Hydraulics and advanced hydraulics _open channels

Lecture 16: Weirs and spillways Associate Prof. Sameh Kantoush
WEUEF12: Hydraulics/ Advanced Hydraulics
Associate Prof. Dr. Sameh Kantoush
Disaster Prevention Research Institute DPRI, Kyoto University
Email of the course instructor: sameh.kantoush@stl.pauwes.dz,
kantoush@yahoo.com
Module III: Hydraulic Structures and
Flow Measurements
March 31, 2022
Lecture 16: Weirs and spillways

Slide 2
Three gorges dam, Yangtze river, China
The world’s recent largest dam
Some weirs in the world
https://www.wsj.com/articles/floodi
ng-again-threatens-chinas-three-
gorges-dam-11597847951

Slide 3
Hoover arch dam, Colorado river, USA
Largest dam at the time
of its completion in 1935
https://graylinelasvegas.com/blog/yo
ur-guide-visiting-hoover-dam/

Slide 4
Son La dam, Da river, Vietnam
Largest hydropower in Vietnam
https://vietnamnet.vn/vn/kinh-
doanh/dau-tu/giai-ma-bi-an-ve-qua-
trinh-tim-noi-dat-dap-thuy-dien-son-la-
466564.html

Slide 5
Wivenhoe dam, Southeast Queensland, Australia

P1
P1
P
Slide 6
Basic parameters of weir-spillway
- Weir height: P and P1
- Length of weir: b (∑b = nb)
- n: number of chambers
- Width of upstream channel: B
- Weir thickness: δ
- Overfall head: H, Ho=H+Vo
2/2g
- Approaching velocity
- Water depth of downstream channel: hd.
- Submerge depth: hs
- Difference between upstream and downstream heads: Z
Note: Critical depth: hk = yc

H
= δ0
δ
Slide 7
Classification of weir-spillway
1. Respect to weir width δ.
 Sharp-crested weir: δ < δ0
Resistance of the weir against the flow is
negligible. The flow is under effect of gravity
δ0 = 0.67H
 Ogee weir: δ0 < δ < (2-3)H
There is resistance of the spillway to
the flow but the flow is still controlled
by gravity
 Broad-crest weir: (2-3)H < δ < (8-10)H
Weir thickness strongly affect the
flow. Gravitational flow occurs only
at the entrance
hd

Slide 8
2. Respect to the shape.
perpendicular
Curved (arch)
Circular (well)
Rectangular
Trapezoidal
Triangle (V-Notch)
circular
3. Respect to location of crest in plane.
oblique
straight
Lateral (side)

Slide 9
5. Respect to overfall condition above the crest
Free overfall weir Submerged overfall weir
hh
hn
hs
4. Respect to the location in river
Suppressed (without
side contraction) weir
Contracted (side
contraction) weir

Slide 10
Sharp-crested weir – rectangular shape
1. Standard sharp-crested weir: rectangular, free overfall,
without side contraction
3/2
0 (21.1)
2
Q m b gH
=
2
1
0.003
0.405 1 0.55( )
H
o
H
m
H P
 
 
= + +
 
 
+
  
m0 is coefficient of discharge
H: overfall head
P1: weir height in the upstream side
P1
P

Slide 11
2. Sharp-crested weir: rectangular, free overfall, with side
contraction
3/2
(21.2)
2
c
Q m b gH
=
2
2
1
0.003
0.405 0.03 1 0.55 .
c
B b b H
m
H B B H P
 
 
−
   
 
= + − +  
 
  +
   
 
 
 
mc: is coefficient of discharge
B: channel width
b: weir length

Slide 12
3. Submerged overfall weir
The flow over the weir is submerged
when satisfied 2 conditions:
hh
hn
hs
hd
P
H
critical critical
Z Z Z
0.65 0.7
P P P
   
< ≈ →
   
   
hd > P ( hs >0 )
and
 Submerged overfall flow, without side contraction
3
2
s o
Q m b 2gH (21.3)
= σ
 Submerged overfall flow, with side contraction
3
2
s c
Q m b 2gH (21.4)
= σ
s 3
s
h Z
1.05 1 0.2
P H
 
