There is a simplest explaination of fluid mechanics

1. CVE 215
2016 - 2017

2. The water pump in the figure maintains a pressure of 6.5 psi
(gage) at point 1. There is a filter, a half-open valve (k = 2.8),
and two elbows (k = 0.64) and there are 80 ft of 4-inch
diameter commercial steel pipe (f = 0.019).
I. If the flow rate is 0.4 ft3 /s, what is the loss coefficient of
the filter?
II. If the valve is wide open (k = 0) and Kfilter = 7, what is
the resulting flow rate?
III. What is the percentage of increase or decrease of the
flow rate?
IV. Draw the TEL and HGL for the system.

3. Pump
Valve
Filter
Elbow

9.0 ft
(1)

4. Taking 𝜌𝑤 = 1.94 𝑠𝑙𝑢𝑔/𝑓𝑡3, the energy equation is written- from point
1 to point 2 “the surface of the tank”- as
𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 1
= 𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 2
+ ℎ𝐿 + ℎ𝑣𝑎𝑙𝑣𝑒 + ℎ𝐿 𝑓𝑖𝑙𝑡𝑒𝑟 + 2ℎ𝐿 𝑒𝑙𝑏𝑜𝑤 + ℎ𝑒𝑥𝑖𝑡
Enter the given values into the above equation gives,
0 +
6.5 × 122
1.94 × 32.2
+
𝑉2
2 𝑔
= 0 + 9 + 0 +
𝑉2
2 𝑔
𝑓 ∙ 𝐿
𝐷
+ 2.8 + 𝑘𝑓𝑖𝑙𝑡𝑒𝑟 + 2 × 0.64 + 1
Since the pipe has a constant diameter, the continuity equation
𝑄 = 𝐴1 ∙ 𝑉1 = 𝐴2 ∙ 𝑉2
𝑜𝑟 𝑉1= 𝑉2 =
𝑄
𝐴1
=
0.4
𝜋 × 4 12 2
4
= 4.58 𝑓𝑡/𝑠
Solution
….. (1)

5. 14.98 − 9 =
4.582
2 × 32.2
8.64 + 𝑘𝑓𝑖𝑙𝑡𝑒𝑟
Substituting into “1” gives,
14.98 − 9 =
4.582
2 × 32.2
0.019 × 80
4 12
+ 2.8 + 𝑘𝑓𝑖𝑙𝑡𝑒𝑟 + 2 × 0.64 + 1 − 1
If 𝑘𝑓𝑖𝑙𝑡𝑒𝑟 = 7;
𝑘𝑓𝑖𝑙𝑡𝑒𝑟 = 9.73 (𝑎)
𝑉1 = 𝑉2 = 𝑉𝑝𝑖𝑝𝑒 = 𝑢𝑛𝑘𝑜𝑤𝑛
Substituting into “1”, with the valve wide opening 𝑘𝑣𝑎𝑙𝑣𝑒 =
0 𝑎𝑛𝑑 𝑘𝑓𝑖𝑙𝑡𝑒𝑟 = 7 and assuming f = 0.019
14.98 − 9 =
𝑉2
2 × 32.2
0.019 × 80
4 12
+ 0 + 7 + 2 × 0.64 + 1 − 1
∴ 14.98 − 9 =
𝑉𝑝𝑖𝑝𝑒
2
2 × 32.2
12.84
….. (2)

6. 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑉
𝑝 𝑔𝑖𝑣𝑒𝑠  𝑉
𝑝= 5.48 ft/s and the flow rate Q = 𝑉
𝑝 × 𝐴𝑝
𝑄 = 𝐴1 ∙ 𝑉1 = 𝐴2 ∙ 𝑉2 =
𝜋
4
×
4
12
2
× 5.48 = 0.48 𝑓𝑡3 𝑠
The flow rate is increased by a percentage of
=
0.48−0.40
0.40
= 20%

7. A sub-marine moves horizontally in sea and has its axis 15m
below the surface of water. A Pitot-tube properly placed just
in front of the sub-marine and along its axis is connected to
the two limbs of a U- tube containing mercury. The difference
of mercury level is found to be 170mm. Find the speed of the
sub-marine in kilometers per hour, knowing that the sp. Gr. of
mercury is 13.6 and that of sea – water is 1.026 with respect
to fresh water.

