chapter v- open channel non-uniform flow
DESCRIPTION
a brief discussion about non uniform flow and some sample problems for solving.TRANSCRIPT
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
∆x
1 2
Chapter VSTEADY NON-UNIFORM FLOW OR VARIED (S≠ SO)FLOW IN OPEN CHANNELS
∆𝑥 =𝐸2 − 𝐸1
𝑆𝑜 − 𝑆
From the above figure, energy equation between section 1-2
𝑉12
2𝑔+ 𝑦1 + 𝑆𝑜 ∆𝑥 =
𝑉22
2𝑔+ 𝑦2 + 𝑆 ∆𝑥
𝑆𝑜 ∆𝑥 − 𝑆 ∆𝑥 = 𝑉2
2
2𝑔+ 𝑦2 −
𝑉12
2𝑔+ 𝑦1
∆𝑥 𝑆𝑜 − 𝑆 = 𝐸2 − 𝐸1
where :
𝑆 =𝑛2𝑉𝑚
2
𝑅𝑚
43
𝑉𝑚 = 𝑉2 + 𝑉1
2 (𝑚𝑒𝑎𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 𝑎𝑛𝑑 2)
𝑅𝑚 = 𝑅2 + 𝑅1
2 (𝑚𝑒𝑎𝑛 𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑅𝑎𝑑𝑖𝑢𝑠)
𝑅1 = 𝐴1
𝑃1
𝑅2 = 𝐴2
𝑃2
Slope can be determined by Manning’s 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 slope’s 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 .
Δx
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
Δx2 Δx2 Δx1
1.10 1.0
Δx
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
311
mSS
EEx
o
73.1971
432
mSS
EEx
o
52.1681
243
,Therefore
52.16873.19709.276 x
∆𝒙 = 𝟔𝟒𝟐. 𝟑𝟒 𝒎
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, 𝑅𝑚
𝑆𝑓 =𝑓𝑉𝑚
2
8𝑔𝑅𝑚
∆𝑥 =
𝑦1 − 𝑦2 +(𝑉1
2−𝑉22)
2𝑔
𝑆𝑓 − 𝑆𝑜
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
∴ 𝑡𝑒 𝑝𝑟𝑜𝑓𝑖𝑙𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑡𝑜 𝑡𝑒 𝑎𝑏𝑟𝑢𝑝𝑡 𝑑𝑟𝑜𝑝 80 𝑚 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚
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:
𝐿 = 𝑉2
2
2𝑔+𝑦2 −
𝑉12
2𝑔+𝑦1
𝑆𝑜− 𝑆
𝑉2
2
2𝑔+ 𝑦2 −
𝑉12
