DIMENSIONAL ANALYSIS
Dimensional analysis is a tool in fluid mechanics and engineering that helps simplify physical problems by
reducing the number of variables. It is especially useful when dealing with complex systems that do not have
direct theoretical solutions. Dimensional analysis is a method used to:
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•
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Simplify physical problems by reducing the number of variables.
Present and interpret experimental data efficiently.
Verify the correctness of equations using dimensional homogeneity.
Identify the relative importance of different physical parameters.
Assist in physical modeling and scaling of experiments.
Dimensions vs. Units
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Dimension: The fundamental nature of a physical quantity (e.g., Length, Time, Mass).
Unit: The measurement scale used (e.g., meters, seconds, kilograms).
Fundamental Dimensions in Fluid Mechanics
The most used fundamental dimensions are, Mass (M) → kilogram (kg), Length (L) → meter (m), Time (T) →
second (s) and Temperature (𝜽) → kelvin (K).
Certain physical quantities are combinations of these fundamental dimensions:
Quantity
Area
Volume
Velocity
Force
Pressure
Density
Dynamic Viscosity
Kinematic Viscosity
Symbol
𝐴
𝑉
𝑈
𝐹
𝑃
𝜌
𝜇
𝑣
Dimensions
𝐿2
𝐿3
𝐿𝑇 −1
𝑀𝐿𝑇 −1
𝑀𝐿−1 𝑇 −2
𝑀𝐿−3
𝑀𝐿−1 𝑇 −1
𝐿2 𝑇 −1
Principle of Dimensional Homogeneity
This principle states that all terms in an equation must have the same dimensions for the equation to be
physically valid. This principle ensures consistent relationships between different physical quantities.
However, it does not determine numerical constants.
Buckingham’s Pi Theorem
Buckingham’s Pi theorem is a mathematical tool used in dimensional analysis to reduce the number of
variables in a physical problem. It helps simplify complex equations and is particularly useful when
conducting experiments or modeling real-world systems.
If a physical problem involves n variables and m fundamental dimensions, the problem can be expressed using
n - m dimensionless groups (called Pi groups, 𝝅). Instead of dealing with many variables, we can reduce the
problem to a few non-dimensional parameters. Mathematically:
𝑓(𝑣1 , 𝑣2 𝑣3 , … 𝑣𝑛 ) = 0
Is rewritten as:
𝜙(𝜋1 , 𝜋2 , … , 𝜋𝑛−𝑚 ) = 0
How to Construct Non-Dimensional Groups?
STEP 1) Identify All Variables in the Problem: List all the variables that affect the phenomenon you are studying.
STEP 2) Determine the Fundamental Dimensions: Identify the fundamental dimensions (M,L,T) of each variable
STEP 3) Find the Number of Pi Groups: Number of Pi groups = 𝑛 − 𝑚
STEP 4) Select the Repeating Variables: Pick m variables that contain all fundamental dimensions (M, L, T).
These are called scaling variables.
STEP 5) Form the Pi Groups: Each remaining variable forms a Pi group with the repeating variables.
𝜋𝑖 = (𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒)(𝑠𝑐𝑎𝑙𝑒1 )𝑎 (𝑠𝑐𝑎𝑙𝑒2 )𝑏 (𝑠𝑐𝑎𝑙𝑒3 )𝑐
Where powers, 𝑎, 𝑏, 𝑐 are determined algebraically so that each Pi group is dimensionless (setting exponents
equal to zero)
EXAMPLE 1
Suppose we are analysing the drag force (𝐹𝐷 ) on a sphere moving through a fluid. The relevant variables are:
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•
•
•
•
Drag Force, 𝐹𝐷 (N)
Fluid density, 𝜌 (kg/m3)
Velocity, V (m/s)
Sphere diameter, D (m)
Fluid viscosity, 𝜇 (Pa.s)
Thus, we have n = 5 variables: 𝐹𝐷 , 𝜌, 𝑉, 𝐷, 𝜇. We then identify the fundamental dimensions of each variable:
Variable
Drag Force
Density
Velocity
Diameter
Viscosity
Symbol
𝐹𝐷
𝜌
𝑉
𝐷
𝜇
Dimension
𝑀𝐿−2
𝑀𝐿−3
𝐿𝑇 −1
𝐿
−1 −1
𝑀𝐿 𝑇
The fundamental dimensions used here are M (mass), L (length), and T (time), so we have m = 3 fundamental
dimensions. The number of dimensionless Pi groups is:
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑖 𝑔𝑟𝑜𝑢𝑝𝑠: 𝑛 − 𝑚 = 5 − 3 = 2
We now pick 𝑚 variables (3 in this case) FROM THE RELEVANT VARIABLES that contain ALL the fundamental
dimensions (M,L,T). A good choice would be, fluid density (𝜌), velocity (𝑉) and diameter (𝐷).
