V AV AQ !!

2

2

2 gA

Q y E

g

V y E

2

2

where A=f(y)

Consider rectangular channel (A = By) and Q = qB

2

2

2 g

q E

A

B

y

3 roots (one is negative)

q is the discharge per unit width of channel

How many possible depths given a specific energy? _____ 2



0

1

3

4

5

6

7

8

9

10

0 1 3 4 5 6 7 8 9 10

E

y

Specific Energy: Sluice Gate

gy

q y

1

2

21 E E !

sluice gate

y ¡

y2

EGL

y1 and y2 are ___________ depths (same specific energy)

Why not use momentum conservation to find y1?

q = 5.5 m2/s

y2 = 0.45 m

V2 = 12.2 m/s

E2 = 8 m

alternate

Given downstream depth and discharge, find upstream depth.

vena contracta



0

1

2

3

4

0 1 2 3 4

E

y

Specific Energy: Raise the Sluice

Gate

2

2

2 gy

q y E

1 2

E 1! E

2

sluice gate

y ¢

y2

EGL

as sluice gate is raised y approaches y2

and E is minimized:

Maximum discharge for given energy.



NO! Calculate depth along step.

0

1

2

3

4

0 1 2 3 4

y

Step Up with Subcritical Flow

Short, smooth step with rise h in channel

h

1 2 E E h= +

Given upstream depth and discharge find y2

Is alternate depth possible? __________________________

0

1

2

3

4

0 1 2 3 4

y

Energy conserved



0

2

3

4

0 2 3 4

E

Max Step Up

Short, smooth step with maximum rise h in channel

h2

E E h= +

What happens if the step is

increased further ? ___________

0

2

3

4

0 2 3 4

E

y

Choked flow



0

2

3

4

0 2 3 4

E

y

Step Up with Supercritical flow

Short, smooth step with rise h in channel

h

2 E E h= +

Given upstream depth and discharge find y2

What happened to the water depth? ______________________________ Increased! Expansion! Energy Loss

0

2

3

4

0 2 3 4

E

y



Hydraulic Jump: Dimensional

Analysis

y1

y2

To what depth will the water rise (y2) given y1?

What forces are important?



Hydraulic Jump

in in out out V y V y!

y1

in out in out LV V y y h

g g

y2

Energy

Mass Per unit width

Unknown losses

cs1

cs2



Hydraulic Jump

y p y p yV yV V V

y1

y2

22

22212

2

21

2

1 y y yV yV K K V V

g

yV y y y 1

2

1

2

11

2

2

22¹

º

¸©

ª

¨

Momentumx ssx p p x x F F F M M x x

2121

2

1

1

y p K

2

12

1

1 1 8

2

Fr y

y

or

x

r V

gl !



Summar y

Open channel flow equations can beobtained in a similar fashion to the Darcy-

Weisbach equation (based on dimensionalanalysis)

The dimensionally incorrect Manningequation is the standard in English speakingcountries

The free surface (an additional unknown)makes the physics more interesting!



Turbulent Flow Losses in Open

Conduits

Maximum

shear stress

No shear

stress



Example

Qn A R S

h o

1 2 3 1 2/ /

Q m m1

0 0129 0 927 0 0004

2 2 3 1 2

.. .

/ /c ha f a f

R

A

P mh 0 927.

A m m m m m 3 15 3 15 92a fa f a fa f . .

P m m m m 3 2 3 15 9 712 2 2a f a f a f . .

n 0 01.

Q m s14. /

S m

m0

1 2

300000004

..



Grand Coulee Dam

http://users.owt.com/chubbard/gcdam/html/galler y.html





Pumps

At the time of original construction the pumping plant contained six 65,000horsepower pumps. In 1973 work began on extending the plant. The pump bay was doubled in length to the south and six 67,500 horsepower pump/generators were added (the last in 1983) providing 12 pumps in all.

Each pump lifts water from Lake Roosevelt up through a 12 foot diameter discharge pipe to the feeder canal above. For most of their length thedischarge pipes are buried in the rock y cliff to the west but at the top of thehill they emerge and can be seen as 12 silver pipes leading to the headworksof the feeder canal. The original pumps can supply water to the feeder canalat a rate of 1,600 cubic feet of water a second while the newer units cansupply 2,000 cubic feet of water a second. They also have the advantage of being reversible. During times of peak power need the new pumps can bereversed thus turning them into generators. Water flows back down throughthe outlet pipes, through the generators and into Lake Roosevelt. Whenoperating in this mode each pump can produce 50 megawatts of electrical power.







Grand Coulee Feeder Canal

The Grand Coulee Feeder Canal is a concretelined canal which runs from the outlet of the

pumping plant discharge tubes to the north end of Banks Lake. The original canal was completed in1951 but has since been widened to accommodatethe extra water available from the six new pump/generators added to the pumping plant. The

canal is 1.8 miles in length, 25 feet deep and 80feet wide at the base. It has the capacity to carr y 16,000 cubic feet of water per second.



Columbia Basin Irrigation Project



Unsteady Hydraulics!

The base width of the feeder canal was increased

from 50 to 80 feet; however, the operating

capacity remained at 16,000 cubic feet per second.Water depth was reduced from 25 to about 20 feet

to safely accommodate wave action when the

water flow is reversed as the pump-generators are

changed from pumping to generating and vice-versa.



Gates



Gates



Banks Lake

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10 Viscous Flow Open Channel