2𝑔+ 𝑦1 = 𝑆𝑜𝐿 − 𝑆𝐿
𝑉2
2
2𝑔−
𝑉12
2𝑔+ 𝑆𝐿 = 𝑦1− 𝑦2 + 𝑆𝑜𝐿 ..Eq.1
@section 1
𝐴1 = 4.5 𝑥 1.5 = 6.75 𝑚2
𝑅1 = 𝐴1 𝑃1 = 6.75 4.5 + (2𝑥1.5) = 0.90𝑚
𝑉1 = 𝑄 𝐴1 = 𝑄/6.75 = 0.148𝑄
𝑉1
2
2𝑔= 0.00112𝑄2
@section 2
𝐴2 = 4.5 𝑥 1.17 = 5.265 𝑚2
𝑅2 = 𝐴2 𝑃2 = 5.265 4.5 + (2𝑥1.17) = 0.77𝑚
𝑉2 = 𝑄 𝐴2 = 𝑄/6.75 = 0.19𝑄 𝑉2
2
2𝑔= 0.0018𝑄2
𝑅𝑚 =𝑅1+𝑅2
2=
0.9+0.77
2= 0.835 𝑚
𝑉𝑚 =𝑉1+𝑉2
2=
0.148𝑄+0.19𝑄
2= 0.169𝑄
𝑆 = 𝑛𝑉𝑚
𝑅𝑚2 3
2=
0.013𝑥0.169𝑄
0.8352 3 2
= 0.00000614𝑄2
From Eq.1
0.0018𝑄2 − 0.00112𝑄2 + 0.00000614𝑄2 𝑥300 = 1.5 − 1.17 + 0.001(300)
∴ 𝑸 = 𝟏𝟓. 𝟖𝟏 𝒎𝟑/𝒔
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 = γ y1 P2 = γ y2 N
y2
y1 V1
y2
y1 Q
V2
Considering the Impulse-Momentum Equation
𝛴𝐹 = (𝜌𝑄𝑉)𝑜𝑢𝑡 − (𝜌𝑄𝑉)𝑖𝑛
𝛴𝐹𝑥 = (𝜌𝑄𝑉𝑥)𝑜𝑢𝑡 − (𝜌𝑄𝑉𝑥)𝑖𝑛
𝐹1 − 𝐹2 − 𝐹𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1
where: Ef = neglected (if distance between sections is relatively small)
𝐹1 = 1
2𝑃1𝑦1𝑏 =
1
2 γ𝑦1𝑦1𝑏 =
1
2 γ𝑦1
2𝑏
𝐹2 = 1
2𝑃2𝑦2𝑏 =
1
2 γ𝑦2𝑦2𝑏 =
1
2 γ𝑦2
2𝑏
Then, 1
2 γ𝑦1
2𝑏 −1
2 γ𝑦2
2𝑏 − 0 = γ
𝑔 𝐴2𝑉2 𝑉2 −
γ
𝑔 𝐴1𝑉1 𝑉1
1
2 𝑔𝑏 𝑦1
2 − 𝑦22 = 𝐴2𝑉2
2 − 𝐴1𝑉12
1
2 𝑔𝑏 𝑦1
2 − 𝑦22 = ( 𝑏𝑦2 )𝑉2
2 − ( 𝑏𝑦1 )𝑉12
1
2 𝑔 𝑦1
2 − 𝑦22 = 𝑦2𝑉2
2 − 𝑦1𝑉12
From continuity equation
𝑄1 = 𝑄2
𝐴1𝑉1 = 𝐴1𝑉1
𝑏𝑦1𝑉1 = 𝑏𝑦2𝑉2
𝑽𝟐 = 𝒚𝟏𝑽𝟏
𝒚𝟐
Substitute values
1
2 𝑔 𝑦1
2 − 𝑦22 = 𝑦2
𝑦12𝑉1
2
𝑦22 − 𝑦1𝑉1
2
1
2 𝑔 𝑦1
2 − 𝑦22 = 𝑉1
2𝑦1 𝑦1
𝑦2− 1
1
2 𝑔 𝑦1
2 − 𝑦22 =
𝑉12𝑦1
𝑦2 𝑦1 − 𝑦2
1
2 𝑔 𝑦1 − 𝑦2 𝑦1 + 𝑦2 =
𝑉12𝑦1
𝑦2 𝑦1 − 𝑦2
1
2 𝑔 𝑦1 + 𝑦2 =
𝑉12𝑦1
𝑦2
𝑽𝟏𝟐 =
𝟏
𝟐 𝒈
𝒚𝟐
𝒚𝟏 𝒚𝟏 + 𝒚𝟐
But
𝑉1 = 𝑄
𝐴=
𝑏𝑞
𝑏𝑦1=
𝑞
𝑦1
𝑞2
𝑦12 =
1
2 𝑔
𝑦2
𝑦1 𝑦1 + 𝑦2
