worked examples

Worked Example (2014) 2
Worked Example
Dam
Strut
1.5 m
1.5 m
60o
45o
2.3 m
2.0 m

A
B
C
Using force components,
calculate the force in the strut
“AB” if these struts have 1.5m
spacing along the small dam
“AC”. Consider all joints to be
pin connected. Plan
Hydrostatic Forces (Plane Surface)
Neglecting distance
Solution
60o
45o
2.3 m
2.0 m

A
B
C
FH
FV
D
Fstrut
Fstrut. cos 15o
Fstrut. sin 15o
2 / 3 m
2 / 3 m
E
Fstrut.
15o
①

We can write the horizontal component of the hydrostatic force as,
N
g
3
)
5
.
1
2
(
)
2
/
2
(
g
A
h
g
FH 








The horizontal component acts through the center of pressure at a distance of
2/3 m (for this condition) above the bottom “point C” . It acts perpendicular to
the dam surface projection as shown in the figure..
The vertical component of the hydrostatic force,
N
g
3
5
.
1
2
2
2
g
w
)
CDE
(area
g
FV 








 





To find the force in the strut, summing the moments about “C” and equating = 0
…… (1)
…… (2)
3
.
2
15
sin
F
3
.
2
15
cos
F
3
2
F
3
2
F o
strut
o
strut
V
H 








)
15
sin
15
(cos
F
3
.
2
)
F
F
(
3
2 o
o
strut
V
H 




or
Substituting the computed values of FH and FV, we get
KN
93
.
13
)
15
sin
15
(cos
3
.
2
10
/
)
81
.
9
10
(
4
)
15
sin
15
(cos
3
.
2
)
g
3
2
(
3
2
F o
o
3
3
o
o
strut 











…… (3)
…… (4)

Worked Example
h
A


Water
Static Forces
A
B
Hinge
ℓ
P
Gate “AB” has length ℓ and width w, is hinged at B and has
negligible weight (see the attached figure). The liquid level h remains
at the top of the gate for any ,. Find: the analytical expression for the
force P, perpendicular to AB, required to keep the gate in equilibrium.
Hydrostatic Forces (Inclined Plane Surface)
h


Water
A
B
Hinge
ℓ
P
⅔ ℓ
F
C.P
The depth to the C.G of the gate is (ℓ /2) and that to the C.P is (⅔ ℓ )
along the gate , thus the hydrostatic force is
C.G
Solution
②

)
w
(
)
2
/
h
(
g
A
h
g
F 





 
Summing moments about the hinge at “B” then gives,
)
3
/
(
F
P 
 


or
3
)
w
(
)
2
/
h
(
g
)
3
/
F
(
P







From the geometry, 
 sin
/
h

………. (1)
………. (2)
Substituting into Eq. (2) gives,
 





 sin
h
w
g
6
1
P 2
Solution
Worked Example
Gate “AB” in the shown figure
is 5ft wide, hinge at “A”, and
restrained by a stop at B. The
stop will break if the water force
on it equals 9200 Ibf. Compute:
→ For what water depth h is
this condition reached?
→Compute the force on the stop at “B” and the reactions at “A” if
the water depth h = 9.5 ft.
h
4.0 ft
A
B

Water
Static Forces
Stop
♨
♨
③

h
4.0 ft
A
B

Water
9200 Ibf
F
C.G
C.P
)
2
h
(
)
4
5
(
)
12
/
4
5
(
h
A
I
y
3
g
.
c







2
h
)
2
/
4
(
h
h 



f
Ib
36
.
1249
)
4
5
(
)
2
4
h
(
2
.
32
94
.
1
A
h
g
F 








The depth to the C.G of the gate is
(h - 2), thus the hydrostatic force is
………. (1)
Solution
)
y
2
(
F
4
Fstop 



The line of action of the force is below the center of the gate
“AB”,(C.G), is given by
ft
)
2
h
(
33
.
1
)
2
h
(
)
4
5
(
)
12
/
4
5
(
h
A
I
y
3
g
.
c








 ………. (2)
Summing moments about the hinge at “A” then gives,
or
)
2
h
33
.
1
2
(
)
2
h
(
36
.
1249
4
9200





 ………. (3)
Solving for h gives → h = 16.06 ft

Worked Example
Gate “AB” in the shown figure
is 5ft wide, hinge at “A”, and
restrained by a stop at B. The
stop will break if the water force
on it equals 9200 Ibf. Compute:
→ For what water depth h is
this condition reached?
→Compute the force on the stop at “B” and the reactions at “A” if
the water depth h = 9.5 ft.
h
4.0 ft
A
B

Water
Static Forces
Stop
♨
♨
h
4.0 ft
A
B

Water
9200 Ibf
F
C.G
C.P
)
2
h
(
)
4
5
(
)
12
/
4
5
(
h
A
I
y
3
g
.
c







2
h
)
2
/
4
(
h
h 



f
Ib
2
.
9370
)
4
5
(
5
.
7
2
.
32
94
.
1
A
h
g
F 







The depth to the C.G of the
gate is (9.5 – 4/2) = 7.5 ft,
thus the hydrostatic force is
………. (1)
Solution
④

