g
v
2
2
1
y1
SoL
Figure 4.6. Non-uniform gradually varied flow
Uniform flow is found only in artificial channels of constant shape, slope, although under these conditions the flow for some distances may be non-uniform, as shown in Figure 4.1. However, with natural stream the slope of the bed and the shape and size of the cross-section usually vary to such an extent that true uniform flow is rare. Hence, the application of Manning equation for uniform flow can be applied to non-uniform flow with accuracy dependent on the length of reach L taken. In order to apply these equations at all, the streams must be divided into several reaches within which the conditions are approximately the same.
E.G.L
. hL=SL
g
v
2
2
2
y2
EGL Slope= S
Channel bed, slope = So
βx
1 2
Chapter V
STEADY NON-UNIFORM FLOW OR VARIED (Sβ SO)FLOW IN OPEN CHANNELS
βπ₯ =πΈ2 β πΈ1
ππ β π
From the above figure, energy equation between section 1-2
π12
2π+ π¦1 + ππ βπ₯ =
π22
2π+ π¦2 + π βπ₯
ππ βπ₯ β π βπ₯ = π2
2
2π+ π¦2 β
π12
2π+ π¦1
βπ₯ ππ β π = πΈ2 β πΈ1
where :
π =π2ππ
2
π π
43
ππ = π2 + π1
2 (ππππ π£ππππππ‘π¦ ππ π πππ‘πππ 1 πππ 2)
π π = π 2 + π 1
2 (ππππ βπ¦ππππ’πππ π ππππ’π )
π 1 = π΄1
π1
π 2 = π΄2
π2
Slope can be determined by Manningβs Equation
There are two types of non uniform flow depending upon the change of depth of flow over the length of the channel. If the depth of flow in a channel changes a gradually over a length of the channel, the flow is said to be Gradually Varied Flow (GVF). If depth of flow changes abruptly over a small length of the channel, the flow is said to be a local non-uniform phenomenon or Rapidly Varied Flow (RVF). Gradually varied flow can occur with either subcritical or supercritical flow, but the transition from one condition to the other is ordinarily abrupt, as between D and E in Figure 4.1. Other cases of local non-uniform flow occur at the entrance and exit of a channel, at channel at changes in cross sections, at bends and at on obstruction such as dams, weirs or bridge piers. See Figure 4.7 for steady non-uniform flow in a channel.
Depth of flow for non-uniform flow conditions varies with longitudinal distance. It occurs upstream and downstream control sections.
Rapid varied flow of occurs on the following condition: 1. Occurrence of hydraulic jump 2. Flow entering a steep channel from lake or a reservoir 3. Flow close to a free out fall from a channel 4. Flow in a vicinity of an obstruction such as bridge pier or sluice gate
Gradual varied flow occurs on the following condition:
1. Backwater created by a dam place in a river 2. Drawdown of a water surface as flow approaches a falls
RVF GVF RVF
RVF GVF
GVF
RVF
GVF
Figure 4.7. Steady Non-uniform flow in a channel.
WATER-SURFACE PROFILES IN GRADUALLY VARIED FLOW
Water surface profiles are classified two different ways: according to the slope of the channel (mild, steep, critical, horizontal, or adverse) and according to the actual depth of flow in relation to the critical and normal depths (zone 1, 2, or 3). The first letter of the type of slope (M, S, C, H or A) in combination with 1, 2, or 3 defines the type of surface profile. If the slope is so small that the normal depth (uniform flow depth) is greater than critical depth for the given discharge, then the slope of the channel is mild, and the water surface profile is given an M classification. Similarly, if the channel slope is so steep that a normal depth less than critical is produced, then the channel is steep, and the water surface profile is given an S designation. If the slopeβs normal depth equals its critical depth, then we have a critical slope, denoted by C. Horizontal and adverse slopes, denoted by H and A, respectively, are special categories because normal depth does not exist for them. An adverse slope is characterized by a slope upward in the flow direction. The 1, 2, and 3 designations of water surface profiles indicate if the actual flow depth is greater than both normal and critical depths (zone 1), between the normal; and critical depths (zone 2) or less than both normal and critical depths (zone 3). The basic shape of the various possible profiles are shown in the Table 4.4.
