IMP_Fluid_Mechanics_lectures_and_Tutorials.pdf

Fluid Mechanics lectures and Tutorials 1
Unit1: Introduction
What we are meaning by Fluids?
Fluid may be defined as a substance which deforms continuously (flows)
when subjected to shearing forces, or
A fluid is a substance which capable of flowing
A fluid has no definite shape unless it is supported (conforms to the shape of the
containing vessel)
Mechanicsis the field of science focused on the force, energy, motion,
deformation interactions of material bodies based on their properties.
What are we meaning by Fluid Mechanics?
Fluid mechanics is the study of fluids, how they move, how they mix, how they
interact with or how they effect on the bodies submerged within, and how they
interact with and effect on the bodies that attached them and their reflections on
human activities.
Fluid mechanics may be defined also as that branch of engineering science that
deals with the behavior of fluid under the condition of rest and motion
Fluid mechanics may be divided into three parts: Statics, Kinematics, and
Dynamics
Statics Deals with fluid at rest in equilibrium state, no force no acceleration
Kinematics Deals With flow behaviors of fluid like velocity, acceleration and flow
patterns.
Dynamics Deals with the effects of flow behaviors on fluid surroundings like
forces and momentum exchange

The matter states
The matter or substance is classified on the bases of the spacing between the
molecules of the matter as follows:
 In solids, the molecules are very closely spacing and then inter-molecules
cohesive forces is quite large, and then possess compactand rigid form.
 Whereas in liquids these spacing are relatively large, and then less inter-
molecules cohesive forces between them, and then can move freely, but it still
has a definite volume (no definite shape, has free interface).
 While these forces is extremely small in gasses, and then have greater
freedom of movement so that the gas fill the container completely in which
they are placed (no definite volume, no definite shape, and no free interface).
Matter or
Substance
Fluid State
Liquid
State
Gaseous
State
Solid State

Attribute Solid Liquid
Gas
Typical
Visualization
Macroscopic
Description
Solids hold their shape;
no need for a container
Liquids take the
shape of the
container and will
stay in open
container
Gases expand to fill a
closed container
Mobility of
Molecules
Molecules have low
mobility because they
are bound in a structure
by strong intermolecular
forces
Liquids typically
flow easily even
though there are
strong
intermolecular forces
between molecules
Molecules move around
freely with little interaction
except during collisions;
this is why gases expand to
fill their container
Typical Density
Often high; e.g., density
of steel is 7700 kg/m3
Medium; e.g.,
density of water is
1000 kg/m3
Small; e.g., density of air at
sea level is 1.2 kg/m3
Molecular
Spacing
Small—molecules are
close together
Small—molecules
are held close
together by
intermolecular forces
Large—on average,
molecules are far apart
Effect of Shear
Stress
Produces deformation Produces flow Produces flow
Effect of Normal
Stress
Produces deformation
that may associate with
volume change; can
cause failure
Produces
deformation
associated with
volume change
Produces deformation
associated with volume
change
Viscosity NA
High; decreases as
temperature
increases
Low; increases as
temperature increases
Compressibility
Difficult to compress;
bulk modulus of steel is
160 × 109 Pa
Difficult to
compress; bulk
modulus of liquid
water is 2.2 × 109 Pa
Easy to compress; bulk
modulus of a gas at room
conditions is about
1.0 × 105 Pa

System of units
MKS system of units
This is the system of units where the metre (m) is used for the unit of length,
kilogram (kg) for the unit of mass, and second (s) for the unit of time as the
base (primary) units.
CGS system of units
This is the system of units where the centimetre (cm) is used for length, gram
(g) for mass, and second (s) for time as the base (primary) units.
International system of units (SI)
SI, the abbreviation of La Systeme International d’Unites, is the system
developed from the MKS system of units. It is a consistent and reasonable
system of units which makes it a rule to adopt only one unit for each of the
various quantities used in such fields as science, education and industry.
There are seven fundamental SI units, namely: metre (m) for length,
kilogram (kg) for mass, second (s) for time, ampere (A) for electric
current, kelvin (K) for thermodynamic temperature, mole (mol) for mass
quantity and candela (cd) for intensity of light. Derived units consist of these
units.
BASIC (PRIMARY) DIMENSIONS
Dimension Symbol Unit (SI)
Length L meter (m)
Mass M kilogram (kg)
Time T second (s)
Temperature θ kelvin (K)
Electric current i ampere (A)
Amount of light C candela (cd)
Amount of matter N mole (mol)

Fluid properties
General fluid (liquid) properties:
1. Mass Density:the density (also known as specific mass or density)of a
liquid defined as the mass per unit volume at a standard temperature and
pressure. It is usually denoted by Latin character ρ (rho). Its unit are Kg/m3
𝜌 =
𝑚
𝑉
𝜌 of water = 1000 kg/m3
at 4o
C and 1 Atm. (see tables A.2-5)
𝜌 = 𝑓(𝑃, 𝑇)
2. Weight Density: (also known as specific weight) is defined as the weight per
unit volume at the standard temperature and pressure, it is usually denoted as
γ. its unit ere N/m3
.
𝛾 =
𝑊
𝑉
=𝜌×𝑔
Where g gravitational acceleration=9.81 m/s2
γ of water = 9810 N/m3
at 4o
C and 1 Atm. (see tables A.2-5)
𝛾 = 𝑓(𝑃, 𝑇, 𝑔)
3. Specific Volume:It is defined as a volume per unit mass of fluid, It is denoted
by v
𝑣 =
𝑉
𝑚
=
1
𝜌
Its unit are m3
/Kg.
4. Specific Gravity:It is defined as the ratio of the specific weight of the fluid to
the specific weight of a standard fluid
For liquids the standard fluid is pure water at the specified temperature, and
denoted by Sg
i.e. 𝑆𝑐 =
𝛾𝑙𝑖𝑞𝑢𝑖𝑑
𝛾𝑤𝑎𝑡𝑒𝑟
)
𝑇
For Gasses the standard fluid is air
- As identical to specific gravity, Relative Density may come as the ratio of the
density of the fluid to the density of a standard fluid
For liquids the standard fluid is pure water at the specified temperature, and
denoted by rd
i.e. 𝒓𝒅 =
𝜌𝑙𝑖𝑞𝑢𝑖𝑑
𝜌𝑤𝑎𝑡𝑒𝑟
)
𝑇

Example2

Example 3:CalculatetheSpecificweight,specificmass,specificvolum and the
total weight of the crude oil truk having a volume of 36m3
and S = 0.85 at 4 o
C and
the empty truck weigh = 16 T

5. Viscosity:it is a property of a real fluid (an ideal fluid has no viscosity) which
determine its resistance to shearing stresses. It is primarily due to cohesion,
adhesion and molecular momentum exchange between fluid layers.
1 - For solids, shear stress reflect on magnitude of
angular deformation (τ ~ angular deformation, θ)
2 – For many fluids shear stress is proportional to
the time rate of angular deformation (τ ~ dθ/dt)
When tow layer of fluid at the distance of δy apart, move one over the other at
different velocities, say u and u+δu, the viscosity together with relative velocity
δx
τ
δθ δθ
u = 0
u = δu
δu δt
τ α
𝒅𝜽
𝒅𝒕
δy
du
dy
u(y)
No slip at wall
Velocityprofile
0
τ = μ
𝒅𝒖
𝒅𝒚

causes shear stress acting between layers. With respect to the distance between
these two layers δy, the shear stress, τ, proportional to angular deformation
𝜏 ∝
𝛿𝜃
𝛿𝑡
From the geometry of Fig. we see that
tan 𝛿𝜃 =
𝛿𝑢 𝛿𝑡
𝛿𝑦
In the limit of infinitesimal changes, this becomes a relation between shear strain
rate and velocity gradient:
𝑑𝜃
𝑑𝑡
=
𝑑𝑢
𝑑𝑦
𝜏 ∝
𝛿𝑢
𝛿𝑦
Newton’s law of viscosity:the shear stresses on a fluid element layers is directly
proportional to the velocity gradient (rate of shear strain). The constant of
proportionality is called the coefficientof viscosity (absolute viscosity, dynamic viscosity, or
simply viscosity) and denoted as μ (mu).
i.e. 𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
Coefficientof Dynamic Viscosity: 𝜇 =
𝜏
𝑑𝑢
𝑑𝑦
Units: N s/m2
or Pa s or kg/m s
The unit Poise (p) is also used where 10P = 1 Pa·s (1P = 0.1 Pa·s)
Water μ = 8.94 × 10−4
Pa s at 25 o
C
Water μ = 1.00 × 10−3
Pa s at 20 o
C
Mercury μ = 1.526 × 10−3
Pa s
Olive oil μ = .081 Pa s
Kinematic Viscosity,ν = the ratio of dynamic viscosity to mass density
𝑣 =
𝜇
𝜌
Units m2
/s and Called kinematic viscosity because it involves no
force (dynamic) dimensions .
The unit Stoke (St) is also used where 1St = 10-4
m2
/s (1St
=cm2
/s)
Water ν = 1.7 × 10−6
m2
/s. at 0 o
C
Water ν = 1.00 × 10−6
m2
/s. at 20 o
C
Air ν = 1.5 × 10−5
m2
/s.

 The fluid is non-Newtonian if the relation between shear stress and shear
strain rate is non-linear
 Typically, as temperature increases, the viscosity will decrease for a liquid, but
will increase for a gas.

: Example4
In figure if the fluid is oil at 20oC (µ = 0.44 Pa.s). What shear stress is required to move the
upper plate at 3.5 m/s?
Solution:
𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
= 0.44 𝑃𝑎.𝑠 ×
3.5 𝑚/𝑠
7
1000𝑚
= 220 𝑃𝑎
Example 5
A board 1 m by 1 m that weighs 25 N slides down an inclined ramp (slope = 20°) with a
velocity of 2.0 cm/s. The board is separated from the ramp by a thin film of oil with a
viscosity of 0.05 N.s/m2. Neglecting edge effects, calculate the space between the board
and the ramp.
Problem Definition
Situation: A board is sliding down a ramp, on a thin film of oil.
Find: Space (in m) between the board and the ramp.
Assumptions: A linear velocity distribution in the oil.
Properties: Oil, μ = 0.05 N・ s/m2.
Sketch:
D=7mm

Plan
1. Draw a free body diagram of the board, as shown in “sketch.”
· For a constant sliding velocity, the resisting shear force is equal to the component of weight parallel to the
inclined ramp (equilibrium condition must be exist).
· Relate shear force to viscosity and velocity distribution.
2. With a linear velocity distribution, dV/dy can everywhere be expressed as ΔV/Δy, where ΔV is the velocity of
the board, and Δy is the space between the board and the ramp.
3. Solve for Δy.
Solution
1. Free-bodyanalysis
𝐹𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 = 𝐹𝑠ℎ𝑒𝑎𝑟
𝑊. sin20𝑜
= 𝜏 × 𝐴𝑟𝑒𝑎
𝑊.sin 20𝑜
= 𝜇
𝑑𝑉
𝑑𝑦
𝐴
2. Substitution of dV/dy as ΔV/Δy
𝑊. sin20𝑜
= 𝜇
∆𝑉
∆𝑦
𝐴
3. Solution for Δy
∆𝑦 = 𝜇
∆𝑉
𝑊. sin20𝑜
𝐴
∆𝑦 = 0.05
0.02
25.sin20𝑜
1 = 0.000117 𝑚 = 0.117 𝑚𝑚
Example 6
Oil has dynamic viscosity (μ = 1.0 × 10-3 Pa.s) filled the space between two concentric
cylinders, where the inner one is movable and the outer is fixed. If the inner and outer
cylinders has diameters 150mm and 156mm respectively and the height of both cylinders
is 250mm, determine the value of the torque (T) that necessary to rotate the internal
cylinder with 12 rpm?
Solution:
m/s
09425
.
0
075
.
0
2
60
12
2
60





 
r
rpm
v

Chr
T 

1
2
1
)
2
(
)
2
( 




 hr
T
r
h
r
T 


𝜏 = 10−3
×
0.09425
0.003
= 31.41667 × 10−2 𝑁
𝑚2
𝑇 = 31.41667 × 10−2
× 2 × .075 × 𝜋 × .25 × .075 = 2.7 × 10−4
𝑁.𝑚
Example 7
Oil has a density of 580 kg/m3 flow through a pipe its diameter 200mm. If it is known from
the pressure calculations for a certain length of the pipe that the shear stress at the pipe
wall equal to 0.07 N/m2, and its known from the velocity calculations through a certain
cross section of the pipe that the velocity profile equation is :
2
100
1 r
v 

Where the velocity dimension is in m/s and the distance from the center of pipe r in m. If
the flow is laminar, calculate the kinematic viscosity for the oil?
Solution:
For laminar flow:
dy
dv

 
the distance from the center of pipe is
y
R
r 

Where R is the radius of the pipe, y is the distance from the pipe wall toward the pipe
centre.
Where dy
dr 
 and then the above equation become:
dr
dv

 

y
v
dy
dv



 