)
y
2
(
F
4
Fstop 



The line of action of the force is below the center of the gate
“AB”,(C.G), is given by
ft
178
.
0
5
.
7
)
4
5
(
)
12
/
4
5
(
h
A
I
y
3
g
.
c






 ………. (2)
Summing moments about the hinge at “A” then gives,
or
)
178
.
0
2
(
2
.
9370
4
Fstop 


 ………. (3)
Solving Eq. (3) for Fstop gives the force on the stop at “B”
→ Fstop = 5102.8 Ibf
9370.2 Ibf
5102.07 Ibf
4268.13 Ibf
A
B
The reactions at “A”

Worked Example
Water
Submerged
concrete block
ℓ
¼ℓ
Stop
Hinge

Determine the minimum volume
of concrete (s.g = 2.40) needed
to keep the gate in a closed
position.
i. If : the gate wide = 2 ft and ℓ = 5 ft
ii. If: the gate wide = 1.0 m ℓ = 2 m.
W
Water
Submerged
concrete block
ℓ
¼ℓ
Stop
Hinge

⅓ ℓ
FH
Ftension
FB
Ftension
0
F
F
W
m
equilibrui
For
tension
B 




Solution
The forces act on the
submerged concrete block
Hydrostatic forces on the gate
⑤

The hydrostatic force due to water =
w
g
5
.
0
)
w
(
)
2
/
(
g
A
h
g
F 2
w
w
w
H 

 








…… (1)
It acts at a distance of ℓ /3 above the hinge (see the attached figure)
The forces act on the concrete block are:
 The weight
gravity
of
center
the
through
acts
and
volume
g
F concrete
concrete 



 The buoyant force FB
gravity
of
center
the
through
acts
and
volume
g
F w
B 



…… (2)
The buoyant force has a magnitude equal to the weight of the fluid displaced by
the concrete body ad is directed vertically upward (Archimedes’ principle)
To satisfy equilibrium
 The tensile force Ftension
75
.
3
w
g
5
.
0
75
.
3
F
25
.
1
)
3
/
(
F
F
2
w
H
H
tension


 





Substituting into Eq. (2), we get
0
F
F
W tension
B 



Thus,
B
tension F
W
F 

To compute Ftension, summing moments about the hinge
…… (3)
















1
g
Volume
75
.
3
w
g
5
.
0
w
concrete
w
2
w 
= (S.g) concrete

)
1
g
.
s
(
75
.
3
w
5
.
0
Volume
.
con
2




…… (4)
i. For, ℓ = 5 ft, w = 2 ft, and s.g conc.= 2.36 in Eq. (4) gives
3
2
ft
76
.
4
)
1
40
.
2
(
75
.
3
2
5
5
.
0
block
concrete
of
volume
Minumum 




 3
2
ft
76
.
4
)
1
40
.
2
(
75
.
3
2
5
5
.
0
block
concrete
of
volume
Minumum 




 3
2
ft
76
.
4
)
1
40
.
2
(
75
.
3
2
5
5
.
0
block
concrete
of
volume
Minumum 





ii. For, ℓ = 2 m, w = 1.0 m, and s.g conc.= 2.36 in Eq. (4) yields
3
2
m
38
.
0
)
1
40
.
2
(
75
.
3
0
.
1
2
5
.
0
block
concrete
of
volume
Minimum 




 3
2
m
38
.
0
)
1
40
.
2
(
75
.
3
0
.
1
2
5
.
0
block
concrete
of
volume
Minimum 




 3
2
m
38
.
0
)
1
40
.
2
(
75
.
3
0
.
1
2
5
.
0
block
concrete
of
volume
Minimum 





Worked Example

0.6 m
2.1 m O
Calculate the moment about
“O” of the resultant forces
exerted by the water:
i. On the end of the cylinder,
and
ii. On the curved surface of
the cylinder.
Hydrostatic Forces (Curved Surface)
⑥


0.6 m
2.1 m O
FH
wide
m
/
m
.
kN
58
.
7
223
.
0
34
y
F
O"
"
about
moments
the
of
sum
THe H 



 wide
m
/
m
.
kN
58
.
7
223
.
0
34
y
F
O"
"
about
moments
the
of
sum
THe H 



 wide
m
/
m
.
kN
58
.
7
223
.
0
34
y
F
O"
"
about
moments
the
of
sum
THe H 




y
where:
wide
.
m
/
kN
34
1000
)
1
1
.
2
(
)
2
/
1
.
2
6
.
0
(
81
.
9
1000
A
h
g
FH









and
m
223
.
0
65
.
1
)
1
.
2
0
.
1
(
12
/
1
.
2
0
.
1
h
A
I
y
3
g
.
c







m
65
.
1
h 
C.P
C.G
The moment of the resultant forces
exerted by the water on the end of the
cylinder about “O” , can be determine as,
y
F
moments
0



Solution
n m

0.6 m
2.1 m O
0
223
.
0
34
446
.
0
17
y
F
x
F
O"
"
about
moments
the
of
sum
THe H
V 







