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Title: HYDRO 1 - Fluid Flow Measurement - Venturi Meter and Nozzle-
Description: Identify devices that can be used to determine discharges in closed conduits. b. Determine discharge in closed conduits using Venturi meter and nozzle. c. Solve problems related to the use of energy and mass conservation law.
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HYDRO 1 (HYDRAULICS)

Topic Learning Outcomes (TLO)
After these topics, the students can
a
...

b
...

c
...


MEASUREMENT OF FLOW THROUGH PIPES
1
...
al
...

Piezometer
s

Q

1

Inle
t

2

Throat

Z
Z

Q

Datum
Line

Write BEE from point 1 to point 2, neglecting head loss between the two points
...

𝑉 12

1

2𝑔
𝑉 22
2𝑔

+

𝑀

𝑉
21
βˆ’

𝑉 22

𝑝

2𝑔

+ 𝑍1 =
𝑝
1

2𝑔

𝑝2
+

𝑀

+ 𝑍2 + 𝐻 𝐿

𝑝2

= ( + 𝑍 1 ) βˆ’ ( + 𝑍 2 ) βˆ’ 𝐻𝐿
𝑀
𝑀

𝑉2 = π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ π‘œπ‘“ π‘“π‘™π‘œπ‘€ π‘Žπ‘‘ π‘π‘œπ‘–π‘›π‘‘ 2
1 |1 4

HYDRO 1 (HYDRAULICS)

The equation
𝑉2𝑑2
2𝑔

𝑝1

𝑉
21
βˆ’

𝑝2

= ( + 𝑍1 ) βˆ’ ( + 𝑍2 )
𝑀
𝑀

2𝑔

shows the Venturi Principle which states that
...

𝑉2𝑑2
2𝑔

𝑉1
2

βˆ’

= π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’ 𝑖𝑛 π‘˜π‘–π‘›π‘’π‘‘π‘–π‘ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦

2
𝑔

𝑝1

𝑝2
(

𝑀

+ 𝑍1 ) βˆ’ (

𝑀

+ 𝑍2) = π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘’ 𝑖𝑛 π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦

Considering a horizontal orientation of the Venturi meter
...
1

Write BEE from point 1 to point 2 (Fig
...

𝑉12
2𝑔

+

𝑉2
2𝑑

𝑝1

𝑝2

+
+ 𝑍1 = 2𝑔
𝑀
𝑉2

𝑀

𝑝
1

𝑉
2

2𝑑

2𝑔

βˆ’

+ 𝑍2 ; 𝑍1 = 𝑍2 = 0

𝑝2

1

=
βˆ’
2𝑔 𝑀 𝑀

The increase in the kinetic energy
...


Considering the pressure head in meters of fluid from point 1 to point 2, let h and Y in
meters, using principles of manometers and Pascal’s Law
...
1)
𝑝1
𝑠𝑝
...
π‘”π‘Ÿπ‘“π‘™π‘’π‘–π‘‘
𝑀
𝑀
𝑝1 βˆ’ 𝑝2
𝑀

𝑀

= β„Ž [(

𝑠𝑝
...
𝑓𝑙𝑒𝑖𝑑

Substitute:

𝑉2
2𝑑

2𝑔

𝑉2
βˆ’

𝑝1

𝑝2

1

=
βˆ’
2𝑔 𝑀 𝑀

𝑠𝑝
...
𝑓𝑙𝑒𝑖𝑑
2𝑔

𝑉2

𝑠𝑝
...
𝑓𝑙𝑒𝑖𝑑
𝐷4
2
[1 βˆ’ (𝐷1) ]

2π‘”β„Ž [(
𝑉2𝑑 = √

The β€œtheoretical discharge”, Qt, is the discharge in the pipe neglecting the loss of head
between the inlet and throat
...
It causes the actual flow, Q, to be less than the theoretical flow, Qt
...
96 to 0
...
A Venturi meter having a throat 15 cm in diameter is installed in a horizontal 30 cm
pipeline carrying a light oil (sp
...
= 0
...
A mercury U-tube connected in the inlet
and the throat shows a difference in height of Hg columns of 20 cm, the remainder
of the tube being filled with oil
...
97, find (a) the actual
discharge in CMS, (b) the loss of head between the inlet and the throat
...
97

