RVF
RVF
RVF
Flow is called rapidly varied flow (RVF) if the flow depth has a large
change over a short distance
Sluice gates
Weirs
Waterfalls
Abrupt changes in cross section
Hydraulic Jump
A hydraulic jump is an abrupt change from a shallow, high-speed flow to a
deep, low-speed flow of lower energy.
Under certain conditions it is possible that the fluid depth will change very
rapidly over a short length of the channel without any change in the channel
configuration.
Such changes in depth can be approximated as a discontinuity in
the free surface elevation (dy/dx=). This discontinuity is called
hydraulic jump
A simplest type of hydraulic jump in a horizontal, rectangular channel
Hydraulic Jump-Contd
Hydraulic Jump-Contd
Assume that the flow at sections (1) and (2) is nearly uniform, steady,
and one-dimensional.
From Continuity Equation
The volume flow rate is the same at each section. For a rectangular channel,
per unit width
q = V1Y1 = V2Y2
Hydraulic Jump-Contd
From Momentum Equation:
Net pressure force = Rate of change of momentum
The average pressure is given by:
Pressure force per unit width by:
Pav =
gy
Pav y =
gy 2
2
gy1 2
2
gy 2 2
2
= q(V2 V1 )
q
1
q
2
2
g y1 y 2 = q
2
y2 y2
1
2 y1 y 2
g ( y1 y 2 )( y1 + y 2 ) = q
2
y1 y 2
Hydraulic Jump-Contd
Divide through by g(y1-y2)(non-zero by assumption) and then multiply by
y1y2
q2
1
y1 y 2 ( y1 + y 2 ) =
g
2
Since we are looking for the depth ratio y2/y1 divide through by: y13
y2 q 2
1 y2
1 + =
2 y1
y1 gy13
The RHS is just V12/gy1 or Fr12 . Hence,
1 y2
y2
1 + = Fr1 2
2 y1
y1
2
y2
2
y
+ 2 2 Fr1 = 0
y1
y1
Hydraulic Jump-Contd
This is a quadratic equation for the depth ratio y2/y1 and its positive root by
the quadratic-equation formula gives the downstream depth in terms of
upstream quantities:
y1
2
y2 = 1 + 1 + 8 Fr1
y2
2
y1 =
1 + 1 + 8 Fr 2
v1
v2
y1 +
= y2 +
+ hL
2g
2g
v1
v2
= y2 +
+ hL
y1 +
2g
2g
v1 v2
hL = ( y1 y2 ) +
2g
2
1
q2 1
2 2
h L = ( y1 y2 ) +
2 g y1
y2
Hydraulic Jump-Contd
Since
q2
1
y1 y 2 ( y1 + y 2 ) =
g
2
q2 1
1
2 2
h L = ( y1 y2 ) +
2 g y1
y2
y1 y2 ( y1 + y2 ) 1
1
2
hL = ( y1 y2 ) +
2
4
y2
y1
Simplifying, the above equation, we have the formula for the energy loss as:
(
y2 y1 )
=
hL
4 y1 y2
Example
# Water on the horizontal apron of the 100-ft-wide spillway shown in Fig. has
a depth o 0.60 ft and a velocity of 18 ft/s. Determine the depth, y2, after the
jump, the Froude numbers before and after the jump, Fr1 and Fr2, and the
energy loss.
Example
Solution
A)
B)
Example-Contd
Y2
1
=
1 +
2
Y1
1 + 8 F1
Y2
1
=
1+
2
Y1
2
1 + 8 * (4.10 )
Y2
1
=
1+
2
Y1
1 + 134.48
Y2
1
= ( 1 +
2
Y1
135.48 )
Y2
1
= ( 1 + 11.639)
2
Y1
Y2
1
= (10.639)
2
Y1
Y2
= 5.32
Y1
Y2 = 5.32 0.60 ft
Y2 = 3.19 ft
Example-Contd
C)
V1Y1 18 0.60
V2 =
=
= 3.39 ft
sec
3.19
Y2
D)
E)
hL
3
(
y2 y1 )
=
4 y1 y2
Example-Contd
(3.19 0.6)
hL =
4 3.19 0.6
3
(
1.59 )
hL =
7.656
4.02
hL =
7.656
hL = 0.525
Weirs
Weirs are elevated structures in open channels that are used to measure
flow and/or control outflow elevations from channels.
There are two types of weirs in common use: Sharp-crested weirs and
the broad-crested weirs.
Sharp-crested or thin plate, weirs consist of a plastic or metal plate that
is set vertically across the width of the channel.
A sharp-Crested weir is essentially a vertical-edged flat plate placed
across the channel.
Weirs-Contd
Sharp-crested weir plate geometry: (a) rectangular, (b) triangular, (c)
trapezoidal.
Weirs-Contd
The falling sheet of water springing from the weir plate is called the
nappe
Weirs-Contd
Broad Crested Weirs
Broad-crested weirs have a horizontal crest with a finite length, Lb, in
the flow direction
A weir is classified as broad-crested if-
2
1
2
2
V
V
= y2 +
y1 +
2g
2g
and
y2=0
2
1
2
1
V
y1 +
=H
2g
2
V2
H=
2g
V = 2 gH
V2 = 2 gH
dQ = VdA = 2 gH LdH
Q = 2g L H
1/ 2
2
dH =
2 g LH 3 / 2
3
2
Q = Cd
2 g LH 3 / 2 = CLH 3 / 2
3
Losses due to the advent of the drawdown of the flow immediately upstream
of the weir as well as any other friction or contraction losses; To account for
these losses, a coefficient of discharge Cd is introduced.
2
Q = Cd
2 g LH 3 / 2 = CLH 3 / 2
3
C d = 0.611 + 0.08H / Z
C d = 0.611 + 0.08H / Z
where,
H- is the head on the weir crest,
Z- is the height of the weir.
L- Crest Width
Example
#A broad-crested weir has a crest length of 0.75m, crest width of 1.0 m, and
crest height of 0.30 m. The water surface at the approach section is 0.20m
above the crest . Determine the Discharge.
Solution
L = 1m
H = 0.2m
Z = 0.3m
2
Q = Cd
2 g LH 3 / 2 = CLH 3 / 2
3
C d = 0.611 + 0.08H / Z
Example-Contd
0.2
C d = 0.611 + 0.08
0.3
C d = 0.611 + 0.08(0.667 )
C d = 0.611 + 0.053
C d = 0.664
Q = CLH
2
C = 0.664 2 9.81
3
C = 0.664 0.667 4.43
C = 1.96
3
Q = 1.96 1 0.21.5
Q = 1.96 1 0.089
Q = 0.175 m 3 s