There are 2 classes of fluid-> liquid & gasses
Chapter 1: PHYSICAL PROPERTIES OF FLUID
Fluid mechanics:
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branch of science which deals with behavior of fluid at
rest as well as motion
Physical properties of liquid
Types of fluid mechanics:
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Hydrodynamic (theoretical aspect)
Hydraulics (practical aspect)
Study of fluid at rest is called fluid statics.
Study of fluid in motion without considering motion is
called fluid kinematics and by considering force is called
fluid dynamics.
Basic concept of fluid mechanics
The basic fluid mechanics principles are the continuity
equation (i.e., Conservation of mass), the momentum
principle (or conservation of momentum) and the
energy equation.
Application of fluid mechanics in civil engineering:
The problem arises in the field of water supply,
irrigation, water power, navigation etc. resulted in
development of fluid mechanics. For the design of civil
related structures, the properties of fluid play vital role.
Hence, fluid mechanics is applicable in various field of
civil engineering such as:
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Water supply
Irrigation
Sanitation
Water distribution
Dam construction
Ground water flow
Flood flow analysis
Water retaining structure
Matter as solid, fluid or gas / shear stress in fluid & diff
btw solid & fluid
From fluid mechanics matter consists of only 2 states:
fluid or solid
The diff btw of solid & fluid is with the reaction of solid
and fluid to shear stress applied a solid can resist shear
stress by static deformation, a fluid cannot. Any shear
stress applied to fluid results in motion of fluid.
A liquid is fluid tends to retain its definite volume
A gas is fluid which is compressible and possess no
definite volume
1. Density/ mass density
mass od fluid (m)
density(ρ) =
volume (V)
m
ρ=
V
unit = kg/m3
2. Specific weight/ weight density
weight
specefic weight (γ) =
volume
mass × gravity
=
volume
m×g
γ=
V
γ=ρ×g
unit = N/m3
3. Specific volume
volume (V)
specefic volim(v) =
mass (m)
1
v=
ρ
unit = m3 /kg
4. Relative density/ specific gravity
density of liquid
δliquid =
density of water at 4°C
5. Vapor pressure and cavitation
The partial pressure of vapor molecule is called
vapor pressure.
If vapor pressure become equal or greater than
liquid surface the liquid starts boiling.
Cavitation is a phenomenon in which the static
pressure of a liquid reduces to below the
liquid's vapor pressure, leading to the
formation of small vapor-filled cavities in the
liquid.
6. Cohesion and adhesion
The attraction of fluid molecules to other fluid
molecules of same kind is called cohesion.
Adhesion is an attraction of two or more
different kind of molecules
7. Surface tension
The surface tension is a phenomenon that
arises due to cohesive interactions between
the molecules in the liquid.
Surface tension on a liquid droplet:
Force acting on droplet of water
1. Tensile force due to surface tension:
= σ × circumference
= σ × (π × d)
2. Pressure force in the area=
πd2
P×
4
8. Capillarity
The phenomenon of raise or fall of liquid in a
small tube relative to hype adjacent level of a
liquid
Causes of viscosity in liquid and gas:
It is due to cohesive force and molecular
momentum transfer. In liquid cohesive force is
pre dominant and in gas is molecular
momentum transfer is pre dominant.
When temperature is increased cohesive force
decreases so viscosity of liquid decreases
with the increase in temp but in case of gases
molecular momentum transfer increases when
temp increases which increases viscosity of gas.
Newton’s
law
of
viscosity:
Shear stress on fluid element layer is directly
proportional to the rate of shear strain
Here
σ = surface tension of fluid
θ
= angle of contact beetween fluid and tube
h = height of liquid in tube
force acting in this phenomenon
1. Weight of liquid in a tube: area of tube x
height x γ
πd2
=
×h×γ
4
2. Vertical component of tensile force
= σ × (πd)cosθ
At equilibrium condition
πd2
σ × (πd)cosθ =
×h×γ
4
4σcosθ
h=
rd
9. Viscosity:
the property of fluid which offers resistance to
the movement of one layer of fluid over
another
adjacent
layer
of
fluid.
It is primarily due to cohesion and molecular
momentum
μdu
dy
Where, µ is the proportionality constant known
as
dynamic
viscosity.
Fluid that obeys newtons law of viscosity is
Newtonian fluid e.g., water, air, alcohol
Fluid that doesn’t obeys newtons law of
viscosity is non-Newtonian fluid e.g. shampoo,
paint etc.
