Fluid Mechanics I

Third Year, Mechanical Department.

Course Coordinator:
Dr. Mohamed Sayed Soliman
Mechanical Power Eng. Dept., Faculty of Engineering,
Port-Said University.
1
Port-Said 2023-2024
 Lecture 11

Chapter 3 Fluid Dynamics

2
A venturimeter of 20 mm throat diameter is used to measure the velocity of water in a horizontal
pipe of 40 mm diameter. The venturimeter equipped with a differential pressure gauge between
the pipe and throat sections is found to be 30 kPa, neglecting frictional losses. Determine the
flow velocity in the pipe.
Data: 𝐷𝐷2 = 20 𝑚𝑚𝑚𝑚 𝐷𝐷1 = 40 𝑚𝑚𝑚𝑚 ∆𝑃𝑃 = 𝑃𝑃1 − 𝑃𝑃2 = 30 kPa water ℎ𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 0.0 𝑚𝑚
Req: 𝑉𝑉1
1- And the continuity equation 𝑄𝑄 = 𝑉𝑉1 𝐴𝐴1 = 𝑉𝑉2 𝐴𝐴2 1 2
𝜋𝜋 2
𝐴𝐴1 . 𝐷𝐷1 0.16
𝑉𝑉2 = . 𝑉𝑉 = 4 . 𝑉𝑉 = 𝑉𝑉 = 4𝑉𝑉1
𝐴𝐴2 1 𝜋𝜋 . 𝐷𝐷 2 1 0.04 1
4 2
2- Apply Bernoulli's equation between the inlet pipe, 1, and the throat, 2.

𝑝𝑝1 𝑢𝑢1 2 𝑝𝑝2 𝑢𝑢2 2

+ + 𝑧𝑧1 = + + 𝑧𝑧2
𝜌𝜌𝜌𝜌 2𝑔𝑔 𝜌𝜌𝜌𝜌 2𝑔𝑔

𝑝𝑝1 − 𝑝𝑝2 𝑉𝑉2 2 − 𝑉𝑉1 2 30 × 1000 16𝑉𝑉1 2 − 𝑉𝑉1 2 15𝑉𝑉1 2

= = =
𝜌𝜌𝜌𝜌 2𝑔𝑔 1000 2 2
60 𝒎𝒎
𝑉𝑉1 2 = =4 𝑽𝑽𝟏𝟏 = 𝟐𝟐 𝑄𝑄 = 𝑉𝑉1 𝐴𝐴1 3
15 𝒔𝒔
3- Weir and Notch

V-notch Rectangular weir

4
3- Weir and Notch
A notch may be defined as a sharp-edged obstruction over which flow of a liquid occurs. As the
depth of flow above the base of the notch is related to the discharge, the notch forms a useful
measuring device. The area of flow is most commonly either rectangular or V-shaped. A large
rectangular notch is more often termed a sharp-crested weir.

(b) V-notch

Consider a sharp-edged, rectangular notch, the analysis requires the following assumptions:
1. Upstream of the notch, the velocities of particles in the stream are uniform and parallel
2. The free surface remains horizontal.
3. The pressure throughout the nape is atmospheric.
4. The effects of viscosity and surface tension are negligible. 5
 2 3�
𝑄𝑄𝑡𝑡𝑡 = . 𝑏𝑏. 2𝑔𝑔. 𝐻𝐻 2 Rectangular weir
3
8 𝜃𝜃 5�
𝑄𝑄𝑡𝑡𝑡 = . tan . 2𝑔𝑔. 𝐻𝐻 2 V-notch
15 2

𝑄𝑄𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐶𝐶𝑑𝑑 . 𝑄𝑄𝑡𝑡𝑡

6
4- Pitot-Static Tube
A point in a fluid stream where the velocity is reduced to
zero is known as a stagnation point. The velocity at X is
therefore zero: X is a stagnation point.
By Bernoulli’s equation (between O and X)