And from the section of velocity distribution, the strain in any point equal to:
r
dr
dv
200


then the shear stress in any distance from the pipe center is expressed as:
r

 200

whereas τ at the wall equal to 0.07 N/m2 then by substitution this value in the last equation
we obtain follows:
Pa.s
0035
.
0
.
20
07
.
0
2


m
s
N

and the kinematic viscosity is :
s
m
kg
m
s
N
m
kg
m
s
N 2
8
8
3
2
10
1176
.
4
.
.
10
1176
.
4
.
850
.
.
0035
.
0 












Example 8
The velocity distribution for flow over a plate is given by u= 2y+y2 where u is the velocity in
m/s at a distance y meters above the plate surface. Determine the velocity gradient and
shear stresses at the boundary and 1.5m from it. Take dynamic viscosity of fluid as 0.9
N.s/m2

ans 1.015 pa.s
H.W
Example 9
Example 10
Example 11

6. Surface Tension: Surface tension is a property of liquids which is making what is
like a thin tensioned membrane at the interface between the liquid and another fluid
(typically a gas). Surface tension has dimensions of force per unit length and
denoted as, σ (Sigma), and its unit is N/m.
 It is a fluid (liquid)-fluid (gas) interface property
Surface tension is a properties of certain fluid-fluid interface
water-air ….. 0.075 N/m at 20oC Water-air …. 0.056 N/m at 100oC
mercury-air … 0.1 N/m
Pressure inside water droplet:
let P= The pressure inside the drop
d= Diameter of droplet
σ= Surface tension of the liquid (water-air interface)
From sectional free body diagram of water droplet we have
Fp
2πRσ

1. ΔP between inside and outside = P-0 =P
2. Pressure force =𝑃 ×
𝜋
4
𝑑2
,𝑎𝑛𝑑
3. Surface tension force acting around the circumference= 𝜎 × 𝜋𝑑,
under equilibrium condition these two forces will be equal and opposite, i.e.
𝑃 ×
𝜋
4
𝑑2
= 𝜎 × 𝜋𝑑
𝑃 =
𝜎 × 𝜋𝑑
𝜋
4 𝑑2
=
4𝜎
𝑑
From this equation we show that (with an increase in size of droplet the pressure
intensity is decreases)
 Derive P for air bubble with the help of figure below
Example 1:
If the surface tension of water-air interface is 0.069 N/m, what is the pressure inside the
water droplet of diameter 0.009 mm?
Solution:
Given d= 0.009 mm; σ= 0.069 N/m
The water droplet has only one surface, hence,
𝑃 =
4𝜎
𝑑
=
4 × 0.069
0.009 × 10−3
= 30667
𝑁
𝑚2
= 𝟑𝟎.𝟔𝟔𝟕
𝒌𝑵
𝒎𝟐
𝒐𝒓 𝒌𝑷𝒂
Surface Tension - Capillarity
 Property of exerting forces on fluids by fine tubes and porous media, due to both
cohesion and adhesion (surface tension)
 Cohesion < adhesion, liquid wets solid, rises at point of contact

 Cohesion > adhesion, liquid surface depresses at point of contact, non-wetting fluid
 The contact angle is defined as the angle between the liquid and solid surface.
 Capillarity is a fluid (liquid)-surface property
 Meniscus: curved liquid surface that develops in a tube
weight of fluid column = surface tension pulling force
𝜌𝑔(𝜋𝑅2
ℎ) = 2𝜋𝑅𝜎 𝑐𝑜𝑠∅
𝒉 =
𝟐𝝈 𝒄𝒐𝒔∅
𝝆𝒈𝑹
● Expression above calculates the approximate capillary rise in a small tube
● The meniscus lifts a small amount of liquid near the tube walls, as r increases this amount may
become insignificant
● Thus, the equation developed overestimates the amount of capillary rise or depression,
particularly for large r.
● For a clean tube, = 0o for water, = 140o for mercury
● For r > ¼ in (6 mm), capillarity is negligible ● Its effects are negligible in most engineering
situations.
● Important in problems involving capillary rise, e.g., soil water zone, water supply to plants
● When small tubes are used for measuring properties, e.g., pressure, account must be made for
capillarity
Example 2: (Example 2.4 Textbook)
To what height above the reservoir level will water (at 20°C) rise in a glass tube, such as
that shown in Figure below, if the inside diameter of the tube is 1.6 mm?
Problem Definition
Situation: A glass tube of small diameter placed in an open reservoir of water induces
capillary rise.
Find: The height the water will rise above the reservoir level.

Properties: Water (20 °C), Table A.5, σ = 0.073 N/m; γ = 9790 N/m3.
Plan
1. Perform a force balance on water that has risen in the tube.
2. Solve for Δh.
Solution
1. Force balance: Weight of water (down) is balanced by surface tension force (up).
Because the contact angle θ for water against glass is so small, it can be assumed to be
0°;
therefore cos θ ≈ 1. Therefore:
2. Solve for Δh
Example 3:
A clean tube of diameter 2.5 mm is immersed in a liquid with a coefficient of surface
tension = 0.4 N/m. the angle of contact of the liquid with the clean glass can be assumed to
be 135o. The density of the liquid= 13600 kg/m3. What would be the level of the liquid in
tube relative to free surface of the liquid inside the tube?
Solution:
Given d= 2.5 mm, σ= 4 N/m, = 135o; ρ = 13600 kg/m3
Level of the liquid in the tube, h:

𝒉 =
𝟐𝝈 𝒄𝒐𝒔∅
𝝆𝒈𝑹
ℎ =
4 × 0.4 × 𝑐𝑜𝑠135
(9.81 × 13600) × 2.5 × 10−3
= −3.3910−3
𝑚 𝑜𝑟 − 3.39𝑚𝑚
Negative sign indicates that there is a capillary depression (fall) of 3.39 mm.
Example 4: Derive an expression for the capillary height change h, as shown, for a
fluid of surface tension σ and contact angle between two parallel plates W apart.
.Evaluate h for water at 20°C (σ=0.0728 N/m) if W = 0.5 mm
Solution: With b the width of the plates into the paper, the capillary forces on each wall
together balance the weight of water held above the reservoir free surface:
ℎ𝑏𝜌𝑔𝑊 = 2(𝜎𝑏𝑐𝑜𝑠∅)
𝒉 =
𝟐(𝝈𝒄𝒐𝒔∅)
𝝆𝒈𝑾
for water at 20°C (σ=0.0728 N/m, 𝛾 = 9790 𝑁/𝑚3
) and W = 0.5 mm.
𝒉 =
𝟐 × (𝟎. 𝟎𝟕𝟐𝟖 × 𝒄𝒐𝒔(𝟎))
𝟗𝟕𝟗𝟎 × 𝟎.𝟎𝟎𝟎𝟓
= 𝟎.𝟎𝟑𝒎 = 𝟑𝟎𝒎𝒎
H.W

The (fluid at rest), fluid statics, or hydrostatics
Pressure is defined as the ratio of normal force to area at a point, or may be
defined as the normal force that’s applied toward the unit area, and
denoted by P. Its units are N/m2
or what is called Pascal, Pa.
Highlights
 Fluid exerted, in general, both normal (due to their weights) and
shearing forces (primary due to their viscosity) on surfaces (any
plane) that are contacted with (or submerged in) it.
 The normal forces that are exerted by fluid weights is called the
fluid pressure force and fluid pressure or intensity of fluid pressure.
So the pressure can be defined also as the weight of fluid
column intensity above a certain area.
 The source of pressure and its effects and its variation of a fluid at
rest is due only to the weight of the fluid.
 Pressure is a scalar quantity; that is, it has magnitude only.
 Pressure is not a force; rather it is a scalar that produces a
resultant force by its action on an area.
 The resultant force is normal to the area and acts in a direction
toward the surface (compressive).
 Fluids at rest cannot resist a shear stress; in other words, when
a shear stress is applied to a fluid at rest, the fluid will not remain
at rest, but will move (flow) because of the shear stress.
 Hydrostatics is the study of pressures throughout a fluid at rest
 The controlling laws are relatively simple, and analysis is based on
a straight forward application of the mechanical principles of force
and moment.

Pressure Units
 Some units for pressure give a ratio of force to area. Newtons per
square meter of area, or pascals (Pa), is the SI unit. The
traditional units include psi, which is pounds-force per square inch,
and psf, which is pounds-force per square foot.
 Other units for pressure give the height of a column of liquid.
Engineers state that the pressure in the balloon is 20 cm of water:
When pressure is given in units of “height of a fluid column,” the
pressure value can be directly converted to other units using Table
below.
Pressure Units
Pascal
(Pa)
bar (bar)
technical
atmosphere
(at)
atmospher
e
(atm)
torr (Torr)
mmHg
pound-
force /in2
(psi)
m of water
1 Pa 1 N/m2
10−5 1.0197×10−
5
9.8692×10−
6
7.5006×10−
3
145.04×10−
6
10.19×10−5
1 bar 100,000
106
dyn/cm2 1.0197 0.98692 750.06 14.5037744
10.1936
1 at 98,066.5 0.980665 1 kgf/cm2
0.96784 735.56 14.223 9.9966
1 atm 101,325 1.01325 1.0332 1 atm 760 14.696 10.33
1 torr 133.322 1.3332×10−3 1.3595×10−
3
1.3158×10−
3
1 Torr;
≈ 1 mmHg
19.337×10−
3
13.59×10−3
1 psi 6.894×103
68.948×10−3 70.307×10−
3
68.046×10−
3 51.715 1 lbf/in2 0.703
1 m
water
9813.54 0.0981 0.10003 0.0968 73.584 1.4225
1 m water
Example 1: 1 Pa = 1 N/m2 = 10−5 bar = 10.197×10−6 at = 9.8692×10−6
atm , etc.
A pressure of 1 atm (Standard atmospheric pressure) can also be stated as:

1.013 25 bar
1013.25 hectopascal (hPa)
1013.25 millibars (mbar, also mb)
760 torr [B]
≈ 760.001 mm-Hg, 0 °C≈ 1.033 227 452 799 886 kgf/cm²
≈ 1.033 227 452 799 886 technical atmosphere
≈ 1033.227 452 799 886 cm–H2O, 4 °C
Pressure at Point
 At a point, fluid at rest has the same
pressure in all direction.
To prove this, a small wedge-shaped free
body element is taken at the point
(x,y,z) in a fluid at rest.
∑ 𝑓𝑥 = 𝑃𝑥 ∙ 𝛿𝑦𝛿𝑧 − 𝑃𝑠 ∙ 𝛿𝑠𝛿𝑧 ∙ 𝑠𝑖𝑛𝜃
= 0 … … … 1
∑ 𝑓𝑦 = 𝑃𝑦 ∙ 𝛿𝑥𝛿𝑧 − 𝑃𝑠 ∙ 𝛿𝑠𝛿𝑧 ∙ 𝑐𝑜𝑠𝜃 −
1
2
𝛿𝑥𝛿𝑦𝛿𝑧 ∙ 𝛾 = 0 … … … … 2
For unit width of element in z direction, and from the geometry of wedge we
have the follows:
𝛿𝑠 ∙ 𝑠𝑖𝑛𝜃 = 𝛿𝑦 , 𝑎𝑛𝑑 𝛿𝑠 ∙ 𝑐𝑜𝑠𝜃 = 𝛿𝑥 … … … … 3
Substitute of eq.3 in eqs. 1 and 2 and rearrange the terms yields:
𝑃𝑥 = 𝑃𝑠
𝑃𝑦 ∙ 𝛿𝑥 = 𝑃𝑠 ∙ 𝛿𝑥 +
1
2
𝛿𝑦𝛿𝑥 ∙ 𝛾
At a point the element limits to have an infinitesimal dimensions and then we
can eliminate the term (
1
2
𝛿𝑦𝛿𝑥 ∙ 𝛾) from the above equation because of it’s a
higher order of differential values. Thus we have at final that:
𝑃𝑥 = 𝑃𝑠 = 𝑃𝑦
Where 𝜃 is an arbitrary angle, these results gives an important first principle of
hydrostatics:
At a point, fluid at rest has the same pressure in all direction.
θ