30 cm diam pipe
15 cm

Q
Ø

Fluid, light oil
Sp
...
= 0
...
gr
...
6)

Required: Q and HL
Solution:
Write BEE from point 1 to point 2 neglecting head loss
𝑉12 𝑝1
𝑉2
2𝑑
+
= +
2𝑔 𝑀 + 𝑍1
2𝑔
𝑉2

𝑝
1

𝑉2

𝑝2
𝑀

+ 𝑍2

𝑝2

+𝑍 βˆ’π‘
βˆ’ 1= βˆ’
1
2
2
𝑀 𝑀
𝑔
𝑝1 𝑝2
𝐷2 4

2𝑑

2𝑔
𝑉2
2𝑑

2𝑔

[1 βˆ’ (

) ]=
βˆ’
𝐷1
𝑀 𝑀

Considering the pressure head in meters of light oil from point 1 to point 2, using
principles of manometers and Pascal’s Law
...

20
0

𝑝1

𝑀

+π‘Œ+

10
0

𝑝1

βˆ’

𝑀

2
βˆ’

13
...
8
𝑀

𝑝2
𝑀 = 3
...
2

𝑉2
5 |1 4

𝑝2

HYDRO 1 (HYDRAULICS)

2𝑔

𝑉2𝑑 = √

30

2𝑔(3
...
183 π‘šβ„π‘  (π‘‘β„Žπ‘’π‘œπ‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ π‘Žπ‘‘ π‘π‘œπ‘–π‘›π‘‘ 2)
15 4
1 βˆ’ (30)

Then
πœ‹

𝑄 = 𝐢𝐴 𝑉 = (0
...
183) = 0
...
140)

= 7
...
152)

2

πœ‹π·
2

𝑉 = 4𝑄 =
1

1

4(0
...
981 π‘šβ„ (π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ π‘Žπ‘‘ π‘π‘œπ‘–π‘›π‘‘ 1)
𝑠

πœ‹(0
...
981)2

𝑝1
+(

2𝑔

𝑀

βˆ’

(1
...
922)2

𝑝2
)+0βˆ’
𝑀

+ 3
...
922)2
2𝑔

= 0
...
A 30 cm by 15 cm Venturi meter is installed in a vertical pipe carrying water
...
A differential manometer containing a gage liquid
(sp
...
= 1
...

Determine the actual flow if C = 0
...


Given:

water

2

30 cm

15 cmØ

30 cmØ

1

80 cm

Datum line

Liquid
Sp
...
= 1
...

𝑝1
80
𝑠𝑝
...
π‘”π‘Ÿπ‘“π‘™π‘’π‘–π‘‘

𝑀

𝑠𝑝
...

π‘”π‘Ÿ
...
30 + π‘Œ = 0
...
50 π‘š
𝑝1
𝑀

( 80

𝑝2
βˆ’

=

𝑀

Substitute:

10
0

1
...
50 = 0
...
70 + 0 βˆ’ 0
...
40
30

𝑉 =
2𝑑

2𝑔(0
...
235 π‘š
1 4
√
1 βˆ’ (2)

Then:

=

𝑄=𝐴 𝑉
𝑑

2 2𝑑

πœ‹
4

(𝐷2)𝑉 =
2

2𝑑

πœ‹

(0
...
235) = 0
...
97(0
...
055 π‘š ⁄𝑠
3
...
5 mm Venturi
meter the total weight of water passing through the meter in 5
...
A mercury-water differential gage connected to inlet and throat of the meter
showed an average mercury difference during that time of 38 cm
...