8. Given:
Diff. of mercury level, x = 170 mm = 0.17m
Sp. G. of mercury, S.G =13.6
Sp. Gr. of sea water, ` Sw=1.026
The difference of mercury level is ℎ =
𝑥 ∙
𝑆. 𝐺𝑚
𝑆. 𝐺𝑤
− 1 = 0.17 ×
13.60
1.026
− 1 = 2.08 𝑚
∴ The speed of the sub−marine in kilometers per hour 𝑉 =
𝐶𝑣 2 𝑔 ℎ = 2 𝑔 ℎ = 2 × 9.81 × 2.08 = 6.39 𝑚
=
6.39 × 60 × 60
1000
= 23 𝑘𝑚/ℎ𝑟
Where CV is a coefficient which is approximately unity for large pipes
and smooth Pitot but is appreciably less for low Reynolds number flow

9. 3m
12m
Invert
Water
O
≈
Hinge
The figure below shows a vertical circular gate in a 3-m diameter tunnel
with water on one side and air on the other side.

10. I. Findthe horizontal reaction at the hinge?
A. 412 kN B. 408 kN
C. 410 kN D. 414 kN
II. How far fromthe invert on the of the tunnel is the net hydrostatic
force acting gate?
A. 1.45 m B. 1.43 m
C. 1.47 m D. 1.41 m
III. Where will the hinge support be located (measuredfromthe invert)to
hold the gate in position?
A. 1.48 m B. 1.46 m
C. 1.44m D. 1.41 m

11. Force on the gate due to air pressure “the diameter of the cylindrical
gate D = 3 m”:
𝐹𝑎𝑖𝑟 = 𝐹𝑎𝑖𝑟 × 𝐴 = 45 × 𝜋 × 32 4 = 318.1 𝑘𝑁
Force on the gate due to water:
𝐹𝑤𝑎𝑡𝑒𝑟 = 𝜌 𝑔 ℎ ∙ 𝐴 =
1000×9.81× 3 2 𝜋×32 4
1000
= 728.1 𝑘𝑁
The net force on the gate:
𝐹𝑤𝑎𝑡𝑒𝑟 − 𝐹𝑎𝑖𝑟 = 728.1 − 318.1 = 410 𝑘𝑁 →
The horizontal reaction at the hinge = net force and opposite to its
direction:
F horizontal = 410 kN  ©

12. The horizontal water force on the gate acts at:
𝑒 =
𝐼𝑐.𝑔
𝐴 ∙ ℎ
=
𝜋 𝐷4 64
𝜋 𝐷2 4 × 12 − 3 2
=
𝐷2
16 × 10.5
=
32
16 × 10.5
= 0.054 𝑚
“e” is a distance down the center line:
The net hydrostatic force acts on the gate at a distance
1.5 − 𝑒 = 1.50 − 0.054 = 1.45 𝑚
above the invert of the tunnel (A)
To find the locationof the hinge support
𝑀𝑜𝑚𝑒𝑛𝑡 𝑂 = 0
∴ 𝐹𝑤𝑎𝑡𝑒𝑟 × 1.45 − 𝑧 = 𝐹𝑎𝑖𝑟 × 1.50 − 𝑧

13. 3m
12m
Water
O
≈
Hinge
Fair
1.5m
Fwater
Z
e
ℎ𝑐𝑝 = 10.5 + 𝑒
≈
F0
1.45 𝑚

14. To find the locationof thehinge support
𝑀𝑜𝑚𝑒𝑛𝑡 𝑂 = 0
∴ 𝐹𝑤𝑎𝑡𝑒𝑟 × 1.45 − 𝑧 = 𝐹𝑎𝑖𝑟 × 1.50 − 𝑧
∴ 728.1 × 1.45 − 𝑧 = 318.1 × 1.50 − 𝑧
The hingesupport will be locatedto hold the gate in position at
𝑧 = 1.41 𝑚
“measured from the invert” (D)

15. Starting with the Bernoulli’s equation, show that the volumetric
flow rate of an incompressible fluid through a horizontal Venturi-
meter where the throat and the upstream positions are connected
by a U-tube containing a mercury of density m can be given by:
𝑄𝑎𝑐𝑡. = 𝐶𝑑 𝐴𝑃
2 ∆𝑃
𝜌 𝛽4 − 1
Where:
𝐶𝑑 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡,
𝐴 𝑖𝑠 𝑝𝑖𝑝𝑒 𝑓𝑙𝑜𝑤 𝑎𝑟𝑒𝑎,
𝜌 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦,
𝛽 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑡ℎ𝑟𝑜𝑎𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟,
∆𝑃 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑉𝑒𝑛𝑡𝑢𝑟𝑖 − 𝑚𝑒𝑡𝑒𝑟,
𝑔 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑖𝑛 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑟𝑦.

