Unit - 5
Dimensional Analysis and Similarities
A dimension is a non-numerical measure of a physical quantity, whereas a unit is a method of assigning a numerical value to that dimension.
Mass, length, time, temperature, electric current, amount of light, and amount of matter are the seven primary dimensions (sometimes known as fundamental or basic dimensions).
All nonprimary dimensions can be formed by some combination of the seven primary dimensions.
Dimensional analysis is a method of dimensions. It is a mathematical technique that is used in research to create and test models. It is concerned with the physical quantities involved in the phenomenon's dimensions. All physical quantities are measured by comparing them to a predetermined value. In Fluid Mechanics, length L, mass M, and time T are three fixed dimensions that are important. If heat is included in any fluid mechanics problem, temperature is also treated as a fixed dimension. Fundamental dimensions or fundamental quantity refers to these fixed dimensions.
Secondary or derived quantities:
Secondary or derived quantities are those which possess more than one fundamental dimension.
Example:
We’ve all heard the old saying, you can’t add apples and oranges.
This is a simplified expression of a far more global and fundamental
mathematical law for equations, the law of dimensional homogeneity,
stated as
“Every additive term in an equation must have the same dimensions.”
Example:
Now LHS,
Now RHS,
If we find ourselves in a situation where two additive terms in an equation have different dimensions at some point throughout the analysis, this is a clear indication that we made an error earlier in the process.
Calculations are valid only when the units are also homogeneous in each additive term, in addition to dimensional homogeneity.
Non-dimensionalization of equations:
Every additive term in an equation has the same dimensions, according to the law of dimensional homogeneity. As a result, dividing each term in the equation by a set of variables and constants whose product has the same dimensions renders the equation nondimensional.
The equation is said to be normalised if the nondimensional terms in it are also of order unity. Even though the two concepts are occasionally (incorrectly) used interchangeably, normalisation is more limiting than nondimensionalization.
“ Each term in a nondimensional equation is dimensionless “
In a problem, dimensional variables are defined as dimensional quantities that change or vary.
Nondimensional (or dimensionless) variables are numbers that alter or vary in a situation but have no dimensions; an example is angle of rotation, which is measured in degrees or radians, which are dimensionless units. While dimensional, the gravitational constant g remains constant and is referred to as a dimensional constant.
The total set of dimensional variables, nondimensional variables, and dimensional constants in the problem is referred to as parameters.
If the number of variables involved in a physical phenomenon is known, the following two methods can be used to discover the relationship between the variables.
1. Rayleigh's method
2. Buckingham's n-theorem.
Rayleigh's method
This method is used to find the expression for a variable that is dependent on no more than three or four variables. When the number of independent variables exceeds four, finding an expression for the dependent variable becomes quite difficult.
Let X is a variable, which depends on
Mathematically it is written as,
It can also be written as,
Where K = constant
a, b, c, are arbitrary Powers
Example: The length (L) of a pendulum and the acceleration due to gravity determine the time-period (t) of the pendulum ( g). Determine a time-period expression.
Solution:
Substituting the dimensions on both sides,
Equating the powers of M, L and T on both sides, we have
Substituting the values,
The value of K is determined from experiments which is given as,
Key Takeaway:
When the number of variables exceeds the number of fundamental dimensions, Rayleigh's approach of dimensional analysis becomes more difficult (M, L, T). Buckingham's pie theorem states, "This difficulty can be overcome by using Buckingham's pie theorem." When a physical phenomenon has n variables (independent and dependent variables) and these variables have m fundamental dimensions (M, L, T), the variables are arranged into (n -m) dimensionless terms. Each term is referred to as a "pie-term."
Mathematically it is written as,
Each of π-terms is dimensionless and is independent of the system. Division or multiplication by a constant does not change the character of the π-term. Each π-term contains m + 1 variables, where m is
The number of fundamental dimensions and is also called repeating variables.
Each π term is written as,
Each equation is solved by the principle of dimensional homogeneity and values of 
etc., are obtained. Obtained values are substituted in above equation.
The values of π obtained is then substituted in function equation.
The final equation for the phenomenon is obtained by expressing any one of the π-terms as a function of others as
Method of Selecting Repeating Variables:
The number of repeated variables is the same as the problem's fundamental dimensions. The following factors influence the selection of repeating variables:
- The dependent variable should not be chosen as a recurring variable as much as possible.
- The recurring variables should be chosen so that one variable has geometric properties, another has flow properties, and the third variable has fluid properties.
Variables with Geometric Property are
Variables with flow Property are
Variables with fluid property
- The selected repeated variables must not form a dimensionless group.
- The number of fundamental dimensions in the recurring variables must be the same.
- There should be no recurring variables with the same dimensions.