Given:

9 |1 4

HYDRO 1 (HYDRAULICS)
50 cm diam pipe

Q

1

12
...
gr
...
0

Y
h = 38 cm

Substance, Hg (sp
...
= 13
...
001162
) (60 𝑠𝑒𝑐 )

π‘š3⁄𝑠 = π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘‘π‘–π‘ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’

1 π‘šπ‘–π‘›

π‘š3
Required:
C = meter coefficient
Solution:

π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘‘π‘–π‘ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’

𝐢=

π‘‘β„Žπ‘’π‘œπ‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ π‘‘π‘–π‘ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’

Write BEE from point 1 to point 2 neglecting head loss between the two points
𝑉12
2𝑔

+

𝑉2

𝑉2
2𝑑
=
+
+ 𝑍1
2𝑔
𝑀

𝑝1

2𝑑

2𝑔

𝐷2 4

2𝑔

𝑀

+ 𝑍2

𝑝2

+𝑍 βˆ’π‘
βˆ’ 1= βˆ’
1
2
2𝑔 𝑀 𝑀

𝑉2
2𝑑

𝑝
1

𝑉2

𝑝2

[1 βˆ’ (

𝑝1

𝑝2

) ]=
βˆ’
+0βˆ’0
𝐷1
𝑀 𝑀

Considering the pressure head in meters of water from point 1 to point 2, let Y in
meters, using principles of manometers and Pascal’s Law
...
π‘šπ‘’π‘Ÿπ‘π‘’π‘Ÿ
𝑝2
)(
)
βˆ’
π‘Œ
=
10
𝑠𝑝
...
6

βˆ’
=
[(
) βˆ’ 1] = 4
...
788
𝑀 𝑀

HYDRO 1 (HYDRAULICS)

𝑉 =
2𝑑

2𝑔(4
...
711 π‘šβ„
)
𝑠
12
...
0
...
711) = 0
...
00116
𝐢=2
= 0
...
00119
2

12 |1 4

𝑠

HYDRO 1 (HYDRAULICS)

4
...
The
vertical distance between the inlet and throat is 30 cm
...
A differential manometer containing a gage liquid (sp
...
= 2
...
Determine the
actual flow if C = 0
...

Q

1

Datum Line

15 cm ΓΈ
2

30 cm
x

ΞΈ

50 cm

Q

Gage liquid (sp
...
= 2
...

𝑉12 𝑝1
𝑉2
2𝑑
+
=
+
2𝑔 𝑀 + 𝑍1
2𝑔
𝑉2

𝑝
1

𝑉2

𝑝2
𝑀

+ 𝑍2

𝑝2

+𝑍 βˆ’π‘
βˆ’ 1= βˆ’
1
2
2𝑔 𝑀 𝑀

2𝑑

𝑉2
2𝑑

2𝑔

2
𝑔

𝐷2 4

[1 βˆ’ (

𝑝1

𝑝2

) ]=
βˆ’
+ 𝑍1 βˆ’ 𝑍2
𝐷1
𝑀 𝑀

Considering the pressure head in meters of water from point 1 to point 2, let X in
meters, using principles of manometers and Pascal’s Law
...
π‘”π‘Žπ‘”π‘’ π‘™π‘–π‘žπ‘’π‘–π‘‘ ) βˆ’ π‘₯ = 𝑝2

10
0

𝑠𝑝
...
π‘”π‘Žπ‘”π‘’ π‘™π‘–π‘žπ‘’π‘–π‘‘
) [(
10
𝑠𝑝
...
30
0

50

2
...
30 = 0
...
45 + 0 βˆ’ (βˆ’0
...
75

𝑉2

2𝑔

14 |1 4

30

HYDRO 1 (HYDRAULICS)

𝑉 =
2𝑑

2𝑔(0
...
962 π‘š
1 4
√
1 βˆ’ (2)

Then:

=

𝑄=𝐴 𝑉
𝑑

2 2𝑑

πœ‹
4

(𝐷2)𝑉 =
2

2𝑑

πœ‹

(0
...
962) = 0
...
96(0
...
067 π‘š ⁄𝑠

15 |1 4

HYDRO 1 (HYDRAULICS)

2
...
al pages 107 - 108
A nozzle is a converging tube connected to the end of a pipe or hose and is used in
engineering practice for the creation of jets and streams for a variety of purposes as well
as for metering
...