Pressure variation:
 For static fluid, pressure varies only with elevation (depth) change
within fluid.
To prove this real, we take a cubic fluid element as shown
While fluid at rest, applying the
equations of equilibrium on the
element. That’s yield:
1. In vertical direction-y:
∑ 𝑓𝑦 = 𝑃𝑦 ∙ 𝛿𝑥𝛿𝑧 − (𝑃𝑦 +
𝛿𝑃𝑦) ∙ 𝛿𝑥𝛿𝑧 − 𝛿𝑥𝛿𝑦𝛿𝑧 ∙ 𝛾 =
0
𝑃𝑦 ∙ 𝛿𝑥𝛿𝑧 − 𝑃𝑦 ∙ 𝛿𝑥𝛿𝑧 − 𝛿𝑃𝑦
∙ 𝛿𝑥𝛿𝑧
− 𝛿𝑥𝛿𝑦𝛿𝑧 ∙ 𝛾
= 0
→ 𝛿𝑃𝑦 ∙ 𝛿𝑥𝛿𝑧 = −𝛿𝑥𝛿𝑦𝛿𝑧 ∙ 𝛾
→ 𝛿𝑃𝑦 = −𝛾 ∙ 𝛿𝑦
For certain fluid surface elevation, when the direction of 𝛿𝑦 downward
away from surface (means in the negative direction of y), this called the
depth difference and denoted as 𝛿h, so the last above equation
become:
𝛿𝑃𝑦 = 𝛾 ∙ 𝛿ℎ
these results gives an important second principle of hydrostatics:
 For static fluid, pressure varies only with elevation (depth) change
within fluid by rate equal to specific weight 𝛾 of that fluid.
 In a fluid, pressure decreases linearly with increase in elevation (height,
y or z) and versa visa.
 In most textbooks and reference applications, they are use z-coordinate
instead of y-coordinate as vertical direction axis so:
∆𝑃𝑦 = −𝛾 ∙ ∆𝑦 becomes ∆𝑃𝑧 = −𝛾 ∙ ∆𝑧
 Second principle of hydrostatics means that for any two point in a same
continuous fluid A and B:
∆𝑃𝐴−𝐵 = −𝛾 ∙ ∆𝑧𝐴−𝐵
𝑃𝐵 − 𝑃𝐴 = −𝛾 ∙ (𝑧𝐵 − 𝑧𝐴)
𝑃𝐵
𝛾
+ 𝑧𝐵 =
𝑃𝐴
𝛾
+ 𝑧𝐴 = 𝐻
 This is the hydrostatics equation
and H called the hydrostatics
H

pressure head or what is called piezometric head. With liquids we
normally measure from the surface.
 Open free surface pressure in liquids mostly is atmospheric, Patmospheric.
 For constant density fluids, and if taking the free surface pressure
(atmospheric pressure, Patmospheric) as zero, the pressure at any depth h
becomes:
𝑃ℎ = 𝛾 ∙ ℎ
 Thus ℎ =
𝑃ℎ
𝛾
 Pressure related to the depth, h, of a fluid column referred to as the
pressure head, h.
2. In horizontal direction-x:
∑ 𝑓𝑥 = 𝑃𝑥 ∙ 𝛿𝑦𝛿𝑧 − (𝑃𝑥 + 𝛿𝑃𝑥) ∙ 𝛿𝑦𝛿𝑧 = 0
𝑃𝑥 ∙ 𝛿𝑦𝛿𝑧 − 𝑃𝑥 ∙ 𝛿𝑦𝛿𝑧 − 𝛿𝑃𝑥 ∙ 𝛿𝑦𝛿𝑧 = 0
→ 𝛿𝑃𝑥 = 0
This equation means there is no change in horizontal pressure with horizontal
direction.
These results gives an important third principle of hydrostatics:
 For certain continuous static fluid, there is no pressure change in
horizontal direction (explain!)
The above mentioned principles is called Pascal principles.

Example 2:
A freshwater lake, has a maximum depth of 60m, and the mean atmospheric
pressure is 91 kPa. Estimate the absolute pressure in kPa at this maximum
depth.
Solution
Take γ = 9790 N/m3
. With pa = 91 kPa and h = 60 m, the pressure at this
depth will be
p = 91 kN/m2
+ (9790 N/m3
)(60 m) 1kN/1000N
= 91 kPa + 587 kN/m2
= 678 kPa Ans.
By omitting Patm we could state the result as p = 587 kPa (gage).
Example 3: (EXAMPLE 3.1 LOAD LIFTED BY A HYDRAULIC JACK)
A hydraulic jack has the dimensions shown. If one exerts a force F of 100 N
on the handle of the jack, what load, F2, can the jack support? Neglect lifter
weight.
Problem Definition
Situation: A force of F = 100 N is applied to the handle of a jack.
Find: Load F2 in kN that the jack can lift.
Assumptions: Weight of the lifter component (see sketch) is negligible.
Plan
1. Calculate force acting on the small piston by applying moment equilibrium.
2. Calculate pressure p1 in the hydraulic fluid by applying force equilibrium.
3. Calculate the load F2 by applying force equilibrium.
Solution
1. Moment equilibrium
2. Force equilibrium (small piston)

Thus
3. Force equilibrium (lifter)
· Note that p1 = p2 because they are at the same elevation.
· Apply force equilibrium:
 The jack in this example, which combines a lever and a hydraulic machine,
provides an output force of 12,200 N from an input force of 100 N. Thus,
this jack provides a mechanical advantage of 122 to 1 !
Example 4
Solution

Example 5
For the Cruid Oil Storage tank of 40 m Dia. shown with Floating steel cover of weight about 5000Kn
what would be the oil height from the tank base if maximum presure at the center of the valve not
exeed 1 Bar

H.W
ans
P2 = 35x62.4 = 2184 psf = 15.2psi

Pressure Measurements.
●Pressure measurement reads as follows:
1. Relative to absolute zero (perfect vacuum): called absolute pressure
2. Relative to atmospheric pressure: called gage (gauge) pressure
●If P < Patm , we call it a vacuum (or negative or suction) pressure ,
its gage value = how much below atmospheric
●Absolute pressure values are all positive
●While gage pressures may be either:
– Positive: if above atmospheric, or
– Negative (vacuum, suction): if below atmospheric
 Relationship between absolute, gage and atmospheric pressure
reading:
Pabs = Patm + Pgage
Example 4: (EXAMPLE 3.3 PRESSURE IN TANK WITH TWO FLUIDS)
Oil with a specific gravity of 0.80 forms a layer
0.90 m deep in an open tank that is otherwise
filled with water. The total depth of water and oil
is 3 m. What is the gage pressure at the bottom
of the tank?
Problem Definition
Situation: Oil and water are contained in a tank.
Find: Pressure (kPa gage) at the bottom of the
tank.
Properties:
1. Oil (10°C), S = 0.8.
2. Water (10°C), Table A.5: γ = 9810 N/m3.
Solution:

Example 5:
Example 6:
Example 7:
In Fig. the tank contains water and immiscible
oil at 20°C. What is h in cm if the density of
the oil is 898 kg/m3
?
at elev 1 P1=P2 at water
=

A closed circular tank filled with water and
connected by a U-piezometric tube as shown
in figure. At the beginning the pressure
above the water table in the tank is
atmospheric, then the gauge that connected
with tank read an increasing in pressure that
caused falling in the water level in the tank
by 3 cm. a) calculate the deference in height
that accrued between water levels inside the
tank and in the external tube leg. b)
Determine the final pressure that was
.reading by the gauge
Example 9 H.W
3cm
h
ᶲ50cm
ᶲ1cm
Example 8
Example 9

Pressure measurement devices:
Absolute pressure measurement
Barometers: The instrument used to measure atmospheric pressure is called
barometer
1. Mercury Barometer: which is illustrated in figure below, which consist of
a one meter length tube filled with mercury and inverted into a pan
that’s filled partially with mercury. The height difference of mercury in
inverted tube respect to outside them reads the atmospheric pressure
value (first was invented by E. Torricelli, 1643).
Values of standard sea-level
atmospheric
pressure=101.325 kPa abs
=1013.25 mbar abs
= 760 mm Hg, Torr
=10.34 m H20
2. Aneroid Barometer: uses elastic diaphragm to measure atmospheric
pressure

Gage pressure measurement:
1. Manometry
1.1. Piezometer
For measuring pressure inside a vessel or pipe in
which liquid is there, a tube open at the top to
atmosphere may be attached, tapped, to the walls of
the container (or pipe or vessel) containing liquid at a
pressure (higher than atmospheric) to be measured,
so liquid can rise in the tube. By determining the
height to which liquid rises and using the relation P1 =
ρgh, gauge pressure of the liquid can be determined.
Such a device is known as piezometer. To avoid
capillary effects, a piezometer's tube should be about
12mm or greater.
1.2. Manometers
The using of piezometers for high pressures
measurement become impractical and it is useless
for pressure measurement in gases and negative
pressure. The manometers in its various forms is an
extremely useful type of pressure measuring
instrument for these cases.
 Professor John Foss (Michigan State
University) Procedure for manometers
pressure calculation:
𝑷𝒅𝒐𝒘𝒏 = 𝑷𝒖𝒑
+ 𝜸𝒉
𝑷𝒖𝒑
= 𝑷𝒅𝒐𝒘𝒏 − 𝜸𝒉
Or for successive manometers we can use the
following formula:
where 𝛾𝑖 and hi are the specific weight and
deflection in each leg of the manometer
When liquids and gases are both involved in a manometer problem, it is well
within engineering accuracy to neglect the pressure changes due to the columns
of gas. This is because

Manometers limitations: manometers suffers from a number of limitations.
1. While it can be adapted to measure very small pressure differences, it
cannot be used conveniently for large pressure differences - although
it is possible to connect a number of manometers in series and to use
mercury as the manometric fluid to improve the range. (limitation)
2. A manometer does not have to be calibrated against any standard; the
pressure difference can be calculated from second and third principles
in hydrostatics. ( Advantage)
3. Some liquids are unsuitable for use because they do not form well-
defined interface. Surface tension can also cause errors due to
capillary rise; this can be avoided if the diameters of the tubes are
sufficiently large - preferably not less than 12 mm diameter.
(limitation)
4. A major disadvantage of the manometer is its slow response, which
makes it unsuitable for measuring fluctuating pressures
5. It is essential that the pipes connecting the manometer to the pipe or
vessel containing the liquid under pressure should be filled with this
liquid and there should be no air bubbles in the liquid.(limitation).
2. Bourdon gage
Curved tube of elliptical cross-section changes curvature with changes
in pressure. Moving end of tube rotates a hand on a dial through a
linkage system. Pressure indicated by gage graduated in kPa or kg/cm2
(=98.0665 kPa) or psi or other pressure units.

Example1: (Piezometers)
In figure pressure gage A reads 1.5 kPa. The
fluids are at 20o
C. Determine the elevations z,
in meters, of the liquid levels in the open
piezometer tubes B and C.
Example 2: (U-manometers)
Water at 10°C is the fluid in
the pipe of Fig. 3.11, and
mercury is the manometer
fluid. If the deflection Δh is 60
cm and l is 180 cm, what is the
gage pressure at the center of
the pipe?

Example 3: (U-manometers)
In Figure fluid 1 is oil (Sg=0.87) and fluid 2 is
glycerin at 20o
C (𝛾=12360 N/m3
). If Patm=98 kPa,
determine the absolute pressure at point A
Example 4:
(Differential-Manometers)
Pressure gage B in figure is to
measure the pressure at
point A in a water flow. If the
pressure at B is 87 kPa
estimate the pressure at A, in
kPa. Assume all fluids at
20oC.

Example 5: (Differential-Manometers)
In figure all fluids are at 20o
C. Determine the pressure difference (Pa) between
points A and B.

Example 6: (Successive Differential-Manometers)
Example 7: Inverted-Manometers
,For inverted manometer of figure
,all fluids are at 20o
C. If pB - pA = 97 kPa
what must the height H be in cm

Example 8
Example 9
Example 10

Pressure Forces and Pressure Distributions on Surface
Hydrostatic Force (Force due to the pressure of a fluid at rest)
(e.g Force exerted on the wall of storage tanks, dams, and ships)
Q. How is Hydrostatic Force on the vertical or inclined planes
determined?
Basic conditions for a Plane surface submerged in a fluid
- Force on the surface: Perpendicular to the surface (No  )
- Pressure: Linearly dependent only to the vertical depth
1. On a Horizontal surface (e.g. the bottom of a tank)
Pressure at the bottom, h
p 

: Uniform on the entire plane
 Resultant force hA
pA
FP 


(A: the bottom area of container)

2. On an Inclined surface
Consider a plane shown
- At surface: p = patm
- Angle  between free surface
& the inclined plane
y axis: Along the surface
x axis: Out of the plane
 Along the vertical depth h
 Pressure linearly changes  Hydrostatic force changes
Differential Force acting on the differential area dA of plane,
)
(
)
(
)
Area
(
)
Pressure
( dA
h
dF 


  (Perpendicular to plane)
Then, Magnitude of total resultant force FP
 
 

 A A
P dA
y
hdA
F 

 sin where 
sin
y
h 

 sin

A
ydA
y A
c


where yc: y coordinate of the center of area (Centroid)
 A
y
ydA c
A


1st moment of the area about x-axis
- Related with the center of
area
x
y


Example 2

The location of point of action of FP (Center of pressure, CP)
- Not passing though Centroid!! (Why?)
- Related with the balance of torques due to of FP
i) Position of FP on y-axis
𝒚𝒑: y coordinate of the point of action of FP
Moment about x axis:
The moment of resultant force = The moment of its components

 A
P
P ydF
y
F
(𝛾𝐴𝑦𝑐𝑠𝑖𝑛𝜃)𝑦𝑝 = 
 
A
A
dA
y
dA
y 2
2
sin
sin 




A
y
dA
y
y
c
A
P


2
=
A
y
I
c
x
where 
 A
x dA
y
I 2
: 2nd moment of area (Moment of inertia, +ve
always )
or, by using the parallel-axis theorem, 2
c
xc
x Ay
I
I 

 yP = c
c
xc
y
A
y
I
 (Always below the centroid !)