τ=
Chapter 2: FLUID PRESSURE AND ITS MEASUREMENT
Pressure is defined as the force per unit area.
pressure(P) =
force(f)
area(A)
unit = N/m2 or Pascal
1 bar = 105 pascal
standered atomospjeric pressure = 101.325 KPa
Application of pressure concept in civil engineering
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Dams
Gates in hydraulic structures
Water tanks
SOME TERMINOLOGIES AND POINTS FOR PRESSURE
COMPUTATION
1. Atmospheric Pressure
• The pressure exerted by the atmospheric air known
as atmospheric pressure.
• The atmospheric pressure varies with the altitude
and can be measured by Barometer.
• It is also called barometric pressure.
• Atmospheric pressure = 760mm Hg or 10.3 m of
water or 101.325 KPa
2. Gauge pressure
The pressure measured with reference of atmospheric
pressure is called gauge pressure.
Pressure measuring instrument measures the gauge
pressure. In such instrument, atmospheric pressure on
the scale is marked zero.
3. Vacuum Pressure
• The pressure below the atmospheric pressure is
known as vacuum pressure.
• It is also called negative gage pressure or suction
pressure.
4. Absolute Pressure
• When there is perfect vacuum, the pressure is zero.
• The pressure measured above a perfect vacuum is
called absolute pressure.
Relationship
Absolute pressure= atmospheric pressure+ gauge
pressure
Absolute pressure= atmospheric pressure- vacuum
pressure
The vertical height of a column of given fluid of above
any point at rest is called head. The gauge pressure can
be expressed in head.
P
h=
ρfluid xg
PASCALS LAW FOR PRESSURE AT A POINT
Statement:
Pressure at ant point in a fluid at rest is small in all
direction.
Proof
Considering X-direction
∑ Fx = 0
Px dydz − Ps dsdz sinθ = 0
In triangle ABF
p
sinθ =
h
dy
Px dydz − Ps dsdz = 0
ds
Px = Ps … … … (a)
Considering Y direction
∑ Fy = 0
Py dxdz − Ps dsdz cosθ − weight of element = 0
In triangle ABF
b
cosθ =
h
dx
1
Py dydz − Ps dsdz − ρg dxdydz = 0
ds
2
Since dx dy and dz are very small so neglected, we get
Py = Ps … … … (b)
from (a) and (b)
Px = Py = Ps
PRESSURE DEPTH RELATIONSHIP
P = ρgh
PRESSURE MEASUREMENT
1. Simple manometer as Piezometer
• Piezometer is the simplest form of manometer.
• The liquid rises to a height depending on the
pressure.
• As the top end is open to the atmosphere, the
pressure measured is gage pressure.
Pressure at A = 𝛾ℎ
Problems with the Piezometer
• It can only be used for liquids.
• Pressure must be above atmospheric
• Liquid height must be convenient, i.e. not be
too small or too large
2. U-tube manometer
• This device consists of a glass tube bent into the
shape of a U, which is filled with manometric
fluid (e.g. mercury) and connected to pipe or
tank.
• This manometer can be used to measure the
pressure of both liquids and gases.
• In this type of manometer, the manometric
fluid density should be greater than that of the
fluid measured and the two fluids should not be
able to mix.
Pressure at B = Pressure at C
P A + γh1 = γman h2
𝐏 𝐀 = 𝛄𝐦𝐚𝐧 𝐡𝟐 − 𝛄𝐡𝟏
In case of air as the fluid, we can generally neglect
𝛾ℎ1 (very small).
3. Differential U-tube manometer
The U-tube manometer can be connected at both
ends to measure pressure difference between
these two points.
Pressure at B = Pressure at C
PA + γa = PD + γb + γman h
𝐏𝐀 − 𝐏𝐃 = 𝛄𝐛 + 𝛄𝐦𝐚𝐧 𝐡 − 𝛄𝐚
4. Inverted U-tube differential manometer.
Pressure at C = Pressure at D
P A − γa − γman h = PB − γ(b + h)
𝐏 𝐀 − 𝐏𝐁 = 𝛄𝐚 + 𝛄𝐦𝐚𝐧 𝐡 − 𝛄(𝐛 + 𝐡)
5. Single column vertical manometer
When the manometer is not connected to the
container, the mercury in the reservoir is at original
level (XX) and at level B in the tube.
Equating pressure at XX
γ1 y = γmercury h1 … … … (a)
Due to pressure, manometric liquid in the reservoir
drops by dy and it will travel a distance of h in the tube.