We form the first Pi group using drag force (𝑭𝑫) with the repeating variables:
𝜋1 = 𝐹𝐷 × 𝜌𝑎 × 𝑉 𝑏 × 𝐷 𝑐
Substituting the dimensions and expanding
𝜋1 = (𝑀𝐿𝑇 −2 )(𝑀𝐿−3 )𝑎 (𝐿𝑇 −1 )𝑏 (𝐿)𝑐
𝜋1 = (𝑀1+𝑎 )(𝐿1−3𝑎+𝑏+𝑐 )(𝑇 −2−𝑏 )
We now solve for exponents, 𝑎, 𝑏, 𝑐 by setting exponents to zero:
𝑀0 𝐿0 𝑇 0 = (𝑀1+𝑎 )(𝐿1−3𝑎+𝑏+𝑐 )(𝑇 −2−𝑏 )
For 𝑴: 1 + 𝑎 = 0 → 𝑎 = −1
For 𝑳: 1 − 3𝑎 + 𝑏 + 𝑐 = 0 → 𝑏 + 𝑐 = −4
For 𝑻: −2 − 𝑏 = 0 → 𝑏 = −2, ∴ 𝑐 = −2
Thus, the first Pi group is:
𝜋1 = 𝐹𝐷 × 𝜌−1 × 𝑉 −2 × 𝐷 −2
∴ 𝜋1 =
𝐹𝐷
𝜌𝑉 2 𝐷2
This is the drag coefficient, commonly written as:
𝐶𝐷 =
𝐹𝐷
𝜌𝑉 2 𝐷2
Now, form the second Pi group using viscosity (𝜇):
𝜋2 = 𝜇 × 𝜌 𝑥 × 𝑉 𝑦 × 𝐷 𝑧
Substituting the dimensions and expanding and setting the exponents equal to zero:
𝜋2 = (𝑀𝐿−1 𝑇 −1 )(𝑀𝐿−3 )𝑥 (𝐿𝑇 −1 )𝑦 (𝐿) 𝑧
𝜋2 = (𝑀1+𝑥 )(𝐿−1−3𝑥+𝑦+𝑧 )(𝑇 −1−𝑦 )
For 𝑴: 1 + 𝑥 = 0 → 𝑥 = −1
For 𝑳: −1 − 3𝑥 + 𝑦 + 𝑧 = 0 → 𝑦 + 𝑧 = −2
For 𝑻: −1 − 𝑦 = 0 → 𝑦 = −1 ∴ 𝑧 = −1
The second Pi group is now:
𝜋2 = 𝜇 × 𝜌−1 × 𝑉 −1 × 𝐷 −1
∴ 𝜋2 =
𝜇
𝜌𝑉𝐷
𝜋2 =
1
𝑅𝑒
This is the inverse of the Reynold’s number:
By Buckingham’s Pi theorem, the original function 𝐹𝐷 = 𝑓(𝜌, 𝑉, 𝐷 𝜇) is now rewritten in terms of two
dimensionless groups:
𝐶𝐷 = 𝑓(𝑅𝑒)
This shows that drag force depends only on the Reynolds number, making it much easier to analyse and apply
experimentally.
EXAMPLE 2
𝑑𝑝
Obtain an expression in non-dimensional form for the pressure gradient, 𝑑𝑥 in a horizontal pipe with flow
velocity V, pipe diameter D, fluid density ρ, fluid viscosity µ and pipe wall roughness e.