ii) Position of FP on x-axis
𝒙𝒑: x coordinate of the point of action of FP
By the similar manner,
The moment of resultant force = The moment of its components


A
P
P xdF
x
F
P
c x
Ay )
sin
( 
 
 
 A
A
xydA
xydA 


 sin
sin
 xP =
A
y
xydA
c
A
 =
A
y
I
c
xy
where A
xydA = Ixy: Area product of inertia (+ve or –ve)
or, by using the perpendicular-axis theorem, c
c
xyc
xy y
Ax
I
I 

 xP = c
c
xyc
x
A
y
I

CP For a symmetric submerged area, xP = xc (Ixyc = 0)

Example 3: A tank of oil has a right-
triangular panel near the bottom, as in
,Figure. Omitting Pa, find: a) hydrostatic force and
2) CP on the panel

Example5
An inclined, circular gate with water on one side as shown in figure.
Determine the total resultant force acting on the gate and the
location of the centre of pressure.
Solution:

Example 6
H.W
ans

 Pressure Prism (Graphical interpretation of pressure distribution)
- Especially applied for a rectangular surfaces (areas)
- Simple method for finding the force and the point of action
Consider the situation shown
 Information from the diagram
- Vertical wall of width b and height h
- Contained liquid with specific weight 
- Pressure: ptop  0 & h
pbottom  
From the last section,
P c av
F  (h )  (A)  p (at the centroid)area = 

 h 

 2
 A
Let’s define a pressure-area space. (See the right figure above]
1. Horizontal axis: Magnitude of the pressure
2. Vertical axis: Height of the area
3. Axis toward the plane: Width of the area
: Resultant volume (Pressure prism)
 How to find the resultant force FR from the pressure prism

 How to find the point of action of FR (the point of action)
From the last section,
yP = h
bh
h
I
y
A
y
I xc
c
c
xc
2
1
)
(
2
1









(In case of rectangular plate, 3
2
12
1
12
1
bh
Ah
Ixc 
 )
yP = h
h
h(bh
bh
6
1
2
1
)
2
1
12
1 3

 + h
2
1
= h
3
2
(from the top)
From the pressure prism
YP = Centroid of the pressure prism
= h
3
2 1
(from the top) = h
3
(above the base)
h/3
h
XP = Horizontal center
 Special case of a plane surface not extending up to the fluid
surface
- Completely submerged plane (See Figure)
Consider the situation shown
Pressure prism
- Trapezoidal cross section
(1) Resultant force FP
= Volume of the shadow region

FP = Volume of hexahedron + Volume of prism
 F1(ABDE)  F2(BCD)
2
1
 (h1)A [ (h2  h1)]A
(2) The location of FR(yA): Consider the moments again
Moment by F acting at yA
P
= Moment by F1 at y1 + Moment by F2 at y2
2
2
1
1 F y
F y
P
F y
P 
 where
2
1
h
y  for rectangle
3
2
2
h
y  for triangle (From the top)
 The effect of the atmospheric pressure patm
: Increasing Volume of hexahedron (F1), NOT the prism (F2)
Example 1
Example 2

Example 3
The dam of figure has a strut AB every 6 m. Determine the compressive force
in the strut, neglecting the weight of the dam.
2?
2?
4?
γ
γ
γ

:Example 4
By using try and error technique (or other techniques like Newton
Raphson method) we find that y=2.196m

Example 5:

:Example 6
Example 7:

Forces on Curved Surfaces (Non-planar surfaces)
px
py
P F
F
F





For unit width of surface
  























ydx
dx
y
F
dx
y
dF
ds
dx
y
P
ds
P
dF
py
py
py
cos
.
cos
.
Where ∀ the volume of liquid above the
surface to the zero pressure surface
x
y

FPy
FPx
FP

 By the same way we find the vertical
component of pressure force if the liquid
exist under the surface by taking the sign
of ∀ as –ve to represent the upward
direction of this force
  































ydx
dx
y
F
dx
y
dF
ds
dx
y
P
ds
P
ds
P
dF
py
py
py
cos
.
cos
.
cos
.
2
2
2
sin
.
sin
.
2
1
2
2
2 2
1
2
1
2
1
y
y
F
y
dy
y
dy
y
F
dy
y
dF
ds
dy
y
P
ds
P
dF
px
y
y
y
y
y
y
px
px
px




































Where (
2
2
y

 ) is the volume of pressure prism
on the surface projection on vertical plan

2
2
)
F
(
)
F
(
F py
px
P 
 : Magnitude &
px
py
F
F
tan 
 : Direction &
the line of action can be find from the concept of:
Moment of resultant force = The summation of the moments of its
components, i.e. 
 )
MF
,
MF
(
MF py
px
P

Example 1: A curved surface AB is a circular arc in
its section with radius of 2m and width of 1m into the
paper. The distance EB is 4m. The fluid above surface
AB is water, and atmospheric pressure applied on
free surface of water and on the bottom side of
surface AB. Find the magnitude and line of action of
the hydrostatic force acting on surface AB.

Example 2: A cylindrical barrier in Fig. holds water as
shown. The contact between the cylinder and wall is
smooth. Consider a 1-m length of cylinder; determine (a)
its weight, and (b) the force exerted against the wall.

Example 3
Example 4

Example 5
Example 6

Example 7
H.W

The Buoyant Force
A buoyant force is defined as the upward force that is produced on a body that is
totally or partially submerged in a fluid. Buoyant forces are significant for most
problems as surface ships. In Fig. 3.10a shown, consider a body ABCD submerged
in a liquid of specific weight Ɣ. The pressures acting on the lower portion of the
body create an upward force equal to the weight of liquid needed to fill the volume
above surface ADC.
Fig. 3.10
The upward force is
As shown by Fig. 3.10a, pressures acting on the top surface of the body create a
downward force equal to the weight of the liquid above the body:
Subtracting the downward force from the upward force gives the buoyant force FB
acting on the body:
Hence, the buoyant force (FB) equals the weight of liquid that would be needed to
occupy the volume of the body.
The body that is floating as shown in Fig. 3.10b. Pressure acts on curved surface
ADC causing an upward force equal to the weight of liquid that would be needed to
fill volume VD (displaced volume). The buoyant force is given by

Hence, the buoyant force equals the weight of liquid that would be needed to
occupy the volume VD. We can write a single equation for the buoyant force:
Stability of Immersed Bodies
When a body is completely immersed in a liquid, its stability depends on the
relative positions of the center of gravity of the body and the centroid of the
displaced volume of fluid, which is called the center of buoyancy.
• If the center of buoyancy is above the center of gravity (Fig. 3.11a), any
tipping of the body produces a righting couple, and consequently, the body
is stable.
• If the center of gravity is above the center of buoyancy (Fig. 3.11c), any
tipping produces an increasing overturning moment, thus causing the body
to turn through 180°.
• Finally, if the center of buoyancy and center of gravity are coincident, the
body is neutrally stable—that is, it lacks a tendency for righting or for
overturning, as shown in Fig. 3.11b.
Fig. 3.11
Stability Floating Bodies
The stability for floating bodies than for immersed bodies is very important
because the center of buoyancy may take different positions with respect to the
center of gravity, depending on the shape of the body and the position in which it is
floating. When the center of gravity G is above the center of buoyancy C (center of
displaced volume) for floating body, the body will be stable and equilibrium. The
reason for the change in the center of buoyancy for the ship is that part of the

original buoyant volume, as shown in Fig.3.12 by the wedge shape AOB, is
transferred to a new buoyant volume EOD. Because the buoyant center is at the
centroid of the displaced volume, it follows that for this case the buoyant center
must move laterally to the right. The point of intersection of the lines of action of
the buoyant force before and after heel is called the metacenter M, and the distance
GM is called the metacentric height.
• If GM is positive—that is, if M is above G, the body is stable
• If GM is negative, the body is unstable.
Fig.3.12
Consider the prismatic body shown in Fig. 3.12, which has taken a small angle of
heel 𝛼𝛼. First evaluate the lateral displacement of the center of buoyancy CC’, then
it will be easy by simple trigonometry to solve for the metacentric height GM or to
evaluate the righting moment.
The righting couple =𝑊𝑊𝑊𝑊𝑊𝑊 sin 𝛼𝛼
Where : W is weight of body and 𝛼𝛼 angle of heel.
The metacenter M distance from center of bouncy (C) or MC
Can be found from:
𝑀𝑀𝑀𝑀 =
𝐼𝐼
𝑉𝑉𝑉𝑉
and then 𝐺𝐺𝐺𝐺 = 𝑀𝑀𝑀𝑀 − 𝐺𝐺𝐺𝐺
Where:
I is the Moment of inertia for the shortest submersed bed about the centroid (m4
).
Vd is the submersed volume (m3
).
GC is the distance from center of bouncy C to center of gravity G (m).

Example(1):
Example(2):
Solution:

Example(3):
Solution:
Example(4):
Solution:
Example(5):

𝐶𝐶𝐶𝐶 =
𝐼𝐼
𝑉𝑉𝑉𝑉
=
60𝑥𝑥25
3
/12
60𝑥𝑥25𝑥𝑥6
= 5.55 𝑓𝑓𝑓𝑓

MG=CM-GC = 5.55-3.97 = 1.58 ft.
Example(6):
Solution:

Example(7):
Solution:

Example(8) H.W.:
For the crude oil ship shown: if the empty ship weight =10000T and its length of
200m.
Find the total oil volume of S=0.85 that can be transmitted by the ship in (barrel)
and check the ship stability if SG. at 2 m above water surface.

Equilibrium of accelerated fluid masses
If a body of fluid is moved at a constant velocity, then it obeys the
equations .derived earlier for static equilibrium
If a body of fluid is accelerated such that, after some time, it has adjusted so that
there are no shearing forces, there is no motion between fluid particles, and it
moves as a solid block, then the pressure distribution within the fluid can be
described by equations similar to those applying to static fluids.
Also, If a=0 the pressure will be hydrostatic
:pressure
∴ 𝑝𝑝 = 𝛾𝛾ℎ �1 +
0
𝑔𝑔
� = 𝛾𝛾ℎ

Example: 1
∴ 𝑝𝑝 = 𝛾𝛾ℎ �1 −
𝑎𝑎
𝑔𝑔
�) .....3.23
In case of a mass of fluid is decelerated uniformly vertically downward with
a=g, the pressure at any point:
𝑔𝑔
∴ 𝑝𝑝 = 𝛾𝛾ℎ �1 −
𝑔𝑔
� = 0
𝑔𝑔
So, at any point of free falling mass of fluid the pressure equal to zero.
In case of a>g; ∴ 𝑝𝑝 = −𝛾𝛾ℎ �1 −
𝑎𝑎
�) .....3.24
If the acceleration downward the pressure will decrease in with ratio of
𝑎𝑎
𝑔𝑔
:
Solution

Pressure Variation
Consider the forces acting on a small horizontal element, area A and length x,
with a uniform acceleration ax in the x direction:
A
x
x
p
=
]
A
)
x
x
p
+
p
(
A
p
[
=
Fx 










But Newton's 2nd law gives: F = m a
a
=
x
p
:
hence
and
a
x
A
=
x
A
x
p
x
x 




 






Now looking at the forces acting on a small vertical element, area A and length
z, with a uniform acceleration ay in the z direction:
a
A
z
=
g
M
]
A
)
z
z
p
+
p
(
A
p
[
=
F z
z 




 



)
a
+
g
(
=
z
p
:
hence
and
a
z
A
=
g
z
A
z
A
z
p
z
z 






 







ax
y P P+P/x
x
P
az
x
P
P+P/z
y
Mg

A

A
A

A
z
x
Ɵ
Uniform horzental accelerationِِ

Free Surface Definition
Here, pressure p, is a function of x and y, so, by definition:
dz
z
p
+
dx
x
p
=
dp
z
z
p
+
x
x
p
=
p











C
z
a
g
x
a
=
p
dz
a
g
+
dx
a
=
dp
z
x
z
x









)
(
)
(
(




This is the pressure distribution in the fluid. The value of C is found by specifying
the pressure at one point. If we apply the point of the center of inclination where
p = Patm at (x, z) = (0, 0), then C = Patm. The final desired distribution is:
readings
pressure
gage
in
0
)
(
or
)
(










z
a
g
x
a
=
p
p
z
a
g
x
a
=
p
z
x
atm
z
x




For free surface p=0 and if we applied this value in the last pressure formula
we find the equation for the free surface as follows:
z
a
g
x
a
= z
x )
(
0 


 

surface
free
of
equation
)
(
x
a
g
a
=
y
z
x



And the inclination of this surface find by the derivative of this equation to find
its slope (or tan θ)

tan
)
(




z
x
a
g
a
=
dx
dy
It is clear that if ax ≠ 0, then the isobars will not be horizontal in this case.
Summary of Uniform linear acceleration:
If a container of fluid is accelerated uniformly and horizontally with az = 0, then
the slope of the isobars within the fluid is given by: g
a
= x
1
tan


But in general when there are uniform accelerations in both vertical and
G

horizontal directions, the slope of the isobars within the fluid is given by
)
(
tan
z
x
1
a
g
a
=