16. An orifice – meter and a Venturi-meter are connected in parallel in a
horizontal pipe of diameter 50mm. The orifice has a throat diameter of 25
mm and discharge coefficient of 0.65, while the Venturi has a throat diameter
of 38 mm and discharge coefficient of 0.95. “The pressure drop across both
instrumentsmust be the same”
Determine the proportionof flowthrougheither meter.
Q Venturi
Q Orifice
Q total Q total

17. It is assumed that:
 The liquid flows along a frictionless horizontal pipe and that
losses due to fittings can be neglected,
The drop across both instruments is the same. Thus for both
meters, the flow is:
𝑄𝑎𝑐𝑡. 𝑉𝑒𝑛𝑡. = 𝐶𝑑 (𝑉𝑒𝑛𝑡.) 𝐴𝑝
2 ∆𝑃
𝜌 𝛽𝑉𝑒𝑛.
4
− 1
𝑄𝑎𝑐𝑡. 𝑂𝑟𝑖𝑓𝑓𝑖𝑐𝑒 = 𝐶𝑑 (𝑂𝑟𝑖𝑓.) 𝐴𝑝
2 ∆𝑃
𝜌 𝛽𝑂𝑟𝑖𝑓.
4
− 1
Where the ratio of the pipe to the Venturi area in terms of
diameter is
1
2

18. 𝐴𝑃
𝐴𝑉𝑒𝑛𝑡.
=
𝐷𝑃
𝐷𝑉𝑒𝑛𝑡.
2
and the ratio of the pipe to the orifice area in terms of diameter
𝐴𝑃
𝐴𝑂𝑟𝑖𝑓.
=
𝐷𝑃
𝐷𝑂𝑟𝑖𝑓.
2
Since the differential pressure across both meters is the same
Eq. (1) and (2) give
𝑄𝑎𝑐𝑡. 𝑉𝑒𝑛𝑡.
𝑄𝑎𝑐𝑡. 𝑂𝑟𝑖𝑓.
=
𝐶𝑑 𝑉𝑒𝑛𝑡.
𝐶𝑑 𝑂𝑟𝑖𝑓.
×
𝐴𝑃 ∙
2 ∆𝑃
𝜌 𝛽𝑉𝑒𝑛𝑡.
4
− 1
𝐴𝑃
2 ∆𝑃
𝜌 𝛽𝑂𝑟𝑖𝑓.
4
− 1

19. For the given data,
𝑄𝑎𝑐𝑡. 𝑉𝑒𝑛𝑡𝑢𝑟𝑖
𝑄𝑎𝑐𝑡. 𝑂𝑟𝑖𝑓𝑓𝑖𝑐𝑒
=
𝐶𝑑 𝑉𝑒𝑛𝑡.
𝐶𝑑 𝑂𝑟𝑖𝑓.
×
𝛽𝑂𝑟𝑖𝑓.
4
− 1
𝛽𝑉𝑒𝑛.
4
− 1
𝑄𝑎𝑐𝑡. 𝑉𝑒𝑛𝑡𝑢𝑟𝑖
𝑄𝑎𝑐𝑡. 𝑂𝑟𝑖𝑓𝑓𝑖𝑐𝑒
=
0.95
0.65
×
0.05 0.025 4 − 1
0.05 0.038 4 − 1
3
4
𝑄𝑎𝑐𝑡. 𝑉𝑒𝑛𝑡.
𝑄𝑎𝑐𝑡. 𝑂𝑟𝑖𝑓.
= 4 𝑜𝑟 𝑄𝑎𝑐𝑡. 𝑉𝑒𝑛𝑡. = 4 𝑄𝑎𝑐𝑡. 𝑂𝑟𝑖𝑓.
Eq. (4) reduces to
𝑄𝑇 = 𝑄𝑎𝑐𝑡. 𝑉𝑒𝑛𝑡𝑢𝑟𝑖 + 𝑄𝑎𝑐𝑡. 𝑂𝑟𝑖𝑓𝑓𝑖𝑐𝑒 = 4 + 1 𝑄𝑎𝑐𝑡. 𝑂𝑟𝑖𝑓𝑓𝑖𝑐𝑒
From continuity, the total flow QT is the sum of the flow through
the two meters. That is

20. Thus 20% of flow passes through the orifice-meter while
the rest (80%) of the flow passes through the Venturi-
meter.

