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VTU – e-Shikshana Programme
Applied Hydraulics
B.E., IV Semester
Course Code 17 CV43
By
Dr. H. B. Balakrishna Professor & Head
Department of Civil Engineeering
Bangalore Institute of Technology
K R Road, V V Pura
Bangalore 560004
Mobile: 9845395535
E-mail: [email protected]
TITLE OF THE COURSE: APPLIED HYDRAULICS
B.E., IV Semester, Civil Engineering [As per Choice Based Credit System (CBCS) scheme]
Course Code 17 CV43 CIE Marks 40
Number of Lecture Hours/Week 04 SEE Marks 60
Total Number of Lecture Hours 50 (10 Hours per Module) Exam Hours 03
CREDITS – 04
COURSE OBJECTIVES: The objectives of this course is to make students to learn:
1. Principles of dimensional analysis to design hydraulic models and Design of various models.
2. Design the open channels of various cross sections including design of economical sections.
3. Energy concepts of fluid in open channel, Energy dissipation, Water surface profiles at
different conditions.
4. The working principles of the hydraulic machines for the given data and analysing the
performance of Turbines for various design data.
MODULE-1
Dimensional analysis: Dimensional analysis and similitude: Dimensional homogeneity, Non
Dimensional parameter, Rayleigh methods and Buckingham Π theorem, dimensional analysis,
choice of variables, examples on various applications.
Model analysis: Model analysis, similitude, types of similarities, force ratios, similarity laws,
model classification, Reynolds model, Froude’s model, Euler’s Model, Webber’s model, Mach
model, scale effects, Distorted models. Numerical problems on Reynold’s, and Froude’s Model
Buoyancy and Flotation: Buoyancy, Force and Centre of Buoyancy, Metacentre and
Metacentric height, Stability of submerged and floating bodies, Determination of Metacentric
height, Experimental and theoretical method, Numerical problems L1, L2, L3, L4
MODULE -2
Open channel flow hydraulics: Uniform Flow: Introduction, Classification of flow through
channels, Chezy’s and Manning’s equation for flow through open channel, Most economical
channel sections, Uniform flow through Open channels, Numerical Problems. Specific Energy
and Specific energy curve, Critical flow and corresponding critical parameters, Metering
flumes, Numerical Problems
L3, L4
MODULE – 3
Non-uniform flow: Non-Uniform Flow: Hydraulic Jump, Expressions for conjugate depths
and Energy loss, Numerical Problems Gradually varied flow, Equation, Back water curve and
afflux, Description of water curves or profiles, Mild, steep, critical, horizontal and adverse
slope profiles, Numerical problems, Control sections L2, L3, L4
MODULE-4
Hydraulic machines: Introduction, Impulse-Momentum equation. Direct impact of a jet on a
stationary and moving curved vanes, Introduction to concept of velocity triangles, impact of
jet on a series of curved vanes- Problems
Turbines – Impulse Turbines: Introduction to turbines, General lay out of a hydroelectric
plant, Heads and Efficiencies, classification of turbines. Pelton wheel components, working
principle and velocity triangles. Maximum power, efficiency, working proportions –
Numerical problems. L1, L2, L3, L4
MODULE-5
Reaction Turbines and Pumps: Radial flow reaction turbines: (i) Francis turbine-
Descriptions, working proportions and design, Numerical problems. (ii) Kaplan turbine-
Descriptions, working proportions and design, Numerical problems. Draft tube theory and unit
quantities. (No problems)
Centrifugal pumps: Components and Working of centrifugal pumps, Types of centrifugal
pumps, Work done by the impeller, Heads and Efficiencies, Minimum starting speed of
centrifugal pump, Numerical problems, Multi-stage pumps. L1, L2, L3, L4
Course outcomes:
After a successful completion of the course, the student will be able to:
1. Apply dimensional analysis to develop mathematical modelling and compute the parametric
values in prototype by analysing the corresponding model parameters
2. Design the open channels of various cross sections including economical channel sections
3. Apply Energy concepts to flow in open channel sections, Calculate Energy dissipation,
4. Compute water surface profiles at different conditions
5. Design turbines for the given data, and to know their operation characteristics under different
operating conditions
TEXT BOOKS:
1. P N Modi and S M Seth, “Hydraulics and Fluid Mechanics, including Hydraulic Machines”,
20th edition, 2015, Standard Book House, New Delhi
2. R.K. Bansal, “A Text book of Fluid Mechanics and Hydraulic Machines”, Laxmi
Publications, New Delhi
3. S K SOM and G Biswas, “Introduction to Fluid Mechanics and Fluid Machines”, Tata
McGraw Hill, New Delhi
REFERENCES BOOKS
1. K Subramanya, “Fluid Mechanics and Hydraulic Machines”, Tata McGraw Hill Publishing
Co. Ltd.
2. Mohd. Kaleem Khan, “Fluid Mechanics and Machinery”, Oxford University Press
3. C.S.P. Ojha, R. Berndtsson, and P.N. Chandramouli, “Fluid Mechanics and Machinery”,
Oxford University Publication – 2010
4. J.B. Evett, and C. Liu, “Fluid Mechanics and Hydraulics ”, McGraw-Hill Book Company.-
2009.
