siphon&HTI 2

Minia University
Civil Eng. Dept.
‫ـــعة‬‫م‬‫جا‬
‫يا‬‫ن‬‫مل‬‫ا‬
Flow through Gate or Siphon
By: Dr. Ezzat El sayed G. SALEH

”
‫داء‬ ‫للك‬
‫به‬ ‫سطاب‬ُ‫ي‬ ‫اء‬‫و‬‫د‬ ‫هل‬
..........
‫ال‬‫ا‬
‫يداوهيا‬ ‫من‬ ‫يت‬‫ع‬‫أ‬ ‫امحلاقة‬
“

Sluice Gate – Old Assuit Barrage

 (2)
Sluice Gate Geometry
Sluice gate of width “W”
 (1)
y1
y2
HGL
EGL
yG
Sluice gates are often used to
regulate and measure the flow
rate in open channel
yG is the gate
opening
 (1)
Sectional Elevation Through a Sluice Gate

 (2)
Sluice gate of width “W”
 (1)
y1
y2
HGL
EGL
yG
yG is the
gate opening
 (1)
Vena-contracta
Taking the bed of the channel as a reference datum

y2
y1
yG
yG is the gate opening
 (1)
 (2) y3
y

y2
y1
yG

y3
Sectional Elevation Through a Radial Gate
 (1)
 (2)
lip

y2
y1
yG

y3
Sectional Elevation Through a Drum Gate
 (1)
 (2)

y2
y1
yG

Sectional Elevation Through a Drum Gate
 (1)
 (2) y3
y

 (1)
 (2)
HGL
EGL
V2/2g
V2/2g
Summit
Z
Siphon

d
 (1)
(2)
Vsiphon
H
Reference Datum (3)
VOrifice
h

H - h

d
Vsiphon
( 3)
VOrifice
h
H


(1)
Reference Datum
(2)

Flow Rates
₪ Volume flow rate: is the volume of fluid flowing past a
section per unit time.
₪ Mass flow rate: is the mass of fluid flowing past a
₪ Weight flow rate: is the weight of fluid flowing past a

GB
Units
SI
Units
Definition
Symbol
Name
ft3/s
m3/s
= A  V
Volume flow rate
Slug/s
Kg /s
=  Q
=  (A  V)
M
Mass flow rate
Ibf/s
N /s
=  g Q
=  g (A  V)
W
Weight flow rate
Flow Rates
Q

Atank
H1
H2
dh
H
g
2
A
C
Q orifice
d
out 


0
Qin 
dh
A
dt
.
Q tank
out 


dh
.
Q
A
dt
t
0
H
H
tank
2
1
 


dh
.
h
g
2
A
.
C
A
dt
t
0
H
H
5
.
0
orifice
d
tank
2
1
 




Qout

Atank
H1
H2
dh
g
2
V
D
L
f
h
2
)
f
(
L 

dh
A
dt
Q tank


.
dh
Q
A
dt
t H
H
tank
.
0
2
1
 


dh
.
h
g
2
A
D
L
f
5
.
1
A
dt
t
0
H
H
5
.
0
pipe
tank 2
1
 




0
Qin 
g
2
V2
g
V
2
5
0
2
.












D
L
f
5
.
1
g
2
V
g
2
V
g
2
V
D
L
f
g
2
V
5
.
0
h
2
2
2
2
5
.
0
h
.
D
L
f
5
.
1
g
2
V


5
.
0
.
5
.
1
2
.
. h
D
L
f
g
A
V
A
Q pipe
pipe




Q
F, L, D
V : the velocity through the pipe

5 cm
5 cm
5 cm 0.4 cm dia.
Water flows from a 10-cm diameter container that contains three holes as
shown in the figure. The diameter of each fluid stream is 0.4cm, and the
distance between holes is 5cm. If viscous effects are negligible and quasi-
steady conditions are assumed.
 Determine the time at which the water stops draining from the top hole.
 Compare it with the time required to stop draining from the container.
Quasi-steady flow: means that at any
moment of time, we can solve the
problem as if the flow were steady.