𝒒𝟐 = 𝟏
𝟐 𝒈𝒚𝟏𝒚𝟐 𝒚𝟏 + 𝒚𝟐
ENERGY LOST AND POWER LOST IN A JUMP
Energy Equation 1 – 2
𝑃1
𝛾+ 𝑧1 +
𝑉12
2𝑔=
𝑃2
𝛾+ 𝑧2 +
𝑉22
2𝑔+ 𝐿
𝑦1 + 𝑉1
2
2𝑔= 𝑦2 +
𝑉22
2𝑔 + 𝐿
𝐸1 = 𝐸2 + 𝐿
𝒉𝑳 = 𝑬𝟏 − 𝑬𝟐 energy head lost
Power Lost: 𝑷 = 𝜸𝑸𝒉𝑳
Depth of Hydraulic Jump
Solve for y2: consider the equation:
𝑞2 = 1
2 𝑔𝑦1𝑦2 𝑦1 + 𝑦2
𝑦2 𝑦1 + 𝑦2 = 2𝑞2
𝑔𝑦1
𝑦22 + 𝑦1𝑦2 =
2𝑞2
𝑔𝑦1
𝑦22 + 𝑦1𝑦2 +
1
2𝑦1
2=
2𝑞2
𝑔𝑦1 +
1
2𝑦1
2
𝑦2 + 1
2𝑦1
2 =
2𝑞2
𝑔𝑦1 +
1
4𝑦1
2
𝑦2 + 1
2𝑦1
2 =
1
4𝑦1
2
8𝑞2
𝑔𝑦13 + 1
Extract the square root:
𝑦2 + 1
2𝑦1 =
1
2𝑦1
8𝑞2
𝑔𝑦13 + 1
𝑦2 = − 1
2𝑦1 +
1
2𝑦1
8𝑞2
𝑔𝑦13 + 1
𝑦2 = 1
2𝑦1 −1 +
8𝑞2
𝑔𝑦13 + 1
But 𝑞2
𝑔𝑦13 =
𝑄2
𝑏2
𝑔𝑦13 =
𝐴2𝑉2
𝑏2
𝑔𝑦13 =
𝑏2𝑦12𝑉1
2
𝑏2
𝑔𝑦13 =
𝑉12
𝑔𝑦1= 𝑁𝐹
2
Hence, 𝒚𝟐 = 𝟏
𝟐𝒚𝟏 −𝟏 + 𝟖𝑵𝑭𝟏
𝟐 + 𝟏 𝑁𝐹1> 1
Likewise,
𝒚𝟏 = 𝟏
𝟐𝒚𝟐 −𝟏 + 𝟖𝑵𝑭𝟐
𝟐 + 𝟏 𝑁𝐹2< 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:
𝛴𝐹𝑥 = (𝑝𝑄𝑉)𝑜𝑢𝑡 − (𝑝𝑄𝑉)𝑖𝑛
𝐹1 − 𝐹2 − 𝐸𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1
𝛾𝐴1𝑦𝑐1− 𝛾𝐴2𝑦𝑐2
− 0 = 𝛾
𝑔 𝐴2𝑉2𝑉2 − 𝐴1𝑉1𝑉1
𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2
= 𝐴2𝑉22 − 𝐴1𝑉1
2
Continuity Equation:
𝑄1 = 𝑄2
𝐴1𝑉1 = 𝐴1𝑉1
𝑉2 = 𝐴1𝑉1
𝐴2
𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2
= 𝐴2𝐴1
2𝑉12
𝐴22 − 𝐴1𝑉1
2
𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2
= 𝐴1𝑉12
𝐴1
𝐴2− 1
𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2
= 𝐴1𝑉12
𝐴1−𝐴2
𝐴2
𝑽𝟏𝟐 =
𝒈𝑨𝟐
𝑨𝟏 𝑨𝟏𝒚𝒄𝟏
−𝑨𝟐𝒚𝒄𝟐
𝑨𝟏− 𝑨𝟐 or 𝑽𝟏
𝟐 = 𝒈𝑨𝟐
𝑨𝟏 𝑨𝟐𝒚𝒄𝟐
−𝑨𝟏𝒚𝒄𝟏
𝑨𝟐− 𝑨𝟏
*Another solution
𝛴𝐹𝑥 = (𝜌𝑄𝑉𝑥)𝑜𝑢𝑡 − (𝜌𝑄𝑉𝑥)𝑖𝑛
𝐹1 − 𝐹2 − 𝐸𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1