MODULE-1
Dimensional analysis: Dimensional analysis and similitude: Dimensional homogeneity, Non
Dimensional parameter, Rayleigh methods and Buckingham Π theorem, dimensional analysis,
choice of variables, examples on various applications.
Model analysis: Model analysis, similitude, types of similarities, force ratios, similarity laws,
model classification, Reynolds model, Froude’s model, Euler’s Model, Webber’s model, Mach
model, scale effects, Distorted models. Numerical problems on Reynold’s, and Froude’s Model
Buoyancy and Flotation: Buoyancy, Force and Centre of Buoyancy, Metacentre and
Metacentric height, Stability of submerged and floating bodies, Determination of Metacentric
height, Experimental and theoretical method, Numerical problems L1, L2, L3, L4
Dimensional analysis is a mathematical technique which makes use of study of dynamics as an art
to the solution of engineering problems. Dimensional analysis is a means of simplifying a physical
problem by appealing to dimensional homogeneity to reduce the number of relevant variables. It is
particularly useful for:
presenting and interpreting experimental data;
attacking problems not amenable to a direct theoretical solution;
checking equations;
establishing the relative importance of particular physical phenomena;
Most physical quantities can be expressed in terms of combinations of five basic dimensions. These
are mass (M), length (L), time (T), electrical current (I), and temperature, represented by the Greek
letter theta (𝜃). These five dimensions have been chosen as being basic because they are easy to
measure in experiments. Dimensions aren't the same as units. For example, the physical quantity,
speed, may be measured in units of meters per second, miles per hour etc.; but regardless of the units
used, speed is always a length divided a time, so we say that the dimensions of speed are length
divided by time, or simply L/T. Similarly, the dimensions of area are L2 since area can always be
calculated as a length times a length. For example, although the area of a circle is conventionally
written as πr2, we could write it as π r (which is a length) × r (another length).
Systems of Units:
All the dimensions of physical quantities need to be written in SI units. The ‘SI’ stands for
International System (Système Internationale). All engineering quantities can be defined in
terms of the four basic dimensions M, L, T and θ. We could use the S.I. units of kilogram’s,
meters, seconds and Kelvin, or any other system of units, but if we stick to M,L,T and θ we
free ourselves of any constraints to a particular system of measurements. The SI unit for current
is ampere, and for temperature the Kelvin. Notice that Kelvin is abbreviated as just K. The
degree symbol, °, and the word "degree" are not used with Kelvin. The official international
system of units (System International Units). Strong efforts are underway for its universal
adoption as the exclusive system for all engineering and science, but older systems, particularly
the CGS and fps engineering gravitational systems are still in use and probably will be around
for some time.
As an example, let's look at speed, which has dimensions of length divided by time or L/T. Its SI units
are then meters divided by seconds, represented as m/s or m·s-1.
SI system:
Primary Quantities: Derived Quantities
Quantity Unit Quantity Unit Quantity
Mass in
Kilogram kg
Force in Newton
(1 N = 1 kg.m/s2) N
Force in Newton
(1 N = 1 kg.m/s2)
Length in
Meter m
Pressure in Pascal
(1 Pa = 1 N/m2) N/m2
Pressure in Pascal
(1Pa =1 N/m2)
Time in
Second
s or as
sec
Work, energy in
Joule ( 1 J = 1 N.m) J
Work, energy in
Joule ( 1J = 1 N.m)
Temperature in
Kelvin K
Power in Watt
(1W = 1 J/s) W
Power in Watt (1
W = 1 J/s)
Mole mol
CGS Units:
The older centimeter-gram-second (CGS) system has the following units for derived quantities:
Quantity Unit
Force in dyne (1 dyn = 1 g.cm/s2) dyn
Work, energy in erg ( 1 erg = 1 dyn.cm = 1 x 10-7 J ) erg
Heat Energy in calorie ( 1 cal = 4.184 J) cal
Dimensions: Dimensions of the primary quantities:
Fundamental dimension Symbol
Length L
Mass M
Time t
Temperature T
Dimensions of derived quantities can be expressed in terms of the fundamental dimensions.
Quantity Representative symbol Dimensions
Angular velocity t-1
Area A L2
Density M/L3
Force F ML/t2
Kinematic viscosity L2/t
Linear velocity v L/t
Some quantities have no dimensions. For example, the sine of an angle is defined as the ratio of the
lengths of two particular sides of a triangle. Thus, the dimensions of the sine are L/L, or 1. Therefore,
the sine function is said to be "dimensionless". There are many other examples of "dimensionless"
quantities listed in the following table.
All trigonometric functions
Exponential functions
Logarithms
Angles (but notice the discussion in the next paragraph)
Quantities which are simply counted, such as the number of people in the room, plain old
numbers (like 2, p, etc.)
(Note-1: Note that some quantities which are "dimensionless" have units. For example, angles can be measured in units of radians or degrees, but angles are "dimensionless")
Dimensions of quantities:
Fundamental Dimensions
All physical quantities are measured by comparison which is made with respect to a fixed value. Length
(L), Mass (M) and Time (T) are three fixed dimensions which are of importance in fluid mechanics and
fluid machinery. In compressible flow problems, temperature is also considered as fundamental
dimensions.
Secondary Quantities or Derived Quantities
Secondary quantities are derived quantities or quantities which can be expressed in terms of two
or more fundamental quantities.