Irrigation canal &
siphon tubes

28
Comparison between Absolute and Gage Pressures

Barometer: instruments used for measuring
atmospheric pressure.
At sea level the average atmospheric pressure
is:
101.3 kPa = 760 mm of mercury
= 14.7 psi = 2117 Ib/ft2
Atmospheric Pressure

Absolute Pressure: is the total pressure measured
above zero (perfect vacuum),
Gage Pressure: is the pressure measured above
atmospheric pressure.
Absolute Pressure = gage pressure + atmospheric
pressure
Suppose that the tire gage measure measures 30 psi
(gage),
Absolute pressure = 30 psi + 14.7 psi = 44.7psi,
Absolute and Gage Pressure

4.6 m
A
50mm dia.
25mm dia.
0.9 m
B
water
» The volume flow-rate of
water from the nozzle.
» The pressure at points “A”
and “B”.
For the siphon shown in the
figure, calculate:
Worked Example

4.6 m
A
50mm dia.
25mm dia.
0.9 m
B
water
» Because p1 = p2 = patm. = 0,
the remaining terms are
2
1
2
2
1
2
2
2

















g
V
g
p
Z
g
V
g
p
Z


» Applying Bernoulli’s Equation
between points “1” and “2”
 
 
  s
m
Z
Z
g
V
/
5
.
9
6
.
4
0
81
.
9
2
2 2
1
2








Reference
Datum
» The volume flow rate of water, Q
s
L
V
D
V
A
Q
/
66
.
4
5
.
9
4
025
.
0
4
2
2
2
2
2
2









 














Worked Example

The continuity equation gives:
water
of
m
Z
g
p
g
p A
B
19
.
1
9
.
0
29
.
0 








B
A
g
V
g
p
Z
g
V
g
p
Z 
















2
2
2
2


s
m
D
D
V
V
siph
Siph /
39
.
2
25
50
5
.
9
2
2
.
2
2
. 



















» To determine the pressure at point “A”, apply Bernoulli’s Equation
between points “1” and “A” gives,
water
of
m
g
V
g
p siph
A
29
.
0
81
.
9
2
39
.
2
2
0
2
2









» Similarly, since the siphon cross section is constant (VA =VB), the
pressure at point “B” can be determined as,
2
3
3
/
85
.
2
10
81
.
9
10
29
.
0
m
kN





2
3
3
/
67
.
11
10
81
.
9
10
19
.
1
m
kN





Worked Example

1.0 m
25mm dia.
h
water
» Does changing the
elevation, h, of the siphon
centerline above the water
level in the tank vary the
volume flow-rate through the
siphon? Explain.
» What is the maximum
allowable value of h ?
A siphon is used to draw water
from a large container as
indicated in the figure.
Worked Example

A
7.5cm dia.
water
» Determine the flow rate
involved?
A water siphon having a
constant inside diameter of
7.5cm is arranged as shown in
the figure. If the friction losses
between A and B is 0.8 V2/2g,
where V is the velocity of flow
in the siphon.
1.0 m
1.0 m
3.0 m
B
Worked Example

What throat diameter is needed for a Venturi-meter in a 200cm
horizontal pipe carrying water with a discharge of 10m3/s if the
differential pressure between the throat and upstream section is to
be limited to 200kPa at this discharge? (take Cd = 0.98)
Solution
 4
1
2
2
1
2
2
1
2
4 D
D
g
p
p
g
D
C
Q d
act








 












The basic discharge equation for a Venturi-meter is:
m
g
p
39
.
20
81
.
9
1000
10
200 3






D1 =2.0m
= 0.98
10 m3/s
………….. (1)
Worked Example

Eliminating Eq. (1) gives: 2
2
2
16
8
.
6400
13
D
D


Solving by trial and error:  D2 = 0.80m
………….. (2)

Water flows through a Venturi-meter that has a 30cm throat. The
Venturi-meter is installed in a 60cm pipe. What deflection occur in a
water-mercury manometer connected between the upstream and
throat sections if the discharge of 0.57m3/s. (Cd = 0.98)
Solution
 4
1
2
2
1
2
2
1
2
4 D
D
g
p
p
g
D
C
Q d
act








