h

B

S

OPEN CHANNEL FLOW: In a channel if the flow is steady (no changes in time) and uniform (no changes down stream) we have normal flow

fc

ghSu

n

Shu

21

32

Manning's Formula mks system

21

21

61

g

chn f

cf is dimensionless BUT n is not

using the definition of discharge Q = (B x h)u (the volume flow per time)

53

21

31

2

2

BS

Qn

gSB

cQh fn

This value can be calculated for any streamas a stream characteristic—the value, however will only coincide with the stream surface under the Normal flow condition

Through a balance of the gravity (weight) to friction shear forces-for wide channels

or where

If there is no change in friction or slope as we move down stream Then the depth of the flow remains constant –the normal depth

L

Shear force acting on fluid element Up along slope is

2uc f )1(,sin ghSgh

Gravity force down stream

fc

ghSu

Derivation:--once moreAs a force balance consider fluid element with unit area base

and height of depth h moving down slope

=

=

h

B

S

L

Engineers (Civil) like to measure energy as a vertical height above a datum

Energy in a fluid: In stream settings We have seen how useful balances of Mass (Volume) and Momentum (Manning) can be in understanding basic behavior-.In addition to these conservations we can also consider the balance of Energy

h

A

B

Consider a large fluid filled tank with a hole near the bottom.If the level h (height) of the tank is maintained, the flow near AIs essential still, and the fluid speed at the exit is u. The potential energy per unit volume at A (relative to the datum) is

This is converted into the Kinetic Energy Per unit volume at B

22 /u

These energy forms can be expressed in terms of height (length) dimensions, bydividing by

gh

g

Then an Energy balance between A and B can be expressed as

g

uhh BAloss 2

2

friction head loss

Derive Normal flow withEnergy balance

balossh

friction head loss =

consider fluid element with unit area baseand height of depth h moving down slope

friction force per unit mass X length /g

h

B

S b

a

1,)cos(

1

)cos(

22

gh

LucL

gh

uc

g

L

hff

L

h

B

S

Applying this idea to a stream channel under-normal flow conditions

a

b

lossbbaaa hhB

Q

gh

hB

Q

ghH

22

2

1

2

1

hB

Qu Average Velocity

Specific Energy

ghLuc f /2

But by definition of slope S same as

ghLuc fba /2

So under normal flow Energy balance will recover our previous result

53

21

31

2

2

BS

Qn

gSB

cQh fn

Bed elevation

BhQuuBhQ /

So In uniform flow energyBalance becomes

ghLucSL f /2

2

2

1

hB

Q

ghH

h

Energy Line

Specific Energy:

2

2

1

hB

Q

g

2

2

1

hB

Q

ghE

The specific energy Energy (head) relative to channel bed

Changes Non-linearly with stream depth, h

See Excell worksheet

(make sure to download “open.xls” along with this power point

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.1 0.2 0.3 0.4 0.5

E-Sp Energy

y-w

ater

dep

th

hcr= 0.14147

Critical Depth: The specific energy

2

2

1

hB

Q

ghE

Sp Energy increases with increase in depth

Sub-critical Flow

Sp Energy increases with decreases in depth

Super-critical Flow

Q= 0.5

B= 3

Is non-linear

E

Can show that, for a given discharge and stream width E will reach a minimum when

31

2

2

gB

Qhcr The critical depth

132

2

ghB

QFr

1Fr

1Fr

At critical depth

crcr

crcrcr

ghuor

g

huh

31

22

1cr

cr

gh

uFr

And

Why we want to know about this

Normal flow is all well and good….

What about changes in depth?

Or slope? ???

???

The Question: How does a disturbance in the flow propagate up and down streamHow far is the effect felt

The form of the flow response to changes such as this depends on a number called the Froude Number

gh

uFr

Flow speed

Max wave speed

05.z

Flow Over an Object: What happens as a sub-critical normal flow Goes over a low object? Recognize that this an important SRES question

??