1.3.1 Dimensions of quantities
Quantity Common Symbol(s) Dimensions
Area A L2
Geometry Volume V L3
Second moment of area I L4
Velocity U LT–1
Acceleration a LT–2
Kinematics
Angle θ 1 (i.e. dimensionless)
Angular velocity ω T–1
Quantity of flow Q L3T–1
Mass flow rate m MT–1
Force F MLT–2
Moment, torque T ML2T–2
Dynamics Energy, work, heat E, W ML2T–2
Power P ML2T–3
Pressure, stress p, τ ML–1T–2
Work done W L2MT-2
Density ρ ML–3
Viscosity μ ML–1T–1
Kinematic viscosity ν L2T–1
Fluid properties Surface tension σ MT–2
Thermal conductivity k MLT–3θ–1
Specific heat cp, cv L2T–2 θ–1
Bulk modulus K ML–1T–2
Examples: –
V = u + at
m/s m/s + m/s2× s
LT-1 = (LT-1) + (LT-2) (T)
2
2
1atutS - all terms have the dimensions of length
HZg
Vp
2
2
- all terms have the dimensions of length
The dimensional homogeneity is a useful tool for checking formulae. For this reason it is useful
when analyzing a physical problem to retain algebraic symbols for as long as possible, only
substituting numbers right at the end. However, dimensional analysis cannot determine
numerical factors; e.g. it cannot distinguish between 2
2
1at and 2at in the first formula above.
Dimensional homogeneity is the basis of the formal dimensional analysis that follows
Analysis- Raleigh’s method, Buckingham’s Π theorem- problems
Rayleigh’s method of dimensional analysis
This method of dimensional analysis was originally proposed by Lord Rayleigh in 1899. He used this
method for determining the effect of temperature on gases. In this method, functional relationship of
variables is expressed in the form of an exponential equation. The equation must be dimensionally
homogeneous.
For example if Y is some function of independent variables X1,X2,X3… etc., then functional relationship
may be written as
Y = f [ X1,X2,X3...]
In this equation, Y is dependent variable and X1,X2,X3 are independent variables.
Dependent variable is one about which information is required. Independent variables are those
which govern the variation of dependent variables.
Procedure
Rayleigh’s method is based on the following steps.
a) Write the functional relationship of the given data. Y = f [ X1, X2, X3...]
b) Write equation in the exponential form with exponents a, b, c…
c) Put the dimensions of variables involved using any one system (MLT, FLT).
d) Apply dimensional homogeneity and evaluate the values of exponents a, b, c…,d.
e) Substitute the values of exponents a, b, c in the equation form in step number 2.
f) Simplify the equation for the required physical quantity.
Limitations of Rayleigh Method: Rayleigh’s method of analysis is adopted when numbers of
parameters or variables are less (3 or 4 or 5).
Methodology
X1 is a function of
X2, x3, X4, ……,Xn then it can be written as
X1 = f(X2, x3, X4, ……,Xn)
X1 = K (X2a, x3
b, X4
c, ……)
Taking dimensions for all the quantities [X1] = [X2]a [X3]
b [X4]
c……
Dimensions for quantities on left hand side as well as on the right hand side are written and using
the concept of Dimensional Homogeneity a, b, c ….can be determined.
Then, X1 = K X2a X3
b X4
c……
Solved Examples
Problems 1: Velocity of sound in air varies as bulk modulus of electricity K,
Mass density . Derive an expression for velocity in form C = √K
ρ
Solution: C = f (K, )
C = M Ka
b
M – Constant of proportionality
[C] = [K]a []
b
[LMoT
-1] = [L
-1MT
-2]a [L
3MT
o]b
[LMoT-1] = [L-a+(-36) Ma+bT-2a]
C – Velocity – LMoT
-1
K – Bulk modulus –L-1MT-2
- Mass density – L-3MT
o
- a –3b = 1 a + b = 0
2b = 1, b = -1
a = 1
2
C = 𝑀𝐾1/2-1/2
𝐶 = 𝑀√𝐾
𝜌
If M= 1, Then 𝐂 = √𝐊
𝛒
Problem 2: Find the equation for the power developed by a pump if it depends onbead H
discharge Q and specific weight of the fluid.
Solution:
P = f (H, Q, )
P = K HaQ
b
c
[P] = [H]a [Q]
b []
c
[L2MT
-3] = [LM
oT
o]a [L
-2MT
-2]b [L
-2MT
-2]c
2 = a + 3b – 2c
Power = L2MT
-3
1 = c
Head = LMoT
o
- 3 = - b - 2
Discharge = L3M
oT
-1
- 3 = - b – 2
Specific Weight = L-2
MT-2
b = - 2 + 3
b = 1
2 = a + 3 – 2
a = 1
P = K H1 Q
1
1
P = K H Q
When, K = 1
P = H Q
Problem 3: Find an expression for drag force R on a smooth sphere of diameter ‘D’ moving with
uniform velocity ‘V’ in a fluid of density and dynamic viscosity ..