Table 4.4. Types of Varied Flow
Problem: 1. A rectangular canal is 2.0m wide and carries 2.4m3/s of water. The bed slope is 0.0009 and
the channel roughness n=0.012. At a certain section the depth is 0.90m and at another section the depth is 1.20m. a. Determine which depth is downstream b. Determine the distance between sections with the given depths using one reach. c. Determine the distance between the sections with the given depths using three
reaches.
2. Water discharges from under a sluice gate into a horizontal channel at a rate of 1m3/s per meter of width as shown in the figure. What is the classification of the water surface profile? Quantitatively evaluate the profile downstream of the gate and determine whether or not it will extend all the way to the abrupt drop 80m downstream. Make the simplifying assumption that the resistance factor f is equal to 0.02 and that the hydraulic radius R is equal to the depth y.
3. A rectangular concrete channel 4.50m wide is carrying water. At an upstream point, the
depth of water is 1.50m and a downstream point 300m away, the depth of flow is 1.17m. If the channel bed slope is 0.0010, find the theoretical flow rate. Use n=0.013.
1. A rectangular canal is 2.0m wide and carries 2.4 m3/s of water. The bed slope is 0.0009 and the channel roughness n=0.012. At a certain section the depth is 0.90m and at another section the depth is 1.20m. a. Determine which depth is downstream. b. Determine the distance between sections with the given depths using one reach. c. Determine the distance between the sections with the given depths using three reaches. GIVEN: Q = 2.4 m3/s; So = 0.0009; n = 0.012
0.82
1.20
0.90
yc= 0.523
2m
SOLUTION:
a.)
@ Critical flow
channel ofh meter widtper /sm2.12
4.2 3b
smVc /276.2528.081.9
2056.1528.02 mbYA cc
mYbP cc 056.3528.0222
mRc
c
P
A
c 346.0056.3
056.1
0009.0003071.0346.0
276.2012.03/4
22
3/4
22
o
c
cc S
R
VnS
Flow is subcritical. Actual slope is mild.
mg
qYc 528.0
9.81
2.13
2
3
2
@ Uniform flow
ooo YbYA 2
ooo YYbP 222
o
o
Po
A
oY
YR o
22
2
2/13/21SRA
nQ oo
2/1
3/2
0009.022
22
012.0
14.2
o
oo
Y
YY
3/2
3/5
148.0
o
o
Y
Y
Let 3/2
3/5
1 o
o
Y
YM
By Trial & error:
Assume oY M
1.0 0.630
0.82 0.482
oY =0.82m < 0.90m
Since Y > oY > cY and the slope is mild, the depth 1.20m is downstream of depth 0.90m. Type of
profile is 1M .
Ξx
1.20
0.90
b.) using one reach
@ Section 1:
mY 90.01
2
11 80.190.02 mbYA
mYbP 80.390.0222 11
mRP
A474.0
80.3
80.11
1
1
smA
QV /333.1
80.1
4.2
1
1
Then
mg
E 991.02
333.190.0
2
1
@ Section 2:
mY 20.12
2
22 40.220.12 mbYA
mYbP 40.420.1222 212
mRP
A545.0
40.4
40.22
2
2
smA
QV /1
40.2
4.2
2
2
Then
mg
E 251.12
0.120.1
2
2
Mean Velocity
sm
VVVm /167.1
2
1333.1
2
21
Mean Hydraulic Radius
m
RRRm 5095.0
2
545.0474.0
2
21
Slope
0004813.05095.0
167.1012.03/4
22
3/4
22
m
m