Volume of fluid fallen = Volume of fluid risen
Ady = ah
A = Area of reservoir
a = area of tube
a
dy = h
A
Equating pressure at new level (ZZ)
PA + γ1 (y + dy) = γmercury (h + h 1 + dy). . (b)
From a and b
PA + γmercury h1 − γ1 dy
= γmercury h1 + γmercury (h + dy)
PA − γ1 dy = γmercury h + γmercury dy
PA = γmercury h + (γmercury + −γ1 )dy
𝐚
𝐏𝐀 = 𝛄𝐦𝐞𝐫𝐜𝐮𝐫𝐲 𝐡 + (𝛄𝐦𝐞𝐫𝐜𝐮𝐫𝐲 + −𝛄𝟏 ) 𝐡
𝐀
If A is very large compared to a, a/A is very small and dy
can be neglected. Then,
𝐏𝐀 = 𝛄𝐦𝐞𝐫𝐜𝐮𝐫𝐲 𝐡
6. Single column inclined manometer
PA + γmercury h1 − γ1 dy
= γmercury h1 + γmercury (h + dy)
PA − γ1 dy = γmercury h + γmercury dy
PA = γmercury h + (γmercury + −γ1 )dy
𝐚
𝐏𝐀 = 𝛄𝐦𝐞𝐫𝐜𝐮𝐫𝐲 𝐡 + (𝛄𝐦𝐞𝐫𝐜𝐮𝐫𝐲 + −𝛄𝟏 ) 𝐱𝐬𝐢𝐧𝛉
𝐀
If A is very large compared to a, a/A is very small and dy
can be neglected. Then,
𝐏𝐀 = 𝛄𝐦𝐞𝐫𝐜𝐮𝐫𝐲 𝐱𝐬𝐢𝐧𝛉
Advantages of manometers
• They are very simple and cheap.
• No calibration is required - the pressure can be
calculated from first principles.
Disadvantages of manometers
• Slow response: It is only really useful for very slowly
varying pressures and cannot be used at all for
fluctuating pressures.
• For the U tube manometer, two measurements
must be taken simultaneously to get the h value.
• It is often difficult to measure small variations in
pressure.
• It cannot be used for very large pressures unless
several manometers are connected in series.
• For very accurate work the relationship between
temperature and density must be known.
Mechanical pressure gage
When the manometer is not connected to the
container, the mercury in the reservoir is at original
level (XX) and at level B in the tube.
Equating pressure at XX
γ1 y = γmercury h1 … … … (a)
Due to pressure, manometric liquid in the reservoir
drops by dy and it will travel a distance of h in the tube.
Volume of fluid fallen = Volume of fluid risen
Ady = ah
A = Area of reservoir
a = area of tube
a
dy = xsinθ
A
Equating pressure at new level (ZZ)
PA + γ1 (y + dy)
= γmercury (h + h 1
+ dy) … … … (b)
From a and b
1. Bourdon’s tube Pressure Gauge:
• The pressure above or below the atmospheric
pressure, may be easily measured with the help of
burden’s tube pressure gauge.
• A burden’s tube pressure gauge, in its simplest
form, consists of an elliptical tube AABC, bent into
an arc of a circle as shown. This bent tube is called
Burdens tube.
• Bourdon tube gage is a simple mechanical device
for measuring pressure.
• It consists of a bent tube of elliptical cross-section
fixed at one end through which fluid enters.
• The other end is linked to a pointer which moves
through the scale. When fluid pressure is made to
enter the tube, its cross-section tends to become
circular, causing the tube to straighten and move
the pointer.
2. Diaphragm Pressure Gauge:
• The pressure above or below the atmospheric
pressure is also measured by using diaphragm
pressure gauge.
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A diaphragm pressure gauge, in its simplest form,
consists of a corrugated diaphragm ( instead of
Bourdin’s tube ) as shown above.\
When the gauge is connected to the fluid ( whose
pressure is required to be found out) at C the fluid
under pressure causes some deformation of the
diaphragm. With the help of some opinion
arrangement the elastic deformation of the
diaphragm rotates the pointer. This pointer moves
over a calibrated scale, which directly gives the
pressure as shown in figure.
3. Dead Weight Pressure Gauge:
• It is the most accurate pressure gauge, which is
generally used for the calibration of the other
pressure gauge in a laboratory.
• A dead weight pressure gauge, in its simplest form,
consists of a piston and cylinder of known area,
which is connected to a fluid through a tube as
shown below.