V1

1

2

Liquid jet
V2

Tip of Nozzle

Datum Line

Base of Nozzle

Q

BEE from 1 to 2, neglecting head loss
...


π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦
π‘‘β„Žπ‘’π‘œπ‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦

𝑉12 + 𝑝

2
= 𝑉2 + 𝐻

1

2𝑔

17 |1 4

𝑀

2𝑔

𝐿

HYDRO 1 (HYDRAULICS)

𝑝
𝑉
𝑉2
2
𝑉2 1
𝐻 =( 1 +
) βˆ’ 2 =π»βˆ’ 2
𝐿
2𝑔 𝑀
2
2𝑔
𝑔
𝑉2
𝐻 =
𝐿

𝑉
2

𝑉2

1

βˆ’ 1)
βˆ’2=2 (
2𝑔𝐢𝑉2 2𝑔 2𝑔 𝐢𝑉2
2

Sample problems:
1
...
At a point 30 cm below the
tip, the pressure is maintained at 145 Kpa
...
5
KW, find (a) the height through which the jet of water will rise, neglecting air friction,
(b) the value of CV, and (c) the diameter of the jet 6 m above the tip of the nozzle,
and (d) head loss in the nozzle
...
5 KW
Required:
h, CV, D4
Solution:
Points 2, 3, and 4 are on the liquid jet so the pressure at those points are all
atmospheric
...

𝑉12
2𝑔
𝑉2
18 |1 4

+

𝑉2
2𝑑
=
+
+ 𝑍1
2𝑔
𝑀

𝑝1

𝑉2

𝑝
1

𝑝2

𝑝2
𝑀

+ 𝑍2

HYDRO 1 (HYDRAULICS)
2𝑑

𝑉2
2𝑑

2𝑔

19 |1 4

2
𝑔
[1 βˆ’ (

βˆ’ 1=
2𝑔 𝑀
𝐷2 4

βˆ’

+𝑍 βˆ’π‘
1

𝑀
𝑝1

2

𝑝2

) ]=
βˆ’
+ 𝑍1 βˆ’ 𝑍2
𝐷1
𝑀 𝑀

HYDRO 1 (HYDRAULICS)
4

4
2
𝑉2𝑑
[1 βˆ’ ( ) ] =
2𝑔

10

9
...
3 = 14
...
81

2𝑔(14
...
076 π‘š
4 4
√
1 βˆ’ (10)

Write BEE from point 2 (just after the tip of the nozzle) to point 3 neglecting air
resistance
𝑉22 𝑝2
𝑉32 𝑝3
+
=
+
2𝑔 𝑀 + 𝑍2 2𝑔 𝑀 + 𝑍3 + 𝐻𝐿(2βˆ’3)
𝑉2
+ 0 + 0
...
3 + β„Ž + 0
2𝑔
2

𝑉22
= β„Ž = π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ β„Žπ‘’π‘Žπ‘‘
2𝑔
Power in kilowatts (KW)
πΎπ‘–π‘™π‘œπ‘€π‘Žπ‘‘π‘‘π‘  =

(𝐾𝑁)(
π‘š)
𝑠𝑒𝑐

π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 𝑀𝑄𝐻
π‘π‘œπ‘€π‘’π‘Ÿπ‘—π‘’π‘‘ = π‘€π‘„β„Ž
𝑀 (πœ‹
𝑉2
) (𝐷 )2(𝑉3)
= 𝑀𝐴 𝑉 ( 2 ) =

π‘π‘œπ‘€π‘’π‘Ÿ
𝑗𝑒𝑑

22

2
𝑔

2𝑔 4

2

2

4 2
9
...
5 πΎπ‘Š = 1 (4) ( 100) (𝑉23)
2𝑔

3

𝑉2 = √

4(1002)(2𝑔)(2
...
846 ⁄𝑠
5)
πœ‹(4)2(9
...

