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Momentum Applications in Open Channel Flow
Momentum
ss mvF
Basic relationship in mechanics:
Sum of forces in the s direction
Change in momentum in the s direction
mass
Velocity in the s direction
Momentum cont.
• For a constant mass and a per unit width consideration: (rectangular channel)
12 vvqmvs
Momentum forces – Open Channel Application
v2
v1
P1
P2
W
Wsin
Rf
L
f21s RPsinWPF
Momentum Forces cont.
• Rf is the frictional resistance.
• P1 and P2 are pressure forces per unit width given by:
2yP
2
Momentum contd.
• Combining terms we get:
)vv(qRsinW2y
2y
12f
22
21
Momentum cont.
• Considering a short section so that Rf is negligible and the channel slope is small so that sin is near zero the equation can be written as:
2
22
1
21 qv
2yqv
2y
or
Mg
qv2y
gqv
2y 2
221
21
Momentum cont.
• M is called the momentum function or the specific force plus momentum.
• For a constant q, M can be plotted against depth to create a curve similar to the specific energy curve.
• Under steady conditions, M is constant from point to point along a channel reach.
Specific force plus momentum curve.
Mc
y
yc
MM
y1
y2
q1
q2
y = yc
Momentum cont.
v2
v1
y1
y2
1 2
Mg
qv2y
gqv
2y 2
221
21
q = q1 = q2
Hydraulic Jump as an application of Momentum Equation (p. 458-465 text)
Lab Jump in Flume (Right to Left)
Hydraulic Jump in a Sink?• http://www.eng.vt.edu/fluids/msc/gallery/waves/sink.htm
hydrojump.mov
Hydraulic Jumps
• Occurs when there is a sudden transition from supercritical (y < yc) to subcritical (y > yc) flow.
• Examples of where this may occur are :– At the foot of a spillway– Where a channel slope suddenly turns flat.
• In analyzing hydraulic jumps we assume there is conservation of momentum, i.e. :
2
222
1
221
gyq
2y
gyq
2y
We can algebraically manipulate this to find:
1F8121
yy 2
11
2
1F8121
yy 2
22
1
• y1 is known as the initial depth and is < yc (supercritical flow).
• y2 is known as the sequent depth and is > yc (subcritical flow).
• The energy loss in a hydraulic jump can be found by:
21
312
1 yy4yyE
gyVyy
y 12
12
112
222
Example
Example 3.18, Streeter, et al.
If 12 m3/sec of water per meter of width flows down a spillway onto a horizontal floor and the velocity is 20 m/sec, determine (a) the downstream depth required to cause a hydraulic jump, (b) the loss in energy head, and (c) the losses in power by the jump per meter of width.
Solution for depth, losses, power lost
kWmmmNlossesQmpower
NmNlosses
m
1659)1.14sec)(/12)(/9806()(/
/1.14)7.6)(6.0(4
)6.07.6(
7.6806.9
)6.0)(20(23.03.0
33
3
22