Solution:
R = f (D, V, , )
R = K DaV
b
c,
d
[R] = [D]a [V]
b []
c []d
[LMT-2
] = [LMoT
o]a [LM
oT
-1]b [L
-3MT
o]c [L
-1MT
-1]d
c + d = 1
Force = LMT-2
c = 1 – d
Diameter = LMoT
o
– b – d = – 2 Velocity = LM
oT
-1
Mass density = L3MT
o
b = 2 – d
1 = a + b – 3c – d
1 = a + 2 – d – 3 (1 – d) – d
1 = a + 2 – d – 3 + 3d – d
a = 2 – d
R = K D2-d
V2-d
1-d
, d
VDDVKR
22.
R = V2 D
2 [NRe]
Buckingham’s π- theorem
The Rayleigh’s method of dimensional analysis becomes bulky when more variables are involved. In
order to overcome this, Buckingham’s method may be used. It states that
If there are n variables in a dimensionally homogeneous equation and if these variables contain m
fundamental dimensions such as (M-L-T). They may be grouped into (n minus m) non dimensional
independent π-terms.
Mathematically, if a variable X1 depends upon independent variables X2, X3, X4, Xn the functional
equation may be written as:
X1 = k (X2, X3, X4…Xn).
The equation may be written in its general form as
F(X1, X2, X3, X4…Xn) = C
Where C is a constant and ‘f’ represents some function. In this equation, there are n variables. If there
are ‘m’ fundamental dimensions, then according to Buckingham’s π-theorem.
f1 (π1, π2, π3… πn-m) = constant
The Buckingham’s π- method is based on the following steps.
1. Write the functional relationship with the given data.
2. Then write the equation in its general form.
3. Now choose m repeating variables and write separate expression for each π- term. Every π-term
will contain the repeating variables and one of the remaining variables. The repeating variables are
written in the exponential form.
4. With the help of principle of dimensional homogeneity find out the values of a,b,c… by obtaining
simultaneous equations.
5. Now substitute the values of these exponents in the π-terms.
6. After the π - terms are determined, write the functional relation in the required form.
Selection of repeating variables
Though there is no hard and fast rule for the selection of repeating variables. Following points should
be kept in mind while selecting repeating variables.
Common π- groups:
During dimensional analysis several groups will appear again and again for different problems.
These often have names. You will recognize the Reynolds number ρvd/µ. Some common non-
dimensional numbers (groups) are listed below.
Reynolds number: inertial, viscous force ratio e
udR
Euler number: pressure, inertial force ratio 2
U
pEn
Froude number: inertial, gravitational force ratio
Weber number: inertial, surface tension force ratio
Mach number: Local velocity, local velocity of sound ratio C
uMa
Choice of repeating variables
Repeating variables are those which we think will appear in all or most of the p groups, and are a
influence in the problem. Before commencing analysis of a problem, one must choose the repeating
variables. There is considerable freedom allowed in the choice.
Some rules which should be followed are
From the 2nd theorem there can be n ( = 3) repeating variables.
When combined, these repeating variables variable must contain all of dimensions (M, L, T) (Note-
*That is not to say that each must contain M,L and T).
No two repeating variables should have same dimensions.
A combination of the repeating variables must not form a dimensionless group.
The repeating variables do not have to appear in all π-groups.
The repeating variables should be chosen to be measurable in an experimental investigation. They
should be of major interest to the designer. For example, pipe diameter (dimension L) is more
useful and measurable than roughness height (also dimension L).
Select: Repeating variables can be selected from each of the following properties.
a. Geometric property Length, height, width, area
b. Flow property Velocity, Acceleration, Discharge
c. Fluid property Mass density, Viscosity, Surface tension
Buckingham's π- theorems:
Although there are other methods of performing dimensional analysis, (notably the indicial method)
the method based on the Buckingham π- theorems gives a good generalized strategy for obtaining a
solution. This will be outlined below.
If there are n – variables in a physical phenomenon and those n-variables contain ‘m’ dimensions, then
the variables can be arranged into (n-m) dimensionless groups called π- terms.
Explanation:
If f (X1, X2, X3, ………Xn) = 0 and variables can be expressed using m dimensions then,
f (π1, π2, π3, ………π n - m) = 0
Where, π1, π2, π3, ………are dimensionless groups.
Each π- term contains (m + 1) variables out of which m are of repeating type and one is of non-
repeating type.
Each term being dimensionless, the dimensional homogeneity can be used to get each term.
So if a physical problem can be expressed:
f ( Q1 , Q2 , Q3 ,………, Qm ) = 0
then, according to the above theorem, this can also be expressed
f ( π 1 , π 2 , π 3 ,………, π m-n ) = 0
In fluids, we can normally take n = 3 (corresponding to M, L, T).
Each π- group is a function of n governing or repeating variables plus one of the remaining variables.
Example
Taking the example discussed above of force F induced on a propeller blade, we have the equation
0 = f ( F, d, u, r, N, m )
n = 3 and m = 6
There are m - n = 3 p groups, so
f ( p1 , p2 , p3 ) = 0
The choice of r, u, d as the repeating variables satisfies the criteria above. They are measurable, good
design parameters and, in combination, contain all the dimension M,L and T. We can now form the
three groups according to the 2nd theorem,
As the p groups are all dimensionless i.e. they have dimensions M0L0T0 we can use the principle of
dimensional homogeneity to equate the dimensions for each p group.