21. A cylindrical tank of diameter 1m mounted on its axis contains a liquid
which drains through a 2 cm diameter orifice in the base of the tank. If
the tank originally contains 1000 liters, determine:
The time taken for total drainage “Take the coefficient for the orifice as
0.60.
A
H
H1
H2
aorifice
dH
Flow out

22. A circular water tank of 4 m diameter contains 5 m deep water. An orifice
of 400 mm diameter is provided at its bottom.
Find the time taken for water level fall from 5 m to 2 m. Take Cd = 0.60.
Solution:
The area of the tank =
𝜋× 𝐷𝑡𝑎𝑛𝑘
2
4
=
𝜋 ×52
4
𝑚2 and
The circular orifice =
𝜋× 𝐷𝑜𝑟𝑖𝑓𝑖𝑐𝑒
2
4
=
𝜋 ×0.42
4
𝑚2
The time “T” taken to fall water level from H1 = 5.0 m to depth H2 = 2.0 m
𝑇 =
2 𝐴𝑡𝑎𝑛𝑘 𝐻1
0.5−𝐻2
0.5
𝐶𝑑 𝐴𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2 𝑔
=
2 ×
𝜋 ×52
4
50.5 − 20.5
0.6×
𝜋 ×0.42
4
2 ×9.81
= 61.9 sec.
Given:
 A circular tank with diameter D = 4.0 m
 Orifice with diameter dorifice = 0.40 m
 Initial water depth in the tank H1 = 5.0 m
 Final water depth in the tank H2 = 2.0 m.

32. A jet of water of diameter d = 50 mm issues (
‫تدفق‬
) with a velocity
of 4.9 m/s from a hole in the vertical side of an open tank which
kept filled with water to a height of 1.5 m above the center of
the hole. Calculate the reaction of the jet on the tank and its
content when:
I. It is stationary,
II. It is moving with a velocity U = 12 m/s in the opposite
direction to the jet while the velocity of the jet relative to the
tank remains unchanged.
Jet Dia. 50 mm &
V = 4.9 m/s
1.5 m
U

33. 1.0 m
Scale
Inflow with an area =
0.01 m2 and V = 8 m/s
Area = 0.1
m2
A container weighting 100 N “empty”
has cross– sectional area of 0.1 m2
and two equal sized openings on
opposite sides as shown in the figure.
The container is kept on a weighting
balance and water flows into it from
the top through a jet of 0.01 m2 area
at 8.0 m/s velocity. If under steady
flow condition the depth of water in
the tank is 1.0 m, determine the scale
reading. Ans. (1720N)

34. For the two orifices shown in the figure, determine y2 such that:
𝑥2 = 3 𝑥1 4
Orifice 2
Orifice 1
X2
X1
ℎ1 = 2.0 𝑚
ℎ2
10 m

35. Consider the jet from the firs orifice, ℎ1 = 10 − 8 = 2𝑚
Since, ℎ1 = 1
2
𝑔 𝑡1
2
∴ 𝑡1= (2 ℎ1) 𝑔= (2 × 8) 9.81 = 1.277 𝑠
𝑎𝑛𝑑 𝑥1 = 𝑉1 × 𝑡1 = 2𝑔 ℎ1= 2 × 9.81 × 2 × 1.277 = 8.0 𝑚
𝐺𝑖𝑣𝑒𝑛, 𝑥2 = 3𝑥1 4 → 𝑥2 = 3 × 8 4 = 6 𝑚
Similarly, for the jet from the second orifice
V 2= 2 g∙ (10 − h2) & t 2 = (2 h2) g
But x2 = V2 ∙ t2 = 2g ∙ (10 − h2)
∴ 6 = 𝑉2 × 𝑡2= 2𝑔 ∙ (10 − ℎ2) × 2 ℎ2 𝑔

36. Rearranging gives, 9 = 10 ℎ2 − ℎ2
2
𝑜𝑟 ℎ2
2
− 10 ℎ2 + 9 = 0
Solving for h2
ℎ2 = 9.0 𝑚 & 1.0 𝑚

39. A weir 36 meters long is divided into 12 equal bays by a vertical posts, each 60
cm wide. Determine the discharge over he weir if the head over the crest is 1.2
m.
Number of end contraction, n = 2 x12 = 24 “Each bay has two end contraction”
Discharge by using Francis Formula with end contraction “neglecting velocity
of approach”
∴ 𝑄 = 1.84 (𝐿 − 01 𝑛 𝐻) ∙ 𝐻3 2
Substituting the given values gives
𝑄 = 1.84 36 − 0.1 × 24 × 1.2 ∙ 123 2 = 80.1 𝑚3/𝑠
Given
Length of the weir ”L” = 36 m
Number of bays “n” = 12
Number of 12 bays, number of vertical post = 11
Width of each post = 60 cm = 0.60 m
∴ Effective length = L – 11x 0.60 = 29.4 m, Head
on weir = 1.2 m