𝑝𝑝0 𝑉𝑉0 2 𝑝𝑝𝑥𝑥 𝑉𝑉𝑥𝑥 2

+ + 𝑧𝑧0 = + + 𝑧𝑧𝑥𝑥
𝜌𝜌𝜌𝜌 2𝑔𝑔 𝜌𝜌𝜌𝜌 2𝑔𝑔
0
𝑝𝑝𝑜𝑜 𝑉𝑉𝑜𝑜 2 𝑝𝑝𝑋𝑋
+ =
�𝛾𝛾 2𝑔𝑔
� �𝛾𝛾
Static Dynamic Stagnation pressure
pressure pressure or total pressure

• Measurement of static pressure, using

piezometer tube.
(a) piezometer tube, (b) pitot tube
7
8
 4- Pitot-Static Tube
 The tubes recording static pressure and stagnation pressure are
frequently combined into one instrument known as a Pitot-
static tube.

 The pitot-static tube can be measured the local velocity

directly.
 A simple Pitot tube can be used to measure, Static head,
Dynamic head Total head

9
10
 3- Conservation of Momentum

It is often important to determine the force produced on a solid body by fluid flowing steadily
over or through it. For example:
(1) the force exerted on a solid surface by a jet of fluid impinging on it;
(2) the aerodynamic forces (lift and drag) on an aircraft wing,
(3) the force on a pipe-bend caused by the fluid flowing within it,
(4) the thrust on a propeller and so on. All these forces are associated with a change in the
momentum of the fluid.

𝐹𝐹 = 𝑚𝑚.̇ 𝑉𝑉

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Nozzle

Nozzle

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 The momentum equation for steady flow, and its applications
In its most general form, Newton’s Second Law states that the net force acting on a body in any fixed
direction is equal to the rate of increase of momentum of the body in that direction.

� 𝐹𝐹 = 𝑚𝑚.̇ 𝑉𝑉 = 𝑚𝑚̇ 2 . 𝑉𝑉2 − 𝑚𝑚̇ 1 . 𝑉𝑉1 = 𝑚𝑚̇ 𝑉𝑉2 − 𝑉𝑉1

1- The force caused by flow round a pipe-bend

𝐹𝐹𝑥𝑥 + 𝑃𝑃1,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴1 − 𝑃𝑃2,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴2 = 𝑚𝑚̇ 𝑉𝑉2𝑥𝑥 − 𝑉𝑉1𝑥𝑥

𝑥𝑥 𝑥𝑥

𝐹𝐹𝑥𝑥 + 𝑃𝑃1 . 𝐴𝐴1 cos 𝜃𝜃 − 𝑃𝑃2 . 𝐴𝐴2 = 𝜌𝜌. 𝑉𝑉. 𝐴𝐴. (𝑉𝑉2 − 𝑉𝑉1 cos 𝜃𝜃)

𝐹𝐹𝑦𝑦 + 𝑃𝑃1,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴1 − 𝑃𝑃2,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴2 = 𝑚𝑚̇ 𝑉𝑉2𝑦𝑦 − 𝑉𝑉1𝑦𝑦

𝑦𝑦 𝑦𝑦

𝐹𝐹𝑦𝑦 + 𝑃𝑃1 . 𝐴𝐴1 sin 𝜃𝜃 − 0 = 𝜌𝜌. 𝑉𝑉. 𝐴𝐴. 0 − 𝑉𝑉1 sin 𝜃𝜃

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 Water flows steadily through a section of expanding elbow shown in the diagram. At the inlet to the
elbow, the absolute pressure is 230 kPa and the cross-sectional area is 0.004 m2. The direction of flow is
45° from horizontal. At the outlet, the absolute pressure is 200 kPa, the cross-sectional area is 0.01 m2,
and the velocity is 2 m/s. Determine the anchoring forces in x and y directions required to hold the elbow
in place.
Data: Water 𝑃𝑃1,𝑎𝑎𝑎𝑎𝑎𝑎 = 230 kPa 𝐴𝐴1 = 0.004 𝑚𝑚2 𝜃𝜃 = 45° 𝑃𝑃2,𝑎𝑎𝑎𝑎𝑎𝑎 = 200 kPa 𝐴𝐴2 = 0.01 𝑚𝑚2 𝑉𝑉2 = 2𝑚𝑚/𝑠𝑠
Req: 𝑅𝑅𝑥𝑥 , 𝑅𝑅𝑦𝑦 to hold the elbow in place