R
VnS
,Therefore
mx 96.6200009.0000481.0
251.1991.0
Ξx2 Ξx2 Ξx1
1.10 1.0
Ξx
1.20
0.90
c.) using three reaches
@ Section 3:
mY 0.13
2
33 0.20.12 mbYA
mYbP 0.40.1222 33
mRP
A50.0
0.4
0.23
3
3
smA
QV /20.1
0.2
4.2
3
3
Then
mg
E 073.12
20.10.1
2
3
@ Section 4:
mY 10.14
2
44 20.210.12 mbYA
mYbP 20.410.1222 44
mRP
A524.0
20.4
20.24
4
4
smA
QV /091.1
20.2
4.2
4
4
Then
mg
E 161.12
091.11.1
2
4
Mean Velocity
sm
VVVm /267.1
2
311
sm
VVVm /145.1
2
432
sm
VVVm /045.1
2
243
Mean Hydraulic Radius
m
RRRm 487.0
2
311
mRR
Rm 512.02
432
mRR
Rm 534.02
243
Slope
000603.0487.0
267.1012.03/4
22
3/4
1
2
1
2
1 m
m
R
VnS
00046.0512.0
145.1012.03/4
22
3/4
2
2
2
2
2 m
m
R
VnS
000363.0534.0
045.1012.03/4
22
3/4
3
2
3
2
3 m
m
R
VnS
,Thus
mSS
EEx
o
09.2761
31
1
mSS
EEx
o
73.1971
432
mSS
EEx
o
52.1681
243
,Therefore
52.16873.19709.276 x
βπ = πππ. ππ π
80 m
10cm
2. Water discharges from under a sluice gate into a horizontal channel at a rate of 1m3/s per meter of
width as shown in the figure. What is the classification of the water surface profile? Quantities
evaluate the profile downstream of the gate and determine whether or not it will extend all the way
to the abrupt drop 80m downstream. Make the simplifying assumption that the resistance factor f is
equal to 0.02 and that the hydraulic radius R is equal to the depth y.
GIVEN:
q=1m3/s per meter width
f=0.02
So=0
R=y
Ys=0.10m (depth of the flow from sluice gate)
SOLUTION:
Critical depth
mYmY
g
qY
sc
c
10.0467.0
9.81
13
2
3
2
(With horizontal bed slope, the water surface profile is classified as type H3, see table 4.4)
Using direct step method
of SS
EEx
21
Where: 3/4
22
22.2 m
mf
R
VnS (Manning Equation English Unit)
m
mf
gR
fVS
8
2
(Darcy-Weisbach Equation)
2
21 VVVm
2
21 RRRm
g
VyE
2
2
111
Sample Computation:
Velocity, @ y=0.10m
smy
qV /10
10.0
1
Using change in depth mooy 4.
mooy 4.
smy
qV /14.7
14.0
1
mx
gx
752.15
0156.0
2
5110014.01.0
Section No.
Depth, y (m)
Velocity @ section, V
(m/s)
Mean Velocity in reach,
Vm
V12
Mean Hydraulics
Radius, π π
ππ =πππ
2
8ππ π
βπ₯ =
π¦1 β π¦2 +(π1
2βπ22)
2π
ππ β ππ
Distance
from gate (m)
1 0.1 10 100 0
8.57 73.4 0.12 0.156 15.7
2 0.14 7.14 51 15.7
6.35 40.3 0.16 0.064 15.3
3 0.18 5.56 30.9 31.0
5.05 25.5 0.2 0.032 15.1
4 0.22 4.54 20.6 46.1
4.195 17.6 0.24 0.019 13.4
5 0.26 3.85 14.8 59.5
3.59 12.9 0.28 0.012 12.4
6 0.3 3.33 11.1 71.9
3.135 9.8 0.32 0.008 10.9
7 0.34 2.94 8.6 82.8
β΄ π‘βπ πππππππ ππ₯π‘ππππ π‘π π‘βπ ππππ’ππ‘ ππππ 80 π πππ€ππ π‘ππππ
0
10
20
30
40
50
60
0 20 40 60 80 100 120x (m)
y (cm)
4.50m
y 2
1
y1 = 1.5 m
y2 = 1.17 m
L
3. A rectangular concrete channel 4.50m wide is carrying water. At an upstream point, the
depth of water is 1.50m downstream point 300m away, the depth of flow is 1.17m. if
the channel bed slope is 0.0010, find the theoretical flow rate. Use n=0.013.