SOLUTION
The relevant variables are:
𝑑𝑝
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Pressure gradient, 𝑑𝑥 (kg/m2.s2)
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Velocity, 𝑉 (m/s)
Diameter, 𝐷 (m)
Fluid density, 𝜌 (kg/m3)
Fluid viscosity, 𝜇 (Pa.s)
Relative roughness, 𝑒 (m)
𝑑𝑝
Thus, we have n = 6 variables: 𝑑𝑥 , 𝑉, 𝐷, 𝜌, 𝜇, 𝑒. We then identify the fundamental dimensions of each variable.
Variable
Pressure Gradient
Symbol
𝑑𝑝
𝑑𝑥
𝑉
𝐷
𝜌
𝜇
𝑒
Velocity
Diameter
Fluid Density
Fluid Viscosity
Relative roughness
Dimension
𝑀𝐿−2 𝑇 −2
𝐿𝑇 −1
𝐿
𝑀𝐿−3
𝑀𝐿−1 𝑇 −1
𝐿
The fundamental dimensions used here are M (mass), L (length), and T (time), so we have m = 3 fundamental
dimensions. The number of dimensionless Pi groups is:
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑖 𝑔𝑟𝑜𝑢𝑝𝑠: 6 − 3 = 3
We now pick 𝑚 variables (3 in this case) FROM THE RELEVANT VARIABLES that contain ALL the fundamental
dimensions (M,L,T). A good choice would be, fluid density (𝜌), velocity (𝑉) and diameter (𝐷).
Forming the first Pi group with pressure gradient:
𝜋1 =
𝑑𝑝
× 𝜌𝑎 × 𝑉 𝑏 × 𝐷 𝑐
𝑑𝑥
𝜋1 = (𝑀𝐿−2 𝑇 −2 )(𝑀𝐿−3 )𝑎 (𝐿𝑇 −1 )𝑏 (𝐿)𝑐
𝜋1 = (𝑀1+𝑎 )(𝐿−2−3𝑎+𝑏+𝑐 )(𝑇 −2−𝑏 )
𝑀0 𝐿0 𝑇 0 = (𝑀1+𝑎 )(𝐿−2−3𝑎+𝑏+𝑐 )(𝑇 −2−𝑏 )
For 𝑴: 1 + 𝑎 = 0 → 𝑎 = −1
For 𝑳: − 2 − 3𝑎 + 𝑏 + 𝑐 = 0 → −2 − 3(−1) + 𝑏 + 𝑐 = 0 → 𝑏 + 𝑐 = −1
For 𝑻: − 2 − 𝑏 = 0 → 𝑏 = −2 ∴ 𝑐 = 1
Thus, the first Pi group is:
𝜋1 =
𝑑𝑝
× 𝜌−1 × 𝑉 −2 × 𝐷1
𝑑𝑥
∴ 𝜋1 =
𝑑𝑝 𝐷
𝑑𝑥 𝜌𝑉 2
Forming the second Pi group (𝜋2 ) using viscosity (𝜇):
𝜋2 = 𝜇 × 𝜌 𝑥 × 𝑉 𝑦 × 𝐷 𝑧
𝜋2 = (𝑀𝐿−1 𝑇 −1 )(𝑀𝐿−3 )𝑥 (𝐿𝑇 −1 )𝑦 (𝐿) 𝑧
𝜋2 = (𝑀1+𝑥 )(𝐿−1−3𝑥+𝑦+𝑧 )(𝑇 −1−𝑦 )
𝑀0 𝐿0 𝑇 0 = (𝑀1+𝑥 )(𝐿−1−3𝑥+𝑦+𝑧 )(𝑇 −1−𝑦 )
For 𝑴: 1 + 𝑥 = 0 → 𝑥 = −1
For 𝑳: −1 − 3𝑥 + 𝑦 + 𝑧 → −1 − 3(−1) + 𝑦 + 𝑧 = 0 → 𝑦 + 𝑧 = −2
For 𝑻: −1 − 𝑦 = 0 → 𝑦 = −1 ∴ 𝑧 = −1
Thus, the second Pi group is:
𝜋2 = 𝜇 × 𝜌−1 × 𝑉 −1 × 𝐷 −1
∴ 𝜋2 =
𝜇
𝜌𝑉𝐷
Forming the third Pi group is:
𝜋3 = 𝑒 × 𝜌 𝑑 × 𝑉 𝑓 × 𝐷 𝑔
𝜋3 = (𝐿)(𝑀𝐿−3 )𝑑 (𝐿𝑇 −1 )𝑓 (𝐿)𝑔
𝑀0 𝐿0 𝑇 0 = (𝐿)(𝑀𝐿−3 )𝑑 (𝐿𝑇 −1 )𝑓 (𝐿)𝑔
For 𝑴: 𝑑 = 0
For 𝑳: 1 − 3𝑑 − 𝑓 + 𝑔 = 0 → 𝑓 − 𝑔 = 1
For 𝑻: −𝑓 = 0 → 𝑓 = 0 ∴ 𝑔 = −1
Thus, the third Pi group is:
𝜋3 = 𝑒 × 𝜌0 × 𝑉 0 × 𝐷 −1
∴ 𝜋3 =
𝑒
𝐷
Similarity and Model Testing
In fluid mechanics, similarity refers to the idea that a scaled-down model can be used to study the behaviour
of a real-life system (called the prototype) as long as certain conditions are met. There are three types of
similarity that must be satisfied to ensure an accurate model test:
1. Geometric Similarity (Shape and Size)
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The model and prototype must have the same shape, but the size can be different. All corresponding