σ
= +
 
 
σs: coefficient of submerged overfall flow

Slide 13
Example
A rectangular sharp-crested weir, having b=0.5m;
P=P1=0.35m; width of upstream channel B=1.2m; water depth in downstream
channel hd=0.55m.
1. Find overfall discharge when H=40cm.
2. Find Hmax if Qma x= 300ℓ/s
Solution:
d s d
s
critical
1/3
s
c
h 0.55m P 0.35m h h P 0.55 0.35 0.2m
H h
Z 0.4 0.2 Z
0.57 0.65 0.7
P P 0.35 P
0.55 0.35 0.2
1.05 1 0.2 0.928
0.35 0.4
0.003 1.2 0.5 0.5
m 0.405 0.03 1 0.55
0.4 1.2 1.2
• = > = → = − = − =
− −  
• = = = < ≈ →
 
 
−
  
σ
= + =
  
  
−
   
= + − +
   
   
2 2
3/2 3/2 3
s c
0.4
0.406
0.4 0.35
Q m b 2gH 0.928 0.406 0.5 4.43 0.4 0.211m / s
Q 211 / s
 
 
=
 
 
+
 
 
 
=
σ = × × × × =
= 
submerged
flow
1.

Slide 14
 Free overfall flow, with
side contraction
Use trial and error method
2. Find Hmax if Qma x= 300ℓ/s
 Hypothesize σs = 1 (free flow); mc = 0.405
2/3 2/3
s c
s
critical
s
2 2
c
Q 0.3
H 0.482m
1 0.405 0.5 4.43
m b 2g
H h
Z 0.48 0.2 z
check 0.8 0.65 0.7
P P 0.35 P
1
0.003 1.2 0.5 0.5 0.48
m 0.405 0.03 1 0.55
0.48 1.2 1.2 0.48 0.35
   
= = =
   
  × × ×
σ  
 
− −  
= = = > ≈ →
 
 
σ =
 
−
     
= + − +

     
+
     


2/3 2/3
s c
0.406
Q 0.3
H 0.481m
1 0.406 0.5 4.43
m b 2g
H 0.481m
=



   
= = =
   
  × × ×
σ  
 
=

Slide 15
Sharp-crested weir – triangle and trapezoidal shapes
1. Triangle shape
5
2
Q 1.4H (21.6)
=
( )
5/2
triangle
Q m 2gH 21.5
=
If θ = 900
Eq. (21.6) is used to measure the discharge very precisely
2. Trapezoidal shape
( )
3/2
trapezoidal
Q m b 2gH 21.7
=
If tgθ = ¼: mtrapezoidal = 0.42 
3
2
Q 1.86bH (21.8)
=

Slide 16
Ogee weir
1. General equation
3/2
s o )
Q m b 2gH (21.9
= σ ε Σ
- Free flow: σs=1. submerged flow: σs<1.
- Without side contraction: ε =1. with side contraction: ε <1.
- Coefficient of discharge (m) depends on H and geometry of
the entrance of the weir
2. Determine ε [ ] o
sa ma
H
1 0.2 (n 1) )
(21.10
nb
ε = − ξ + − ξ
Side abutment middle abutment
sa
ξ
ma
ξ
squared rounded
oblique
rounded
squared angle curved
Side abutment
middle abutment
n = 3 chambers
 2 middle
abutments

Slide 17
Ogee weir
3. Determine σs
The overfall flow is submerged when satisfied both 2 conditions
critical
Z Z
P P
 
<  
 
hd > P ( hs >0 ) and (Z/P)critical ≈ 0.7÷0.75
(Z/P)critical is obtained from here

Slide 18
Ogee weir
3. Determine σs
- When hs ≤ 0 (hd ≤ P): free overfall→ σs = 1
- When hs > 0 and (Z/P)<(Z/P)critical :
submerged overfall
→ σs = f(hs/Ho) <1
σs
0
s
h
H Weir without vacuum Weir with vacuum

Slide 19
Ogee weir
4. Determine m
tan (21.11)
s dard shape H
m m σ σ
=
 mstandard : is the coefficient of discharge when the weir
has a standard shape and H=Hdesign.
- Weir type 1: mstandard = 0.504
- Weir type 2: mstandard = 0.48
e
P1
Coordinates of the standard weir
according to Corigio-Ophixerop
Type 1 Type 2
Type 1 Type 2
Table to determine R

Slide 20
Ogee weir
4. Determine m
tan (21.11)
s dard shape H
m m σ σ
=
 σshape : modified factor when the actual shape of weir is
different from the standard one
σshape = f(α, β, e/P1).
 σH : modified factor of overflow head H ≠ Hdesign
(designed overflow head).
e
P1