Solution:
πΏ = π2
2
2π+π¦2 β
π12
2π+π¦1
ππβ π
π2
2
2π+ π¦2 β
π12
2π+ π¦1 = πππΏ β ππΏ
π2
2
2πβ
π12
2π+ ππΏ = π¦1β π¦2 + πππΏ ..Eq.1
@section 1
π΄1 = 4.5 π₯ 1.5 = 6.75 π2
π 1 = π΄1 π1 = 6.75 4.5 + (2π₯1.5) = 0.90π
π1 = π π΄1 = π/6.75 = 0.148π
π1
2
2π= 0.00112π2
@section 2
π΄2 = 4.5 π₯ 1.17 = 5.265 π2
π 2 = π΄2 π2 = 5.265 4.5 + (2π₯1.17) = 0.77π
π2 = π π΄2 = π/6.75 = 0.19π π2
2
2π= 0.0018π2
π π =π 1+π 2
2=
0.9+0.77
2= 0.835 π
ππ =π1+π2
2=
0.148π+0.19π
2= 0.169π
π = πππ
π π2 3
2=
0.013π₯0.169π
0.8352 3 2
= 0.00000614π2
From Eq.1
0.0018π2 β 0.00112π2 + 0.00000614π2 π₯300 = 1.5 β 1.17 + 0.001(300)
β΄ πΈ = ππ. ππ ππ/π
HYDRAULIC JUMP
A hydraulic jump is a transition flow from supercritical to subcritical flow.
Hydraulic jump is one means of reducing the velocity of flow. It may also be used to separate lighter
solids from heavier ones.
y2>yc
V1 y1 < yc
HYDRAULIC JUMP IN A RECTANGULAR CHANNEL
Consider a freebody of water containing hydraulic jump
F2
F1
W
P1 = Ξ³ y1 P2 = Ξ³ y2 N
y2
y1 V1
y2
y1 Q
V2
Considering the Impulse-Momentum Equation
π΄πΉ = (πππ)ππ’π‘ β (πππ)ππ
π΄πΉπ₯ = (ππππ₯)ππ’π‘ β (ππππ₯)ππ
πΉ1 β πΉ2 β πΉπ = ππ2π2 β ππ1π1
where: Ef = neglected (if distance between sections is relatively small)
πΉ1 = 1
2π1π¦1π =
1
2 Ξ³π¦1π¦1π =
1
2 Ξ³π¦1
2π
πΉ2 = 1
2π2π¦2π =
1
2 Ξ³π¦2π¦2π =
1
2 Ξ³π¦2
2π
Then, 1
2 Ξ³π¦1
2π β1
2 Ξ³π¦2
2π β 0 = Ξ³
π π΄2π2 π2 β
Ξ³
π π΄1π1 π1
1
2 ππ π¦1
2 β π¦22 = π΄2π2
2 β π΄1π12
1
2 ππ π¦1
2 β π¦22 = ( ππ¦2 )π2
2 β ( ππ¦1 )π12
1
2 π π¦1
2 β π¦22 = π¦2π2
2 β π¦1π12
From continuity equation
π1 = π2
π΄1π1 = π΄1π1
ππ¦1π1 = ππ¦2π2
π½π = πππ½π
ππ
Substitute values
1
2 π π¦1
2 β π¦22 = π¦2
π¦12π1
2
π¦22 β π¦1π1
2
1
2 π π¦1
2 β π¦22 = π1
2π¦1 π¦1
π¦2β 1
1
2 π π¦1
2 β π¦22 =
π12π¦1
π¦2 π¦1 β π¦2
1
2 π π¦1 β π¦2 π¦1 + π¦2 =
π12π¦1
π¦2 π¦1 β π¦2
1
2 π π¦1 + π¦2 =
π12π¦1
π¦2
π½ππ =
π
π π
ππ
ππ ππ + ππ
But
π1 = π
π΄=
ππ
ππ¦1=
π
π¦1
π2
π¦12 =
1
2 π
π¦2
π¦1 π¦1 + π¦2