 0
223
.
0
34
446
.
0
17
y
F
x
F
O"
"
about
moments
the
of
sum
THe H
V 







 0
223
.
0
34
446
.
0
17
y
F
x
F
O"
"
about
moments
the
of
sum
THe H
V 








where:
w
)
Qmnr
area
pQmn
area
(
g
Fv 




and
p
Q
r
The moment of the resultant forces
exerted by the water on the curved
surface of the cylinder about “O” , can be
determine as,
x
Fv
x
F
moments
0



wide
m
/
kN
17
1000
0
.
1
)
2
4
1
.
2
(
81
.
9
1000
w
)
pQr
area
(
g
F
2













It acts at a distance “x” right to
the line np.
m
446
.
0
3
05
.
1
4
3
R
4
x






y

Worked Example
120 ft
Submersible
pump Well
air
Storage tank
Air of P =
40 psi (gage)
Well casing
A submersible deep-well pump delivers
745 gallons/h of water through a 1-in pipe
when operating in the system shown in the
figure. An energy loss of 10.5 ft occurs in
the piping system, (1.0 ft3/s = 449 gal./min)
Calculate the power delivered by the
pump to the water, and
If the pump draws 1 hp, calculate the
efficiency.
.
.
s
/
ft
028
.
0
449
1
60
745
Q
discharge
The 3



Applying the energy equation (Bernoulli’s Eq. ) between the points identified
as “1” and “2” in the shown figure, we have
L
2
2
p
1
2
h
g
2
V
g
p
z
H
g
2
V
g
p
z 
























For the given condition, p1= patm. = 0, V1 = V2  0, hL = 10.5ft and
taking the water surface at the well as a reference datum (Z1= 0 & Z2 =
120 ft ) and hL = 10.5 ft, the above equation gives
 0
 0
Bernoulli’s Equation and it’s Applications
Solution
⑦

  50
.
10
0
2
.
32
94
.
1
12
40
120
h
0
0
0
2
p 
















Solving for hp gives  hp = 222.7 ft.
We can now calculate the power delivered by the pump
hp
70
.
0
550
7
.
222
028
.
0
2
.
32
94
.
1
h
Q
g
power p 







To calculate the mechanical efficiency of the pump,
70
.
0
0
.
1
70
.
0
power
pump
H
Q
g
Efficiency
p






Thus, at the stated condition, the pump is 70% efficient.
Bernoulli’s Equation and it’s Applications
.
Worked Example
⑧


5 ft
12 ft
4 in. dia.
h
d
Calculate the maximum “h” and
the minimum “d” that will permit
cavitation-free flow through the
frictionless pipe system shown.
Atmospheric pressure = 14.7 psi;
vapor pressure = 1.5 psi (abs).
water
4 in. dia.
Bernoulli’s equation Applications

5 ft
12 ft
4 in. dia.
h
d
water
4 in. dia.
④
③
②
① Reference datum
Solution

Pipe Flow and Pipe Systems
Solution
Taking ( g)w = 62.4 Ibf / ft2
ft
92
.
33
)
ft
/
Ib
(
4
.
62
)
ft
/
in
(
12
)
in
/
Ib
(
7
.
14
g
P
3
f
2
2
2
2
.
atm 



ft
46
.
3
4
.
62
12
5
.
1
g
P 2
V 



Similarly,
ft
92
.
33
g
/
P .
atm 

ft
46
.
3
g
/
Pv 

ft
46
.
30
46
.
3
92
.
33
g
/
P
&
g
/
P 3
2






P
Datum
Abs. 0
Gage 0
To compute the velocity at exit “V4”
and consequently the flow rate “Q”,
we apply Bernoulli’s equation
between points “1” and “4”
L
4
2
1
2
h
g
2
V
g
P
Z
g
2
V
g
P
Z 























For the given condition, taking a reference datum as shown in the figure,
V4 = ?????
Z4 = 12 ft
V1 = 0 (large tank)
hL = 0 (frictionless pipe)
Z1 = 0 ft
P1 = P4 = Patm. ( = 0 gage)
Now, substituting into Eq. (1),
…………… (1)
  0
g
2
V
0
12
0
0
0
2
4 














or
s
/
ft
80
.
27
12
2
.
32
2
V4 


 …………… (2)

Taking the pressure at the constriction equals the vapor pressure permits us
to calculate the velocity at the constriction,
and the flow rate
s
/
ft
43
.
2
80
.
27
)
12
/
4
(
4
V
A
Q 3
2
4
4 











 s
/
ft
43
.
2
80
.
27
)
12
/
4
(
4
V
A
Q 3
2
4
4 











 s
/
ft
43
.
2
80
.
27
)
12
/
4
(
4
V
A
Q 3
2
4
4 












To compute the minimum diameter “d”, we apply Bernoulli’s equation
between points “1” and “2”
2
2
1
2
g
2
V
g
P
Z
g
2
V
g
P
Z






















 …………… (3)
  














g
2
V
)
46
.
30
(
5
0
0
0
2
2
s
/
ft
80
.
47
46
.
35
81
.
9
2
V2 



…………… (4)
The continuity equation gives
4
d
ft
05
.
0
8
.
47
/
43
.
2
V
/
Q
A
2
2
2
2






Solving for the diameter at the constriction,
in
3
ft
25
.
0
4
05
.
0
d 



 in
3
ft
25
.
0
4
05
.
0
d 



 in
3
ft
25
.
0
4
05
.
0
d 



 …………… (5)
The velocity through the summit equals to that at the exit V3 = V4 (the same
diameter and the same discharge) and the maximum height “h” is computed
by assuming that the pressure at point ③ is limited by vapor pressure.