 The free surface of a liquid is normally taken as a line of constant pressure
- or isobar - and the equation above gives the surface slope of an
accelerated fluid.
 The resultant acceleration perpendicular on the isobar surfaces is:
2
2
)
( z
x a
g
a
=
G 

Example: 2
g
a
=
 1 x
tan

For az=0
And
P=.G.S
and the presure at any point P
where S=h.cos

Example: 3
Example: 4
Solution
Solution

Example: 5

Example: 6 H.W

Uniform rotation about a vertical axis:
If liquid is placed in a container and rotated about a vertical axis at a constant
angular velocity, then after some time it will move as a solid body with no
shearing of the fluid. Under such conditions the liquid is said to be moving as a
"forced vortex" with velocity q =  r at any radius r from the axis. [This contrasts
with "free-vortex" motion in which the fluid velocity varies inversely with
distance from the axis of rotation.]
With  constant, all fluid experiences an acceleration 2
r (centrifugal) directed
towards the axis of rotation, and for equilibrium of a typical small horizontal
element:


p
r
r A = A r . r
     2
r
v
r
=
r
p 2
2


 



In the vertical direction, the usual expression for pressure distribution in a static
fluid holds:
g
=
z
p




Clearly, in such circumstances, the pressure is varying with both r and y: p = f(r,
y),and
dz
z
p
+
dr
r
p
=
dp




Hence:
dp =  2
r. dr - g dz,
P+P/r r
P
A
r
r

v r
θ
ω g
z

and if using a liquid with constant , this can be integrated:
C
+
z
g
r
=
p 2
2


 
2
1
This equation can be written similar to hydrostatic equation terms as follows:
𝑝
𝛾
+ 𝑧 −
𝜔2𝑟2
2𝑔
= 𝐶 (eq. 4.13a in textbook)
However, this is the pressure distribution in the fluid. The value of C is found by
specifying the pressure at one point. If p = Po at (r, z) = (0, 0), then C = Po. The
final desired distribution is:
o
2
2
P
+
z
g
r
=
p 

 
2
1
or z
g
r
=
p 2
2


 
2
1
(in gage reading)
 This reveals that all isobars in such a rotating liquid are paraboloids with the
form:
r2
= k z + ( p - Po )
 and the free surface of the liquid (being an isobar) will also take this form.
 For free surface p=0 and if we applied the point of the center of rotation in
(r = 0, z = 0) location, the constant of equation is concluded as zero and k
become
g
k 2

2

 We can find the equation of each
isobar lines by applying the
boundary condition of the center
of rotation which have the
coordinate of (0, zo).
 From the mass conservation we
have that the volume that
descending near the center of
rotation must be equal to that ascending in opposite side of rotation. In
parapoloid volume (
h
R
V
2
2

 ), this lead to
k
R
h
h rise
fall
2

 …. (Prove that!)
Example:
The coffee cup in previous example is removed from the
drag racer, placed on a turntable, and rotated about its
central axis until a rigid-body mode occurs. Find (a) the
angular velocity which will cause the coffee to just reach
the lip of the cup and (b) the gage pressure at point A for
this condition.
ω
2
Constant for certain isobar line
ω

rad/sec. x (60/2#)= rpm

Solution:

Example
For the open cylinder shown, Find the angular velosity for wihich
halve of the volume will be exposed

anc
H.W

Fluid flow
Motion (flowing) of a fluid mass
accrues when it is subjected to
unbalanced forces that reveal if the
fluid mass was subjected to
hydraulic gradient (e.g. tilting of
free surface by certain angle or
connect two containers have
different levels). This means that
fluid mass lies under an acceleration
toward its flow direction. This
motion continues as long as unbalanced forces are applied.
Flow is defined as the quantity (mass or volume) of fluid (gas, liquid or vapour)
that passes a point (section) per unit time. A simple equation to represent this
is:
𝑭𝒍𝒐𝒘 =
𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚
𝒕𝒊𝒎𝒆
Flow Classification (Flow pattern)
Having introduced the general concepts of flow patterns, it is convenient to
make distinctions between different types of flows. These concepts can be best
introduced by expressing the velocity of the fluid in the form:
𝑽 = 𝑽(𝒔, 𝒕)
where s is the distance traveled by a fluid particle along a path, and t is the
time.
 Uniform or Non-uniform
 A uniform flow is a flow in which the velocity does not change along a
streamline, i.e. 0



s
v
In uniform flows the streamlines are straight and parallel.
 A non-uniform flow is a flow in which the velocity changes along a
streamline, i.e. 0



s
v
 Steady or Unsteady
 In a steady flow the velocity at a given point on a streamline does not
change with time: 0



t
v
h1
h2
Flow
Free surface
FP2 = g.h2. A
FP1= g.h1. A
FP1 ≠ FP2
A
Kinematic of Fluid Motion

 An unsteady flow exists if: 0



t
v
 Combining the above we can classify any flow into one of four types:
 Steady uniform flow. Conditions do not change with position in the stream
or with time. An example is the flow of water in a pipe of constant diameter
at constant velocity or discharge (flow rate).
 Steady non-uniform flow. Conditions change from point to point in the
stream but do not change with time. An example is flow in a tapering pipe
with constant velocity at the inlet - velocity will change as you move along
the length of the pipe toward the exit.
 Unsteady uniform flow. At a given instant in time the conditions at every
point are the same, but will change with time. An example is a pipe of
constant diameter connected to a pump pumping at a constant rate which
is then switched off or in open/close valves.
 Unsteady non-uniform flow. Every condition of the flow may change from
point to point and with time at every point. An example is surface waves in
an open channel.
Flow rate
timetakento accumulatethis mass
  volum flowrate
mass flow rate
mass
mass flowrate 
 Volume flow rate - Discharge.

timetakento accumulatethis volume
Volume
volume flowrate  Discharge(Q)
 More commonly we use volume flow rate
 Also known as discharge.
 The symbol normally used for discharge is Q.
Weight flowrate time taken to accumulatethismass
weight
= =  g.Q

Discharge and mean velocity
Cross sectional area of a pipe is A
Mean velocity is vm.
Q = A . vm
We usually drop the “m” and imply mean velocity
Flow Equations
Equation of Continuity
The application of the principle of conservation of mass to fluid flow in a
stream tube results in the "equation of continuity' expressing the continuity of
the flow from point to point along the stream tube. If the cross-sectional areas
and average velocities at sections 1 and 2 in the stream tube of Fig. 31 are
designated by A1, A2, V1and V2, respectively, the .quantity of fluid passing
section 1 per unit of time will be expressed by A1V1, and the mass of fluid
passing section 1 per unit of time will be 𝐴𝐴1𝑑𝑑1𝜌𝜌1. Similarly, the mass of fluid
passing section 2 will be𝐴𝐴2𝑑𝑑2𝜌𝜌2, Obviously, no fluid mass is being created or
destroyed between sections 1 and 2, and therefore

𝐴𝐴1𝑑𝑑1𝜌𝜌1 = 𝐴𝐴2𝑑𝑑2𝜌𝜌2 ……4.1
Fig. 4.6
Thus the mass of fluid passing any point in a streamtube per unit of time is the
same.
• If this equation is multiplied by g, the acceleration due to gravity, there
results giving the equation of continuity in terms of weight.
𝐴𝐴1𝑑𝑑1𝑊𝑊1 = 𝐴𝐴2𝑑𝑑2𝑊𝑊2 ……4.2
The product will be found to have units of N/s and is termed the "weight rate of
flow" or "weight flow."
• For liquids, and for gases when pressure and temperature changes are
negligible, W1=W2 , resulting in
𝐴𝐴1𝑑𝑑1 = 𝐴𝐴2𝑑𝑑2 = 𝑄𝑄 ……4.3
Energy Equation (Bernoulli’s Equation):
Consider a small element of ideal fluid
(non-viscous and incompressible fluid)
ossraligned along a streamline. It has a c
sectional area ΔA, pressure is assumed
launiform across its ends ΔA, and the loc
velocity is defined v and subject to
ontal)zacceleration in both directions x (hori
.(and z (vertical instead of y

1. First from previous lectures, recall to the pressure difference due to
pressure variation in both directions (x, z)
∆𝑝 =
𝜕𝑝
𝜕𝑥
∆𝑥 +
𝜕𝑝
𝜕𝑧
∆𝑧
or dz
z
p
dx +
x
p
dp =




a
=
x
p
x
 


, and 
=
z
p


( g +az )
Also, we know that
So
∆𝑝 = −𝜌 ∙ 𝑎𝑥 ∆𝑥 + (−𝜌 ∙ (𝑎𝑥 + 𝑔)) ∆𝑧
or ( g +a ) dz
 
a dx
dp = z
x
 
 0








a dz
gdz
ax dx
dp
a dz
gdz
ax dx
dp =
z
z






(1)
2. We look at the acceleration of the fluid element.
 Ignoring the possibility that the flow might be steady,


t
v
 0
 v can change with time t, and also with position s along the
direction of motion.
i.e. v = f (t, s).
 Hence, if the element moves a distance δs in time δt, then the total
change in velocity δv is given by:
(Hydrostatic eq. extension
due to accelerations)
t
t
v
s +
s
v
v= 




and in the limit as δt tends to zero, the "substantive" derivative represent the
acceleration in that direction and is given as:
tempo

rarily
spat

ially
t0
s
t
v
+
s
v
= v
t
v
+
dt
ds
s
v
=
t
v
=
dt
dv
a










Lim
 For a steady flow the local velocity at a point does not vary with time,
so the last term under such conditions (
t
v

) will be zero. And the
acceleration remain as:
ds
dv
= v
dt
dv
as  (i.e.
dx
dv
= v
dt
dv
a x
x
x
x  , and
dz
dv dv
= v
dt
a z z
z
z  )

3. Now substitute the form of horizontal and vertical acceleration in equ's. (1)
we get;
0
.
.
0
.
.









z
z
x
x
z
z
x
x
v dv
gdz
v dv
dp
dz
dz
dv
v
gdz
dx
dx
dv
v
dp


+v dv v dv + gdz =0
dp
x x z z

This is a form of Euler's equation, and relates p, v, and z in flow field.
- it then becomes possible to integrate it - giving:
v  v )+ g z=C
2
1
+
p
z
x
( 2
2

Euler's equation (for ideal,
steady flow)
v + g z=C
2
1
+
p 2

v +  g z=C

2
1
p+ 2
+ z = C
2 g
v
+
g

p 2
The three equations above are valid for incompressible, frictionless
steady flow, and what they state is that total energy is conserved
along a streamline.
The first of these forms of the Bernoulli equation is a measure of energy
per unit mass, the second of energy per unit volume, and the third of
"head", equivalent to energy per unit weight.
In the second equation, the term p is the static pressure, {½ρv2
} is the
dynamic pressure, ρgz is the elevation term, and the SUM of all three is
known as the stagnation (or total) pressure, p0
Bernoulli’s equation
(for ideal, steady
flow)
In the third equation:
 p/ρg is known as the pressure head (or flow work head or flow energy
head), which is the work done to move fluid against pressure,
 z is the potential head (elevation head),
 the summation of two terms (p/ρg + z) is called piezometric head or
hydraulic head,
 v2
/2g as the kinetic head (dynamic energy head or velocity head), and
 the sum of the three terms as the Total Head H. The sum of first and
third tem of 3rd
equation is called the piezometric head respect to
piezometer's tube.

where C is a constant along a streamline.
 For the special case of irrotational flow, the constant C is the same
everywhere in the flow field.
 Therefore, the Bernoulli equation can be applied between any two points in
the flow field if the flow is 1
ideal, 2
steady, 3
incompressible, and 4
irrotational.
 i.e. for two points 1 and 2 in the flow field:
2
2
2
2
1
2
1
1
+ z
2 g
v
+
g

p
+ z =
2 g
v
+
g

p
Equation DERYHLV called Bernoulli’s equation (for frictionless, steady flow).
All of terms of Bernoulli’s equation having dimension of length (L) or
dimension of energy times dimension of weight( FL/F). The elevation head
,represent the potential energy per unit weight as below
∴ 𝑎𝑎𝑎𝑎𝑎𝑎𝑣𝑣𝑚𝑚𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ℎ𝑎𝑎𝑚𝑚𝑑𝑑 = 𝛿𝛿
The velocity head represent the kinetic energy per unit weight as below,
∴ 𝑣𝑣𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑣𝑣 ℎ𝑎𝑎𝑚𝑚𝑑𝑑 =
𝑑𝑑2
2𝑔𝑔
The pressure head represent the pressure energy per unit weight as below,
∴ 𝑝𝑝𝑎𝑎𝑎𝑎𝑚𝑚𝑟𝑟𝑎𝑎𝑎𝑎 ℎ𝑎𝑎𝑚𝑚𝑑𝑑 =
𝑝𝑝
𝛾𝛾
The sum of elevation, velocity and pressure heads for ideal steady
incompressible flow is constant for all point in stream line,
𝑃𝑃
𝛾𝛾
+ 𝛿𝛿 +
𝑑𝑑2
2𝑔𝑔
= 𝐻𝐻 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑚𝑚𝑎𝑎 ℎ𝑎𝑎𝑚𝑚𝑑𝑑
the sum of elevation and pressure heads called piezometric head which
represent the manometric height of liquid from datum,
𝑃𝑃
𝛾𝛾
+ 𝛿𝛿 = ℎ = 𝑝𝑝𝑎𝑎𝑎𝑎𝛿𝛿𝑎𝑎𝑚𝑚𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ℎ𝑎𝑎𝑚𝑚𝑑𝑑

Hydraulic and Energy Grade Lines
The energy grade line (EGL) shows the height of the total Bernoulli constant
𝑃𝑃
𝛾𝛾
+ 𝛿𝛿 +
𝑑𝑑2
2𝑔𝑔
= 𝐻𝐻 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑚𝑚𝑎𝑎 ℎ𝑎𝑎𝑚𝑚𝑑𝑑. The EGL has constant height.
𝛾𝛾
The hydraulic grade line (HGL) shows the height corresponding to elevation
and pressure head
𝑃𝑃
+ 𝛿𝛿 = ℎ = 𝑝𝑝𝑎𝑎𝑎𝑎𝛿𝛿𝑎𝑎𝑚𝑚𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ℎ𝑎𝑎𝑚𝑚𝑑𝑑, that is, the EGL minus the
velocity head V2
/2g. The HGL is the height to which liquid would rise in a
piezometer tube
• In an open-channel flow the HGL is identical to the free surface of the
water.
• The EGL will drop slowly due to friction losses and will drop sharply
due to a substantial loss (a valve or obstruction) or due to work extraction
(to a turbine).
• The EGL can rise only if there is work addition (as from a pump or
propeller).
• The HGL generally follows the behavior of the EGL with respect to
losses or work transfer, and it rises and/or falls if the velocity decreases
and/or increases.