 












The basic discharge equation for a Venturi-meter is:
D1 =0.60 m
= 0.98
0.57 m3/s
………….. (1)
  









 1
/
2
1


 i
h
g
p
p 
D2 =0.30 m
Worked Example

For the given case,
6
.
13
0625
.
0
16
1
60
.
0
30
.
0
0706
.
0
4
3
.
0
4
4
4
1
2
2
2
2
2
2
























w
m
L
i
D
D
m
D
A







Substituting these values into Eq. (1) gives
 
0625
.
0
1
1
6
.
13
81
.
9
2
0706
.
0
98
.
0
57
.
0







h
 ………….. (2)
Solving for h,  h (the deflection in the manometer) = 25.70cm
Worked Example

What is Pitot tube:
Pitot Tube is a device used for measuring the velocity
of flow at any point in a pipe or a channel. The Pitot
tube was invented by the French Engineer Henri Pitot.
The principle of Pitot tube is based on the Bernoulli's
equation, where each term can be interpreted as a
form of pressure.

Pitot Tube
It is used to measure flow rate at a particular point in a flowing
fluid. Such devices are frequently used as air speed indicators in
aircraft
The Pitot tube consists of a tube with a small opening at the
measuring end. This small hole faces the flowing fluid,
When the fluid contacts the Pitot tube, the fluid velocity is zero
and the pressure is maximum.
This small hole or impact opening, provide the higher pressure for
pressure measurement.
The purpose of the static ports is to sense the true static pressure
of the free airstream.

The true static
pressure hole
Impact Opening with
higher pressure hole
Pitot Static tube

Note: the gray circles outlined in dotted red boxes are Pitot-static ports,
which, combined with an aircraft's Pitot tubes

With no flow, the pressure at
both ports is the same and the
level in each “leg” of the U-tube
is the same.
Flow Sensors are typically a Form of Pitot Tube.
 

The Pitot tube is used to find the Velocity of a fluid. The
parameters for the model include density of fluid “L”,
density of manometer liquid (i), differential head (h).

Laminar flow: slow, organized, parallel to vessel
walls, parabolic profile.
Vortex flow: swirling, often countercurrent eddies
seen at bifurcations and after stenosis
Turbulent flow: disorganized, random, with high
velocities.

Types of Flow
Laminar Flow
Low velocity
Particles flow in straight
lines
No Mixing
Rare (
‫اندرة‬
) in water system
Turbulent Flow
High velocity
Particles move in irregular
paths
Mixing
Most common type of flow.

Additional to Weirs
By: Dr. Ezzat El-sayed Gomaa SALEH

 The weir is one of the oldest, cheapest, most straight
forward and reliable structures for the determination of
flow in channels where sufficient water depth is available.
 A simple weir consists of a structure in wood, metal or
concrete placed perpendicular to the flow in a channel.
The structure has a sharp-edged opening or notch of
specific shape and dimension through which the water
can flow. Weirs are identified by shape as indicated
below.
Weir

Rectangular Shape
Trapezoidal Shape
Triangular Shape
Parabolic Shape 
Circular Shape 
Shapes of Canal

Sharp crested weir:
(a)Rectangular,
(b)Cipoletti or trapezoidal and
(c) Triangular or (V- notch).
(a) (b)
(c)

Typical Weir Shapes
Cipolletti
Weir
V- notch
Weir
Rectangular
Weir

Each type of weir has certain advantages under specific
conditions. Generally, for accurate readings, a standard V-
notch or a rectangular suppressed (parallel sides with no
side contractions) weir should be used. The Cipolletti and
rectangular weirs with full end contractions are especially
suited for water division.
 Normally the observer has a reasonable idea of the
quantities to be measured and taking the following into
account, a choice can be made of the suitable weir for
particular circumstances:
Choice of weir

 The maximum expected water height above the weir must be
at least 60 mm to prevent the nape from adhering to the weir
crest. Furthermore it is difficult to take accurate readings on
the measuring scale if (h) is too low.
 The length of the rectangular and Cipolletti weirs must be at
least equal to three times the water height above the weir
crest.
 The V-notch is the most suitable for measurements smaller
than 100 m³/h.