To compute the maximum height “h”, we apply Bernoulli’s equation between
points “1” and “3”
3
2
1
2
g
2
V
g
P
Z
g
2
V
g
P
Z






















 …………… (6)
Substituting the following data for the given condition gives,
V3 = 27.80 ft/s
Z3 = h ft
P3 /  g = -30.48 ft
V1 = 0 (large tank)
Z1 = 0 ft
P1 = Patm (= 0 gage)
  














2
.
32
2
80
.
27
)
46
.
30
(
h
0
0
0
2
Solving for “h” gives,
→ h = 18.46 ft≃ 18.50 ft
Incipient cavitation m ust be assum ed in ideal fluid flow problem s
of this nature in order for head losses to be considered neg lig ible.
Also, w ith little cavitation there is m ore likelihood of the pip e
flow ing full at the exit.
Notice:
Worked Example
⑨

The maximum flow rate in a 250mm water pipeline is expected to be 142
Liters/s. The Venturi-meter is attached a mercury-under-water manometer,
its reading is 90.5cm. Calculate the minimum throat diameter which should
be specified. (take Cd = 0.98).
In this application we use the Venturi-meter equation,
)
1
.
g
.
s
(
h
g
2
A
A
A
A
C
Q
2
2
2
1
2
1
d 



But first we will calculate the areas of the inlet section (pipeline) and the throat
of the Venturi-meter,
…………… (1)
Bernoulli’s equation Applications “Venturi-meter”
Solution
2
2
2
1 m
049
.
0
)
25
.
0
(
4
D
4
A 






2
2
2 m
d
4
A 


For the given application, we have
h = 0.905 m
A2 = ???? m2
Indicating liquid is mercury ( s.g. = 13.6)
Cd = 0.98
A1 = 0.049 m2
Q = 0.142 m3/s
)
1
6
.
13
(
905
.
0
81
.
9
2
A
049
.
0
A
049
.
0
98
.
0
142
.
0
2
2
2
2
2








Substituting into Eq. (1) gives
Solving for A2 gives
2
2
3
2 m
d
4
10
36
.
9
A 



  or mm
110
m
011
.
0
d 2
2 


2
2
2
1 m
049
.
0
)
25
.
0
(
4
D
4
A 






2
2
2 m
d
4
A 


For the given application, we have
h = 0.905 m
A2 = ???? m2
Indicating liquid is mercury ( s.g. = 13.6)
Cd = 0.98
A1 = 0.049 m2
Q = 0.142 m3/s
)
1
6
.
13
(
905
.
0
81
.
9
2
A
049
.
0
A
049
.
0
98
.
0
142
.
0
2
2
2
2
2








Substituting into Eq. (1) gives
Solving for A2 gives
2
2
3
2 m
d
4
10
36
.
9
A 



  or mm
110
m
011
.
0
d 2
2 

Worked Example
Water flows through the shown nozzle at a rate of 15 ft3/s and
discharges into the atmosphere. D1 = 12 in. and D2 = 8 in. Determine
the force required at the flange to hold the nozzle in place. Neglect the
gravitational forces.
①
②
① ②
D1
D2
Flange
⑩

The data given for this case are:
Z2 = 0
P2 = Patm. = 0
P1 = ?????
D2 = 8 in
The nozzle and pipe are horizontal
Z1 = 0
Q = 15 ft3/s
D1 = 12 in
The inlet and outlet velocities of water can be computed using the continuity
equation as:
s
/
ft
10
.
19
)
12
/
12
(
4
15
A
Q
V
2
1
1 




s
/
ft
97
.
42
)
12
/
8
(
4
15
A
Q
V
2
2
2 




and
Solution
①
②
① ②
D1
D2
Flange
Momentum Equation and it’s Applications
C.V
)
V
V
(
Q
A
P
F
A
P 1
2
2
2
1
1 






x
y
F
= 0
)
10
.
19
97
.
42
(
15
94
.
1
0
F
)
4
1
(
16
.
1437
2










Ib
12
.
434
F 
 Ib
12
.
434
F 
 Ib
12
.
434
F 

Therefore, the horizontal force on the flange is 434.12 Ib acting in the
negative x-direction (the nozzle is trying to separate from the pipe). The
connectors (such bolts) used must be strong enough to withstand this force.
It acts In the
Negative x-direction