?

h =0.3B=3Q=.5

h =0.3B=3Q=.5

A B

We can construct an Energy Balance between A and B—neglecting slope changes and bed friction (small distance)

22

2

1

2

1

Bh

Q

gzh

Bh

Q

gh

BB

AA

zEE BA

Bh

Implies that BA hzh

05.z

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.1 0.2 0.3 0.4 0.5

E-Sp Energy

y-w

ater

dep

th

h =0.3B=3Q=.5

A B

We can construct an Energy Balance between A and B—neglecting slope changes and bed friction (small distance)

22

2

1

2

1

Bh

Q

gzh

Bh

Q

gh

BB

AA

zEE BA

Bh

implies that BA hzh

See Excell worksheet

05.z

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.1 0.2 0.3 0.4 0.5

E-Sp Energy

y-w

ater

dep

th

h =0.3B=3Q=.5

A B

What happens when we increase the “bump” heightThe level of the flow over the bump will continue to decrease--- until the critical height hcr is reached.

1410.hcr

See Excell worksheet

1037.z

If the bump height increases more: The level over the bump is caught betweenA “rock and a hard place” going both up or down will only increase the energy

--so with an increasing bump the the level over the bump is fixed at hcr

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.1 0.2 0.3 0.4 0.5

E-Sp Energy

y-w

ater

dep

th

--so with an increasing bump the the level over the bump is fixed at hcr

A B

1037.z

The only way to establish an energy balance is for the height at A toIncrease

22

2

1

2

1

Bh

Q

gzh

Bh

Q

gh

crcr

AA

This will mean that down-stream of the bump the flow Will not be normal

nA hh As we move further back friction will reestablishNormal flow—the distance to reestablish normal flow is the

Back Water Distance—will be calculated

1410.hcr 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.1 0.2 0.3 0.4 0.5

E-Sp Energy

y-w

ater

dep

th

Back water

A B

crhh

What about down stream of our high bump ?

Since fluid has lost the pot. energy The energy is more easily recoveredBy the level dropping below critical

Further down streamThe bottom frictionWill drive it back up to the normalDepth

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.1 0.2 0.3 0.4 0.5

E-Sp Energy

y-w

ater

dep

th

A B

crhh

If normal depth is sub-critical hn > hcr something interesting happens

The flow level can notPass smoothly Thorough the critical depthAnd you get a hydraulic jump

How Far down stream in an other wise normal flow is an obstacle felt ?

AF1

SS

dx

dh2r

f

32

2

hgB

QcS ff

32

2

ghB

QFr

The Stress slope

nf

nf

hh,SS

hh,SS

when

when

See derivation on Board

? This elevation change is important

h

B

S

Now consider head balance behind an obstacle-not in uniform flow

a

b

lossbbaa hhB

Q

gh

hB

Q

gh

22

2

1

2

1

hB

Qu Average Velocity Bed elevation

h

B

S

Now consider head balance behind an obstacle-not in uniform flow

a

b

22

2

1

2

1bbbaaa u

ghu

gh

hB

Qu Average Velocity

ghLuc f /2

LSSug

hug

h fbbaa )(2

1

2

1 22

slopefriction // slope, bed, 2322 BghQcghucSL

S fffba

For a small distance deltax

)(2

2

SShg

u

dx

df

)( SS

dx

dh

dx

du

g

uf

Result follows on subbing for uand noting that

dx

dh

Bh

Q

dx

du2

AF1

SS

dx

dh2r

f

32

2

hgB

QcS ff

32

2

ghB

QFr

The Stress slope

nf

nf

hh,SS

hh,SS

when

when

critical

normal

Fr=1

S=Sf

Fr > 1, Sf>S, A > 0

Mild slope, normal flow depth > critical flow depth

Fr < 1, Sf>S, A < 0

Fr < 1, Sf<S, A > 0

critical

normal

Fr=1

S=Sf

Fr > 1, Sf >S, A > 0

Fr >1, Sf <S, A < 0

Fr < 1, Sf <S, A > 0

Steep slope, normal flow depth < critical flow depth

Behavior of water surface near an object depends on normalflow level in relation to critical flow level

Homework: find a web page that shows examples of water level change for obstacles In streams/channels—send web-address to Voller

Assumed to beSlope at normal flow