Slide 21
3 basic problems of Ogee weir
1. Type 1: find Q.
Known: H, b, n, P, P1, hd, Bupstream (V0)
2. Type 2: find H.
Known: Q, b, n, P, P1, hd, Bupstream (V0)
3. Type 3: find b.
Known: H, Q, n, P, P1, hd, Bupstream (V0)

Slide 22
Ogee weir
1. Type 1: find Q
Ogee weir type 2 without vacuum, having α=45o; β=60o; e/P1=0.9 ;
Hdesign=2m. The weir has 5 chambers, width of each b=10m,
rounded side and middle abutments.
Calculate Q when a) H = Hdesign and b) H = 2.6m; Downstream
water head is below the weir crest, ignore approaching velocity
Vo.
Solution:
Free overfall flow with side contraction.
3/2
0
s tan dard shape H shape s tan dard
sa ma 0
sa ma
0 0
Q m b 2gH
m m ; 0.978; m 0.48
(n 1) H
1 0.2 . with 0.7; 0.45
n b
H H because V 0
=
ε Σ
= σ σ σ = =
ξ + − ξ
ε = − ξ = ξ =
= =

Slide 23
Ogee weir
1. Type 1: find Q
H
sa ma 0
3/2
0
3/2 3
1 m 0.978 1 0.48 0.4694
(n 1) H
1 0.2 .
n b
0.7 (5 1) 0.45 2
1 0.2 0.98
5 10
Q m b 2gH
0.98 0.4694 5 10 2 9.81 2 288.16m / s
σ = ⇒ = × × =
ξ + − ξ
ε = −
+ − ×
=− × =
⇒ =
ε Σ
= × × × × × × =
H
design
3/2 3
H 2.6
1.3 1.0275
H 2
m 0.978 1.0275 0.48 0.4823
0.7 (5 1) 0.45 2.6
1 0.2 0.974
5 10
Q 0.974 0.4823 5 10 2 9.81 2.6 435.95m / s
= = ⇒ σ=
⇒ = × × =
+ − ×
ε = − × =
⇒= × × × × × × =
a) H = Hdesign=2m
b) H = 2.6m

Slide 24
Ogee weir
2. Type 2: find H
Ogee weir type 1 without vacuum, having: α = 15o, β=60o, e/P1=0.6;
n=8, b = 10m, rounded side and middle abutment. Hdesign = 3m.
When upstream water level Zupstream = +31m, the discharge Q =
1200 m3/s. Downstream channel’s bed level is +18m and
downstream water level is +25m (ignore approaching velocity).
Determine the crest level Zcrest
Solution: Trial and error method
Zcrest = Zupstream- H and hd = Zdownstream- Zbed= 25 – 18 = 7m
Assume hd < P  free overfall flow 
In which ε, m depend on H
1 round: assume m = 0.458; ε = 0.96
2/3
0
Q
H ( )
m b 2g
=
ε Σ
2/3 2/3
0
Q 1200
H ( ) ( ) 3.9m
m b 2g 0.96 0.458 8 10 2 9.81
⇒
= = =
ε Σ × × × ×

Slide 25
Ogee weir
2. Type 2: find H
Check assumption: α = 15o, β=60o, e/P1=0.6 ⇒ σshape=0.902; mstandard=0.504
m = 0.465 ≠ massume = 0.458; ε = 0.962 ≠ εassume = 0.96
2 round: assume m = 0.465; ε = 0.962
m ≈ massume ; ε ≈ εassume .
 H0=3.85m. Ignore V0  H=H0=3.85m
⇒ Zcrest = Zupstream – H = +31 – 3.85 = +27.15m (Zcrest >Zdownstream  free flow).
H
design
H 3.9
1.3 1.023 m 0.902 1.023 0.504 0.465
H 3
0.7 (8 1) 0.45 3.9
1 0.2 0.962
8 10
= = ⇒ σ= ⇒ = × × =
+ − ×
ε = − × =
2/3 2/3
0
Q 1200
H ( ) ( ) 3.85m
m b 2g 0.962 0.465 8 10 2 9.81
⇒
= = =
ε Σ × × × ×
H
design
H 3.85
1.285 1.022 m 0.902 1.022 0.504 0.465
H 3
0.7 (8 1) 0.45 3.85
1 0.2 0.963
8 10
= = ⇒ σ
= ⇒
= × × =
+ − ×
ε = − × =