Example 1: A flow of water from a reservoir to a pipe of different diameters
shown in Figure below. Calculate 1) the discharge and velocity at each pipe, 2)
the pressure in each pipe and 3) the energy and hydraulic grade lines.
Example 2: A flow of water from a closed reservoir with interior pressure of 50
kPa to a pipe of different diameters shown in figure below. Calculate 1) the
discharge and velocity at each pipe, 2) the pressure in each pipe and 3) the
.energy and hydraulic grade lines
Example 3: A pipe gradually tapers from 0.6m at A to 0.9m at point B. the
elevation difference between A and B is 3m. Find pressure head and pressure at
point B if the pressure head at A is 15m and velocity at A is 2m/s. Assume the
frictionless flow.

Example.4
Example. 5

Applications of Bernoulli’s equation:
1) Flow through orifice
a) With constant head
When an open tank fill with liquid and drains through a port at the
bottom of the tank. The elevation of the liquid in the tank is constant
above the drain. The drain port is at atmospheric pressure. The flow is
steady, viscous effects are unimportant and velocity at liquid surface is
zero. The Bernoulli equation between points 1 and 2 on streamline:
2
2
2
2
1
2
1
1
z
+
g
2
v
+
g
p
=
z
+
g
2
v
+
g
p
ρ
ρ
p1 = p2 because the pressure at the outlet and the tank surface are the
same (atmospheric).
The velocity at the tank surface zero, then:
g
V
)
z
z
(
2
0
0
0
2
2
2
1 +
=
−
+
+
gH
z
z
g
V 2
)
(
2 2
1
2 =
−
=
𝑄𝑄𝑡𝑡ℎ𝑒𝑒𝑒𝑒 = 𝑉𝑉2𝐴𝐴2
𝑄𝑄𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐶𝐶𝑑𝑑 × 𝑄𝑄𝑡𝑡ℎ𝑒𝑒𝑒𝑒
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐶𝐶𝑑𝑑 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (0.64)
• If the tank is closed with interior pressure of Po then Bernoulli’s
equation can be expressed as:
g
V
z
z
Po
2
0
)
(
0
2
2
2
1 +
=
−
+
+
g
Fig.4.8
)
2g(
2
Po
+ H
V =
g
then

b) With variable head (time for the tank to empty)
For cylindrical tank, the tank cross sectional area is A. In a time dt the
level falls by dH
We have an expression for the discharge from the tank
gH
A
C
Q d
act 2
2
=
This discharge out of the orifice is the same as the flow in the tank so,
gH
A
C
dt
dH
A d 2
2
=
−
Integrating between the initial level, h1, and final level, h2, gives the
time it takes to fall this height:
∫
∫
−
=
−
2
1
2
1
2
2
5
.
0
t
t
d
H
H
dt
A
g
A
C
dH
H
[ ] t
A
g
A
C
H d
H
H ∆
−
=
2
2 2
5
.
0 2
1
t
A
g
A
C
H
H d
∆
=
−
2
2
2
1

2) Syphon
34.3
-
Example 1

3) Pitot Tube
Two piezometers, one as normal and one as a Pitot tube within the pipe
can be used as shown below to measure velocity of flow
By applying Bernoulli’s eq.,
2
2
2
2
1
2
1
1
+ z
2 g
v
+
g

p
+ z =
2 g
v
+
 g
p
we have the equation for p2 ,
0
We now have an expression for velocity from two pressure measurements
and the application of the Bernoulli equation.

The holes on the side connect to one side of a manometer, while the
central hole connects to the other side of the manometer
Using the theory of the manometer,
The Pitot/Pitot-static is:
 Simple to use (and analyse)
 Gives velocities (not discharge)

Example1

4) Venturi Meter
The Venturi meter is a device for measuring discharge in a pipe. It is a
rapidly converging section which increases the velocity of flow and hence
reduces the pressure. It then returns to the original dimensions of the pipe by
a gently diverging ‘diffuser’ section.
Apply Bernoulli along the streamline from point 1 to point 2
2
2
2
2
1
2
1
1
+ z
2 g
+
v
g
ρ
+ z =
p
2 g
+
v
g
ρ
p
Substituting and rearranging gives
Or 𝑄𝑄𝑑𝑑ℎ𝑒𝑒𝑒𝑒 = 𝐴𝐴1�
2𝑔𝑔(
𝑃𝑃1−𝑃𝑃2
𝛾𝛾
+𝑧𝑧1−𝑧𝑧2)
(
𝐴𝐴1
𝐴𝐴2
)2−1
= 𝐴𝐴1�
2𝑔𝑔(
𝑃𝑃1−𝑃𝑃2
𝛾𝛾
+𝑧𝑧1−𝑧𝑧2)
(
𝐷𝐷1
𝐷𝐷2
)4−1

Actual discharge takes into account the losses due to friction, include a
coefficient of discharge (Cd ≈ 0.9)
𝑄𝑄𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐶𝐶𝑑𝑑𝐴𝐴1�
2𝑔𝑔(
𝑃𝑃1 − 𝑃𝑃2
𝛾𝛾
+ 𝑧𝑧1 − 𝑧𝑧2)
(
𝐷𝐷1
𝐷𝐷2
)4 − 1
In terms of the manometer readings
𝑃𝑃1 + 𝛾𝛾𝑧𝑧1 − 𝛾𝛾𝑚𝑚∆ℎ − 𝛾𝛾(𝑧𝑧2 − ∆ℎ) = 𝑃𝑃2
𝑃𝑃1 − 𝑃𝑃2 + 𝛾𝛾𝑧𝑧1 − 𝛾𝛾𝑧𝑧2 = 𝛾𝛾𝑚𝑚∆ℎ − 𝛾𝛾∆ℎ = ∆ℎ(𝛾𝛾𝑚𝑚 − 𝛾𝛾)
𝑃𝑃1 − 𝑃𝑃2
𝛾𝛾
+ 𝑧𝑧1 − 𝑧𝑧2 = ∆ℎ(
𝛾𝛾𝑚𝑚
𝛾𝛾
− 1)
Giving 𝑄𝑄𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐶𝐶𝑑𝑑𝐴𝐴1�
2𝑔𝑔∆ℎ(
𝛾𝛾𝑚𝑚
𝛾𝛾
−1)
(
𝐷𝐷1
𝐷𝐷2
)4−1
• This expression does not include any elevation terms. (z1 or z2) When used with a
.manometer, the Venturimeter can be used without knowing its angle
Example 1
Piezometric tubes are tapped into a Venturi
section as shown in the figure. The liquid is
incompressible. The upstream piezometric head is
1 m, and the piezometric head at the throat is 0.5
m. The velocity in the throat section is twice large
as in the approach section. Find the velocity in the
.throat section
Solution
The Bernoulli equation with v2 = 2v1 gives
2
2
2
2
1
2
1
1
+ z
2 g
v
+
 g
p
+ z =
2 g
v
+
 g
p
3
1.81 3.62 m/s
v
0.5 1.81m/s
2 g
v
v
2 g
2 g
v

2 g
v
2 g
v
 z )=
 g
p
+ z (
 g
p
3
1.0  0.5
3
2
1
2
1
2
1
2
1
2
2
2
2
1
1
 2







5) Notches and Weirs
A weir is a vertical barrier in the side of a tank or reservoir. The liquid is flow
over the weir with free surface. It is a device for measuring discharge. It is used
as both a discharge measuring device and a device to raise water levels. There
are many different designs of weir depending on the shape of weir opening. It
may be rectangular, trapezoidal or triangular weir. The weirs also classified
according its crest width to sharp crested weir, broad crested weir and ogee
weir. The sharp crested rectangular weir may be contracted when the length of
weir opening is less than the channel width or suppress when the length of weir
is equal to channel width.
Fig. 4.9
Weir Assumptions
• velocity of the fluid approaching the weir is small so we can ignore kinetic
energy.
• The velocity in the flow depends only on the depth below the free surface.
A General Weir Equation
Consider a horizontal strip of width b, depth h below the free surface

velocity through the strip,
discharge through the strip,
Integrating from the free surface, h = 0, to the weir crest, h = H, gives the total
theoretical discharge
This is different for every differently shaped weir or notch. We need an
expression relating the width of flow across the weir to the depth below the free
surface
Rectangular Weir
The width does not change with depth so,
Substituting this into the general weir equation gives
To get the actual discharge we introduce a coefficient of discharge, Cd, to
account for losses at the edges of the weir and contractions in the area of flow,

‘V’ Notch Weir
The relationship between width and depth is dependent on the angle of the “V”.
The width, b, depth, h, from the free surface relationshipis
So the discharge is
The actual discharge is obtained by introducing a coefficient of discharge

Example 1: Water enters the Millwood flood storage area via a rectangular
weir when the river height exceeds the weir crest. For design purposes a flow
rate of 0.163 m3
/s over the weir can be assumed
1. Assuming a height over the crest of 200mm and Cd=0.2, what is the necessary
width, B, of the weir?
2. What will be the velocity over the weir at this design?
Example 2: Water is flowing over a 90o ‘V’ Notch weir into a tank with a is
cross-sectional area of 0.6m2. After 30s the depth of the water in the tank 1.5m.
If the discharge coefficient for the weir is 0.8, what is the height of the water
above the weir
Solution:
Given: Q=0.163 m3
/s., H=200mm = 0.2 m., Cd=0.2.
Required B and V.
𝑄𝑄 = 𝐶𝐶𝑑𝑑.
2
3
3
2
. 𝐵𝐵�2𝑔𝑔. 𝐻𝐻 → 0.163 = 0.2𝑥𝑥
2
3
𝑥𝑥𝐵𝐵𝑥𝑥√19.62𝑥𝑥0.2
3
2
∴ 𝐵𝐵 = 3.08 𝑚𝑚
𝑉𝑉 = �2𝑔𝑔ℎ → 𝑉𝑉 = √19.62𝑥𝑥0.2 = 1.98
𝑚𝑚
𝑠𝑠
Solution:
Given: ᶿ=90o
, At=0.6m2, T=30s, Ht=1.5m, Cd=0.8.
Required H.
𝑄𝑄 =
𝑉𝑉𝑉𝑉
𝑇𝑇
=
0.6𝑥𝑥1.5
30
= 0.03
𝑚𝑚3
𝑠𝑠
𝑄𝑄 = 𝐶𝐶𝑑𝑑.
8
15
. �2𝑔𝑔. tan(
𝜃𝜃
2
5
2
). 𝐻𝐻 → 0.03 = 0.8𝑥𝑥
8
15
𝑥𝑥√19.62𝑥𝑥 tan(
90
2
)𝑥𝑥 𝐻𝐻
3
2
∴ 𝐻𝐻 = 0.06 𝑚𝑚