Procedure for solving:
During flight, the resistive force R of a supersonic plane is determined by the length of the aircraft l, velocity V, air viscosity, air density p, and bulk modulus of air K. The functional relationship between these variables and the resistive force should be expressed.
Solution:
Total number of variables, n = 6
Out of six variables 
three variables are to be selected as repeating variable.
is a dependent variable and should not be selected as a repeating variable.
One of the remaining five variables should have a geometric property, the second should have a flow property, and the third should have a fluid property.
The recurring variables should not create a dimensionless term, and their fundamental dimensions should be equal to m.
Each π-term is solved by the principle of dimensional homogeneity
Key Takeaway:
Inspectional analysis for nondimensionalization of an equation is only useful if the equation is known to begin with. In many circumstances, however, in real-life engineering, the equations are either unknown or too complicated to solve; in these cases, testing is often the only way to gather reliable data. Most investigations are carried out on a geometrically scaled model rather than a full-scale prototype to save time and money.
In such instances, it's crucial to scale the results properly. We'll go through dimensional analysis, which is a powerful tool.
The three primary purposes of dimensional analysis are:
- To obtain scaling laws so that prototype performance can be predicted from model performance .
- To (sometimes) predict trends in the relationship between parameters .
- To generate nondimensional parameters that aid in the design of experiments (physical and/or numerical) and the reporting of experimental results.
In several domains of engineering, dimensional less numbers are collections of variables that provide order-of-magnitude estimations of a system's behaviour.
- In engineering, dimensionless numbers are frequently employed to capture underlying phenomena and scale up.
- Dimensionless numbers lower the number of variables that represent a system, minimising the amount of experimental data needed to make scalable system correlations from physical occurrences. The Reynolds number is the most common dimensionless group in fluid dynamics (Re).
- It is feasible to determine whether particular effects or forces are important or can be safely ignored in the model by studying dimensionless numbers.
- It demonstrates whether certain effects or forces may be safely ignored.
- It enables us to adapt the force correlations discovered in a specific (experimentally convenient) system to a considerably broader class of multiphase flow issues.
Now let's look at how to make the nondimensional parameters, or π. There have been various methods devised for this goal, but the most popular (and easiest) one is Edgar Buckingham's (1867–1940) method of repeating variables. In 1911, Russian physicist Dimitri Riabouchinsky (1882–1962) described the approach for the first time. This method can be thought of as a "recipe" or step-by-step procedure for producing nondimensional parameters.
Step 1: List the parameters in the problem and count their total number 
.
Step 2: List the primary dimensions of each of the 
parameters.
Step 3: Set the reduction 
as the number of primary dimensions. Calculate 
, the expected number of II's,
Step 4: Choose 
repeating parameters.
Step 5: Construct the 
II's, and manipulate as necessary.
Step 6: Write the final functional relationship and check your algebra.
We first describe the basic notion of dimensional analysis—the principle of similarity—before moving on to the practise of dimensional analysis.
For perfect likeness between a model and a prototype, three conditions must be met. The first requirement is geometric similarity—the model must have the same shape as the prototype, but it can be scaled by a constant amount. The second criteria is kinematic similarity, which requires that the velocity at each location in the model flow be proportional to the velocity at the corresponding position in the prototype flow (by a constant scale factor).
The velocity at corresponding sites must scale in magnitude and point in the same relative direction for kinematic similarity. Geometric similarity can be thought of as length-scale equivalent, whereas kinematic similarity can be thought of as time-scale equivalence. Kinematic similarity is dependent on geometric similarity. The velocity scale factor, like the geometric scale factor, can be less than, equal to, or greater than one.
The geometric scale factor for image above is less than one (model is smaller than prototype), while the velocity scale is bigger (velocities around the model are greater than those around the prototype).
Dynamic similarity is the third and most restricted similarity condition. When all forces in the model flow scale by a constant factor to equivalent forces in the prototype flow, dynamic similarity is obtained (force-scale equivalence). Forces have a scale factor that can be less than, equal to, or greater than one, just like geometric and kinematic similarity.
For image above, the force-scale factor is less than one since the force on the model building is less than that on the prototype.
Dynamic similarity requires, but is not sufficient for, kinematic similarity. A model flow and a prototype flow can therefore attain geometric and kinematic similarity, but not dynamic similarity. For complete similarity to be guaranteed, all three-similarity criterion must be met.
“Complete resemblance between a model and prototype in a general flow field is only accomplished when there is geometric, kinematic, and dynamic similarity.”
The ratio of the respective forces acting at the corresponding sites in the model and prototype should be identical for dynamic similarity between the two. The forces' ratios are dimensionless numbers. It indicates that the dimensionless numbers for the model and the prototype should be the same for dynamic similarity.