 The V-notch is as accurate as the other weir types for
flows between 100 and 1 000 m³/h provided that
submergence does not occur.
 The weir crest must be as high above the channel floor
as possible so that free overflow (of the nape) will take
place.
 The flow depth over the weir should not exceed 600 mm.

When the channel feeding the water to the weir is wider than
the weir crest, the sides of the stream will narrow where it
crosses the weir. The width of the stream flowing over the
weir will be slightly narrower than crest width. The
phenomenon is known as end contraction.
 To make allowance for end contractions, the overflow
width in the basic equation must be modified as follows:
L' = L – 0.1 n h (8.4)
where:
L' = effective length of weir [m]
L = measured length of weir [m]
n = number of end contractions
h = overflow depth [m].
End Contractions

A rectangular weir that does not span the whole channel is called a weir
with end contractions. The effective length of the weir will be less than the
actual weir length due to contraction of the flow jet caused by the
sidewalls.
where,
N = the number of contractions, and
L‘ = measured length of the crest.
L = L' – 0.1 n h
L'

A lined channel is provided with a lining of impervious material
on its bed and banks to prevent the seepage of water.
Lined Canal

Advantages of Canal Lining:
It reduces the loss of water due to seepage and hence the
duty is enhanced,
It controls the water logging and hence the bad effects of
water-logging are eliminated,
It provides smooth surface and hence the velocity of flow
can be increased.
Advantage of Canal Lining

Disadvantages of Canal Lining:
 Higher initial investment,
 Repair is costly,
 Shifting of outlet is costly because it involve dismantling
and relaying of lining,
(
‫يلها‬‫ص‬‫تو‬‫و‬ ‫نة‬‫ا‬‫البط‬ ‫تفكيك‬ ‫يشمل‬ ‫نه‬‫ل‬ ‫ملكف‬ ‫اخملرج‬ ‫نقل‬ ‫ن‬‫ا‬
)
 Longer construction period,
 Sophisticated construction equipment and labor is
needed. { ‫هناك‬
‫حاجة‬
‫ىل‬‫ا‬
‫معدات‬
‫بناء‬‫ل‬‫ا‬
‫املتطورة‬
‫العامةل‬‫و‬ }
Dis - advantage of Canal Lining

Pumps in Series
Volume Flow- rate V
.
Performance Curve “one pump”
Performance Curve “two pump”
System Curve
A
B
Head
H
p
When two pumps are placed in series, the resulting pump performance curve is
obtained by adding heads at the same flow-rate. As shown in the figure, for two
identical pumps in series, both the actual head gained by the fluid and the flow-
rate increased, but neither will be doubled if the system curve remains the
same. The operating point is at “A” for one pump and moves to “B” for two
pumps in series.
For two pumps in series,
added heads

Pumps in Parallel
Volume Flow- rate V
.
Performance Curve “one pump”
Performance Curve “two pump”
System Curve
A
B
Head
H
p
For two identical pumps in parallel, the combined performance curve is
obtained by adding flow-rates at the same head. As shown in the figure above,
the flow-rate for the system will not be doubled with the addition of two pumps
in parallel (if the same system curve applies). However, for a relatively flat
system curve (see the shown figure), a significant increase in flow-rate can be
obtained as the operating point moves from point “A” to point “B”.
For two pumps in parallel,
added flow-rates


y
D

 

 sin
8
,
2


D
A
Area
2
,

D
P
perimeter
Wetted 
 



4
sin
,



D
P
A
R
raduis
hydraulic
The h
Where  is in radians
If the section is laid on a constant slope “So”, and the Manning coefficient “n”
 





 

 3
2
3
5
3
2
3
8
2
1 sin
4
8
,



D
S
n
k
Q
discharge
The o
The above equation can be written in terms of the flow depth, y, using
   