Worked Example
①
②
③
④
Negligible
20 ft
3 ft
The pressure in pipe at section 1 is 30 psi
and the cross-sectional areas of the pipe at
sections 1 and 2 are 0.20 ft2 and 0.05 ft2.
a) Compute the flow velocity at section 3,
b) What is the vertical component of the
force necessary to hold the deflecting
vane in position?. Neglecting friction,
weight of the fluid in contact with the
vane,
c) Show the control volume with all forces
and velocities labeled.
Deflecting vane
①
②
③
④
Negligible
20 ft
3 ft
Deflecting vane
Reference datum
C.V
x
y
Solution
⑪

 Using Bernoulli’s equation between points ① and ② to calculate
the velocity at the exit of the nozzle:
2
2
1
2
g
2
V
g
P
Z
g
2
V
g
P
Z























 Taking the center of the nozzle, point ②, as the reference datum, we
have
),
gage
(
0
P
P
,
0
Z
,
psi
30
P
,
ft
3
Z .
atm
2
2
1
1 




and
2
2
2
1 ft
05
.
0
A
,
ft
20
.
0
A 

…………. (1)
 The continuity equation gives V1 = Q/ A1 & V2 = Q/ A2 and
substituting the known values into Eq. ① becomes
2
2
2
1
2
2
2
05
.
0
2
.
32
2
Q
0
0
20
.
0
2
.
32
2
Q
2
.
32
94
.
1
12
30
3



























Solving for Q gives,
s
/
ft
52
.
3
375
16
.
72
2
.
32
2
Q 3




Thus, V2 = Q/A2 =3.52/0.05 = 70.40 ft/s
 Applying Bernoulli’s equation between points ② and ③
= 0 = 0 = 0
…………. (2)
V3 = ?????
Z3 = -20 ft
V2 = 70.40 ft / s
???
3
2
2
2
g
2
V
g
P
Z
g
2
V
g
P
Z
























 Solving Eq. (2) for V3 and substituting give,
s
/
ft
79
2
.
32
2
)
20
97
.
76
(
V3 



 The vertical component of the force necessary to hold the
deflecting vane in position;






 Ib
540
79
52
.
3
94
.
1
V
Q
F 3 





 Ib
540
79
52
.
3
94
.
1
V
Q
F 3 





 Ib
540
79
52
.
3
94
.
1
V
Q
F 3
It acts
upward
Worked Example
 (70 m)
K = 0.5
The three pipes have the same
length L = 300 m and the same
diameter = 30 cm
(1)
(2)
(3)
 (100 m)
Open gate valve k = 5
Neglect bend losses
i. What horsepower must the pump add to the water to
pump 250 Lit / s to the upper reservoir? (Include the
local losses except at bends).
ii. Sketch the energy line approximately to scale and
label the changes in slope.
⑫

HGL
 (1) (70 m)
(1)
(2)
(3)
 (2) (100 m)
Valve
hL
Inlet
hL
p
H
HGL
Solution
 (1) (70 m)
K = 0.5
The three pipes have the same
length L = 300 m and the same
diameter = 30 cm
(1)
(2)
(3)
Open gate valve k = 5
Neglect bend losses
 (2) (100 m)
Applying Bernoulli equation between points (1) and (2),

























 L
2
2
p
1
2
h
g
2
V
g
P
Z
H
g
2
V
g
P
Z
Note that the choice of points (1) and (2) at the surfaces of the upstream lower
reservoir and the downstream upper reservoir means that
………… (1)
0
V
V
&
)
gage
(
0
P
P
P 2
1
.
atm
2
1 





The velocity of the flow through all the pipes shown in the system is the same
and equals, (the pipes have the same diameter 0.30 cm)
s
/
m
54
.
3
4
/
)
3
.
0
(
1000
/
250
A
Q
V 2





For f = 0.20, L = 3x 300 m and D = 0.3 m, the head losses can be computed as,
Main losses
m
32
.
38
81
.
9
2
54
.
3
3
.
0
300
3
02
.
0
g
2
V
D
L
f
2
2







Minor losses
m
32
.
0
81
.
9
2
54
.
3
5
.
0
g
2
V
5
.
0
2
2












Inlet losses
Valve Losses
Total
losses
m
2
.
3
81
.
9
2
54
.
3
5
g
2
V
5
2
2












………… (2)
Substituting in Eq. (2) gives
    )
20
.
10
02
.
1
36
.
122
(
0
0
300
H
0
0
200 p 








ft
58
.
233
Hp 
 ft
58
.
233
Hp 
 ft
58
.
233
Hp 

The horsepower must the pump add to the water to pump 9 ft3/ s to the
upper reservoir can now be calculated
hp
240
550
58
.
233
0
.
9
2
.
32
94
.
1
H
Q
g
power p 







Worked Example
 (constant)
Knozzle = 0.04
L = 1500 m (total), and f = 0.02
Kinlet = 0.1
Nozzle of
150 mm dia.
0.30 m dia.
The flow rate through this pipeline and nozzle when the
pump is not running is 0.28 m3/s. How much power
must be supplied by the pump to produce the same flow
rate with a 100 mm nozzle at the end of the line?
Pump
 (constant)
Knozzle = 0.04
L = 1500 m (total), and f = 0.02
Kinlet = 0.1
Nozzle of
150 mm dia.
0.30 m dia.
Pump
1
2
Reference
datum
Pipe Flow and pipe system
H
Solution
⑬