lengths, areas, and volumes must maintain the same ratio. This ensures that flow patterns remain
the same.
2. Kinematic Similarity (Velocity and Acceleration)
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The model and prototype must have the same ratios of velocities and accelerations at
corresponding points. This ensures that fluid particles move in the same way in both the model and
the prototype.
3. Dynamic Similarity (Forces)
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The forces acting on the fluid particles in the model and prototype must be in the same ratio. This
ensures that forces like gravity, viscosity, pressure, and inertia behave in the same way.
To maintain similarity, we use dimensionless numbers that compare the importance of different forces. The
most common ones are:
Non-Dimensional Number
Reynold’s Number (Re)
Froude Number (Fr)
Cauchy Number (Ca)
Mach Number (Ma)
Euler Number (Eu)
Weber Number (We)
Strouhal Number (St)
Formula
𝜌𝑉𝐷
𝑅𝑒 =
𝜇
𝐹𝑟 =
𝑉
√𝑔𝐿
𝜌𝑉 2
𝐸𝑣
𝑉
𝑀𝑎 =
𝑐
𝐶𝑎 =
𝑃
𝜌𝑉 2
𝜌𝑉 2 𝐿
𝑊𝑒 =
𝜎
𝐸𝑢 =
𝑆𝑡 =
𝜔𝐿
𝑉
Significance
Compares inertia forces to
viscous forces. Used in viscous
flows.
Compares inertia forces to
gravitational forces. Used in free
surface flows (e.g., ships, rivers).
Compressible Flows
Compares inertia forces to
compressibility effects. Used in
high-speed flows (e.g., aircraft,
rockets).
Compares pressure forces to
inertia forces.
Compares inertia forces to
surface tension forces. Used in
bubble/droplet flows.
Oscillating Flows
To ensure that a model experiment accurately represents the prototype, we ensure that the relevant nondimensional numbers are the same for both.
EXAMPLE
A prototype gate valve which will control the flow in a pipe system conveying paraffin is to be studied in a
model. Perform dimensional analysis to obtain the relevant nondimensional groups. A 1/5 scale model is built
to determine the pressure drop across the valve with water as the working fluid. DATA:
𝜌𝑝 = 800 kg/m3, 𝜇𝑝 = 0.002 kg/m.s, 𝑣𝑤𝑎𝑡𝑒𝑟 = 1.12 × 10−5 m2/s
a) For a particular opening, when the velocity of paraffin in the prototype is 3.0 m/s what should be the
velocity of water in the model for dynamic similarity?
b) What is the scale ratio of the quantity of flow?
c) Find the pressure drop in the prototype if it is 60 kPa in the model.
SOLUTION
The pressure drop is expected to depend upon the gate opening h, the overall depth H, the velocity V, density ρ
and viscosity µ.