41. A
B
C
Tank full of water
1.5 m
0.2 kgf/ m2

42. A
B
C
Tank full of water
1.5 m
2.0 m
G
H
O
E F


46. Extreme suction head (for centrifugal pump)
Centrifugal pump suction occurs due to pressure differential between vessel from which
pumped medium is taken, and impeller blades. Excessive increase of pressure differential
may result in the occurrence of cavitation – process when pressure drops down to value at
which fluid boiling temperature lowers below pumped medium temperature and it starts
to evaporate in flow space forming multiple bubbles. Bubbles are carried away by stream
further downstream where under action of building up pressure they are condensed and
collapse, accompanied with multiple hydraulic shocks that negatively tell on pump service
life. In order to avoid negative influence of cavitation the suction head of centrifugal pump
has to be limited.

48. Pump
Suction pipeline
Pump impeller
axis
Pressure
pipeline
Receiving tank
Suction
head
Total suction
head
Pressure
head
Intake
tank

49. Geometric suction head can be determined by the formula:
hг = (P0-P1)/(ρ·g) - hсв - w²/(2·g) - σ·H
hГ – geometric suction head, m
P0 – intake tank pressure, Pa
P1 – pressure of the impeller blades, Pa
ρ – pumped medium density, kg/m3
g – gravity acceleration, m/s2
hсв – losses due to overcoming suction pipeline hydraulic resistance, m
w²/(2·g) – suction pipeline velocity head, m
σ*H – added resistance losses proportional to head, m
where σ – cavitation factor, H – head created by pump
Cavitation factor can be calculated by the empirical formula:
σ = [(n·√Q) / (126H4/3)]4/3
σ – cavitation factor
n – impeller rotation speed, s-1
Q – pump performance capacity, m3/s
Н – created head, m
There is also formula for centrifugal pumps for calculation of head margin providing
absence of cavitation:
Hкв = 0,3·(Q·n²)2/3
Hкв – head margin, m
Q – centrifugal pump performance capacity, m3/s
n – impeller rotation speed, s -1

51. y
40o

66. The theory is to build a dam on a large river that has a large drop in.
The dam stores lots of water behind it in the reservoir. Near the
bottom of the dam wall there is the water intake.
Gravity causes it to fall through the penstock inside the dam. At the
end of the penstock there is a turbine, which is turned by the
moving water.
The shaft from the turbine goes up into the generator, which
produces the power. Power lines are connected to the generator that
carry electricity to your home and mine.
The water continues past the propeller through the tailrace into the
river past the dam. By the way, it is not a good idea to be playing in
the water right below a dam when water is released!

69. Egypt forms part of the world's most water-scarce region; more than 95% of the
country is desert. This satellite photograph shows Lake Nasser (centre) and the
Toshka Lakes (centre left).

70. Toshka project

73. The shown figure shows a free overfall in a horizontal frictionless
rectangular channel. Assuming the flow to be horizontal at section “1” and
the pressure at the brink of section “2” to be atmospheric throughout the
depth, prove that:
𝑦2
𝑦1
=
2 𝐹1
2
2 𝐹1
2
+ 1
Where q = discharge per unit width, and 𝐹1
2
=
𝑞2
𝑔 𝑦1
3
y2
y1
q
Brink
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80. Open channel flow is covered in essentially all civil and environmental engineering
programs, usually by final-year undergraduate or graduate students studying water
resources.
Developed by an author who has been teaching open channel flow to university
students for the past fifteen years, Fundamentals of Open Channel Flow provides the
students with a detailed explanation of the basics of open channel flow using examples
and animation, and offers expert guidance on the practical application of graphical and
computational tools.
It highlights the practical computational tools students can use to solve problems, such
as spreadsheet applications.
It assumes a foundation in fluid mechanics, then adopts a deliberately logical sequence
through )
‫من‬ ‫عمدا‬ ‫المنطقي‬ ‫التسلسل‬ ‫تتبنى‬
‫خالل‬
( energy, momentum, friction, gradually varied
flow (first qualitative, then quantitative), and the basics of sediment transport.