ππ = π
π πππππ ππ + ππ
ENERGY LOST AND POWER LOST IN A JUMP
Energy Equation 1 β 2
π1
πΎ+ π§1 +
π12
2π=
π2
πΎ+ π§2 +
π22
2π+ βπΏ
π¦1 + π1
2
2π= π¦2 +
π22
2π + βπΏ
πΈ1 = πΈ2 + βπΏ
ππ³ = π¬π β π¬π energy head lost
Power Lost: π· = πΈπΈππ³
Depth of Hydraulic Jump
Solve for y2: consider the equation:
π2 = 1
2 ππ¦1π¦2 π¦1 + π¦2
π¦2 π¦1 + π¦2 = 2π2
ππ¦1
π¦22 + π¦1π¦2 =
2π2
ππ¦1
π¦22 + π¦1π¦2 +
1
2π¦1
2=
2π2
ππ¦1 +
1
2π¦1
2
π¦2 + 1
2π¦1
2 =
2π2
ππ¦1 +
1
4π¦1
2
π¦2 + 1
2π¦1
2 =
1
4π¦1
2
8π2
ππ¦13 + 1
Extract the square root:
π¦2 + 1
2π¦1 =
1
2π¦1
8π2
ππ¦13 + 1
π¦2 = β 1
2π¦1 +
1
2π¦1
8π2
ππ¦13 + 1
π¦2 = 1
2π¦1 β1 +
8π2
ππ¦13 + 1
But π2
ππ¦13 =
π2
π2
ππ¦13 =
π΄2π2
π2
ππ¦13 =
π2π¦12π1
2
π2
ππ¦13 =
π12
ππ¦1= ππΉ
2
Hence, ππ = π
πππ βπ + ππ΅ππ
π + π ππΉ1> 1
Likewise,
ππ = π
πππ βπ + ππ΅ππ
π + π ππΉ2< 1
2
y2
y1
1
yc2
y2
y1
yc1
HYDRAULIC JUMP IN A NON-RECTANGULAR SECTION
section thru 1 - 1 section thru 2 - 2
Impulse-Momentum Equation:
π΄πΉπ₯ = (πππ)ππ’π‘ β (πππ)ππ
πΉ1 β πΉ2 β πΈπ = ππ2π2 β ππ1π1
πΎπ΄1π¦π1β πΎπ΄2π¦π2
β 0 = πΎ
π π΄2π2π2 β π΄1π1π1
π π΄1π¦π1β π΄2π¦π2
= π΄2π22 β π΄1π1
2
Continuity Equation:
π1 = π2
π΄1π1 = π΄1π1
π2 = π΄1π1
π΄2
π π΄1π¦π1β π΄2π¦π2
= π΄2π΄1
2π12
π΄22 β π΄1π1
2
π π΄1π¦π1β π΄2π¦π2
= π΄1π12
π΄1
π΄2β 1
π π΄1π¦π1β π΄2π¦π2
= π΄1π12
π΄1βπ΄2
π΄2
π½ππ =
ππ¨π
π¨π π¨ππππ
βπ¨ππππ
π¨πβ π¨π or π½π
π = ππ¨π
π¨π π¨ππππ
βπ¨ππππ
π¨πβ π¨π
*Another solution
π΄πΉπ₯ = (ππππ₯)ππ’π‘ β (ππππ₯)ππ
πΉ1 β πΉ2 β πΈπ = ππ2π2 β ππ1π1
Ξ³π΄1π¦π1 β Ξ³π΄2π¦π2 =Ξ³
π π΄2π2
2 β π΄1π12
π΄1π¦π1 β π΄2π¦π2 =1
π π΄2
π2
π΄22 β π΄1
π2
π΄12
π΄1π¦π1 β π΄2π¦π2 =π2
π
1
π΄2β
1
π΄1
πΈπ
π=
π¨ππππβπ¨ππππ
π
π¨πβ
π
π¨π
1.2 m
2.0 m
1.2 m
1. Water flows in a rectangular channel with a width of 4.0 m at a uniform depth of 1.2 m. Adjustment is made downstream to raise water level to 2.0 m. consequently causing hydraulic jump. a. Calculate the discharge in the canal. b. Determine the power lost in a jump.