 Using Bernoulli’s equation between points ① and ② to calculate
the velocity at the exit of the nozzle:

























 L
2
2
p
1
2
H
g
2
V
g
P
Z
H
g
2
V
g
P
Z
 Taking the center of the nozzle, point ②, as the reference datum,
we have
running)
not
is
pump
(the
0
H
and
,
0
Z
),
gage
(
0
P
P
P
,
m
H
Z
p
2
.
atm
2
1
1






s
/
m
84
.
15
)
4
/
15
.
0
(
28
.
0
A
Q
V
,
s
/
m
96
.
3
)
4
/
30
.
0
(
28
.
0
A
Q
V 2
n
n
2
p
p 









…………. (1)
 Using the continuity equation gives VP = Q/ AP & Vn = Q/ An
For f = 0.020, L = 1500 m and Dp = 0.3 m, dn = 0.15 m, the head losses can
be computed as,
Main losses
m
93
.
79
81
.
9
2
96
.
3
3
.
0
1500
02
.
0
g
2
V
D
L
f
2
2






Minor losses
m
08
.
0
81
.
9
2
96
.
3
1
.
0
g
2
V
1
.
0
2
2
p












Inlet losses
Valve Losses
Total
losses
m
51
.
0
81
.
9
2
84
.
15
04
.
0
g
2
V
04
.
0
2
2
n 











and substituting the known values into Eq. ① becomes

  )
51
.
0
08
.
0
93
.
79
(
81
.
9
2
84
.
15
0
0
0
0
0
H
2

















 ……. (2)
Solving for H gives  H = 93.31 m.
If the nozzle diameter at the end of the pipe changed to be 0.10 m with
producing the same flow rate and keeping the pipe diameter constant, the
exit velocity will be
s
/
m
65
.
35
)
4
/
10
.
0
(
28
.
0
A
Q
V
V 2
n
n
exit 





N.B. : The velocity through the pipe will not change (Vp = 3.96 m /s), the
water surface in the tank is constant (H = 93.31 m).
For f = 0.020, L = 1500 m and Dp = 0.3 m, dn = 0.10 m and H = 93.31 m, the
head losses can be computed as,
Main losses
m
93
.
79
81
.
9
2
96
.
3
3
.
0
1500
02
.
0
g
2
V
D
L
f
2
2







Minor losses
m
08
.
0
81
.
9
2
96
.
3
1
.
0
g
2
V
1
.
0
2
2
p












Inlet losses
Valve Losses
Total
losses
m
59
.
2
81
.
9
2
65
.
35
04
.
0
g
2
V
04
.
0
2
2
n 











Substituting the known values into Eq. ① gives

  )
59
.
2
08
.
0
93
.
79
(
81
.
9
2
65
.
35
0
0
H
0
0
31
.
93
2
p 

















Solving for Hp gives  Hp = 54.07 m.
 The power must be supplied by the pump to produce the same flow rate
with a 100 mm nozzle at the end of the line,
kw
150
kW
52
.
148
1000
07
.
54
28
.
0
81
.
9
1000
H
Q
g
Power p 






 kw
150
kW
52
.
148
1000
07
.
54
28
.
0
81
.
9
1000
H
Q
g
Power p 






 kw
150
kW
52
.
148
1000
07
.
54
28
.
0
81
.
9
1000
H
Q
g
Power p 







Worked Example
25 ft
L = 600 ft, D = ?? , f = 0.02
Main
A
B
Reference datum
The figure shows a pipe delivering water to the putting green on a golf
course. The pressure in the main is 80 p.s.i and it is necessary to
maintain a minimum of 60 p.s.i at point B to adequately supply a
sprinkler system. Specify the required size of steel pipe ( f = 0.02) to
supply 0.50 ft3/s.
Q = 0.5 ft3/s
⑭

 Using Bernoulli’s equation between points Ⓐ and Ⓑ to calculate
the pressure at Ⓐ :
























 L
B
2
A
2
H
g
2
V
g
P
Z
g
2
V
g
P
Z
 Taking the line passing through, point Ⓐ, as a reference datum, we
have
/s
ft
0.5
Q
and
0.02,
f
???,
D
ft,
600
L
and
,
i
.
s
.
p
60
P
and
V
V
V
,
m
25
Z
i
.
s
.
p
80
P
???,
diameter)
pipe
same
the
(
V
V
V
,
m
0
Z
3
p
p
B
p
A
B
B
A
p
B
A
A













s
/
ft
)
4
/
D
(
50
.
0
A
Q
V 2
p
p
p




…………. (1)
 To calculate Vp, we use the continuity equation VP = Q/ AP
…………. (2)
Solution
The head losses can be computed as,
ft
D
0755
.
0
2
.
32
5
.
0
D
600
02
.
0
8
g
Q
D
L
f
8
g
2
V
D
L
f 5
p
2
2
5
2
2
5
2













5
p
2
B
2
2
A
2
D
0755
.
0
g
2
V
2
.
32
94
.
1
12
60
25
g
2
V
2
.
32
94
.
1
12
80
0 


