81. The Hydraulic Institute defines NPSH as the total suction head in feet
absolute, determined at the suction nozzle and corrected to datum, less the
vapor pressure of the liquid in feet absolute. Simply stated, it is an analysis of
energy conditions on the suction side of a pump to determine if the liquid will
vaporize at the lowest pressure point in the pump.
The pressure which a liquid exerts on its surroundings is dependent upon its
temperature. This pressure, called vapor pressure, is a unique characteristic of
every fluid and increased with increasing temperature. When the vapor
pressure within the fluid reaches the pressure of the surrounding medium, the
fluid begins to vaporize or boil. The temperature at which this vaporization
occurs will decrease as the pressure of the surrounding medium decreases.
A liquid increases greatly in volume when it vaporizes. One cubic foot of water
at room temperature becomes 1700 cu. ft. of vapor at the same temperature.

82. It is obvious from the above that if we are to pump a fluid effectively, we must
keep it in liquid form. NPSH is simply a measure of the amount of suction head
present to prevent this vaporization at the lowest pressure point in the pump.
NPSH Required is a function of the pump design. As the liquid passes from the
pump suction to the eye of the impeller, the velocity increases and the pressure
decreases. There are also pressure losses due to shock and turbulence as the
liquid strikes the impeller. The centrifugal force of the impeller vanes further
increases the velocity and decreases the pressure of the liquid. The NPSH
Required is the positive head in feet absolute required at the pump suction to
overcome these pressure drops in the pump and maintain the majority of the
liquid above its vapor pressure. The NPSH Required varies with speed and
capacity within any particular pump. Pump manufacturer's curves normally
provide this information.
Net Positive Suction Head (NPSH) NPSH Available is a function of the system in
which the pump operates. It is the excess pressure of the liquid in feet absolute
over its vapor pressure as it arrives at the pump suction. Fig. 4 shows four
typical suction systems with the NPSH Available formulas applicable to each. It
is important to correct for the specific gravity of the liquid and to convert all
terms to units of "feet absolute" in using the formulas.

83. Fig. 4 Calculation of system Net Positive Suction Head Available
for typical suction conditions.

84. PB = Barometric pressure in feet absolute.
VP =Vapor pressure of the liquid at maximum pumping temperature,
in feet absolute.
P = Pressure on surface of liquid in closed suction tank, in feet
absolute.
Ls = Maximum static suction lift in feet.
LH = Minimum static suction head in feet.
hf= Friction loss in feet in suction pipe at required capacity
In an existing system, the NPSHAvailable can be determined by a gauge on
the pump suction.The following formula applies:
NPSHA =PB –VB  Gr + hV
Where
Gr = Gauge reading at the pump suction expressed in feet (plus if above
atmospheric, minus if below atmospheric) corrected to the pump
centerline.
hv =Velocity head in the suction pipe at the gauge connection,
expressed in feet.

85. Cavitation is a term used to describe the phenomenon, which occurs in a pump
when there is insufficient NPSH Available. When the pressure of the liquid is
reduced to a value equal to or below its vapor pressure the liquid begins to boil
and small vapor bubbles or pockets begin to form. As these vapor bubbles
move along the impeller vanes to a higher pressure area above the vapor
pressure, they rapidly collapse.
The collapse, or "implosion" is so rapid that it may be heard as a rumbling
noise, as if you were pumping gravel. In high suction energy pumps, the
collapses are generally high enough to cause minute pockets of fatigue failure
on the impeller vane surfaces. This action may be progressive, and under
severe (very high suction energy) conditions can cause serious pitting damage
to the impeller.

86. Capacity
Capacity (Q) is normally expressed in gallons per minute (gpm). Since liquids
are essentially incompressible, there is a direct relationship between the
capacity in a pipe and the velocity of flow.This relationship is as follows:
Where
A = area of pipe or conduit in square feet.
V = velocity of flow in feet per second.
Q = Capacity in gallons per minute
NOTE:On vertical pumps the correction should be made to the eye of the
suction or lowest impeller.

87. Centrifugal Pump Fundamentals - Head Section
The pressure at any point in a liquid can be thought of as being caused by a vertical
column of the liquid which, due to its weight, exerts a pressure equal to the pressure at
the point in question. The height of this column is called the static head and is
expressed in terms of feet of liquid.
The static head corresponding to any specific pressure is dependent upon the weight of
the liquid according to the following formula.
A Centrifugal pump imparts velocity to a liquid. This velocity energy is then transformed
largely into pressure energy as the liquid leaves the pump. Therefore, the head
developed is approximately equal to the velocity energy at the periphery of the impeller
This relationship is expressed by the following well-known formula:

88. Where H = Total head developed in feet.
v = Velocity at periphery of impeller in feet per sec.
g = 32.2 Feet/Sec2
We can predict the approximate head of any centrifugal pump by calculating the
peripheral velocity of the impeller and substituting into the above formula. A handy
formula for peripheral velocity is:
D = Impeller diameter in inches
V = Velocity in ft./sec
The above demonstrates why we must always think in terms of feet of liquid rather
than pressure when working with centrifugal pumps. A given pump with a given
impeller diameter and speed will raise a liquid to a certain height regardless of the
weight of the liquid, as shown in Fig. 1.