• The pressure on the fluid in the pipe is calculated
from the relation
• P = Weight/ Area of the piston
• A pressure gauge, to be calibrated is fitted on the
other end of the tube as shown above. By changing
the weight, on the piston, the pressure on the fluid
is calculated and marked on the gauge at the
respective points, indicated by the pointer. A small
error due to frictional resistance to the motion of
the piston may come into play. But the same may
be avoided by taking adequate precaution.
Chapter 4: EQUILIBRIUM STABILITY
Condition of equilibrium of a floating and submerged
body
Equilibrium Floating body
Submerged
M is above G
B is above G
Stable
M is below G
B is below G
Unstable
M and G coincide B and G coincide
Neutral
Meta center
The point about which body starts to oscillate is called
meta center
Meta centric height
Distance between meta center and center of gravity
Metacenter
𝐺𝑀 =
𝐼𝑐𝑔
− 𝐵𝐺
𝑉𝑠𝑢𝑏
Experimentally
𝐺𝑀 =
𝑤1 𝑥
𝑊𝑡𝑎𝑛𝜃
Linear equilibrium
𝑡𝑎𝑛𝜃 = −
Chapter 3: HYDROSTATIC FORCE
The force exerted by static fluid when fluid come in
contact with the surface is called total pressure
The point of application of total pressure on the surface
is called center of pressure.
Horizontal plane
Total pressure= 𝜚𝑔ℎ̅𝐴
Vertical plane
Total pressure= 𝜚𝑔ℎ̅𝐴
Center of pressure
𝐼𝑐𝑔
ℎ𝑝 =
+ℎ
𝐴ℎ̅
Inclined plane surface
Total pressure = 𝜚𝑔ℎ̅𝐴
Center of pressure=
𝐼𝑐𝑔 sin2 𝜃
ℎ𝑝 =
+ℎ
𝐴ℎ̅
𝑎𝑥
𝑔 + 𝑎𝑧
Cases
1. Horizontal acceleration
𝑡𝑎𝑛𝜃 = −
𝑎𝑥
𝑔
2. Vertical acceleration
𝑎𝑧
)ℎ ↑
𝑔
𝑎𝑧
𝑃 = 𝛾 (1 − ) ℎ ↓
𝑔
3. Angular acceleration
𝜕𝑃
= 𝜚𝑟𝜔2
𝜕𝑟
𝜔2 𝑟 2
Δ𝑧 =
2𝑔
z = vertical height
𝑃 = 𝛾 (1 +
Chapter 5: FLUID KINEMATICS
Lagrangian approach
Studying individual fluid particles and history of motion
of particle. Equation can be derived
Eulerian approach
Based on fixed space. This approach is used for
engineering problem
Classification of fluid motion
1. Steady and unsteady flow
If velocity, depth, density doesn’t change with
time then steady otherwise unsteady
2. Uniform and non uniform flow
If velocity, depth, density doesn’t change with
distance then uniform otherwise non uniform
3. Laminar vs turbulent flow
Reynolds number
𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝑅𝐸 =
𝑣𝑖𝑠𝑐𝑖𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒
If RE< 2000 → laminar
If RE> 4000 → turbulent
If RE 2000 ≤ 𝑅𝐸 ≤ 4000 → transition phase
4. Subcritical, critical and super critical force
Fraud number
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝐹𝑟 =
√𝑔𝑟𝑎𝑣𝑖𝑡𝑦 × ℎ𝑢𝑑𝑟𝑢𝑙𝑖𝑐 𝑑𝑒𝑝𝑡ℎ (𝐷)
𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝐷=
𝑡𝑜𝑝 𝑤𝑖𝑑𝑡ℎ
F(r)=1 →critical
F(r)<1 → sub-critical
F(r)>1 → super critical
5. Compressible and incompressible fluid
If density is constant then incompressible
otherwise in compressible.
6. Rotational and irrotational flow
Particle moving along a steam line rotates
about its own axis the it is called rotational
otherwise irrotational.