 
2
sin
1
2 
 

 D
y
Uniform Flow in a Circular Section


V2 = 6m/s

 
V1 = 4m/s
V3

10 cm
12 cm
90o
The water jets collide and form one
homogenous jet as shown in the
figure. Determine:
» The speed, V, and direction,
, of the combined jet.
» The loss for a fluid particle
flowing from “1” to “3” and
from “2” to “3”. Gravity is
negligible.
Linear Momentum Equation

V2 = 6m/s

 
V1 = 4m/s
V3

10 cm
12 cm
90o
V3 cos 
V3 sin 
s
m
Q /
0314
.
0
4
1
.
0
4
3
2
1 










s
m
Q /
0679
.
0
6
12
.
0
4
3
2
2 










s
m
Q
Q
Q /
1
.
0
0679
.
0
0314
.
0 3
2
1
3 




Applying momentum equation in x-direction gives,
2
2
3
3
3
3
2
2
)
( cos
cos
0 V
Q
V
Q
A
p
A
p
F x 


 





Substituting p2 = p3 = patm (gage) and incompressible flow ( is constant),
….. (1)
2
2
3
3 cos V
Q
V
Q 

 
2
2
2
3
2
3
cos
A
Q
A
Q


or
For Q3 = 0.1 m3/s , Q2 = 0.0679 m3/s




















2
2
3
2
2
2
3
2
3
2
2
4
1
.
0
0679
.
0
cos
D
A
A
A
Q
Q

 (2)

Applying momentum equation in y-direction gives,



















2
1
3
2
2
1
3
2
3
2
1
4
1
.
0
0314
.
0
sin
D
A
A
A
Q
Q


….. (4)
1
1
3
3
3
3
1
1
)
( sin
sin
0 V
Q
V
Q
A
p
A
p
F x 


 




 ….. (3)
Similarly,
Solving Eq. (2) and Eq. (4) for ,
2
2
2
1
2
2
10
.
0
12
.
0
0679
.
0
0314
.
0
0679
.
0
0314
.
0
tan 





























D
D

308
.
0
10
.
0
12
.
0
0679
.
0
0314
.
0
0679
.
0
0314
.
0
tan
2
2
2
1
2
2
































D
D
 and


 12
.
17


y2
y1
yG


y3
 (1)
 (2)
Sectional Elevation Through a Radial Gate

o
5m 35m
20m
5m
Hydrostatic Forces “HTI Dec. 2012”

20m
o
5m 35m
20m
5m
F1
2
cp )
y
(
2
CG )
y
(
CG
cp
5 m
m
77
25
20
25
62
20
y 2
CG .
.
)
( 


m
14
27
62
20
200
77
25
12
62
20
200
77
25
y
3
2
cp .
)
.
(
.
)
/
.
(
.
)
( 





o
1
96
.
75
5
20
tan 






 



N
10
525
.
24
)
5
200
(
)
2
/
5
(
81
.
9
10
A
h
g
F 6
3
1
1
1 






 
The hydrostatic force on the vertical face of the dam, F1
It acts at a vertical distance = ⅔  5 =3.33 m below the water surface
The hydrostatic force on the inclined face of the dam, F2
N
10
85
.
606
)
62
.
20
200
(
)
2
/
20
5
(
81
.
9
10
A
h
g
F 6
3
2
2
2 







 

cos
F
F
2
F
F
F 2
1
2
2
2
1
t 





 = 165.96

Weight
= 250kg
Gate
Hinge
1.0 m
100 kg
45o


45o
500 mm dia-
pinlet = 1000 KN/m2
Linear Momentum Equation “HTI Dec. 2012”

45o
500 mm dia-
pU.S = 1000 KN/m2
2
0
0
m
m
d
i
a
-
Linear Momentum Equation “HTI Dec. 2012)
 A reducing pipe bend is shown in the figure. If the discharge is 500L/s and the
upstream pressure is1000kN/m2, find the magnitude and direction of the force
on the bend.