89. Fig. 1 Identical Pumps Handling Liquids of
Different Specific Gravities.

90. All of the forms of energy involved in a liquid flow system can be expressed in
terms of feet of liquid. The total of these various heads determines the total
system head or the work which a pump must perform in the system. The
various forms of head are defined as follows.
SUCTION LIFT exists when the source of supply is below the center line of the
pump. Thus the STATIC SUCTION LIFT is the vertical distance in feet from the
centerline of the pump to the free level of the liquid to be pumped.

91. Fig. 2-a Suction Lift ? Showing Static Heads in a Pumping System Where the Pump is
Located Above the Suction Tank. (Static Suction Head)
SUCTION HEAD exists when the source of supply is above the centerline of the pump.
Thus the STATIC SUCTION HEAD is the vertical distance in feet from the centerline of the
pump to the free level of the liquid to be pumped.

92. STATIC DISCHARGE HEAD is the vertical distance in feet between the pump centerline
and the point of free discharge or the surface of the liquid in the discharge tank.
TOTAL STATIC HEAD is the vertical distance in feet between the free level of the source of
supply and the point of free discharge or the free surface of the discharge liquid.
FRICTION HEAD (hf) is the head required to overcome the resistance to flow in the pipe
and fittings. It is dependent upon the size, condition and type of pipe, number and type
of pipe fittings, flow rate, and nature of the liquid. Frictional tables are included in Water
Data.
VELOCITY HEAD (hv) is the energy of a liquid as a result of its motion at some velocity V.
It is the equivalent head in feet through which the water would have to fall to acquire
the same velocity, or in other words, the head necessary to accelerate the water.
Velocity head can be calculated from the following formula:
The velocity head is usually insignificant and can be ignored in most high head systems.
However, it can be a large factor and must be considered in low head systems.

93. PRESSURE HEAD must be considered when a pumping system either begins or terminates
in a tank which is under some pressure other than atmospheric. The pressure in such a
tank must first be converted to feet of liquid. A vacuum in the suction tank or a positive
pressure in the discharge tank must be added to the system head, whereas a positive
pressure in the suction tank or vacuum in the dis-charge tank would be subtracted. The
following is a handy formula for converting inches of mercury vacuum into feet of liquid.
Vacuum, feet of liquid = (Vacuum, in of Hg x 1.13)/ Sp. Gr.
The above forms of head, namely static, friction, velocity, and pressure, are combined to
make up the total system head at any particular flow rate. Following are definitions of
these combined or "Dynamic" head terms as they apply to the pump.
TOTAL DYNAMIC SUCTION LIFT (hs) is the static suction lift minus the velocity head at the
pump suction flange plus the total friction head in the suction line. The total dynamic
suction lift, as determined on pump test, is the reading of a gauge on the suction flange,
converted to feet of liquid and corrected to the pump centerline*, minus the velocity
head at the point of gauge attachment.

94. TOTAL DYNAMIC SUCTION HEAD (hs) is the static suction head plus the velocity head
at the pump suction flange minus the total friction head in the suction line. The total
dynamic suction head, as determined on pump test, is the reading of the gauge on the
suction flange, converted to feet of liquid and corrected to the pump centerline*, plus
the velocity head at the point of gauge attachment.
TOTAL DYNAMIC DISCHARGE HEAD (hd) is the static discharge head plus the velocity
head at the pump discharge flange plus the total friction head in the discharge line.
The total dynamic discharge head, as determined on pump test, is the reading of a
gauge at the discharge flange, converted to feet of liquid and corrected to the pump
centerline*, plus the velocity head at the point of gauge attachment.
TOTAL HEAD (H) or TOTAL Dynamic HEAD (TDH) is the total dynamic discharge head
minus the total dynamic suction head or
TDH = hd + hs (with a suction lift)
TDH = hd - hs (with a suction head)

95. Representation of static discharge head, static suction lift and total static head

98. Cavitation and NPSH
If the suction pressure is only slightly grater than the vapor
pressure, some liquid may flash to vapor inside the pump.
This process is called as “Cavitation” which greatly reduce
the pump capacity and causes severe erosion. If the
suction pressure is actually less than the vapor pressure
there will be vaporization in the suction line and no liquid
can be drawn into the pump.
To avoid the cavitation the pressure at pump inlet must
exceed the vapor pressure by a certain value called as the
Net Positive Suction Head. The required recommended
value of NPSH is about 2 to 3m (5 to 10ft) for small
centrifugal pumps but it increases to 15m (50ft) for larger
pumps.