Solution:
a.) π2 = 1
2 ππ¦1π¦2 π¦1 + π¦2
π2 = 1
2 (9.81)(1.2)(2) 1.2 + 2
π = 6.138 π3 π πππ πππ‘ππ π€πππ‘β ππ π‘βπ πππππ π = ππ = 6.138 4
πΈ = ππ. ππ ππ π b.) Power Lost, π = πΎπβπΏ βπΏ = πΈ1 β πΈ2
where πΈ1 = π¦1 + π1
2
2π= 1.2 +
24.55
1.2 (4)
2
2π= 2.533 π
πΈ2 = π¦2 + π2
2
2π= 2 +
24 .55
2 (4)
2
2π= 2.480 π
βπΏ = 2.533 β 2.480 = 0.053 π
Thus, π = πΎπβπΏ = (9.81)(24.55)(0.053) π· = ππ. πππ ππΎ
2. A hydraulic jump occurs in a 5 m wide rectangular canal carrying 6 m3/s on a slope of 0.005. the depth
after the jump is 1.4m.
a.) Calculate the depth before the jump.
b.) Calculate the power lost in a jump.
y2 = 1.4 m
y1 =?
Given: b= 5 m S= 0.005
Q= 6 m3/s yafter= 1.4 m
Solution:
a.) π΄2 = ππ¦2 = 5 1.4 = 7 π2
π2 = π
π΄2=
6
7= 0.857 π/π
ππΉ2=
π2
ππ¦2=
0.857
9.81 x 1.4= 0.23 < 1
There is a hydraulic jump that occurs. And the depth before the jump is
π¦1 = 1
2π¦2 β1 + 8ππΉ2
2 + 1 =1
2(1.4) β1 + 8(0.23)2 + 1
β΄ ππ = π. πππ π
b.) π1 =π
π΄1=
π
ππ¦1=
6
5 x 0.136= 8.82 π/π ππ
πΈ1 = π¦1 + π1
2
2π= 0.136 +
8.82 2
2π= 4.1 π
πΈ2 = π¦2 + π2
2
2π= 1.4 +
0.857 2
2π= 1.44 π
Therefore,
βπΏ = πΈ1 β πΈ2
βπΏ = 4.1 β 1.44 = 2.66 π
Thus,
π = πΎπβπΏ
π = 9.81(6)(2.66)
π· = πππ. ππ ππΎ
3. A rectangular canal has a width of 4.0m and carries water at the rate of 12m3/s. its bed slope is 0.0003 and roughness is 0.02. To control the flow, a sluice gate is provided at the entrance to the canal.
a. Determine whether a hydraulic jump would occur when the sluice gate is adjusted so that minimum depth after the gate is 0.40 m.
b. If a hydraulic jump would occur in letter (a), how far from the sluice gate will it occur?
y1=depth req. to
cause a jump
yo
ys=0.4
0m
βπ₯
Given: Q = 2m3/s; b = 4m; n = 0.02 ; So = 0.0003 Solution:
π =π
π=
12
4= 3 π3 π πππ π π€πππ‘β
@critical flow
π¦π = π2
π
1 3
= 32
9.81
1 3
= 0.972 π
π΄π = ππ¦π = 4 0.972 = 3.887 π2 ππ = π + 2π¦π = 4 + 2 0.972 = 5.944 π
π π =π΄π
ππ= 0.654 π
ππ =π
π΄π ππ ππ = ππ¦π = 3.087 π π
ππ =π2ππ
2
π π4 3 =
0.00223.0872
0.654 4 3 = 0.006715 > So = 0.0003 β΄ π ππππ ππ ππππ.