However, satisfying the condition that all dimensionless numbers for the model and prototype are the same is rather challenging. As a result, models are created based on the force ratio that is dominant in the occurrence. Model laws or laws of similarity are the laws on which the models are based for dynamic similarity.
Types of model laws are
- Reynolds model laws
In flow situations where in addition to inertia, viscous force is the other predominant force, the similarity of flow in the model and its prototype can be established if Reynolds number is same for both the systems.
This is known as Reynolds law and according to this law,
Where, ρm = density of fluid in model
Vm = velocity of fluid in model,
Lm = length of linear dimension of the model, and
µm = viscosity of fluid in model
And ρ P, VP, LP and µP are the corresponding values of density, velocity, linear dimension and viscosity of fluid in prototype.
Following are some of the phenomena for which Reynolds model law can be a sufficient criterion for similarity of flow in the model and the prototype:
- Motion of airplanes,
- Flow of incompressible fluid in closed pipes,
- Motion of submarines completely under water, and
- Flow around structures and other bodies immersed completely under moving fluids.
2. Froude model laws
When the gravitational force can be the only predominant force which controls the motion in addition to the inertia force, the similarity of the flow in any two such systems can be established if the Froude’s number of both the systems is the same.
This is known as Froude’s Model Law.
Some of the phenomenon for which the Froude Model Law can be sufficient criterion for dynamic similarity to be established in the model and the prototype are:
- Free surface flows such as flow over spillways, sluices, etc.
- Flow of jet from an office or nozzle.
- Where waves are likely to be formed on the surface.
- Where fluids of different mass densities flow over one another
Let, Vm = velocity of fluid in model
Lm = length (or linear dimension) of the model
gm = acceleration due to gravity (at a place where model is tested)
And VP, LP and gP are the corresponding values of the velocity, length and acceleration due to gravity for the prototype.
Then according to Froude Law,
(Fr)m = (Fr)P
As the value of g at the site of mode testing will be practically the same as at site of the proposed prototype, therefore gm = gP and the equation 1 values as
Where Lr = scale ratio for length
The law of Froude's model is applied
- Free local flow such as streaming to spill, heirs, sluices, channels etc.
- Flow of the jet from the orifice or nozzle
- Where waves may form more
- When a flow of liquid of different strength passes one anther
3. Euler model laws
Euler's model law states that models should be designed on Euler's number, which means that the Euler number for the model and prototype should be same for dynamic similarity. When the pressure forces alone predominate in addition to the inertia force, Euler's model law applies.
4. Weber model laws
Used in:
- Capillary rise in narrow passages
- Capillary movement of water in soil
- Capillary waves in channel
- Flow over weirs for small heads
5. Mach model laws
Models are designed on Mach number, which is the square root of the ratio of a flowing fluid's inertia force to its elastic force. The dynamic similarity is obtained by:
Used in:
- Flow of aeroplane and projectile through air at supersonic speed
- Aerodynamic testing
- Under water testing of torpedoes
- Water hammer
Key Takeaway:
Undistorted Model:
Undistorted models are ones that are geometrically identical to their prototypes, or in other words, if the scale ratio between the model's linear dimensions and its prototype is the same. The results of the undistorted model can easily anticipate the behaviour of the prototype.
Distorted Model:
If a model is not geometrically identical to its prototype, it is said to be distorted. Different scale ratios for the linear dimensions are used for a distorted model. For rivers, harbours, and reservoirs, for example, two separate scale ratios are used, one for horizontal dimensions and the other for vertical dimensions. As a result, models of rivers, harbours, and reservoirs will be warped. If the horizontal and vertical scale ratios for the river are taken to be the same so that the model is undistorted, the depth of water in the river model will be extremely small, making it difficult to measure precisely.
Advantages:
- The model's vertical dimensions can be precisely measured.
- The model's cost can be decreased.
- The model's turbulent flow can be maintained.
Although the distorted model has several advantages, the outcomes of the distorted model cannot be simply translated to the prototype. However, occasionally highly relevant information can be gleaned from distorted models.
Scale Ratio for Distorted Models:
Scale ratio for Velocity:
Scale ratio for area of flow:
Scale ratio for discharge:
Key Takeaway:
Examples:
Q.1. An oil of specific gravity 0.9 and viscosity 0.9 poise is to be transported at the rate of 1000 l/s through a. 1.2 m diameter pipe. Tests were conducted on a 10 cm diameter pipe using water at .20°C. Viscosity of water at 20°C is 0.01 poise. Find the rate of flow in the model.
Solution:
Prototype
D = 1.2 m.
Sp = 0.9,
Model
Dm = 10 cm = 0.1 m
Sm = 1
μp = 0.9 poise. μm = 0.01 poise = 1 x 10-² poise
To find: Velocity and flow rate.
Since it is pipe flow, Reynolds’s number must be applied.