99. The figure above shows a vertical circular gate in a 3-m
diameter tunnel with water on one side and air on the
other side.

101. A container weighting 100
N “empty” has cross –
sectional area of 0.1 m2
and two equal sized
openings on opposite sides
as shown in the figure. The
container is kept on a
weighting balance and
water flows into it from the
top through a jet of 0.01
m2 area at 8.0 m/s velocity.
If under steady flow
condition the depth of
water in the tank is 1.0 m,
determine the scale

106. Discharge striking the vane = 0.0554 m3/s

109. A Pelton turbine develops 3000 kw under a head of 300m. The overall
efficiency of the turbine is 83%. If the speed ratio = 0.46, Cv = 0.98 and
specificspeed is 16.5, thenfind:
I. Diameter of the turbine, and
II. Diameter of the jet.
Solution
Given:
Power, 𝑃 = 3000 𝑘𝑤 ,
Net head, 𝐻 = 300 𝑚,
Overall efficiency 𝑜 = 83%,
Speed ratio = 0.46,
Cv value, = 0.98, 𝑎𝑛𝑑
Specific speed 𝑁𝑠 = 16.5

110. Using equation,
𝑁𝑠 =
𝑁 𝑃
𝐻5 4
∴ 𝑁 =
𝑁𝑠 ∙ 𝐻5 4
𝑃
=
16.5 ∙ 3005 4
3000
= 375 𝑟. 𝑝. 𝑚.
The velocity “Vn” at the outlet of nozzle is given by:
𝑉
𝑛 = 𝐶𝑉 2 𝑔 𝐻 = 𝐶𝑉 2 × 9.81 × 300 = 75.1 𝑚 𝑠
Since the speed ratio 𝜑 =
𝑉𝑏
2 𝑔 𝐻
∴ 𝑉𝑏 = 𝜑 2 𝑔 𝐻 = 0.46 2 × 9.81 × 300 = 34.95 𝑚 𝑠
The speed of the bucket ratio 𝑉𝑏 =
𝜋 𝐷 𝑁
60
 Diameter of the wheel “D” =
60 𝑉𝑏
𝜋∙𝑁
=
60 ×34.95
𝜋×375
= 1.78 𝑚

111. To determine the nozzle diameter, let
Q = Discharge through the turbine in m3/s, which can be calculated
using,
𝑜 =
𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
𝑖𝑢𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
=
𝑃
𝑊. 𝐻. 𝑃.
=
𝑃
𝜌 𝑔 𝑄 𝐻
𝑐
Substituting the given values
0.83 =
3000
1000 × 9.81 × 𝑄 × 300
1000
∴ 𝑄 =
10
0.83 × 9.81
= 1.228 𝑚3/𝑠
To find the jet diameter,
𝐴𝑗 =
𝜋 𝑑𝑗
2
4
=
𝑄
𝑉𝑗
∴ 𝑑𝑗 =
4 𝑄
𝜋 ×𝑉𝑗
0.5
=
4 ×1.228
𝜋 ×75.1
0.5
= 0.144 𝑚 = 14.4 𝑐𝑚 (𝐴𝑠𝑠𝑢𝑚𝑒 𝑉
𝑛 = 𝑉
𝑗)

112. In turbines, water enters from very high head i.e. very high pressure. As the water
passes through turbine vanes, it gives its energy to vanes and pressure of water
decreases to sub-atmospheric values so that we can have high energy output because of
higher pressure difference across turbine. Now we just can't make turbine open to atm.
because of low pressures otherwise air entrapment could cause problems like
cavitation.
Now in order to again have water exiting at atmospheric pressure, we need to raise
water pressure. To achieve this, we use draft tube and because of its tapering
outward shape, it acts as diffuser since area of flow increases. Thus we have water
comingout fromturbine at atm. pressure.
Another function of draft tube is that it also acts as collector for water coming out
of turbineand directs it to reservoir.

115. Surge Tank
Closed aqueduct to absorb sudden pressure rise and
provide during pressure drop
It is mostly installed when distance between dam and
power house is large
Functions
When load decreases water moves backward & get stored,
Addition supply of water will be provided, when load
increases,
Avoid vacuum during water supply closed.