γ𝐴1𝑦𝑐1 − γ𝐴2𝑦𝑐2 =γ
𝑔 𝐴2𝑉2
2 − 𝐴1𝑉12
𝐴1𝑦𝑐1 − 𝐴2𝑦𝑐2 =1
𝑔 𝐴2
𝑄2
𝐴22 − 𝐴1
𝑄2
𝐴12
𝐴1𝑦𝑐1 − 𝐴2𝑦𝑐2 =𝑄2
𝑔
1
𝐴2−
1
𝐴1
𝑸𝟐
𝒈=
𝑨𝟏𝒚𝒄𝟏−𝑨𝟐𝒚𝒄𝟐
𝟏
𝑨𝟐−
𝟏
𝑨𝟏
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.) 𝑞2 = 1
2 𝑔𝑦1𝑦2 𝑦1 + 𝑦2
𝑞2 = 1
2 (9.81)(1.2)(2) 1.2 + 2
𝑞 = 6.138 𝑚3 𝑠 𝑝𝑒𝑟 𝑚𝑒𝑡𝑒𝑟 𝑤𝑖𝑑𝑡 𝑜𝑓 𝑡𝑒 𝑐𝑎𝑛𝑎𝑙 𝑄 = 𝑞𝑏 = 6.138 4
𝑸 = 𝟐𝟒. 𝟓𝟓 𝒎𝟑 𝒔 b.) Power Lost, 𝑃 = 𝛾𝑄𝐿 𝐿 = 𝐸1 − 𝐸2
where 𝐸1 = 𝑦1 + 𝑉1
2
2𝑔= 1.2 +
24.55
1.2 (4)
2
2𝑔= 2.533 𝑚
𝐸2 = 𝑦2 + 𝑉2
2
2𝑔= 2 +
24 .55
2 (4)
2
2𝑔= 2.480 𝑚
𝐿 = 2.533 − 2.480 = 0.053 𝑚
Thus, 𝑃 = 𝛾𝑄𝐿 = (9.81)(24.55)(0.053) 𝑷 = 𝟏𝟐. 𝟕𝟔𝟒 𝒌𝑾
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.) 𝐴2 = 𝑏𝑦2 = 5 1.4 = 7 𝑚2
𝑉2 = 𝑄
𝐴2=
6
7= 0.857 𝑚/𝑠
𝑁𝐹2=
𝑉2
𝑔𝑦2=
0.857
9.81 x 1.4= 0.23 < 1
There is a hydraulic jump that occurs. And the depth before the jump is
𝑦1 = 1
2𝑦2 −1 + 8𝑁𝐹2
2 + 1 =1
2(1.4) −1 + 8(0.23)2 + 1
∴ 𝒚𝟏 = 𝟎. 𝟏𝟑𝟔 𝒎
b.) 𝑉1 =𝑄
𝐴1=
𝑄
𝑏𝑦1=
6
5 x 0.136= 8.82 𝑚/𝑠𝑒𝑐
𝐸1 = 𝑦1 + 𝑉1
2
2𝑔= 0.136 +
8.82 2
2𝑔= 4.1 𝑚
𝐸2 = 𝑦2 + 𝑉2
2
2𝑔= 1.4 +
0.857 2
2𝑔= 1.44 𝑚
Therefore,
𝐿 = 𝐸1 − 𝐸2
𝐿 = 4.1 − 1.44 = 2.66 𝑚
Thus,
𝑃 = 𝛾𝑄𝐿
𝑃 = 9.81(6)(2.66)
𝑷 = 𝟏𝟓𝟔. 𝟓𝟕 𝒌𝑾
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
∆𝑥
Given: Q = 2m3/s; b = 4m; n = 0.02 ; So = 0.0003 Solution:
𝑞 =𝑄
𝑏=
12
4= 3 𝑚3 𝑠 𝑝𝑒𝑟 𝑚 𝑤𝑖𝑑𝑡
@critical flow
𝑦𝑐 = 𝑞2
𝑔
1 3
= 32
9.81
1 3
= 0.972 𝑚
𝐴𝑐 = 𝑏𝑦𝑐 = 4 0.972 = 3.887 𝑚2 𝑃𝑐 = 𝑏 + 2𝑦𝑐 = 4 + 2 0.972 = 5.944 𝑚
𝑅𝑐 =𝐴𝑐
𝑃𝑐= 0.654 𝑚
𝑉𝑐 =𝑄
𝐴𝑐 𝑜𝑟 𝑉𝑐 = 𝑔𝑦𝑐 = 3.087 𝑚 𝑠
𝑆𝑐 =𝑛2𝑉𝑐
2
𝑅𝑐4 3 =
0.00223.0872
0.654 4 3 = 0.006715 > So = 0.0003 ∴ 𝑠𝑙𝑜𝑝𝑒 𝑖𝑠 𝑚𝑖𝑙𝑑.
