OPEN CHANNEL FLOW: In a channel if the flow is steady (no changes in time)
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Transcript of OPEN CHANNEL FLOW: In a channel if the flow is steady (no changes in time)
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
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Manning's Formula mks system
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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)
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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
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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
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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
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2
2
gB
Qhcr The critical depth
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2
ghB
QFr
1Fr
1Fr
At critical depth
crcr
crcrcr
ghuor
g
huh
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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)
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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
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0.25
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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
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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
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E-Sp Energy
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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
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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
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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
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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
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2
hgB
QcS ff
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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
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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
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2
hgB
QcS ff
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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