=
.
in
4
ft
324
.
0
D
or
D
0755
.
0
1
.
21 p
5
p


 .
in
4
ft
324
.
0
D
or
D
0755
.
0
1
.
21 p
5
p


 .
in
4
ft
324
.
0
D
or
D
0755
.
0
1
.
21 p
5
p



Solving for Dp,

Worked Example
Main
A
Reference datum
Water is being delivered to a
tank on the roof of a factory
building. The system consists
of a pipe (1½ in. diameter of
f = 0.02), valve and elbow as
shown in the figure.
 What pressure must exist at
point “A” for 200 L/min to be
delivered?
Flow of
200 L/s
25 m
2.5 m
Elbow
k = 1.0
Valve
k = 2.0
Factory
building
Main
A
Reference datum
Flow of
200 L/s
25 m
2.5 m
Elbow
k = 1.0
Valve
k = 2.0
Factory
building
B
m
5
.
27
5
.
2
25
length
pipe
total
The 


s
/
m
10
33
.
3
60
1000
200 3
3




The flow rate Q =
The pipe diameter Dp =
m
0381
.
0
100
54
.
2
5
.
1


Solution
⑮

 To compute the pressure at point Ⓐ, we apply Bernoulli’s equation
between points Ⓐ and Ⓑ :
























 L
B
2
A
2
H
g
2
V
g
P
Z
g
2
V
g
P
Z
 Taking the line passing through, point Ⓐ, as a reference datum, we
have
and
),
gage
(
0
P
P
and
V
V
V
,
m
25
Z
???
P
,
s
/
m
92
.
2
V
V
,
m
0
Z
.
atm
B
exit
p
B
B
A
p
A
A









s
/
m
92
.
2
)
4
/
0381
.
0
(
10
33
.
3
A
Q
V 2
3
p
p 






…………. (1)
 To calculate Vp (velocity through pipe, we use the continuity
equation - VP = Q/ AP
…………. (2)
For f = 0.020, L = 27.5 m and Dp = 0.0381 m, Kvalve = 2.0 and Kelbow = 1. 0
and V =2 92 m /s, the head losses can be computed as,
Main losses
m
27
.
6
81
.
9
2
92
.
2
0381
.
0
5
.
27
02
.
0
g
2
V
D
L
f
2
2
p







Minor losses
m
87
.
0
81
.
9
2
92
.
2
0
.
2
g
2
V
0
.
2
2
2
p












Valve losses
Elbow Losses
Total
losses
m
43
.
0
81
.
9
2
92
.
2
0
.
1
g
2
V
0
.
1
2
2
B 












Substituting in Eq. (2) gives
)
43
.
0
87
.
0
27
.
6
(
g
2
V
0
25
g
2
V
g
P
0
2
p
2
p
A 
























kw
320
1000
81
.
9
1000
58
.
32
P
or
m
58
.
32
g
P A
A 





 kw
320
1000
81
.
9
1000
58
.
32
P
or
m
58
.
32
g
P A
A 





 kw
320
1000
81
.
9
1000
58
.
32
P
or
m
58
.
32
g
P A
A 






=
Worked Example
Main
line A
The figure shows a pipe delivering water from a main line to a factory.
The pressure at the main is 415 KPa. Compute the maximum
allowable flow rate if the pressure at the factory must be no less than
415 KPa.
100 m & Dp = 10 cm & f = 0.02
415 KPa 200 KPa
B
Valve k = 5
Q = ???
⑯

 To compute the pressure at point Ⓐ, we apply Bernoulli’s equation
between points Ⓐ and Ⓑ :
























 L
B
2
A
2
H
g
2
V
g
P
Z
g
2
V
g
P
Z
 Taking the line passing through point Ⓐ, as a reference datum, we
have
m
0.1
D
,
02
.
0
f
m,
100
L
and
,
i
.
s
.
p
200
P
,
i
.
s
.
p
415
P
),
pipe
same
the
(
V
V
V
,
m
0
Z
Z
p
p
B
A
p
B
A
B
A









…………. (1)
 Substituting into, Eq. (1) gives
…………. (2)













L
B
A
H
g
P
g
P
Solution
For f = 0.020, L = 100 m and Dp = 0. 1 m, and Kvalve = 5.0
Main losses
m
Q
4
.
16525
Q
1
.
0
g
100
02
.
0
8
Q
D
g
L
f
8 2
2
5
2
2
5
p
2





