@Normal depth π΄π = ππ¦π = 4π¦π ππ = π + 2π¦π = 4 + 2π¦π = 2(2 + π¦π)
π π =π΄π
ππ=
4π¦π
2(2+π¦π )=
2π¦π
(2+π¦π )
ππ =1
ππ΄ππ π
2
3ππ
1
2
12 =1
0.024π¦π
2π¦π
(2+π¦π )
2
30.0003
1
2
2.812 =π¦π
5 3
(2+π¦π )2 3 let π =π¦π
5 3
(2+π¦π )2 3
Trial and error: Assume y M 1.0 0.481 3.053 2.182 β΄ π¦π = 3.053 π
yc
yo
Depth required to cause a jump:
π¦1 =1
2π¦π β1 + 8ππΉπ
2 + 1
where ππ =π
π΄π=
12
ππ¦π=
12
4(3.053)= 0.983 π/π ππ
ππΉπ2 =
ππ2
ππ¦π=
0.983 2
9.81 π₯ 3.053= 0.032
π¦1 =1
2
3
053 β1 + 8 0.032 2 + 1 = 0.184 π < π¦π = 0.40π
β΄ "π»ππππβ²π ππ πππ ππππππ ππππ ππππππ"
π¦π > π¦π > π¦ π π’πππππππ‘ππππ ππππ€
y1
yo=1.60
m ys=0.50m
Ξx
4. A rectangular channel has a width of 5m, so=0.0009 and n=0.012. its uniform flow depth is 1.60m. if a sluice gate is adjusted such that a min. depth immediately downstream of the gate is 0.50m.
a. Determine whether a hydraulic jump would occur, and if it occurs b. how far downstream will it occur c. type of profile
Solution: a.) @ normal depth, yo=1.60 m π΄π = ππ¦π = 5 1.60 = 8 π2 ππ = π + 2π¦π = 4 + 2 1.60 = 8.20π
π π =π΄π
ππ=
8
8.20= 0.976 π
ππ =1
ππ π
2 3 π1 2 =1
0.012(0.976)2 3 (0.0009)1 2 = 2.46 π/π ππ
ππΉπ2 =
ππ2
ππ¦π=
2.46 2
9.81 π₯ 1.60= 0.385
Depth required to cause a jump
π¦1 =1
2π¦π β1 + 8ππΉπ
2 + 1 =1
2 1.60 β1 + 8 0.385 2 + 1 = 0.816 π
π¦1 = 0.816 π > π¦π = 0.5 β΄ ππππ ππππ πππππ @ critical flow π = 8 2.46 = 19.68 π3/π π = 19.68 5 = 3.936 π3/π πππ π π€πππ‘β
π¦π = π2
π
1 3
= 3.9362
9.81
1 3
= 1.16 π < π¦π = 1.60 π β΄ πππππ ππ ππππ
b.) distance between ys=0.50 m to y1=0.816 m using one reach
βπ₯ =πΈπ βπΈ1
πβππ
Mean velocities: ππ =π
π΄π =
19.68
5(0.5)= 7.872 π/π
π1 =π
π΄1=
19.68
5(0.816)= 4.824 π/π
ππ =1
2 ππ + π1 = 6.348 π/π
Mean hydraulic radius: π π =π΄π
ππ =
5(0.5)
5+2(0.5)= 0.417 π
π 1 =π΄1
π1=
5(0.816)
5+2(0.816)= 0.615 π
π π =1
2 π π + π 1 = 0.516 π
Then, π =π2ππ
2
π π4 3 =
(0.012)2(6.348)2
(0.516)4 3 = 0.014
πΈπ = π¦π + ππ
2
2π= 0.5 +
(7.872)2
2π= 3.658 π
πΈ1 = π¦1 + π1
2
2π= 0.816 +
(4.824)2
2π= 2.002 π
Thus, βπ =π¬πβπ¬π
πΊβπΊπ=
π.πππβπ.πππ
π.πππβπ.ππππ= πππ. πππ π
c.) Type of profile
Since π¦π > π¦π > π¦ ,
β΄ π»ππ πππππππ ππ π΄π
5. Examine the flow conditions in a very long 10ft wide open rectangular channel of rubble
masonry with n=0.017 when the flow rate is 400cfs. The channel slope is 0.020 and an ogee weir
5ft high with Cw=3.80 are located at the downstream end of the channel.
yn
yn
Solutions:
Normal depth of flow in the channel,
π =1.48
ππ΄π 2/3π1/2
400 =1.49
0.017x10π¦π
10π¦π
10+2π¦π
2
3(0.020)
1
2
By trial and error:
π¦π = 2.36 ππ‘.