𝑃𝑃1,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 230−101=129 kPa 𝑃𝑃2,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 200−101= 99 kPa

𝐴𝐴2 0.01
𝜌𝜌 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑄𝑄 = 𝑉𝑉1 𝐴𝐴1 = 𝑉𝑉2 𝐴𝐴2 𝑉𝑉1 = . 𝑉𝑉2 = . 𝑉𝑉 = 5 𝑚𝑚/𝑠𝑠
𝐴𝐴1 0.004 2
� 𝐹𝐹 = 𝑚𝑚.̇ 𝑉𝑉 = 𝑚𝑚̇ 𝑉𝑉2 − 𝑉𝑉1

𝐹𝐹𝑥𝑥 + 𝑃𝑃1,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴1 − 𝑃𝑃2,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴2 = 𝑚𝑚̇ 𝑉𝑉2𝑥𝑥 − 𝑉𝑉1𝑥𝑥

𝑥𝑥 𝑥𝑥

𝐹𝐹𝑥𝑥 + 𝑃𝑃1 . 𝐴𝐴1 cos 𝜃𝜃 − 𝑃𝑃2 . 𝐴𝐴2 = 𝜌𝜌. 𝑉𝑉2 . 𝐴𝐴2 . (𝑉𝑉2 − 𝑉𝑉1 cos 𝜃𝜃)

𝐹𝐹𝑥𝑥 + 129 ∗ 103 ∗ 0.004 ∗ cos 45 − 99 ∗ 103 ∗ 0.01 = 1000 ∗ 0.01 ∗ 2 (2 − 5 ∗ cos 45)
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𝐹𝐹𝑥𝑥 = [𝑁𝑁]
𝐹𝐹𝑦𝑦 + 𝑃𝑃1,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴1 − 𝑃𝑃2,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴2 = 𝑚𝑚̇ 𝑉𝑉2𝑦𝑦 − 𝑉𝑉1𝑦𝑦
𝑦𝑦 𝑦𝑦

𝐹𝐹𝑦𝑦 + 129 ∗ 103 ∗ 0.004 ∗ sin 45 − 0 = 1000 ∗ 0.01 ∗ 2 ∗ (0 − 5 ∗ sin 45)

𝐹𝐹𝑦𝑦 = [𝑁𝑁]

∴ 𝑡𝑡𝑡𝑡𝑡 𝑅𝑅𝑥𝑥 , 𝑅𝑅𝑦𝑦 to hold the elbow in place

∴ 𝑡𝑡𝑡𝑡𝑡 𝑅𝑅𝑥𝑥 = −𝐹𝐹𝑥𝑥 = 𝑁𝑁

𝑅𝑅𝑦𝑦 = − 𝐹𝐹𝑦𝑦 = [𝑁𝑁]

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  Application of Momentum Equation
1- The force caused by flow round a pipe-bend
1- The net momentum change in x-direction

𝐹𝐹𝑥𝑥 + 𝑃𝑃1,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴1 − 𝑃𝑃2,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴2 = 𝑚𝑚̇ 𝑉𝑉2𝑥𝑥 − 𝑉𝑉1𝑥𝑥

𝑥𝑥 𝑥𝑥

𝐹𝐹𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉2𝑥𝑥 − 𝑉𝑉 = 𝑚𝑚.̇ 𝑉𝑉2 cos 𝜃𝜃 − 𝑉𝑉1

2- Similarly, the net momentum change in Y-direction

𝐹𝐹𝑦𝑦 + 𝑃𝑃1,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴1 − 𝑃𝑃2,𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 𝐴𝐴2 = 𝑚𝑚̇ 𝑉𝑉2𝑦𝑦 − 𝑉𝑉1𝑦𝑦