Minor losses
m
Q
3
.
4131
)
4
/
1
.
0
(
81
.
9
2
Q
5
g
2
V
0
.
5 2
2
2
2
2
p







Valve losses
Total
losses
Substituting the known and computed values into Eq. ② gives
)
3
.
4131
4
.
16525
(
Q
81
.
9
1000
1000
)
200
415
( 2











s
/
L
6
.
32
s
/
m
0325
.
0
Q 3


 s
/
L
6
.
32
s
/
m
0325
.
0
Q 3


 s
/
L
6
.
32
s
/
m
0325
.
0
Q 3




Worked Example
Water backs up behind a concrete dam (see the shown figure). Leakage
under the foundation gives a pressure distribution under the dam as
indicated. If the water depth, h, is too great, the dam will topple over ( )
about its toe (point A). For the dimensions given, determine the maximum
water depth for the following widths of the dam, B = 6, 9, 12, and 15 m.
Base your analysis on a unit length of the dam. The specific gravity of the
concrete dam is 2.44.
h
g

m
24
m
3
h
3
g

44
.
2
G
S 
B
Water
Water
▼
▼
A
Solution
The forces acts on the concrete dam are:
 “W” – Self-weight load =
B
g
28
.
29
)
B
24
2
/
1
(
g
44
.
2
volume
g
.)
G
.
S
( 










 …… (1)
It acts at a distance = 2B/3 left to “A”
 FU.S - U.S. water load
= )
2
/
h
(
g
)
0
.
1
h
(
)
2
/
h
(
g
A
h
g 2










 ……………………… (2)
⑰

It acts at a distance = h/3 above “A”
 FD.S - D.S. water load
= g
5
.
1
)
0
.
1
(
)
2
/
3
(
g
A
h
g 








 
 ……… (3)
where 
 cos
/
3
 (N.B.  changes as B change and can be
calculated with the aid of the sketch shown below).
 Fuplift- Uplift load = Fuplift, 1 + Fuplift, 2
 g
B
3
)
0
.
1
)
B
(
g
3
F 1
,
uplift 





 ………………...… (4)
(it acts at a distance= B/2 left to “A”)
 g
)
2
/
B
(
)
h
3
(
)
2
/
B
(
g
)
h
3
(
F 2
,
uplift 








 .(5)
(it acts at a distance = 2B/3 left to “A”)
i- For B =6m
m
6
B 
m
24


m
3
m
09
.
3
cos
/
3 



3

o
1
14
)
24
/
6
(
tan 

 
x
o Eq. (1) gives: “W” = g
68
.
175
g
6
28
.
29
B
g
28
.
29 







o Eq. (2) gives: FU.S )
2
/
h
(
g
)
0
.
1
h
(
)
2
/
h
(
g 2









o Eq. (3) gives: FD.S g
64
.
4
g
09
.
3
5
.
1
g
5
.
1 





 
o Eq. (4) gives: g
18
g
6
3
F 1
,
uplift 




 and
g
)
h
3
(
3
g
)
2
/
6
(
)
h
3
(
F 2
,
uplift 









The calculated values can be summarized in the following table
It
acts
at:
W FU.S FD.S Fuplift
Fuplif t, 1 Fuplif t, 2
175.68  g  g x (h2
/2) 4.64  g 18  g 3 (h-3)  g
4m left to “A” h/3 above “A” 1.03 m above
“A” along the
inclined face
of the dam
B/2 =3m
left to “A”
2B/3 =4m
left to “A
We can now sum moments about “A”  0
M "
A
" 
 and therefore
0
4
F
3
F
)
3
/
h
(
F
F
4
W 2
,
uplif
1
,
uplif
S
.
U
S
.
D 








  ……… (5)
or
0
4
g
)
3
h
(
3
3
g
18
)
3
/
h
(
2
h
g
03
.
1
g
64
.
4
4
g
66
.
175
2

















Rearranging and canceling gives
 0
)
3
h
(
12
6
h
4
.
653
3



 …………………… (6)
Solving by trial gives h = 14.56 m
By repeating the calculations for various values of B “B = 9.0, 12, and
15m), the results are shown in Table ( ) and Fig. ( ).

W FU.S FD.S Fuplift
Fuplif t, 1 Fuplif t, 2
B = 9 m
263.52  g  g x (h2/2) 4.80  g 27  g 4.5 (h-3)  g
It acts at: 6m left to
“A”
h/3 above “A” 1.06 m above “A”
along the inclined
face of the dam
B/2 =4.5m
left to “A”
2B/3 =6m
left to “A
0
)
3
h
(
12
6
/
h
4
.
653 3




B = 12 m 351.36  g  g x (h2/2) 5.03  g 36  g 6 (h-3)  g
“A”
h/3 above “A” 1.12m above “A”
along the inclined
face of the dam
B/2 =6m
left to “A”
2B/3 =8m
left to “A
0
)
3
h
(
48
6
/
h
63
.
25599 3




B = 15 m 439.20  g  g x (h2/2) 5.31  g 45  g 7.5 (h-3)  g
“A”
h/3 above “A” 1.18 m above “A”
along the inclined
face of the dam
B/2 =7.5m
left to “A”
2B/3 =4m
left to “A
0
)
3
h
(
75
6
/
h
44
.
4060 3




Substituting the above values into Eq. (5) and solving by trial and error we
obtain:
B = 6 m = 9 m = 12 m = 15 m
h = 14.56 m = 18.45 m = 21.70 m = 24.90 m

10.00
15.00
20.00
25.00
30.00
6 9 12 15
Dame Base "B" in meters
Water
Depth
"h"
in
meters
Note: As the dam base width ”B” increases, the U.S water depth
increases.

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