7. 1D, 2D and 3D flow
1D→ velocity depend on one space
2D→ velocity depend upon two space
Example: flow over weir
3D→ depend upon three space variable and
time
Chapter 7: FLOW MEASUREMENT
Hydrostatic coefficient
1. Coefficient of velocity (Cv)=
𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
2. Coefficient of contraction (Cc)=
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑣𝑒𝑛𝑎 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑎
𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑎𝑟𝑒𝑎
3. Coefficient of discharge (Cv)=
𝑎𝑐𝑡𝑢𝑎𝑙 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒
𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒
Venturi meter
It is determined by using Bernoulli’s equation
𝑷𝟏 𝒗𝟏 𝟐
𝑷𝟐 𝒗𝟐 𝟐
+
+ 𝒛𝟏 =
+
+ 𝒛𝟐
𝜸
𝟐𝒈
𝜸
𝟐𝒈
When head loss is given
𝑷𝟏 𝒗𝟏 𝟐
𝑷𝟐 𝒗𝟐 𝟐
+
+ 𝒛𝟏 =
+
+ 𝒛𝟐 + 𝒉𝒆𝒂𝒅 𝒍𝒐𝒔𝒔
𝜸
𝟐𝒈
𝜸
𝟐𝒈
For horizontal pipe
z1=z2
Continuity equation
A1V1=A2V2
Theoretical discharge (Q th) =
𝐴1 𝐴2 √2𝑔ℎ
√𝐴1 2 −𝐴1 2
Actual discharge (Q actual) =
𝐶𝑑 𝐴1 𝐴2 √2𝑔ℎ
√𝐴1 2 −𝐴1 2
For vertical
Z1 and z2 are different
ℎ=
𝑃1 − 𝑃2
+ (𝑧1 − 𝑧2 )
𝛾
ℎ = 𝑥(
𝛾𝐻𝑔
− 1)
𝛾
here x= distance between two limb
ORIFICE
End effect
Cases
Discharge over rectangular weir considering end effect
1. Flow through small orifice
Theoretical velocity (v th) = √2𝑔ℎ
Actual velocity (v actual) = 𝐶𝑑 √2𝑔ℎ
2. Flow through large orifice
3
3
2
𝑄 = 𝐶𝑑 𝑏√2𝑔 (𝐻2 2 − 𝐻1 2 )
3
Here
b= breadth of a vessel
H1, H2= height of vessel up to orifice
𝑄=
3
2
𝐶𝑑 (𝐿 − 0.2𝐻)√2𝑔 (𝐻 2 )
3
Discharge over triangular weir considering end effect
𝑛𝑜 𝑒𝑛𝑑 𝑒𝑓𝑓𝑒𝑐𝑡
Considering both end effect and approach velocity in
rectangle
3
3
2
𝑄 = 𝐶𝑑 𝐿√2𝑔 ((𝐻 + ℎ𝑎 )2 − ℎ𝑎 2 )
3
Cipolletti weir/ trapezoidal weir
3. Flow through fully submerge orifice
v2=√2𝑔ℎ
4. Flow through partially submerge orifice
3
3
2
𝑄 = 𝐶𝑑 𝑏√2𝑔 (𝐻2 2 − 𝐻1 2 )
3
+ 𝐶𝑑 𝐵(𝐻2 − 𝐻1 )√2𝑔ℎ
It is a trapezoidal weir with side slope (1:4) in such a way
that discharge over it
Proof
Discharge over rectangular weir considering end effect
Q= actual discharge – losses
Notches and weir
𝑄=
1. Flow over rectangular notches/ weir
𝑄=
3
2
𝐶𝑑 𝑏√2𝑔 (𝐻2 )
3
2. Flow over triangular notches
𝑄=
5
8
𝜃
𝐶𝑑 √2𝑔 𝑡𝑎𝑛 (𝐻 2 )
15
2
3. Flow through trapezoidal notch/ weir