Nozzle flow
Fig. 4.9
The pressure of all points of liquid jet outside the nozzle is equal to
atmospheric pressure. So, it will be equal to zero when the atmospheric
pressure is the reference pressure. Then, the Bernoulli’s equation between
points 1 and 2 in Fig. 4.9 can be written as:
The velocity components for Vn are:
𝑉𝑉
𝑥𝑥 = 𝑉𝑉
𝑛𝑛 cos 𝜃𝜃 and 𝑉𝑉
𝑦𝑦 = 𝑉𝑉
𝑛𝑛 sin 𝜃𝜃
Therefore, for point 1: 𝑉𝑉1𝑥𝑥 = 𝑉𝑉
𝑛𝑛 cos 𝜃𝜃 and 𝑉𝑉1𝑦𝑦 = 𝑉𝑉
𝑛𝑛 sin 𝜃𝜃
For point 2: 𝑑𝑑2𝑥𝑥 = 𝑑𝑑
𝑛𝑛 cos 𝜃𝜃 and 𝑑𝑑2𝑦𝑦 = 0, so, 𝑑𝑑2 = 𝑑𝑑
𝑛𝑛 cos 𝜃𝜃,
which mean that the horizontal velocity component is constant along the
nozzle jet path.
∴
𝑑𝑑1𝑥𝑥
2
2𝑔𝑔
+
𝑑𝑑1𝑦𝑦
2
2𝑔𝑔
= ℎ +
𝑑𝑑2
2
2𝑔𝑔
(𝑑𝑑
𝑛𝑛 𝑎𝑎𝑎𝑎𝑚𝑚 𝜃𝜃)2
2𝑔𝑔
+
(𝑑𝑑
𝑛𝑛 𝑚𝑚𝑎𝑎𝑎𝑎 𝜃𝜃)2
2𝑔𝑔
= ℎ +
Therefore the maximum height of nozzle jet will be: ℎ =
(𝑑𝑑𝑛𝑛 sin 𝜃𝜃)2
2𝑔𝑔
The vertical component of velocity Vy is varied along the nozzle jet path
as below:
𝑑𝑑
𝑦𝑦 = 𝑑𝑑
𝑛𝑛 sin 𝜃𝜃 − 𝑔𝑔. 𝑎𝑎
The horizontal and vertical axes(x,y) of any point along the nozzle jet
path as below:
(6
(𝑑𝑑
𝑛𝑛 𝑎𝑎𝑎𝑎𝑚𝑚 𝜃𝜃)2
2𝑔𝑔

𝑦𝑦 = 𝑉𝑉
𝑛𝑛 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃 . 𝑡𝑡 −
1
2
𝑔𝑔𝑡𝑡2
𝑥𝑥 = 𝑉𝑉
𝑥𝑥. 𝑡𝑡 = 𝑉𝑉
𝑛𝑛 cos 𝜃𝜃. 𝑡𝑡
at the highest point of jet Vy reaches to zero, so:
ℎ =
1
2
𝑔𝑔𝑡𝑡2
→ 𝑡𝑡2 = �
2ℎ
𝑔𝑔
and the horizontal distance of highest point determined as:
𝑋𝑋 = 𝑉𝑉
𝑥𝑥. 𝑡𝑡2 = 𝑉𝑉
𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃. 𝑡𝑡2
Example 1: Determine the vertical and horizontal distance of highest point of
water jet from nozzle with velocity of 20m/s. Also, find the diameter of the jet
at the highest point if the diameter of nozzle is 2cm. The jet is inclined at 60o
with horizontal. Neglect air resistance.
Solution: ℎ =
(𝑑𝑑𝑛𝑛 𝑎𝑎𝑒𝑒𝑠𝑠 𝜃𝜃)2
2𝑔𝑔
=
(20 sin 60)2
2×9.81
=15.29m
𝑎𝑎2 = �
2ℎ
𝑔𝑔
2 × 15.29
= �
9.81
= 1.766𝑚𝑚𝑎𝑎𝑎𝑎
𝑑𝑑
𝑥𝑥 = 𝑑𝑑
𝑛𝑛 𝑎𝑎𝑎𝑎𝑚𝑚 𝜃𝜃 = 20 cos 60 = 10m/s
𝑋𝑋 = 𝑑𝑑
𝑥𝑥. 𝑎𝑎2 = 10 × 1.766 = 17.66𝑚𝑚
Q=Vn ×Anozzle=20.π/4(0.02)2
=6.28×10-3
m3
/s
A2=Q/Vx=6.28/10=6.28×10-4
m2
D=0.028m=2.8cm

Solution
𝑒𝑒 = 𝑑𝑑
𝑥𝑥. 𝑎𝑎 = 𝑑𝑑
𝑛𝑛 cos 𝜃𝜃. 𝑎𝑎
𝑑𝑑
𝑛𝑛 =
𝑒𝑒
𝑎𝑎𝑎𝑎𝑚𝑚 𝜃𝜃. 𝑎𝑎
=
42.43
𝑎𝑎
… … … . .1
𝑣𝑣 = 𝑑𝑑
𝑛𝑛 𝑚𝑚𝑎𝑎𝑎𝑎 𝜃𝜃 . 𝑎𝑎 −
1
2
𝑔𝑔𝑎𝑎2
𝑑𝑑
𝑛𝑛 =
𝑣𝑣 +
1
2
𝑔𝑔𝑎𝑎2
𝑚𝑚𝑎𝑎𝑎𝑎 𝜃𝜃. 𝑎𝑎
=
21.21
𝑎𝑎
+ 6.94𝑎𝑎 … … … … … 2
42.43
𝑎𝑎
=
21.21
𝑎𝑎
+ 6.94𝑎𝑎 → 6.94𝑎𝑎 =
21.21
𝑎𝑎
→ 𝑎𝑎 = 1.749𝑚𝑚𝑎𝑎𝑎𝑎
Vn=42.43/1.749=24.26m/s
Q=Vn.Anozzle=0.107m3
/s
Example 2: Point (b) is located on the stream line. Determine the flowrate

Pumps and Turbines
The energy line of liquid flow through turbine drops down directly due to
consumption of energy by turbine which call turbine head (ht) . While the
energy line of liquid flow through pump rises up directly due to adding of
energy to the flow by pump which call pump head (hp). So, the Bernoulli’s
equation will be:
Fig. 4.10
Pumps and turbines power
The head is the energy of unit weight: ℎ =
𝐸𝐸
𝑊𝑊
𝑡𝑡ℎ𝑒𝑒𝑒𝑒, 𝐸𝐸 = 𝑊𝑊 × ℎ
Power is the energy per unit time: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 =
𝑊𝑊ℎ
𝑡𝑡
𝑤𝑤ℎ𝑖𝑖𝑖𝑖𝑖𝑖 𝑄𝑄𝑤𝑤 =
𝑊𝑊
𝑡𝑡
= 𝑄𝑄𝑄𝑄
∴ 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑄𝑄𝑄𝑄ℎ
and 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝑄𝑄𝑄𝑄ℎ𝑝𝑝
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝑄𝑄𝑄𝑄ℎ𝑡𝑡
(7

Example 1: Draw the E.G.L. and H.G.L. of the pipe system in Fig. and
determine the power of pump. The discharge is 0.15m3
/s. neglect the friction of
pipe.
Example 2: The depth of water in tank shown in Fig. is 10m and discharge
required through the system is 0.15m3
/s. Determine the velocity and the
pressure in each pipe, the power of the pump. Plot E.G.L. and H.G.L.
Example 3: Calculate the depth of water in tank shown in Fig. which will
produce a discharge of 85 l/s. The input power of the turbine is 15kW. What
flowrate may be expected if the turbine is removed?

Momentum Equation
The Momentum equation is a statement of Newton’s Second Law. It
relates the sum of the forces to the acceleration or rate of change of
momentum. From solid mechanics you will recognize
F = ma
What mass of moving fluid we should use?
We use a different form of the equation.
Consider a stream tube and assume steady non-uniform flow:
In time δt a volume of the fluid moves from the inlet a distance v1δt, so
volume entering the stream tube = area × distance
=A1v1δt
The mass entering,
mass entering stream tube = volume density
=ρ A1 v1 δt
And momentum
momentum entering stream tube = mass velocity
= ρ A1 v1 δt v1
Similarly, at the exit, we get the expression:
momentum leaving stream tube = ρ A2 v2 δt v2
By another reading of Newton’s 2nd
Law.
where Momentum = m x v,
= rate of change of momentum
=
Force = mass x acceleration =
dt
dmv
dt
m
dv
u1
A2
u2

We know from continuity that
And if we have a fluid of constant density,
The Momentum equation
This force acts on the fluid in the direction of the flow of the fluid
The previous analysis assumed the inlet
and outlet velocities in the same
direction (i.e. a one dimensional system).
What happens when this is not the
case?
We consider the forces by resolving in
the directions of the co-ordinate axes.
The force in the x-direction
And the force in the y-direction
The resultant force can be found by combining
these components
And the angle of this force

This hydrodynamic force exerted on fluid mass due to time rate of change
of the linear momentum of the system is countered with other external
forces exist within control volume (i.e. pressure forces and body forces and
thrust to result an excess action force that exerted on any solid body
touching the control volume, R:
FB = Force exerted due to fluid body (e.g. gravity)
FP = Force exerted on the fluid control volume due to fluid pressure at the
open fluid edges of the control volume
FR = Force exerted on the fluid by any solid body touching the control
volume
So we say that the total force, FT, is given by the sum of these forces:
FT = FR + FB + FP
The force exerted by the fluid on the solid body touching the control volume
is opposite to FR (action force).
So the reaction force, R, is given by
R = -FR

Applications of Momentum Equation:
1) Pipe reducer and nozzle
In pipe reducer and nozzle, the inlet and outlet velocities are in the same
direction as shown in Fig. 4.8 which represent reducer fitting in pipe line,
equation 4.12 written as:
𝑃𝑃1𝐴𝐴1 − 𝑃𝑃2𝐴𝐴2 − 𝑅𝑅𝑥𝑥 = 𝜌𝜌. 𝑄𝑄. (𝑉𝑉2 − 𝑉𝑉1) , then Rx can be found.
Fig. 4.8
Application – Force exerted by a firehose
Example 1
A firehose discharges 5 l/s. The nozzle inlet and outlet diameters are 75 and 25 mm
respectively. Calculate the force required to hold the hose in place.
Solution
The control volume is taken as shown:

There are three forces in the x-direction:
• The reaction force R
F provided by the fireman;
• Pressure forces P
F : 1 1
p A at the left side and 0 0
p A at the right hand side;
• The momentum force M
F caused by the change in velocity.
So we have: M P R
F F F
= +
The momentum force is:
( )
2 1
M = ρ
F Q −
v v
Therefore, we need to establish the velocities from continuity:
(0.075)
3
1 2
1
5
4
1.13 m/s
v
A π
×10−
Q
= = =
And
(0.025)
3
2 2
5
4
10.19 m/s
v
π
×10−
= =
Hence:
( ) 10 5 10
3 3
( ) 10.19 1.
− 13
( )
2 1 45 N
M ρ
F Q v v
−
= − = × =
The pressure force is: 1 1 0 0
P =
F p A − p A
0
If we consider gauge pressure only, the p = 0 and we must only find 1
p . Using
Bernoulli’s Equation between the left and right side of the control volume:
2 2
1 1 0 0
0
2 2
⎛ ⎞
p
p
+
v v
g
ρ ⎝ ⎠ g
=
g
= ⎜ ⎟
ρg
+
Thus:
v ) ( 1.13 )
2 2
1 1 0
3
2 2 2
10
10.19
2
51.28 kN/m
p
ρ
⎛ ⎞
⎜ ⎟
= −
(v
2
⎝ ⎠
⎛ ⎞
⎜ ⎟
= −
⎝ ⎠
=
Hence
( ×
⎜
⎜
)
π
⎛ (0.075)
1 1 0 0
2
3
51.28 10
4
226 N
P
F p
= −
A p A
⎞
= −
⎟
⎟
⎝ ⎠
=
0
Hence the reaction force is:
226 181 N
R M P
=
F F − F 45
= − = −
This is about a fifth of an average body weight – not inconsequential.