For the first p group,
In terms of SI units
And in terms of dimensions
For each dimension (M, L or T) the powers must be equal on both sides of the equation, so
for M: 0 = a1 + 1
a1 = -1
for L: 0 = -3a1 + b1 + c1 + 1
0 = 4 + b1 + c1
for T: 0 = -b1 - 2
b1 = -2
c1 = -4 - b1 = -2
Giving p1 as
And a similar procedure is followed for the other p groups. Group
For each dimension (M, L or T) the powers must be equal on both sides of the equation,
for M: 0 = a2
for L: 0 = -3a2 + b2 + c2
0 = b2 + c2
for T: 0 = -b2 - 1
b2 = -1
c2 = 1
Giving p2 as
And for the third,
For each dimension (M, L or T) the powers must be equal on both sides of the equation,
for M: 0 = a3 + 1
a3 = -1
for L: 0 = -3a3 + b3 + c3 -1
b3 + c3 = -2
for T: 0 = -b3 - 1
b3 = -1
c3 = -1
Giving p3 as
Thus the problem may be described by the following function of the three non-dimensional p groups,
f ( p1 , p2 , p3 ) = 0
This may also be written:
Solved Problems: Dimensional Analysis
Example 1: The discharge Q through an orifice is a function of the diameter d, the pressure
difference p, the density ρ, and the viscosity µ, show that, where f is some unknown function.
Solution:
Write out the dimensions of the variables
ρ: ML-3 u: LT-1
d: L m: ML-1T-1
p:(force/area) ML-1T-2
We are told from the question that there are 5 -variables involved in the problem: d,
p, ρ, µ and Q.
Choose the three recurring (governing) variables; Q, d, r.
From Buckingham's p theorem we have m-n = 5 - 3 = 2 non-dimensional groups.
For the first group, π1:
M] 0 = c1 + 1
c1 = -1
L] 0 = 3a1 + b1 - 3c1 - 1
-2 = 3a1 + b1
T] 0 = -a1 - 1
a1 = -1
b1 = 1
And the second group π2 :
(note p is a pressure (force/area) with dimensions ML-1T-2)
M] 0 = c2 + 1
c2 = -1
L] 0 = 3a2 + b2 - 3c2 - 1
-2 = 3a2 + b2
T] 0 = -a2 - 2
a2 = - 2
b2 = 4
the physical situation is described by this function of non-dimensional numbers,
The question wants us to show:
Take the reciprocal of square root of π2: ,
Convert p1 by multiplying by this new group, p2a
Then
Example-2 Show that Reynold number, vd/, is non-dimensional. If the discharge Q through
an orifice is a function of the diameter d, the pressure difference p, the density , and the
viscosity , show that Q = Cp1/2d2/1/2 where C is some function of the non-dimensional group
(d1/2p1/2/).
The dimensions of these following variables are
ML-3
u LT-1
d L
ML-1T-1
Re = ML-3 LT-1L (ML-1T-1)-1 = ML-3 LT-1 L M-1LT = 1
i.e. Re is dimensionless.
We are told from the question that there are 5 variables involved in the problem: d,
p, , and Q.
Choose the three recurring (governing) variables; Q, d,
From Buckinghams theorem we have m-n = 5 - 3 = 2 non-dimensional groups.
As each group is dimensionless then considering the dimensions, for the first group, 1:
M] 0 = c1 + 1
c1 = -1
L] 0 = 3a1 + b1 - 3c1 - 1
-2 = 3a1 + b1
T] 0 = -a1 - 1
a1 = -1
b1 = 1
And the second group 2:
(note p is a pressure (force/area) with dimensions ML-1T-2)
M] 0 = c2 + 1
c2 = -1
L] 0 = 3a2 + b2 - 3c2 - 1
-2 = 3a2 + b2
T] 0 = -a2 - 2
a2 = - 2
b2 = 4
The physical situation is described by this function of non-dimensional numbers,
The question wants us to show :
Take the reciprocal of square root of 2: ,
Convert 1 by multiplying by this number
then
Example 3 The resistance to motion, ‘R’, of a sphere through a fluid is a function of the
density ρand viscosity m of the fluid, and the radius r and velocity u of the sphere, show that R
is given by
Show that if at very low velocities the resistance R is proportional to the velocity u, then
R = kµρu where k is a dimensionless constant.
A fine granular material of specific gravity 2.5 is in uniform suspension in still water of depth
3.3m. Regarding the particles as spheres of diameter 0.002cm find how long it will take for the
water to clear. Take k=6π and µ=0.0013 kg/ms.[218mins 39.3sec]
Choose the three recurring (governing) variables; u, r, ρ, µ, R
From Buckingham’s π- theorem we have m-n = 5 - 3 = 2 non-dimensional groups.
As each p group is dimensionless then considering the dimensions, for the first group, p1:
M] 0 = c1 + 1
c1 = -1
L] 0 = a1 + b1 - 3c1 - 1
-2 = a1 + b1
T] 0 = -a1 - 1
a1 = -1
b1 = -1
i.e. the (inverse of) Reynolds number
And the second group π2 :
M] 0 = c2 + 1
c2 = -1
L] 0 = a2 + b2 - 3c2 - 1
-3 = a2 + b2
T] 0 = -a2 - 2
a2 = - 2
b2 = -1
The physical situation is described by this function of non-dimensional numbers,
or
the question asks us to show or
Multiply the LHS by the square of the RHS: (i.e. π2(1/p12) )
Hence
The question tells us that R is proportional to u so the function f must be a constant, k
The water will clear when the particle moving from the water surface reaches the bottom.