𝑄=
𝑄=
3
3
2
2
𝐶𝑑 𝐿√2𝑔 (𝐻2 ) − 𝐶𝑑 √2𝑔(0.2𝐻)𝐻 2
3
3
If the angle made in slope of trapezoidal weir i.e. angle
of slope is 14.03° then there is no end effect
Discharge over narrow created weir
1. For rectangular
5
3
8
2
𝐶𝑑 √2𝑔 𝑡𝑎𝑛𝜃 (𝐻 2 ) + 𝐶𝑑 𝑏√2𝑔 (𝐻 2 )
15
3
Approach velocity (va)
𝑄=
𝑄=
5
8
𝐶𝑑 √2𝑔 𝑡𝑎𝑛𝜃 (𝐻 2 )
15
Discharge over sharp crested weir:
𝑄=
3
3
2
𝐶𝑑 𝐿√2𝑔 ((𝐻 + ℎ𝑎 )2 − ℎ𝑎 2 )
3
For rectangular
𝑄=
For triangular
𝑄=
3
2
𝐶𝑑 𝐿√2𝑔 (𝐻 2 )
3
2. For triangular
It is a speed of upstream water to approach the
weir/notch
For rectangular
3
2
𝐶𝑑 (𝐿 − 0.2𝐻)√2𝑔 (𝐻 2 )
3
5
5
8
𝜃
𝐶𝑑 √2𝑔 𝑡𝑎𝑛 ((𝐻 + ℎ𝑎 )2 − ℎ𝑎 2 )
15
2
3
2
𝐶𝑑 𝐿√2𝑔 (𝐻 2 )
3
For triangular
𝑄=
5
8
𝐶𝑑 √2𝑔 𝑡𝑎𝑛𝜃 (𝐻 2 )
15
Leaving mass=𝜚2 𝐴2 𝑉2 𝑑𝑡
Discharge over ogee weir
𝑄=
3
2
𝐶𝑑 𝐿√2𝑔 (𝐻 2 )
3
Change in momentum (d(mv))= 𝜚1 𝐴1 𝑉1 𝑑𝑡 − 𝜚2 𝐴2 𝑉2 𝑑𝑡
(𝑑(𝑚𝑣)) = 𝐹𝑑𝑡
Discharge over board crested weir
𝐹 = 𝜚1 𝐴1 𝑉1 𝑑𝑡 − 𝜚2 𝐴2 𝑉2 𝑑𝑡
For incompressible fluid
𝜚1 = 𝜚2 = 𝜚
EMPTYING AND FILLING OF TANK
By continuity equation
Without inflow
𝑄 = 𝐴1 𝑉1 = 𝐴2 𝑉2
For cylindrical tank/ rectangular tank
So
𝑇=−
2𝐴
𝐶𝑑 𝑎 √2𝑔
1
[𝐻1 2
1
− 𝐻2 2 ]
𝐹 = 𝜚𝑄(𝑣2 − 𝑣1 )
Application
1. Pipe reducer
𝐹𝑥 = (𝑃1 𝐴1 − 𝑃2 𝐴2 ) + 𝜚𝑄(𝑣1 − 𝑣2 )
For conical tank
𝑇=
−𝜋𝑅1 2
1
3
2 5
4
[ ℎ2 + 2𝐻0 2 ℎ2 + 𝐻0 ℎ2 ]
3
𝐶𝑑 𝑎 √2𝑔 (𝐻1 + 𝐻0 )2 5
For hemispherical tank
𝑇=
5
14𝜋𝑅 2
15 𝐶𝑑 𝑎 √2𝑔
Chapter 8: MOMENT ANALYSIS OF FLOW
Here impulse momentum
𝐹𝑑𝑡 = 𝑚𝑑𝑣
m= mass
dv= change in velocity
Force based on impulse momentum
𝐹 = 𝜚𝑄(𝑣2 − 𝑣1 )
Proof
Entering mass= mass x velocity
= 𝜚1 𝐴1 𝑉1 𝑑𝑡
𝐹𝑦 = 0
2. In pipe bending
𝐹𝑥 = (𝑃1 𝐴1 − 𝑃2 𝐴2 𝑐𝑜𝑠𝜃) + 𝜚𝑄(𝑣1 − 𝑣2 𝑐𝑜𝑠𝜃)
𝐹𝑦 = 𝑃2 𝐴2 𝑠𝑖𝑛𝜃) + 𝜚𝑄(𝑣2 𝑠𝑖𝑛𝜃)
IMPACT JET
1. Stationary vertical plate
𝐹𝑥 = 𝜚𝐴𝑣 2
𝐹𝑦 = 0
2. Inclined stationary plate
𝐹𝑛 = 𝜚𝐴𝑣 2 sin 𝜃
𝐹𝑥 = 𝜚𝐴𝑣 2 sin2 𝜃
𝐹𝑦 = 𝜚𝐴𝑣 2 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠𝜃
For discharge
By continuity equation
𝑄 = 𝑄1 + 𝑄2 … … … (1)
Mass conserved = output – input
0 = 𝜚𝑄1 𝑉1 − 𝜚𝑄2 𝑉2 − 𝜚𝑄𝑉 cos 𝜃
𝑣1 = 𝑣2 = 𝑣
𝑄𝑐𝑜𝑠𝜃 = 𝑄1 − 𝑄2
Solving 1 and 2
𝑄
𝑄1 = (1 + cos 𝜃)
2
𝑄
𝑄2 = (1 − cos 𝜃)
2
3. Impact on curved surface
a. Impact at center
𝐹𝑥 = 𝜚𝐴𝑣 2 (1 + cos 𝜃)
𝐹𝑦 = −𝜚𝐴𝑣 2 sin 𝜃
b. Impact at one end (symmetrical)
𝐹𝑥 = 2𝜚𝐴𝑣 2 (cos 𝜃)
𝐹𝑦 = 0