Slide 26
Ogee weir
3. Type 3: find b
Ogee weir type 1 without vacuum, having standard shape, α
= 60o, P = P1 = 8m. n=6, rounded side and middle abutment,
Hdesign= 2.5m. hd = 2.8m, V0 = 0.6m.
Find b when H = 3m and Q = 409 m3/s.
Solution:
hd < P  free overfall flow
With H = 3m; V0=0.6m/s ⇒ Ho
3/2
0
Q
b
mn 2gH
ε =
2 2
0
0
V 0.6
H H 3 3.02m
2g 19.62
α
=
+ =
+ =

Slide 27
Ogee weir
3. Type 3: find b
H
design
shape s tan dard s tan dard shape H
H 3
1.2 1.020
H 2.5
1, m 0.504 m m . . 0.504 1 1.02 0.5141
= = → σ
=
σ = = ⇒= σ σ
= × × =
3/2
409
b 5.704
0.5141 6 2 9.81 3.02
⇒ ε =
× × ×
sa ma 0
sa ma
0
sa ma
(n 1) H
1 0.2 .
n b
(n 1)
b b 0.2 .H
n
with 0.7; 0.45 and n 6
ξ + − ξ
∴ε= −
ξ + − ξ
⇒ =
ε +
ξ= ξ = =
0.7 (n 1) 0.45
b 5.704 0.2 3 5.704 0.297 6m
6
+ − ×
⇒
= + ×
= + =

Slide 28
Broad-crested weir
1. Conditions to determine the flow types
hs
s s
k k critical
h h
1.2 1.4
h h
   
> =
÷
   
   
s s
0 0 critical
h h
0.7 0.85
H H
   
> =
÷
   
   
 submerged flow when
or
hd
P
P1
 free flow when
s s
k k critical
h h
1.2 1.4
h h
   
< =
÷
   
   
s s
0 0 critical
h h
0.7 0.85
H H
   
< =
÷
   
   
or

Slide 29
Broad-crested weir
2. Equations of computation: rectangular weir
c
c
o
o
H
hc
c
c
o
o
H
hc
a. Free flow
3/2
o (21.12)
Q mnb 2gH
=
With sill,
without side
contraction: mα
m: coefficient of discharge
Without sill,
with side
contraction:
mβ

Slide 30
2. Equations of computation: rectangular weir
b. Submerged flow
s o
Q bh 2g(H h) (21.13)
=
ϕ −
( )
2
o s o
V
H h 1 k V 2g(H h)
2g
= + + ∑ → = ϕ −
Write energy equation for section (0-0) and (2-2)
The general equation for weir is not suitable!
hs
hs= hd
Broad-crested weir

Slide 31
Broad-crested weir
s
1
1 k
ϕ =
+ ∑
: Coefficient of velocity of submerged flow
d d
2
V (V V )
Z
g
−
=
h = hs - Z2 With
Z2 is relatively small  can be ignore in calculation
n s o s (21.14)
Q bh 2g(H h )
=
ϕ −
→
Z2 = 0
If the weir is not rectangular, replace bh in Eq. (21.13) and (21.14) by A
s o
Q A 2g(H h) (21.13b)
=
ϕ − n o s )
Q A 2g(H h ) (21.14b
=
ϕ −
ϕs

Slide 32
Some examples of sluice gates

Slide 33
Definitions
0
Ho
Vo
0
H
a
I
C
hc
2
α
hk hh
Gate
Gate
Gate opening
hd
H.J
H: upstream water head
V0: approaching velocity
a: gate opening
hc: water head at contraction section
hk: critical depth (yc)
hd: water head at downstream channel

Ho
Vo
2/2g
H
a
c
c
hd
Thân cống Cửa ra
Cửa vào
Gate
hc
Slide 34
Flow characteristics
Ho= H+Vo
2/2g
From experiments, when a/H < 0.75, the flow after the gate
is supercritical.
hc=ε.a
The flow at the contraction section is connected with the
downstream flow via supercritical flow when hd < hk, and via
hydraulic jump when hd > hk
Inlet Outlet
Sluice gate
body/foundation

Slide 35
Flow modes via sluice gates
S3 hh
Gate
Inlet Outlet
1. When hd < hk ↔ no hydraulic jump →
→ Free flow
2. When hd > hk : with hydraulic jump.
a. If hd < h”c→ repelled hydraulic
jump → Free flow
b. If hd = h”c→ free hydraulic jump,
starting at C-C → Free flow
c. If hd > h”c → Submerged hydraulic
jump
→ Submerged flow
M3
hh
Gate
Inlet
Outlet
hd
Gate
Inlet Outlet
hc and h”c: conjugate depths
h”c: sequent depth of hc through H.J