At terminal velocity there is no acceleration - the force R = mg – up thrust.
From the question:
s = 2.5 , r = 2500kg/m3 m = 0.0013 kg/ms k = 6p
r = 0.00001m depth = 3.3m
Model analysis: Model analysis, similitude, types of similarities, force ratios, similarity laws,
model classification, Reynolds model, Froude’s model, Euler’s Model, Webber’s model, Mach
model, scale effects, Distorted models. Numerical problems on Reynold’s, and Froude’s Model
Before constructing or manufacturing hydraulics structures or hydraulics machines tests are
performed on their models to obtain desired information about their performance. Models are small
scale replica of actual structure or machine. The actual structure is called prototype.
Similitude / Similarity
It is defined as the similarity between the prototype and its model.
Types of Similarity There are three types of similarity.
Geometric similarity Kinematic similarity Dynamic similarity
Geometrical Similarity Geometric similarity is said to exist between the model and prototype if the ratio of corresponding linear dimensions between model and prototype are equal.
Kinematic Similarity Kinematic similarity exists between prototype and model if quantities such at velocity and acceleration at corresponding points on model and prototype are same.
(𝑉1)𝑃
(𝑉1)𝑀=
(𝑉2)𝑃
(𝑉2)𝑀=
(𝑉3)𝑃
(𝑉3)𝑀= 𝑉𝑟
where Vr → is velocity ratio
Dynamic Similarity Dynamic similarity is said to exist between model and prototype if ratio of forces at corresponding points of model and prototype is constant.
(𝐹1)𝑃
(𝐹1)𝑀=
(𝐹2)𝑃
(𝐹2)𝑀=
(𝐹3)𝑃
(𝐹3)𝑀= 𝐹𝑟
Where Fr Force ratio
Force Ratios – Dimensionless numbers
i.e. Lp
h p
Hp
............ Lr
Lm h m
Hm
Lr scale ratio / linear ratio
Ap Lr2
Vp
Lr3
Am
V
m
In a system involving flow of fluid, different forces due to different causes may act on a fluid
element. These forces are as follows:
1. Viscous force (due to viscosity) Fv
2. Pressure force (due to difference in pressure) Fp
3. Gravity force (due to gravitational attraction) Fg
4. Capillary force (due to surface tension) Fs
5. Compressibility force (due to elasticity) Fe
Since inertia force always exists when any mass is in motion, the condition for dynamic
similarity are developed by considering the ratio of the inertia force and any one of the
remaining forces mentioned above. Each of these ratios will obviously be non-dimensional
factor. These ratios are discussed below.
Dimensionless Numbers Following dimensionless numbers are used in fluid mechanics.
1. Reynold’s number 2. Froude’s number 3. Euler’s number 4. Weber’s number 5. Mach number
Syllabus:
Buoyancy and Flotation: Buoyancy, Force and Centre of Buoyancy, Metacentre and Metacentric height, Stability of submerged and floating bodies, Determination of Metacentric height, Experimental and theoretical method, Numerical problems
Buoyancy or up-thrust, is an upward force exerted by a fluid that opposes the weight of an immersed object.
In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus the pressure at the bottom of a column of fluid is greater than at the top of the column.
Similarly, the pressure at the bottom of an object submerged in a fluid is greater than at the top of the object. The pressure difference results in a net upward force on the object. The magnitude of the force is proportional to the pressure difference, and (as explained by Archimedes' principle) is equivalent to the weight of the fluid that would otherwise occupy the volume of the object, i.e. the displaced fluid.
For this reason, an object whose average density is greater than that of the fluid in which it is submerged tends to sink. If the object is less dense than the liquid, the force can keep the object afloat. This can occur only in a non-inertial reference frame, which either has a gravitational field or is accelerating due to a force other than gravity defining a "downward" direction.
The centre of buoyancy of an object is the centroid of the displaced volume of fluid.
The forces at work in buoyancy. The object floats at rest because the upward force of buoyancy is equal to the downward force of gravity.
A metallic coin floats in mercury due to the buoyancy force upon it and appears to float higher because of the surface tension of the mercury.
Illustration of the stability of bottom-
heavy (left) and top-heavy (right) ships
with respect to the positions of their centres of buoyancy (CB) and gravity
(CG)
Archimedes' principle is named after Archimedes of Syracuse, who first discovered this law in
212 B.C. For objects, floating and sunken, and in gases as well as liquids (i.e. a fluid),
Archimedes' principle may be stated thus in terms of forces:
“Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to
the weight of the fluid displaced by the object”.
With the clarifications that for a sunken object the volume of displaced fluid is the volume of
the object, and for a floating object on a liquid, the weight of the displaced liquid is the weight
of the object.
More tersely: buoyancy = weight of displaced fluid.
Archimedes' principle does not consider the surface tension (capillarity) acting on the body,
but this additional force modifies only the amount of fluid displaced and the spatial distribution
of the displacement, so the principle that buoyancy = weight of displaced fluid remains valid.
The weight of the displaced fluid is directly proportional to the volume of the displaced fluid
(if the surrounding fluid is of uniform density).