@Normal depth 𝐴𝑜 = 𝑏𝑦𝑜 = 4𝑦𝑜 𝑃𝑜 = 𝑏 + 2𝑦𝑜 = 4 + 2𝑦𝑜 = 2(2 + 𝑦𝑜)
𝑅𝑜 =𝐴𝑜
𝑃𝑜=
4𝑦𝑜
2(2+𝑦𝑜 )=
2𝑦𝑜
(2+𝑦𝑜 )
𝑄𝑜 =1
𝑛𝐴𝑜𝑅𝑜
2
3𝑆𝑜
1
2
12 =1
0.024𝑦𝑜
2𝑦𝑜
(2+𝑦𝑜 )
2
30.0003
1
2
2.812 =𝑦𝑜
5 3
(2+𝑦𝑜 )2 3 let 𝑀 =𝑦𝑜
5 3
(2+𝑦𝑜 )2 3
Trial and error: Assume y M 1.0 0.481 3.053 2.182 ∴ 𝑦𝑜 = 3.053 𝑚
yc
yo
Depth required to cause a jump:
𝑦1 =1
2𝑦𝑜 −1 + 8𝑁𝐹𝑜
2 + 1
where 𝑉𝑜 =𝑄
𝐴𝑜=
12
𝑏𝑦𝑜=
12
4(3.053)= 0.983 𝑚/𝑠𝑒𝑐
𝑁𝐹𝑜2 =
𝑉𝑜2
𝑔𝑦𝑜=
0.983 2
9.81 𝑥 3.053= 0.032
𝑦1 =1
2
3
053 −1 + 8 0.032 2 + 1 = 0.184 𝑚 < 𝑦𝑠 = 0.40𝑚
∴ "𝑻𝒉𝒆𝒓𝒆′𝒔 𝒏𝒐 𝒉𝒚𝒅𝒓𝒂𝒖𝒍𝒊𝒄 𝒋𝒖𝒎𝒑 𝒐𝒄𝒄𝒖𝒓𝒔"
𝑦𝑜 > 𝑦𝑐 > 𝑦 𝑠𝑢𝑝𝑒𝑟𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑓𝑙𝑜𝑤
y1
yo=1.60
m ys=0.50m
Δx
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 𝐴𝑜 = 𝑏𝑦𝑜 = 5 1.60 = 8 𝑚2 𝑃𝑜 = 𝑏 + 2𝑦𝑜 = 4 + 2 1.60 = 8.20𝑚
𝑅𝑜 =𝐴𝑜
𝑃𝑜=
8
8.20= 0.976 𝑚
𝑉𝑜 =1
𝑛𝑅𝑜
2 3 𝑆1 2 =1
0.012(0.976)2 3 (0.0009)1 2 = 2.46 𝑚/𝑠𝑒𝑐
𝑁𝐹𝑜2 =
𝑉𝑜2
𝑔𝑦𝑜=
2.46 2
9.81 𝑥 1.60= 0.385
Depth required to cause a jump
𝑦1 =1
2𝑦𝑜 −1 + 8𝑁𝐹𝑜
2 + 1 =1