Slide 36
Basic equations of computation
1. Free flow
M3
hh
Gate
Energy equation for (1-1) and (c-c):
2 2 2
o c c
c
2
c
o c c
c o c
c o c c o c
V V V
H h k
2g 2g 2g
V
H h (1 k) ; h a
2g
1 1
V 2g(H h ) ;
1 k 1 k
V 2g(H h ); Q A 2g(H h )
+ = + + Σ
= + + Σ = ε
→
= − ϕ
=
+ Σ + Σ
→ =
ϕ − =
ϕ −
1
1
For rectangular shape c c c o c
Q bh 2g )
A (H h )
bh (22.1
= ϕ
= → −
ϕ: coefficient of velocity ( with ∑k ≈ 0.1÷0.2→ ϕ ≈ 0.9 ÷ 0.95 )

Slide 37
Basic equations of computation
2. Submerged flow: hd > hc’’
hz
hd
1
1
2
2
Energy equation for (1-1) and (c-c):
2
c
o z
c o z
c o z
V
H h (1 k)
2g
1 1
V 2g(H h );
1 k 1 k
Q A 2g(H h )
= + + Σ
→
= − ϕ
=
+ Σ + Σ
→ =
ϕ −
For rectangular shape c c c o z
Q bh 2g )
A (H h )
bh (22.2
= ϕ
= → −

Slide 38
Equations to calculate hz
Momentum equation for (c-c) and (2-2):
2
2 d c
z d
d
2
2 2 2
d
z
2 c
d c c
h
h Q Q Q
b b Q(V V ) ; q (22.3)
q
2 2 bh bh
(h h )
2
h h
b g h h
ρ ρ  
γ − γ = ρ −

−
=
=
− →
  −
=

 Eq. (22.3) is used in the problem of finding H (known Q and a)
 Eq. (22.4) is used in the problem of finding Q (known H and a)
 Eq. (22.5) is used in the problem of finding a (known Q and H)
2 2
2 2
2 2 2 2
d c c 0
d
c
2
z
2 c d c
z d
z d z d d
d c d c d
o
(h h ) 2 h 2g(H h )
2q 2q 1 1 1 1
h h h h h
g h h g h h g h
H
h
(2
h (h h )
M 4
M
2.4
M
h h M )
h
2 4
   
− ϕ −
= − → = − − = − −
−
   
   
→

−
 
=
+ −
  =

ϕ
Rearranging Eq. (22.3)
z
2
2
d
2
z o d
h q
A H h
8 2q
A và B=h
g
0 (22
B .5)
gh
 
 
−

+ ϕ
− =
=

+

Slide 39
3 basic problems of sluice gate
1. Type 1: find Q.
Known: H, a, hd, b, Bupstream, ϕ
2. Type 2: find H.
Known: Q, a, hd, b, Bupstream, ϕ
3. Type 3: find a.
Known: Q, H, hd, b, Bupstream, ϕ

Slide 40
Example 1
1. Type 1: find Q. (Known: H, a, hd, b, Bupstream, ϕ)
Calculate the discharge flowing through a rectangular sluice
gate, having: b= 3m; Bupstream= 5m; H= 4m; a= 1m; ϕ=0.9 in two
cases: 1. hd= 2m; 2. hd= 3m. Neglect V0.
Solution
c
a 1
0.25 0.622 h .a 0.622 1 0.622m
H 4
= = → ε = → = ε = × =
Determine flow mode
Assume free flow
2
c c
2 2
'' c
c 3 3
c
q h 2g(H h ) 0.9 0.622 4.43 (4 0.622) 4.56m / s
h q 0.622 4.56
h 1 1 8 1 1 8 2.32m
2 gh 2 9.81 0.622
→ ≈ ϕ − = × × × − =
   
→ = − + + = − + + =
   
   
×
   

Slide 41
Example 1
1. Type 1: find Q. (Known: H, a, hd, b, Bupstream, ϕ)
 Case 1: hd= 2m
hd= 2m < h”
c → free flow → Q=q.b=13.68m3/s
 Case 2: hd= 3m
hd= 3m > h”
c → submerged flow→ find hz
2 2 c d c
z d o
d
2 2
z
2 3
c z
h (h h )
M M
h h M H (22.4) M 4
2 4 h
0.622(3 0.622) 1.6 1.6
M 4 0.9 1.6 h 3 1.6 4 2.6m
3 2 4
q h 2g(H h ) 0.9 0.622 4.43 (4 2.6) 2.93m / s Q 8.8m / s
−
 