In simple terms, the principle states that the buoyancy force on an object is equal to the weight
of the fluid displaced by the object, or the density of the fluid multiplied by the submerged
volume times the gravitational acceleration, g.
Thus, among completely submerged objects with equal masses, objects with greater volume
have greater buoyancy. This is also known as up-thrust.
Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum
with gravity acting upon it. Suppose that when the rock is lowered into water, it displaces water
of weight 3 newtons. The force it then exerts on the string from which it hangs would be 10
newtons minus the 3 newtons of buoyancy force: 10 − 3 = 7 newtons. Buoyancy reduces the
apparent weight of objects that have sunk completely to the sea floor. It is generally easier to
lift an object up through the water than it is to pull it out of the water.
Example: If you drop wood into water, buoyancy will keep it afloat.
Example: A helium balloon in a moving car. During a period of increasing speed, the air mass inside the car moves in the direction opposite to the car's acceleration (i.e., towards the rear). The balloon is also pulled this way. However, because the balloon is buoyant relative to the air, it ends up being pushed "out of the way", and will actually drift in the same direction as the car's acceleration (i.e., forward). If the car slows down, the same balloon will begin to drift backward. For the same reason, as the car goes round a curve, the balloon will drift towards the inside of the curve.
Buoyancy
When a body is immersed in a fluid either wholly or partially it is subjected to an upward
force which tends to lift (buoy) it up. This tendency for an immersed body to be lifted up
in the fluid, due to an upward force opposite to the action of gravity, is known as buoyancy.
The force tending lift up the body under such conditions is known as buoyant force or
force of buoyancy or up-thrust.
Centre of buoyancy The point of application of the force of buoyancy on the body is known as centre of buoyancy.
Ex. A float valve regulates the flow of oil of sp. Gr. 0.8 into a cistern. The spherical float is 15
cm in diameter. AOB is a weightless link carrying the float at one end, and a valve at the other
end which closes the pipe through which oil flows into the cistern. The link is mounted on a
frictionless hinge at O and the angle AOB is 135o. The length of OA is 20 cm, and the distance
between the centre of the float and the hinge is 50 cm. When the flow is stopped AO will be
vertical. The valve is to be pressed on to the seat with a force of 9.81 N to completely stop the
flow of oil into the cistern. It was observed that the flow of oil is stopped when the free surface
of oil in the cistern is 35 cm below the hinge. Determine the weight of the float.
META – CENTRE
It is defined as the point about which a body starts oscillating when the body is tilted by a small
angle. The meta-centre may also be defined as the point at which the line of action of the force
of buoyancy will meet the normal axis of the body when the body is given a small angular
displacement.
Consider a body floating in a liquid as shown
in fig. (a). Let the body is in equilibrium and
G is the centre of gravity and B is centre of
Buoyancy. For equilibrium, both the points
lie on the normal axis, which is vertical.
Let the body is given a small angular displacement in the clockwise direction as shown in Fig.
(b). The centre of buoyancy, which is the centre of gravity of the displaced liquid or centre of
gravity of the portion of the body submerged in liquid, will now be shifted towards right from
the normal axis. Let it is at B1 as shown in Fig. (b). The line of action of the force of buoyancy
in this new position, will intersect the normal axis of the bodyat some point say M. This point
M is called the Meta-centre.
Meta-Centric Height
The distance MG, i.e., the distance between the meta-
centre of a floating body and the centre of gravity of
the body is called meta-centric height.
Analytical Method for determining Meta-Centric height
Fig (a) shows the position of a body in equilibrium. The location of centre of gravity and centre
of buoyancy in this position is at G and B. The floating body is given a small angular
displacement in the clockwise direction as shown fig. (b). The new centre of buoyancy is at B1.
The vertical line through B1 cuts the normal axis at M. Hence M is the meta-centre and GM is
meta-centric height.
Ex. A rectangular pontoon is 5 m long, 3 m wide and 1.2 m high. The depth of immersion of
pontoon is 0.8 m in sea water. If the centre of gravity is 0.6 m above the bottom of the pontoon
, determine the meta-centric height. The density for sea water = 1025 kg/m3.
Solution:
Given, Dimension of pontoon = 5 m x 3 m x 1.2 m
Depth of immersion = 0.8 m
BODY IMMERSED IN TWO DIFFERENT FLUIDS
Fig. shows a body of volume V immersed in two different fluids of specific weights W1
and W2 respectively.
In this case the up-thrust on the body = Weight of volume V1 of liquid of specific weight
W1 + Weight of volume V2 of the liquid of specific weight W2
Where V1 and V2 are the volumes of the two liquids displaced by the body.
Conditions of Equilibrium of a Floating and Sub-merged Bodies
A sub-merged or a floating body is said to be stable if it comes back to its original position
after a slight disturbance. The relative position of the centre of gravity (G or g) and centre of
buoyancy (B) of a body determines the sub-merged body.
Stability of a Sub-merged Body
The position of centre of gravity and centre of buoyancy in case of a completely sub-merged
body are fixed. Consider a balloon, which is completely sub-merged in air. Let the lower
portion of the balloon contains heavier material, so that its centre of gravity is lower than its
centre of buoyancy as shown Fig. (a). Let the weight of balloon is W. The weight W is acting
through , vertically in the downward direction, while the buoyant force FB is acting vertically
up through B.