2 1.60 −1 + 8 0.385 2 + 1 = 0.816 𝑚
𝑦1 = 0.816 𝑚 > 𝑦𝑠 = 0.5 ∴ 𝒋𝒖𝒎𝒑 𝒘𝒊𝒍𝒍 𝒐𝒄𝒄𝒖𝒓 @ critical flow 𝑄 = 8 2.46 = 19.68 𝑚3/𝑠 𝑞 = 19.68 5 = 3.936 𝑚3/𝑠 𝑝𝑒𝑟 𝑚 𝑤𝑖𝑑𝑡
𝑦𝑐 = 𝑞2
𝑔
1 3
= 3.9362
9.81
1 3
= 1.16 𝑚 < 𝑦𝑜 = 1.60 𝑚 ∴ 𝒔𝒍𝒐𝒑𝒆 𝒊𝒔 𝒎𝒊𝒍𝒅
b.) distance between ys=0.50 m to y1=0.816 m using one reach
∆𝑥 =𝐸𝑠−𝐸1
𝑆−𝑆𝑜
Mean velocities: 𝑉𝑠 =𝑄
𝐴𝑠=
19.68
5(0.5)= 7.872 𝑚/𝑠
𝑉1 =𝑄
𝐴1=
19.68
5(0.816)= 4.824 𝑚/𝑠
𝑉𝑠 =1
2 𝑉𝑠 + 𝑉1 = 6.348 𝑚/𝑠
Mean hydraulic radius: 𝑅𝑠 =𝐴𝑠
𝑃𝑠=
5(0.5)
5+2(0.5)= 0.417 𝑚
𝑅1 =𝐴1
𝑃1=
5(0.816)
5+2(0.816)= 0.615 𝑚
𝑅𝑚 =1
2 𝑅𝑠 + 𝑅1 = 0.516 𝑚
Then, 𝑆 =𝑛2𝑉𝑚
2
𝑅𝑚4 3 =
(0.012)2(6.348)2
(0.516)4 3 = 0.014
𝐸𝑠 = 𝑦𝑠 + 𝑉𝑠
2
2𝑔= 0.5 +
(7.872)2
2𝑔= 3.658 𝑚
𝐸1 = 𝑦1 + 𝑉1
2
2𝑔= 0.816 +
(4.824)2
2𝑔= 2.002 𝑚
Thus, ∆𝒙 =𝑬𝒔−𝑬𝟏
𝑺−𝑺𝒐=
𝟑.𝟔𝟓𝟖−𝟐.𝟎𝟎𝟐
𝟎.𝟎𝟏𝟒−𝟎.𝟎𝟎𝟎𝟗= 𝟏𝟐𝟔. 𝟒𝟏𝟐 𝒎
c.) Type of profile
Since 𝑦𝑁 > 𝑦𝑐 > 𝑦 ,
∴ 𝑻𝒉𝒆 𝒑𝒓𝒐𝒇𝒊𝒍𝒆 𝒊𝒔 𝑴𝟑
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,
𝑄 =1.48
𝑛𝐴𝑅2/3𝑆1/2
400 =1.49
0.017x10𝑦𝑛
10𝑦𝑛
10+2𝑦𝑛
2
3(0.020)
1
2
By trial and error:
𝑦𝑛 = 2.36 𝑓𝑡.