𝑦𝑦 𝑦𝑦

𝐹𝐹𝑦𝑦 = 𝑚𝑚.̇ 𝑉𝑉2𝑦𝑦 − 𝑉𝑉1𝑦𝑦 = 𝑚𝑚.̇ 𝑉𝑉2 sin 𝜃𝜃 − 0

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EXAMPLE
A 45𝑜𝑜 reducing pipe-bend (in a horizontal plane) tapers from 600 mm diameter at inlet to 300
mm diameter at outlet. The rate of flow of water through the bend is 0.425 m3/s. Neglecting
pressure force and friction, calculate the net resultant force exerted by the water on the bend.
Data: D1 = 600 mm D2= 300 mm Q = 0.425 m3/s
Req.: FR
Solution
Form continuity equation 𝑄𝑄 = 𝑉𝑉1 𝐴𝐴1 = 𝑉𝑉2 𝐴𝐴2

𝑄𝑄 𝑄𝑄
𝑉𝑉1 = 𝜋𝜋 = 1.503 𝑚𝑚/𝑠𝑠 𝑉𝑉2 = 𝜋𝜋 = 6.01 𝑚𝑚/𝑠𝑠
2 2
. 𝑑𝑑 . 𝑑𝑑
4 1 4 2

𝐹𝐹𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉2𝑥𝑥 − 𝑉𝑉1𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉2 cos 𝜃𝜃 − 𝑉𝑉1 = 1000 ∗ 0.425 6.01 cos 45 − 1.503 = 1167.35 𝑁𝑁
𝐹𝐹𝑦𝑦 = 𝑚𝑚.̇ 𝑉𝑉2𝑦𝑦 − 𝑢𝑢1𝑦𝑦 = 𝑚𝑚.̇ 𝑉𝑉2 sin 𝜃𝜃 − 0 = 1000 ∗ 0.425 6.01 sin 45 − 0 = 1806.13 𝑁𝑁

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  Application of Momentum Equation

2- The force caused by a jet striking a surface

1 − 𝐹𝐹𝐹𝐹𝐹𝐹 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝐹𝐹𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝜌𝜌. 𝐴𝐴. 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 (𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 0)

2
𝐹𝐹𝑥𝑥 = 𝜌𝜌. 𝐴𝐴. 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗

𝐹𝐹𝑦𝑦 = 0.0

Jet striking a plate surface

2 − 𝐹𝐹𝐹𝐹𝐹𝐹 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝐹𝐹𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝜌𝜌. 𝐴𝐴. 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 . (𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 )
2
𝐹𝐹𝑥𝑥 = 𝜌𝜌. 𝐴𝐴. 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝐹𝐹𝑦𝑦 = 0.0 18
A jet of water of 100 mm diameter impinges normally on a fixed plate with a velocity of 30 m/s.
Find the force exerted on the plate.
𝑚𝑚
Data: d = 100 mm = 0.1 m fixed plate 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 = 30 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 0.0
𝑠𝑠
Req: F
𝐹𝐹𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝑚𝑚.̇ 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 0
𝜋𝜋
𝑚𝑚̇ = 𝜌𝜌. 𝐴𝐴. 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 = 1000 ∗ ∗ 0.12 ∗ 30 = 235.62 [𝑘𝑘𝑘𝑘⁄𝑠𝑠]
4
𝐹𝐹𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 0 = 235.62 ∗ 30 − 0.0 = 7068.58 [𝑁𝑁]
𝐹𝐹𝑦𝑦 = 0.0
∴ 𝐹𝐹 = −𝐹𝐹𝑥𝑥 = −7068.58 [𝑁𝑁]

If the plate is moving by 20 m/s, find the force exerted on the plate
𝐹𝐹𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝜋𝜋
𝑚𝑚̇ = 𝜌𝜌. 𝐴𝐴 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 1000 ∗ ∗ 0.12 ∗ (30 − 20) = 78.54 [𝑘𝑘𝑘𝑘⁄𝑠𝑠]
4
𝐹𝐹𝑥𝑥 = 𝑚𝑚.̇ 𝑉𝑉𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 78.54 ∗ 30 − 20 = 785.4 [𝑁𝑁]
19
𝐹𝐹𝑦𝑦 = 0.0
20