c. Impact at one end (un symmetrical)
𝐹𝑥 = 𝜚𝐴𝑣 2 (cos 𝜃 + cos 𝛼)
𝐹𝑦 = 𝜚𝐴𝑣 2 (𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛𝛼)
4. Impact of jet on moving plates
Velocity ko thau ma (v-u)
1. vertical plate
𝐹𝑥 = 𝜚𝐴(𝑣 − 𝑢)2
𝐹𝑦 = 0
2. Inclined y plate
𝐹𝑛 = 𝜚𝐴(𝑣 − 𝑢)2 sin 𝜃
𝐹𝑥 = 𝜚𝐴(𝑣 − 𝑢)2 sin2 𝜃
𝐹𝑦 = 𝜚𝐴(𝑣 − 𝑢)2 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠𝜃
3. Impact on curved surface
a) Impact at center
i. 𝐹𝑥 = 𝜚𝐴(𝑣 − 𝑢)2 (1 +
cos 𝜃)
ii. 𝐹𝑦 = −𝜚𝐴(𝑣 −
𝑢)2 sin 𝜃
b) Impact at one end (symmetrical)
i. 𝐹𝑥 = 2𝜚𝐴(𝑣 −
𝑢)2 (cos 𝜃)
ii. 𝐹𝑦 = 0
c) Impact at one end (un symmetrical)
i. 𝐹𝑥 = 𝜚𝐴(𝑣 −
𝑢)2 (cos 𝜃 + cos 𝛼)
ii. 𝐹𝑦 = 𝜚𝐴(𝑣 −
𝑢)2 (𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛𝛼)
Chapter 9: BOUNDARY LAYER
Displacement thickness
Mass of fluid flowing in strip=mass (m)
𝑚𝑎𝑠𝑠(𝑚) = 𝜚𝑄 = 𝜚𝐴𝑣 = 𝜚(𝑏 × 𝑑𝑦)𝑢
When there is no plate
Mass entering= 𝜚(𝑏 × 𝑑𝑦)𝑈
Difference in flowing fluid
𝜚(𝑏 × 𝑑𝑦)𝑈 − 𝜚(𝑏 × 𝑑𝑦)𝑢
Total difference=
𝛿
∫ 𝜚(𝑏 × 𝑑𝑦)(𝑈 − 𝑢) … … … (1)
0
Let path is displaced by distance
𝛿 ∗ = 𝜚𝑈(𝑏𝛿 ∗ ) … … … (2)
Equating 1 and 2 we get
𝛿
∫
0
Momentum thickness
Momentum = mass x velocity
Angular momentum
(𝜚(𝑏 × 𝑑𝑦)𝑢) × 𝑢
Momentum at end 1= 𝑚1 𝑣1 𝑟1
𝛿
∫ 𝜚𝑏𝑢(𝑈 − 𝑢) 𝑑𝑦 = 𝜚𝑈 2 (𝑏𝜃)
Momentum at end 2= 𝑚2 𝑣2 𝑟2
0
𝛿
Change in momentum= 𝑚(𝑣2 𝑟2 − 𝑣1 𝑟1 )
𝜃=∫
0
Rate of change in momentum=
𝑚
= (𝑣2 𝑟2 − 𝑣1 𝑟1 )
𝑡
𝑢
𝑢
(1 − ) 𝑑𝑦
𝑈
𝑈
Energy thickness
1
𝐾. 𝐸 = 𝑚𝑣 2
2
= 𝜚𝑄(𝑣2 𝑟2 − 𝑣1 𝑟1 )
𝐾. 𝐸 =
Rate of change of momentum is torque so,
𝑇 = 𝜚𝑄(𝑣2 𝑟2 − 𝑣1 𝑟1 )
1
(𝑈 − 𝑢)𝑑𝑦
𝑈
1
(𝜚𝑏𝑑𝑦𝑢)𝑢2
2
𝛿
∫ 𝜚𝑏𝑢(𝑈 2 − 𝑢2 ) 𝑑𝑦 = 𝜚(𝛿 ∗∗ 𝜃)𝑈3
0
𝛿
𝛿 ∗∗ = ∫
0
𝑢
𝑢2
(1 − 2 ) 𝑑𝑦
𝑈
𝑈
From drag force
Dimensional homogeneity:
𝜏0
𝜕 𝛿𝑢
𝑢
=
∫ (1 − ) 𝑑𝑦
2
𝜚𝑈
𝜕𝑥 0 𝑈
𝑈
𝑄 =
3
2
√2𝑔𝐿𝐻 2
3
Dimension of Left-hand side = 𝐿3 𝑇 −1
It is equal to:
Dimension of right-hand side
𝜏0
𝜕𝜃
=
2
𝜚𝑈
𝜕𝑥
1
3
= (𝐿𝑇 2 )2 𝐿 × 𝐿2 = 𝐿3 𝑇 −1
Chapter 10: DIMENSION ANALYSIS
RAYLEIGH’S METHOD
Widely used dimension
𝑋 = 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒
fundamental
Geometric
Kinematic
quantities
kg
Mass = M
m
Length = L
Functional relationship: =f(𝑥1 𝑥2 𝑥3 , . . . . . . . . . , 𝑥𝑛 )
t
Time = T
Equation in exponential form:
𝑥1 𝑥2 𝑥3 , . . . . . . . . . , 𝑥𝑛 = 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝑚2