Critical depth: 𝑦𝑐 = 𝑄2
𝐵2𝑔
1
3=
4002
10232.2
1
3= 3.67 𝑓𝑡
Since yn < yc , the flow is supercritical. The head required on the weir to discharge:
𝑄 = 𝐶𝑤𝐿𝐻3
2
400 = 3.80x10 +
400
5+ x10 2
64.4
3
2
By trial and error, = 4.53 𝑓𝑡 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:
𝑦2 = −2.36
2+
2.362
4+
2 400
23.6
2x2.36
32.2
1/2
= 5.42 𝑓𝑡
The distance from the weir to the jump
𝑦𝐴 = 5.42 𝑓𝑡 𝑉𝐴 = 400/54 = 7.39𝑓𝑡/𝑠𝑒𝑐 𝑉𝐴2 2𝑔 = 0.85𝑓𝑡
𝑦𝐵 = 9.53 𝑓𝑡 𝑉𝐵 = 400/97.4 = 4.20𝑓𝑡/𝑠𝑒𝑐 𝑉𝐵2 2𝑔 = 0.27𝑓𝑡
𝑉𝑎𝑣𝑒 = (VA + VB )/2 = 5.74 ft/sec
𝑅𝑎𝑣𝑒 = 2.95 ft
𝑆 = 0.017x5.79 2 (1.486x2.490)4 3 = 0.00104
𝑥 = 5.42+0.85−9.53−0.27
0.00104−0.02= 186.18 ft
5 ft
4.53 ft
5.40 ft 2.36 ft
EGL
𝑉2
2𝑔= 0.85 𝑓𝑡 𝑉2
2𝑔= 0.26 𝑓𝑡 𝑉2
2𝑔= 4.45 𝑓𝑡
186.18 ft
2
y2 = 1.80m
y1 = 1.20m
1
4m 4m
𝐴23 𝐴22
𝐴13
𝐴12
1.80
1.20 𝐴11
𝐴21
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:
𝐴1 = 𝐴𝑛 + 2𝐴12
𝑥
𝑦=
1
2
𝑥1 = 1
2𝑦1 = 0.60 𝑚
𝐴1 = 4 1.20 + 2(1
2 × 0.60 + 1.20)
𝐴1 = 5.52 𝑚2
𝐴2 = 𝐴21+ 2𝐴22
𝑥
𝑦=
1
2
𝑥2 = 1
2𝑦2 = 0.90 𝑚
𝐴2 = 4 1.80 + 2(1
2 × 0.90 × 1.80)
𝐴2 = 8.82 𝑚2
𝐴1𝑦𝑐1= 𝐴11𝑦𝑐1 1
+ 2𝐴12𝑦𝑐1 2
𝐴1𝑦𝑐1= 1.20𝑥4 0.60 +
2(1
2𝑥 0.60 𝑥 1.2)(0.40)
𝐴1𝑦𝑐1= 2.88 + 0.288
𝐴1𝑦𝑐1= 3.168 𝑚2
𝐴2𝑦𝑐2= 𝐴21𝑦𝑐2 1
+ 2𝐴22𝑦𝑐2 2
𝐴2𝑦𝑐2= 4 1.8𝑥 0.90 + 2
1
2𝑥0.90𝑥1.80 𝑥0.60)
𝐴2𝑦𝑐2= 7.452 𝑚2
𝑉12 =
9.81(8.82)
5.52
7.452−3.168
8.82−5.52 = 20.349
𝑉1 = 4.511 𝑚/𝑠
a.) Flow rate in canal
𝑄1 = 𝐴1𝑉1 = 5.52 4.511
𝑸𝟏 = 𝟐𝟒. 𝟗𝒎𝟑/𝒔
b.) Power Lost, 𝑃 = 𝛾𝑄𝐿
𝐿 = 𝐸1 − 𝐸2
where 𝐸1 = 𝑦1 +𝑉1
2
2𝑔= 1.2 +
4.5112
2𝑔= 2.237𝑚
𝐸2 = 𝑦2 +𝑉2
2
2𝑔= 1.8 +
2.8232
2𝑔= 2.206𝑚
Then, 𝐿 = 2.237 − 2.206 = 0.031𝑚
Therefore,
𝑃 = 𝛾𝑄𝐿
𝑃 = 9.81(24.9)(0.031)
𝑷 = 𝟕. 𝟓𝟕𝟐 𝒌𝑾