Critical depth: π¦π = π2
π΅2π
1
3=
4002
10232.2
1
3= 3.67 ππ‘
Since yn < yc , the flow is supercritical. The head required on the weir to discharge:
π = πΆπ€πΏπ»3
2
400 = 3.80x10 β +
400
5+β x10 2
64.4
3
2
By trial and error, β = 4.53 ππ‘ depth of water upstream weir is 9.53 which is greater than yc.
The flow @ this point is subcritical & hydraulic jump must occur upstream. The depth y2 after
the jump is:
π¦2 = β2.36
2+
2.362
4+
2 400
23.6
2x2.36
32.2
1/2
= 5.42 ππ‘
The distance from the weir to the jump
π¦π΄ = 5.42 ππ‘ ππ΄ = 400/54 = 7.39ππ‘/π ππ ππ΄2 2π = 0.85ππ‘
π¦π΅ = 9.53 ππ‘ ππ΅ = 400/97.4 = 4.20ππ‘/π ππ ππ΅2 2π = 0.27ππ‘
πππ£π = (VA + VB )/2 = 5.74 ft/sec
π ππ£π = 2.95 ft
π = 0.017x5.79 2 (1.486x2.490)4 3 = 0.00104
π₯ = 5.42+0.85β9.53β0.27
0.00104β0.02= 186.18 ft
5 ft
4.53 ft
5.40 ft 2.36 ft
EGL
π2
2π= 0.85 ππ‘ π2
2π= 0.26 ππ‘ π2
2π= 4.45 ππ‘
186.18 ft
2
y2 = 1.80m
y1 = 1.20m
1
4m 4m
π΄23 π΄22
π΄13
π΄12
1.80
1.20 π΄11
π΄21
6. A hydraulic jump occurs in a trapezoidal section with bottom width of 4m and side slope of 1:2. The
depth before the jump is 1.20m and after the jump is 1.80m.
a.) Calculate the flow rate in the canal.
b.) Calculate the power lost.
Solution:
π΄1 = π΄π + 2π΄12
π₯
π¦=
1
2
π₯1 = 1
2π¦1 = 0.60 π
π΄1 = 4 1.20 + 2(1
2 Γ 0.60 + 1.20)
π΄1 = 5.52 π2
π΄2 = π΄21+ 2π΄22
π₯
π¦=
1
2
π₯2 = 1
2π¦2 = 0.90 π
π΄2 = 4 1.80 + 2(1
2 Γ 0.90 Γ 1.80)
π΄2 = 8.82 π2
π΄1π¦π1= π΄11π¦π1 1
+ 2π΄12π¦π1 2
π΄1π¦π1= 1.20π₯4 0.60 +
2(1
2π₯ 0.60 π₯ 1.2)(0.40)
π΄1π¦π1= 2.88 + 0.288
π΄1π¦π1= 3.168 π2
π΄2π¦π2= π΄21π¦π2 1
+ 2π΄22π¦π2 2
π΄2π¦π2= 4 1.8π₯ 0.90 + 2
1
2π₯0.90π₯1.80 π₯0.60)
π΄2π¦π2= 7.452 π2
π12 =
9.81(8.82)
5.52
7.452β3.168
8.82β5.52 = 20.349
π1 = 4.511 π/π
a.) Flow rate in canal
π1 = π΄1π1 = 5.52 4.511
πΈπ = ππ. πππ/π
b.) Power Lost, π = πΎπβπΏ
βπΏ = πΈ1 β πΈ2
where πΈ1 = π¦1 +π1
2
2π= 1.2 +
4.5112
2π= 2.237π
πΈ2 = π¦2 +π2
2
2π= 1.8 +
2.8232
2π= 2.206π
Then, βπΏ = 2.237 β 2.206 = 0.031π
Therefore,
π = πΎπβπΏ
π = 9.81(24.9)(0.031)
π· = π. πππ ππΎ