Area = 𝐿2
𝑚3
Volume = 𝐿3
𝑚/𝑠
Velocity = 𝐿𝑇 −1
𝑥 = 𝐾 𝑥1𝑎 𝑥2𝑏 𝑥3𝑐 , . . . . . . . . . , 𝑥𝑛𝑧
where a, b, c.......zare constants
𝑟𝑎𝑑/𝑠
Angular velocity = 𝑇 −1
𝑚/𝑠 2
Acceleration = 𝐿𝑇 −2
𝑟𝑎𝑑/𝑠 2
𝑚3 /𝑠
Ang acceleration = 𝑇 −2
3 −1
BUCKINGHAM’S Π THEOREM
Discharge = 𝐿 𝑇
Kinematic viscosity = 𝐿2 𝑇 −1
Dynamic
quantities
If there are more than three constants, then three
appropriate constants are expressed in terms of other
constants. The variables of the expression to be derived
are examined, and the powers of these three variables
are usually expressed in terms of other constants.
𝑘𝑔 𝑚/𝑠
Force, Weight = 𝑀𝐿𝑇 −2
𝑘𝑔/𝑚3
Density = 𝑀𝐿−3
Specific weight = 𝑀𝐿−2 𝑇 −2
Dynamic viscosity = 𝑀𝐿−1 𝑇 −1
Pressure = 𝑀𝐿−1 𝑇 −2
Surface tension = 𝑀𝑇 −2
Write down the functional relationship between
dependent and independent variables.
𝑥1 = 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒
𝑥2 𝑥3 𝑥4 , . . . . . . . . . 𝑥𝑛 = 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝑥1 = 𝑓(𝑥2 𝑥3 𝑥4 , . . . . . . . . . 𝑥𝑛 )
𝐹1 (𝑥1 𝑥2 𝑥3 𝑥4 , . . . . . . . . . 𝑥𝑛 ) = 0 … … … . (𝐼)
Count total number of variables (n) and fundamental
dimensions (m)
Work, energy = 𝑀𝐿2 𝑇 −2
𝑁𝑜. 𝑜𝑓 𝜋 𝑡𝑒𝑟𝑚𝑠 = 𝑛 − 𝑚
Shear stress = 𝑀𝐿−1 𝑇 −2
Write down the functional relationship in π terms.
𝜑 (𝜋1 , 𝜋2 , 𝜋3 , . . . . . . . . . . . . , 𝜋𝑚−𝑛 ) = 0 … … . . (𝐼𝐼)
Power = 𝑀𝐿2 𝑇 −3
2 −2
Torque = 𝑀𝐿 𝑇
Momentum = 𝑀𝐿𝑇 −1
For each π term, write down equation in exponential
form with (m+1) variables, where m=3 is called
fundamental dimension and it is also called repeating
variables.
Each π group is a function of n repeating variables plus
one of the remaining variables. The π term is
dimensionless.
Let 𝑥2 , 𝑥3 , 𝑥4 be the repeating variables.
𝜋1 = 𝑥2𝑎1 𝑥3𝑏1 𝑥3𝑐1
𝜋2 = 𝑥2𝑎2 𝑥3𝑏2 𝑥3𝑐2
𝑎(𝑛−𝑚) 𝑏(𝑛−𝑚) 𝑐(𝑛−𝑚)
𝑥3
𝑥3
𝜋 𝑛−𝑚 = 𝑥2
Solve each equation by the principle of dimensional
homogeneity (writing dimensions and equating the
powers of M, L and T to get constants. )
Obtain π1, π2, π3,..........πn-m and substitute in eqn. (II).
Establish functional relationship between π terms.
Reynold number (RE)
𝑅𝐸 =
𝐼𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒
𝑅𝐸 =
𝜚𝑣𝑙
𝜇