45o
500 mm dia-
P inlet = 103 KN/m2
V1
V2
p1
p2
P2 cos 45o
P2 sin 45o
C V

Section “2”
Section “1”
P2
in KN
P1
in KN A2 in m2
D2 ( m)
A1 in m2
D1 ( m)
0.0314
0.20
0.196
0.50
???
103
The data given are:
To calculate the pressure at the outlet section P2, we use the
Bernoulli’s equation. The Bernoulli’s equation for a streamline
passing through the centerline of the pipe bend is expressed as
2
2
1
2
g
2
V
g
p
g
2
V
g
p

















The pipe bend is in a
horizontal position

or
The inlet and outlet velocities, V1 and V2, may be determined
using the continuity equation as











g
2
V
g
2
V
g
p
g
p 2
2
2
1
1
2


………… (1)
s
/
m
55
.
2
196
.
0
5
.
0
A
Q
V 1
1 


s
/
m
92
.
15
0314
.
0
5
.
0
A
Q
V 2
2 


and
Then, substituting the given data in Eq. (1),
  water
of
m
35
.
89
92
.
15
55
.
2
81
.
9
2
1
81
.
9
10
10
10
g
p 2
2
3
3
3
2









or
Hence, the pressure forces on the pipe bend ends are:
2
3
2 m
/
kN
53
.
876
35
.
89
81
.
9
10
35
.
89
g
p 




  ……… (2)
P2 . A2 in KN
P1 .A1 in KN
876.530.0314 = 27.52
1030.196 = 196
The reducing pipe bend, surrounded by a suitable control
volume, is shown in the figure. The forces acting in the bend
are due to the change in direction and the reduction in
diameter.

The momentum equations along the x- and y- directions are
)
V
45
cos
V
(
Q
45
cos
A
p
A
p
F 1
2
2
2
1
1
)
x
( 




 


…… (3)
Substituting the given values, Eq. (3) gives
)
0
45
sin
V
(
Q
45
sin
A
p
0
F 2
2
2
)
y
( 



 


(where all terms are taken positive with x- and y- axis)
)
55
.
2
45
cos
92
.
15
(
5
.
0
10
45
cos
10
52
.
27
10
196
F 3
3
3
)
x
( 





 

kN
19
.
172
F )
x
( 


Similarly,

Hence
)
0
45
sin
92
.
15
(
5
.
0
10
45
sin
10
52
.
27
0
F 3
3
)
y
( 




 

and
kN
09
.
25
F )
y
( 


kN
174
09
.
25
19
.
172
F
F
F 2
2
2
)
y
(
2
)
x
(
t 





'
)
x
(
)
y
(
17
8
19
.
172
09
.
25
F
F 



 From negative x-direction to
negative y-direction
Then the total force of the elbow on the control volume is 174 KN
Fy
Fx
Ft


kN
174
F
F t
flange
on
t 



From Newton’s third law, the force exerts on the bend is the negative
of Ft
Fx
Fy Ft


Air
S.g. = 0.80
water
0.85 ft
1.60 ft
2.50 ft
4.0 ft
6
.
0
f
t
The figure shows a container holding
oil and water. Air at 710Ibf / ft2 below
atmospheric pressure is above the
oil. Calculate:
₪
The pressure and force at the
bottom of the container in Ibf /ft2
(gage).
B
The pressure at the bottom of the container
2
f
w
w
oil
oil
air
B
ft
/
Ib
946.13
)
5
.
2
6
.
1
8
.
0
(
2
.
32
94
.
1
710
h
g
h
g
P
P









 

The force at the bottom of the container f
B
B
B Ib
707
22
)
6
4
(
13
.
946
A
P
F 





Hydrostatic Pressure Forces “HTI Dec. 2012)

o 5m 35m
20m
5m
Hydrostatic Forces “HTI Dec. 2012)
A mass concrete dam has the section shown in the figure and spans a channel
200m wide.
I) Estimate the magnitude of the resultant force,
II) Its angle to the horizontal, and
III) The point at which its line of action passes through the base line.
Hydrostatic Pressure Forces “HTI Dec. 2012)

Air
S.g. = 0.80
water
0.85 ft
1.60 ft
2.50 ft
4.0 ft

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