(Re)model = (R) prototype
Rate of flow through model
Q.2. From the following data, find the scale ratio of model Velocity of water 1 m/'s through circular pipe. Prototype: Velocity of oil 0.12 ms through 50 mm diameter pipe. Assume kinematic viscosity of water 0.01 cms and that of oil 0.008 cm/s. Assume dynamic similarity. Also find the diameter of pipe used for model.
Solution:
Prototype
Vp -012 m/s
Dp = 50mm = 0.05 m
Vp = 0.006 cm²/sec
Model
Vm= 1m/s
Dm=?
Vm= 0.01 cm^2/sec
To find:
Since it is pipe flow, Reynold's ember mast be applied
Model scale ratio is 1:5
Now, diameter of model pipe
Q.3. Water at 15.6°C flows at 3.5 m/s in a 150 mm diameter pipe. At what velocity must a fuel oil at 32.6°C flow in a 75 mm diameter pipe for the flow to be dynamic similar? v.-1.13 x 10 m/s and 2.96 x 10 m/s.
Solution:
Water
Dw = 150 mm = 0.15 m
Oil
To find: Velocity of oil V.
Since it is pipe flow, Reynold's number must be applied
The velocity of cl at 32.6°C is 15.33 m'k
Q.4. A ship has a length of 150 m and wetted area 3000 m² A model of this ship 5 m in length when towed in fresh water (p= 1000 kg/m) at 2 m/s produces a resistance of 40N. Calculate (1) corresponding speed of the ship. (i) the shaft power required to propel the ship at this speed through sea water (p=1030 kg/m). Take the propeller efficiency as 75%
Solution:
Model
Lm= 5 m
Vm= 2 m/s
Fm = 40 N
Prototype
Lp= 150 m
Ap= 3000m^2
Propeller efficiency= n= 75 %
To find:
Speed of ship Vp
Since it is ship motion Reynolds number must be applied
Ratio of drag force
Drag force on prototype
Actual power input
Q.5. A ship 300 m long moves in seawater whose density is 1030 kg/m A1.75 Model of this ship is to be tested in a wind tunnel. The velocity of air in wind tunnel is 29 m/s the resistance of the model is 60 N. Determine velocity and resistance of the ship in seawater. Air density is 1.24 kg/m Kinematic viscosity of air=0.018 stokes Kinematic viscosity of seawater 0,01 stokes.
Solution:
Prototype
Scale 1:75
Lp = 300 m
To find:
Vp and Fp
Since it is ship motion Reynolds number must be applied
Drag force ratio
Q.6. In a geometrically similar model of weir the discharge is 0.15 m².s.. if the scale of the model is 1/50 find the discharge of the prototype.
Solution:
Discharge per meter length = 0.15 m/sec.
Scale ratio Lr = Lp/Lm = 50
To find: qp
Since it is a spillway, Froude Number must be applicable.
The discharge ratio for spillway is given by
Q.7. The performance of a spillway is to be studied by means of a model constructed to a scale of 1:9. Determine: (1) Rate of flow in the model for a prototype discharge of 1400 m/s. (2) Energy lost in the prototype if the energy loss in model is 0.3 kW
Solution:
Scale of model 1:9
Lr = 9
Discharge of prototype Q = 1400 m^3 /s
Since it is a spillway, Froude number must be applicable.
Using discharge scale ratio
Energy loss
Q.8. In the model test of a spillway the discharge per meter length is 1/6 m^3/sec if the scale of the model is 1/36 find the discharge per meter run of the prototype.
Solution:
Qm = 1/6 m^3/sec
Lr = Lp/Lm = 36
To find: QP
Since it is a spillway, Froude Number must be applicable. The discharge ratio for spillway is given by
Q.9. Find the suitable scale for model of a spillway if maximum discharge available in the laboratory is 10 L.P.S. And the prototype discharge is 168.07 m/sec,
Solution:
Q=10 lps = 10 x10³ m³/sec;
Qp = 168.07 m^3/sec
Since it is a spillway, Froude Number must be applicable. Using discharge scale ratio
Model scale should be 1:49.
Q.10. A ship model of scale 1/60 is towed through sea water at a speed of 1:1 m/s A force of 2.1 N is required to tow the model. Determine the speed of ship and the propulsive force on the ship if the prototype is subjected to wave resistance only.
Solution:
Speed of model Vm=1.1 m/s
Resistance to prototype in sea water
References:
- Fluid Mechanics and Hydraulic Machines by R.K. Bansal, Laxmi Publications
- Fluid Mechanics: Fundamentals and Applications by Y. A. Çengel and J. M. Cimbala, by McGraw-Hill
- Fluid Mechanics by Frank .M. White, mcgraw Hill Publishing Company Ltd.
- Fundamentals of Fluid Mechanics by munson, Wiley India Pvt. Ltd