2) Pipe Bends
Calculating the force on pipe bends is important to design the support system. In
pipe bend the inlet and outlet velocities are in different directions. There are two
cases of pipe bend can be illustrated as below:
Case 1: pipe bend in horizontal plan
According Fig. 4.9, equation 4.12 written as:
Fig. 4.9
∑ 𝐹𝐹
𝑥𝑥 = 𝑃𝑃1𝐴𝐴1 − 𝑃𝑃2𝐴𝐴2 cos 𝜃𝜃 − 𝑅𝑅𝑥𝑥 = 𝜌𝜌. 𝑄𝑄. (𝑑𝑑2 cos 𝜃𝜃 − 𝑑𝑑1) , to find Rx
∑ 𝐹𝐹
𝑦𝑦 = 0 − 𝑃𝑃2𝐴𝐴2 sin 𝜃𝜃 − 𝑅𝑅𝑦𝑦 = 𝜌𝜌. 𝑄𝑄. (𝑑𝑑2 sin 𝜃𝜃 − 0) , to find Ry
The resultant can be get by: 𝑅𝑅 = �𝑅𝑅𝑥𝑥
2 + 𝑅𝑅𝑦𝑦
2
𝑅𝑅𝑦𝑦
The resultant inclined with horizontal with angle of: ∅ = tan−1 𝑅𝑅𝑥𝑥

Case 2: pipe bend in perpendicular plan
According Fig. 4.10, the summation of forces in x-direction is same as in case
of horizontal plan. The summations of forces in y-direction include the effect of
fluid weight in pipe bend Wf.
Fig. 4.10
∑ 𝐹𝐹
𝑦𝑦 = 0 − 𝑃𝑃2𝐴𝐴2 cos 𝜃𝜃 − 𝑊𝑊𝑓𝑓 − 𝑅𝑅𝑦𝑦 = 𝜌𝜌. 𝑄𝑄. (𝑑𝑑2 sin 𝜃𝜃 − 0) , to find Ry
The weight of fluid can be founded as below:
𝑊𝑊𝑓𝑓 = 𝛾𝛾∀
the volume of fluid is a cross section area times
the length of center for the pipe bend. The length
of center line can be calculated as below:
For pipe bend of θ ≤ 90ᵒas shown in Fig. 4.11,
the radius for center line of bend given by:
𝐻𝐻 = 𝑅𝑅 − 𝑅𝑅 cos 𝜃𝜃 = 𝑅𝑅(1 − cos 𝜃𝜃)
Fig. 4.11

∴ 𝑅𝑅 =
𝐻𝐻
1 − cos 𝜃𝜃
For pipe bend of θ 90ᵒas shown in Fig. 4.12, the radius for center line of bend
given by:
𝑦𝑦 = 𝑅𝑅 sin 𝜃𝜃
𝑅𝑅 = 𝐻𝐻 − 𝑦𝑦 = 𝐻𝐻 − 𝑅𝑅 sin 𝛼𝛼
𝐻𝐻 = 𝑅𝑅(1 + sin 𝛼𝛼)
𝑅𝑅 =
𝐻𝐻
1 + 𝑅𝑅 sin 𝛼𝛼
Then, length of pipe bend centerline
is given by;
𝐿𝐿 = 𝜋𝜋𝜋𝜋
𝜃𝜃
180
Fig. 4.12
Volume of fluid in pipe bend given by;
∀=
𝜋𝜋
3
𝐿𝐿(𝑎𝑎1
2
+ 𝑎𝑎1𝑎𝑎2 + 𝑎𝑎2
2)

Example-1: Forces on a vertical Bend
The outlet pipe from a pump is a bend of 45o
rising in the vertical plane
(i.e. and internal angle of 135o
). The bend is 150mm diameter at its inlet
and 300mm diameter at its outlet. The pipe axis at the inlet is horizontal
and at the outlet it is 1m higher. By neglecting friction, calculate the force
and its direction if the inlet pressure is 100kN/m2
and the flow of water
through the pipe is 0.3m3
/s. The volume of the pipe is 0.075m3
.
Solution:
12 Draw the control volume and the axis
System
3. Calculate the total force
in the x direction
∑ 𝐹
𝑥 = 𝜌 (∑ 𝑄𝑜𝑢𝑡𝑣𝑥𝑜𝑢𝑡
𝑐𝑠
− ∑ 𝑄𝑖𝑛𝑣𝑥𝑖𝑛
𝑐𝑠
) = 𝐹𝑇𝑥
𝐹𝑇𝑥 = 𝜌 (∑ 𝑄𝑜𝑢𝑡𝑣𝑥𝑜𝑢𝑡
𝑐𝑠
− ∑ 𝑄𝑖𝑛𝑣𝑥𝑖𝑛
𝑐𝑠
)
𝐹𝑇𝑥 = 𝜌𝑄(𝑣2𝑥 − 𝑣1𝑥)
𝐹𝑇𝑥 = 𝜌𝑄(𝑣2 cos 𝜃 − 𝑣1)
by continuity: 𝐴1𝑣1 = 𝐴2𝑣2 = 𝑄
𝑣1 =
0.3
𝜋 (
0.152
4 )
= 16.98
𝑚
𝑠
𝑣2 =
0.3
0.0707
= 4.24
𝑚
𝑠
and in the y-direction 𝐹𝑇𝑦 = 𝜌𝑄(𝑣2𝑦 − 𝑣1𝑦)
𝐹𝑇𝑥 = 𝜌𝑄(𝑣2 sin 𝜃 − 0)

4. Calculate the pressure force.
We know pressure at the inlet, but not at the outlet
we can use the Bernoulli equation to calculate this unknown pressure.
2
2
2
2
1
2
1
1
+ z
v
+
p
+ z =
v
+
p
 g 2 g  g 2 g
The height of the pipe at the outlet is 1m above the inlet.
Taking the inlet level as the datum:
z1 = 0 , z2 = 1m
So the Bernoulli equation becomes:
5. Calculate the body force
The body force is the force due to gravity. That
is the weight acting in the -ve y direction.
N
FBy  735.75
There are no body forces in the x direction,

6. Calculate the resultant force
And the resultant force on the fluid is given by
And the direction of application is
The reaction force on bend is the same magnitude but in the opposite
direction

Example-2
The diameter of a pipe-bend is 300 mm at inlet and 150 mm at outlet and the flow is
turned through 120o in a vertical plane. The axis at inlet is horizontal and the centre of
the outlet section is 1.4 m below the centre of the inlet section. The total volume of
fluid contained in the bend is 0.085 m3. Neglecting friction, calculate the magnitude
and direction of the net force exerted on the bend by water flowing through it at 0.23
m3·s−1 when the inlet gauge pressure is 140 kPa.
solution

3) Momentum force of flow through diversion
For flow through diversion shown in Fig.13, the momentum equation become as
follow:
Fig. 4.13
� 𝐹𝐹
𝑥𝑥 = (𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜𝑜𝑜)𝑥𝑥 − (𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑖𝑖𝑖𝑖)𝑥𝑥
� 𝐹𝐹
𝑦𝑦 = (𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜𝑜𝑜)𝑦𝑦 − (𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑖𝑖𝑖𝑖)𝑦𝑦
The main pipe supply each branching pipe according to its diameter as follow:
𝑄𝑄1 =
𝐷𝐷1
2
𝐷𝐷1
2+𝐷𝐷2
2+𝐷𝐷3
2 × 𝑄𝑄0 , 𝑄𝑄2 =
𝐷𝐷2
2
𝐷𝐷1
2+𝐷𝐷2
2+𝐷𝐷3
2 × 𝑄𝑄0 , 𝑄𝑄3 =
𝐷𝐷3
2
𝐷𝐷1
2+𝐷𝐷2
2+𝐷𝐷3
2 × 𝑄𝑄0, ….etc.
∴ � 𝐹𝐹
𝑥𝑥 = � 𝜌𝜌𝜌𝜌𝑉𝑉
𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜
− 𝜌𝜌𝜌𝜌𝑉𝑉
𝑥𝑥𝑖𝑖𝑖𝑖
𝐹𝐹
𝑜𝑜 − 𝐹𝐹1 cos 𝜃𝜃1 − 𝐹𝐹2 cos 𝜃𝜃2 − 𝐹𝐹3 cos 𝜃𝜃3 − 𝑅𝑅𝑥𝑥 = 𝜌𝜌[𝑄𝑄1𝑉𝑉1 cos 𝜃𝜃1 + 𝑄𝑄2𝑉𝑉2 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃2 +
𝑄𝑄3𝑉𝑉3 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃3] − 𝜌𝜌[𝑄𝑄𝑜𝑜𝑉𝑉
𝑜𝑜] , then Rx can be founded.
In the same way:
∴ � 𝐹𝐹
𝑦𝑦 = � 𝜌𝜌𝜌𝜌𝑉𝑉
𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜
− 𝜌𝜌𝜌𝜌𝑉𝑉
𝑦𝑦𝑖𝑖𝑖𝑖
0 − 𝐹𝐹1 sin 𝜃𝜃1 − 𝐹𝐹2 sin 𝜃𝜃2 + 𝐹𝐹3 sin 𝜃𝜃3 + 𝑅𝑅𝑦𝑦 = 𝜌𝜌[𝑄𝑄1𝑉𝑉1 sin 𝜃𝜃1 + 𝑄𝑄2𝑉𝑉2 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃2 −
𝑄𝑄3𝑉𝑉3 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃3] − 𝜌𝜌[0] , then Ry can be founded.
The resultant can be get by: 𝑅𝑅 = �𝑅𝑅𝑥𝑥
2 + 𝑅𝑅𝑦𝑦
2
𝑅𝑅𝑦𝑦
The resultant inclined with horizontal with angle of: ∅ = tan−1 𝑅𝑅𝑥𝑥

m / s
A
A
V V
V A
V A
V A
A symmetry V V )
(  A
V A V A V A
Q
in out
3.37
2 .0043
0.0095
3.05
2
2
2
2
1
1
2
3
3
2
2
1
1
3
2
3
2
3
3
2
2
1
1







Example-1
A cylindrical metal container 60 cm high with an inside diameter of 27 cm, weights
22N when empty. The container is placed on a scale and water flows in through an
opening in the top and out through the two equal area 45o-deflection openings in the
sides as shown in the diagram. Under steady flow conditions the height of the water
in the tank is h = 58 cm. Your friend claims that the scale will read the weight of the
volume of water in the tank plus the tank weight, i.e., that we can treat this as a
simple statics problem. You disagree, claiming that a flow analysis is required. Who
is right, and what is the scale reading in Newtons
Q 
For CV shown
.83N  scale  reading
N
N
F
F
 p
,p
p
N
/
.
F
F
F
F
F
F
F
F
N
cos
V
QV )
F = ( QV
Ry
py
By
By
y py
Ry
By
Ry
py
y
F =
y
watersurface
in
yin
yout
y




 

 
 
 22 












F 







416
347.77
69.06
0
0
0
347.77
0.58 9810
0 27 4
45 2  0  69.06
1000  3.37
0
3
2
1
2
2

Then scale reading not equal the static loads only, but with addition value
of dynamic effects
jet
Scale
jet
Example-6:
The 6-cm-diameter 20°C water jet in Fig.
strikes a plate containing a hole of 4-cm
diameter. Part of the jet passes through the
hole, and part is deflected. Determine the
horizontal force required to hold the plat
0
980
998 0 0.0314 0.0707  25
25  0.0314
25  0.0707
4
3
2
3
2



 25 
 0 


 

y
x
x
x
x out outx in inx
hole
in
F
N
R
N
980
F =
)
(
F =
Q V )
F = ( Q V
out in
for devided or branched flow
4
F = Q(V V )
m s
(0.04 )
Q
m s
(0.06 )
Q


x43/10000

Example-7:
Water at 20°C exits to the standard sea-level
atmosphere through the split nozzle in Fig.
Duct areas are A1 = 0.02 m2
and A2 = A3 =
0.008 m2
. If p1= 135 kPa (absolute) and the
flow rate is Q2 = Q3 = 275 m3
/h, compute
the force on the flange bolts at section 1.

Example: A water jet of velocity Vj impinges normal to a flat plate which
moves to the right at velocity Vc, as shown in Fig. Find the force required
to keep the plate moving at constant velocity if the jet density is 1000
kg/m3
, the jet area is 3 cm2
, and Vj and Vc are 20 and 15 m/s,
respectively. Neglect the weight of the jet and plate, and assume steady
flow with respect to the moving plate with the jet splitting into an equal
upward and downward half-jet.
For moving control volume with V=Vc we have
m s
in j c
V V V  20 15  5
2
1
2
2
1
2
1
2
2
1
1
2
1
2
1
1
V V
V
2
V , but from symmetry and neglactingthe wight :V V
V
V
A
AV  AV , A  A
AV
By continuity equation we have:
Q  Q
in
1
in
j
j in
out
in





0
7
7.5
7.5 0
7
0 1000 0.0003 5

 
 
 
 F 


5



 

 

y
x
px
x
Rx
Rx
px
x
x
x
in in inx
out out outx
x
in inx
out outx
x
in
out
F
.5N
R
F
F
F
F  F
.5 N
F =
)
(
F =
A V V )
F = A V V
(
Q V )
F = Q V
(
for devided or branched flow
V )
F = Q(V


REAL FLOW IN PIPES
Viscous Flow in Ducts (Flow in Pipes)
The pipe defined as:
1. Same container sectional area along whole length (L)
2. Same container material along whole length (L)
3. Circular sectional area along whole length (L)
4. Straightforward along whole length (L)
5. Full flow (closed conduit) along whole length (L)
CV Analysis: For pipe segment control volume below
Continuity:
.
const
Q
Q 
 2
1 

1 2 1 2
. . sin , ., ave
i e V V ce A A const and V V

   
Energy (Bernoulli’s) equation in real fluid flow conditions:
𝑝1
𝛾
+
𝑉1
2
2𝑔
+ 𝑧1 =
𝑝2
𝛾
+
𝑉2
2
2𝑔
+ 𝑧2 + 𝒉𝑳
where ℎ𝐿represents the head loss between two sections 1 and 2 which
divided into two parts ℎ𝑓(head losses due to friction, called major losses), and
ℎ𝑚(head losses due to fitting, valves, and any other source of losses along
individual pipe, called miner losses)
ℎ𝐿 = ℎ𝑓 + ℎ𝑚

IMP_Fluid_Mechanics_lectures_and_Tutorials.pdf

IMP_Fluid_Mechanics_lectures_and_Tutorials.pdf

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to IMP_Fluid_Mechanics_lectures_and_Tutorials.pdf

Similar to IMP_Fluid_Mechanics_lectures_and_Tutorials.pdf (20)

More from ssuser79cb761

More from ssuser79cb761 (8)

Recently uploaded

Recently uploaded (20)

IMP_Fluid_Mechanics_lectures_and_Tutorials.pdf