g v 2 2 1 y 1 S o L 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 . h L =SL g v 2 2 2 y 2 EGL Slope= S Channel bed, slope = S o ∆x 1 2 Chapter V STEADY NON-UNIFORM FLOW OR VARIED (S≠ S O )FLOW IN OPEN CHANNELS
A brief discussion about Open Channel (non uniform flow) and some sample problems with answers for exercises.
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
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 V
STEADY 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 ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 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
Qq
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
31
1
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)
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+ ℎ𝐿
𝑦1 + 𝑉1
2
2𝑔= 𝑦2 +
𝑉22
2𝑔 + ℎ𝐿
𝐸1 = 𝐸2 + ℎ𝐿
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.
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,
ℎ𝐿 = 𝐸1 − 𝐸2
ℎ𝐿 = 4.1 − 1.44 = 2.66 𝑚
Thus,
𝑃 = 𝛾𝑄ℎ𝐿
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:
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