= + − − =
ϕ
 
 
−  
= × × = → = + − × − =
 
 
→ ≈ ϕ − = × × × − = → =

Slide 42
2. Type 2: find H. (Known: Q, a, hd, b, Bupstream, ϕ)
Example 2
Calculate the upstream water head of a rectangular sluice gate,
having: b= 3m; V0≈ 0m; a= 1m; ϕ=0.9; Q = 14 m3/s in two
cases: 1. hd= 2m; 2. hd= 3m
Solution
- Assume εgt = 0.63 (normally in range ε ≈ 0.62÷ 0.63 )
- Determine flow mode
2
c
2 2
'' c
c 3 3
c
h .a 0.63 1 0.63m; q Q / b 14 / 3 4.67m / s
h q 0.63 4.67
h 1 1 8 1 1 8 2.36m
2 gh 2 9.81 0.63
=
ε = × = = = =
   
→ = − + + = − + + =
   
   
×
   

Slide 43
Example 2
 Case 1: hd= 2m
hd= 2m < h”
c  free flow
2 2
o c 2 2 2 2
c
q 4.67
H H h 0.63 4.09m
2gh 0.9 19.62 0.63
→ ≈ = + = + =
ϕ × ×
''
c c
2
a 1
0.244 h 0.622m h 2.38m
H 4.09
stillfreeflow
a
0.
0.622
0.
2 62
. 4
H
H 4 17m
= = → → =
→
ε =
ε =
→ →
=
→ =
→ =
- Checking the assumption

Slide 44
Example 2
 Case 2: hd= 3m
hd= 3m > h”
c  submerged flow
2 2
2 2
d c
z d
d c
2 2
0 z 2 2 2 2
c
(h h )
2q 2 4.67 (3 0.63)
h h 3 1.85m
g h h 9.81 3 0.63
q 4.67
H H h 1.85 5.3m
2gh 0.9 19.62 0.63
− × −
→ = − = − =
×
≈ = + = + =
ϕ × ×
- Checking the assumption
''
c c
z 2
a 1
0.188 h 0.62m h 2.39m
H 5.3
stillsubm
9
ergedflow
a
h 1.82m 0.185
0.62
0
H 5.3
H
m .6
= = =
ε =
=
→ → = →
→
→ = ε =
→ → = →

Slide 45
Example 3
3. Type 3: find a. (Known: Q, H, hd, b, Bupstream, ϕ)
A rectangular sluice gate, having: b= 3m; Bupstream=5m; ϕ=0.9;
Q = 16 m3/s; H=5m. Determine the gate opening in two cases:
1. hd= 2m; 2. hd= 3m. Neglect V0.
Solution
2
q Q / b 5.33m / s
= =
c c c c
2 2
'' c
c c 3 3
c
q h 2g(H h ) 5.33 0.9.h 2g(5 h )
h q 0.64 5.33
h 0.64m h 1 1 8 1 1 8 2.7m
2 gh 2 9.81 0.64
→ =
ϕ − → = −
   
→ = → = − + + = − + + =
   
   
×
   
- Assume free flow
c
m
h 0.64 a
0.128 1
0.21
H 5 H
a .05
= =
= → = →
 Case 1: hd= 2m
hd= 2m<h”
c → free flow

Slide 46
Example 3
3. Type 3: find a. (Known: Q, H, hd, b, Bupstream, ϕ)
 Case 2: hd= 3m
hd= 3m>h”
c=2.7m → submerged flow→ Find hz.
2
2 2
z o z d
d
2
z z z
c
c
o z
8 2q
h A H h B 0 (22.5) A q và B=h
g gh
8
A 0.9 5.33 4.33 và B 10.93 h 4.33 5 h 10.93 0 h 1.7m
9.81
h
q 5.33 0.736
h 0.736 0.147
H 5
2g(H h ) 0.9 19.62 (5 1.7)
a
0.24 a 1.2m
H
 
+ − − = =
ϕ +
 
 
= × × = = → + − − = → =
= = = → = =
ϕ − × × −
→ = → =

ً‫ا‬‫ﺷﻛر‬
Vielen Dank
Thank you for your attention
ありがとうございます

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