For the equilibrium of the balloon W = FB. If the balloon is given an angular displacement in
the clockwise direction as shown in Fig. (a), then W and FB constitute a couple acting in anti-
clockwise direction and brings the balloon in the original position. Thus the balloon in the
position, shown in Fig. (a) is in stable equilibrium.
a) Stable Equilibrium: When W = FB and point B is above G, the body is said to be in
stable equilibrium.
b) Unstable equilibrium: If W = FB, but the centre of buoyancy (B) is below centre of
gravity (G), the body is in unstable equilibrium as shown Fig. (b). A slight displacement
of the body, in the clockwise direction, leads to formation of a couple due to W and FB
also in the clockwise direction. Thus the body does not return to its original position
and hence the body is in unstable equilibrium.
c) Neutral equilibrium: If W = FB, and B and G are at the same point, as shown in Fig.
(c), then the body is said to be in Neutral equilibrium.
Stability of Floating body
The stability of a floating body is determined from the position of the Meta-centre (M). In case of floating body, the weight of the body is equal to the weight of liquid displaced.
Conditions of Equilibrium of a floating body
i) When a body is floating in a liquid, the weight of the body is equal to the weight of the liquid displaced.
ii) When the metacentre is above the centre of gravity as shown in Fig., we find on giving a small tilt to the floating body, the weight of the body and the up-thrust will form a righting couple i.e., a restoring couple to bring the body back to its position. The body, is therefore, said to be in stable equilibrium.
iii) When the metacentre is below the centre of gravity as in Fig. we find on giving a small tilt
to the floating body, the weight of the body and the upthrust will make the body tilt further and the body will not therefore be restored to its earlier position. In this case the body is said to in unstable equilibrium.
iv) When the metacentre and the centre of gravity coincide, the body will remain in equilibrium in any position in which it is floating as shown in Fig.
If the body is tilted, it will remain in equilibrium in the new position, since in the new position also the weight of the body and the up-thrust will remain in the same vertical. A body floating in this way is said to be in Neutral equilibrium.
Ex. A rectangular pontoon is 5 m long, 3 m wide and 1.2 m high. The depth of immersion of pontoon is 0.8 m in sea water. If the centre of gravity is 0.6 m above the bottom of the pontoon , determine the meta-centric height. The density for sea water = 1025 kg/m3.
Solution:
Given, Dimension of pontoon = 5 m x 3 m x 1.2 m
Depth of immersion = 0.8 m
Ex. A solid cylinder of diameter 1 m and height 1 m floats in fresh water with its axis vertical. The cylinder is made of a material of specific gravity 0.7. Determine the metacentric height and state the condition of its equilibrium.
Solution:
Diameter of cylinder, D = 1 m
Height of cylinder, H = 1 m
Specific gravity of cylinder, S = 0.7
The wooden block is as shown in Fig.
The depth of cylinder in water is found as follows:
Ex. Show that a cylindrical buoy of 1 m diameter and 2.0 m height weighing 7.848 kN will not float vertically in sea water of density1030 kg/m3. Find the force necessary in a vertical chain attached at the centre of base of buoy that will keep it vertical.
As the meta-centric height is –ve, the point M lies below G and hence the cylinder will be in unstable equilibrium and hence cylinder will not float vertically
The equilibrium of floating or submerged bodies requires that the weight of the body acting through its centre of gravity has to be collinear with an equal buoyant force acting through the centre of buoyancy.
A submerged body will be in stable, unstable or neutral equilibrium if its centre of gravity is below, above or coinciding with the centre of buoyance respectively.
Meta-centre of a floating body is defined as the point of intersection of the centre line of cross section containing the centre of gravity and centre of buoyancy with the vertical line through new centre of buoyancy due to any small angular displacement of the body.
Example :
A large iceberg floating in sea water is of cubical shape and its specific gravity is 0.9 If 20 cm proportion of the iceberg is above the sea surface, determine the volume of the iceberg if specific gravity of sea water is 1.025.
Solution :
Let the side of the cubical iceberg be h.
Total volume of the iceberg = h3
volume of the submerged portion is = ( h -20) x h2
Now,
For flotation, weight of the iceberg = weight of the displaced water
The side of the iceberg is 164 cm.
Thus the volume of the iceberg is 4.41m3
Ex. Show that a cylindrical buoy of 1 m diameter and 2.0 m height weighing 7.848 kN will not float vertically in sea water of density1030 kg/m3. Find the force necessary in a vertical chain attached at the centre of base of buoy that will keep it vertical.
As the meta-centric height is –ve, the point M lies below G and hence the cylinder will be in unstable equilibrium and hence cylinder will not float vertically
For the cone, the distance of centre of gravity from the apex is A is
Experimental method of determination of Meta-centric height
The meta-centric height of a floating vessel can be determined, provided we know the centre of gravity of floating vessel. Let w1 is a known weight placed over the centre of the vessel as shown in Fig. (a) and the vessel is floating.
w for sea water = 10104 N/m3 = 10.104 kN/m3
Movable weight, w1 = 343.35 kN
Distance moved by w1, x = 6 m
Centre of buoyancy = 2.25 m below water surface
Find i) Meta-centric height
ii) Position of centre of gravity, G