πŸ–πŸ’πŸ”)𝟐

= 𝟏𝟐
...
πŸ–πŸ’πŸ”
=

πŸπ’•

πŸπŸ•
...
πŸ—πŸπŸ–

Write BEE from point 2 (just after the tip of the nozzle) to point 4 neglecting air
resistance
𝑉22
2𝑔

+

𝑝2
𝑀

𝑉
+ 𝑍2 =

24

2
𝑔

+

𝑝4
𝑀 + 𝑍4 + 𝐻𝐿(2βˆ’4)

𝑉22
𝑉
+ 0 + 0
...
30 + 6 + 0
2𝑔
2
𝑔

21 |1 4

HYDRO 1 (HYDRAULICS)

(15
...
798

𝑉4 = √2𝑔(6
...
549π‘šβ„π‘ 
𝑄 = 𝑄1 = 𝑄2 = 𝑄4
𝐴2𝑉2 = 𝐴4𝑉4
πœ‹
πœ‹
(𝐷 )2𝑉 = (𝐷 )2𝑉
4 2 2 4 4 4
𝐷 =√

𝐷22(𝑉2)

4

𝑉4

=√

(4)2(15
...
549

= 4
...

𝑉12 𝑝1
𝑉22 𝑝2
+
=
+
+ 𝑍1
+ 𝑍2 + 𝐻𝐿(1βˆ’2)
2𝑔 𝑀
2𝑔 𝑀
1

𝐷

2

[(
2𝑔

2

(15
...
30 + 𝐻

9
...
30 = 2
...
81

2
...
The nozzle,
trained vertically up, is 1
...
The
head losses are: pump to hydrant: 3J/N; hydrant: 2 J/N; hydrant to nozzle base: 12
J/N; nozzle: 6% velocity head in the nozzle
...

Given:
Y
h

jet
X 2 cmØ tip
nozzle

8 cmØ
22 |1 4

1
...
π‘š
=3
= 3π‘š
𝐿(π‘π‘’π‘šπ‘βˆ’β„Žπ‘¦π‘‘π‘Ÿπ‘Žπ‘›π‘‘
𝑁
𝑁
𝐽
𝐻𝐿(β„Žπ‘¦π‘‘π‘Ÿπ‘Žπ‘›π‘‘ = 2 ⁄𝑁
𝐻𝐿(β„Žπ‘¦π‘‘π‘Ÿπ‘Žπ‘›π‘‘βˆ’π‘›π‘œπ‘§π‘§π‘™π‘’ π‘π‘Žπ‘ π‘’ = 12 𝐽⁄𝑁
2
𝐻
= 6% (𝑉𝑋 )
𝐿(π‘›π‘œπ‘§π‘§π‘™π‘’)
2𝑔
𝑃𝐴 = 550 πΎπ‘ƒπ‘Ž
= 3 𝐽⁄

Required: h
Solution:
Write BEE from A (a point just after the pump) to X (a point just after the tip of the
nozzle)
𝑉𝐴2 𝑝𝐴
𝑉𝑋2 𝑝𝑋
+
=
+
2𝑔 𝑀 + 𝑍𝐴
2𝑔 𝑀 + 𝑍𝑋 + 𝐻𝐿(π΄βˆ’π‘‹)
1
2𝑔

𝐷
2
[(
)
𝐷𝐴

𝑋

2

550

1

0
...
8
2
]
1
𝑔

2
𝑔

{𝑉𝑋}2]

55
𝑉𝑋2 2 4
0
[( ) βˆ’ 1 βˆ’ 0
...
81
𝑉𝑋 = 22
...
657)
2

+ 0 + 12 = 0 + 0 + 12 + β„Ž + 0 β„Ž =

2𝑔
26
Title: HYDRO 1 - Fluid Flow Measurement - Venturi Meter and Nozzle-
Description: Identify devices that can be used to determine discharges in closed conduits. b. Determine discharge in closed conduits using Venturi meter and nozzle. c. Solve problems related to the use of energy and mass conservation law.
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