Variable
Pressure Drop
Gate opening
Overall Depth
Velocity
Fluid Density
Fluid Viscosity
Symbol
∆𝑃
ℎ
𝐻
𝑉
𝜌
𝜇
Dimension
𝑀𝐿−1 𝑇 −2
𝐿
𝐿
𝐿𝑇 −1
𝑀𝐿−3
𝑀𝐿−1 𝑇 −1
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑖 𝐺𝑟𝑜𝑢𝑝𝑠 = 6 − 3 = 3
Selecting the scale variables, overall depth (𝐻), velocity (𝑉) and fluid property (𝜌)
𝜋1 = ∆𝑃 × 𝐻 × 𝑉 × 𝜌
𝜋2 = ℎ × 𝐻 × 𝑉 × 𝜌
𝜋3 = 𝜇 × 𝐻 × 𝑉 × 𝜌
After performing dimensional analysis, we obtain:
𝜋1 =
∆𝑃
ℎ
𝜇
, 𝜋2 = , 𝜋3 =
2
𝜌𝑉
𝐻
𝜌𝑉𝐷
a) Dynamic similarity requires that all non-dimensional groups be the same in model and prototype; i.e.
𝜋1𝑚𝑜𝑑𝑒𝑙 = 𝜋1𝑝𝑟𝑜𝑡𝑜𝑡𝑦𝑒 , 𝜋2𝑚𝑜𝑑𝑒𝑙 = 𝜋2𝑝𝑟𝑜𝑡𝑜𝑡𝑦𝑝𝑒 , 𝜋3𝑚𝑜𝑑𝑒𝑙 = 𝜋3𝑝𝑟𝑜𝑡𝑜𝑡𝑦𝑝𝑒
∆𝑃
∆𝑃
( 2)
= ( 2)
𝜌𝑉 𝑝𝑟𝑜𝑡𝑜
𝜌𝑉 𝑚𝑜𝑑𝑒𝑙
ℎ
ℎ
( )
=( )
𝐻 𝑝𝑟𝑜𝑡𝑜
𝐻 𝑚𝑜𝑑𝑒𝑙
𝜇
𝜇
𝜌𝑉𝐷
𝜌𝑉𝐷
(
)
=(
)
→(
)
=(
)
𝜌𝑉𝐷 𝑝𝑟𝑜𝑡𝑜
𝜌𝑉𝐷 𝑚𝑜𝑑𝑒𝑙
𝜇 𝑝𝑟𝑜𝑡𝑜
𝜇 𝑚𝑜𝑑𝑒𝑙
Since the problem involves fluid flow through a valve, Reynolds number similarity must be satisfied.
𝜌𝑝 𝑉𝑝 𝐷𝑝 𝜌𝑚 𝑉𝑚 𝐷𝑚
=
𝜇𝑝
𝜇𝑚
Solving for 𝑉𝑚 (because 𝑉𝑝 is given as 3 m/s):
𝜌𝑝 𝜇𝑝 𝐷𝑚
𝑉𝑚 = 𝑉𝑝 ×
×
×
𝜌𝑚 𝜇𝑚 𝐷𝑝
𝑉𝑚 = 3 (
800
0.002
1
)(
)( )
−6
1000 1.12 × 10
5
∴ 𝑉𝑚 = 6.72 𝑚/𝑠
b) Flow (volumetric, 𝑄) can be written as the product of velocity and area (𝐿3 𝑇 −1)
𝑄𝑝
𝑉𝑝 𝐷𝑝 2
3 5 2
=
( ) =
( ) = 11.2
𝑄𝑚 𝑉𝑚 𝐷𝑚
6.72 1
c) We can use the first Pi group to solve for ∆𝑃𝑝
∆𝑃𝑝
∆𝑃𝑚
=
2
𝜌𝑝 𝑉𝑝
𝜌𝑚 𝑉𝑀2
𝜌𝑝 𝑉𝑝 2
( )
𝜌𝑚 𝑉𝑚
800
3 2
∆𝑃𝑝 = (60) ×
(
) = 9.57 𝑘𝑃𝑎
1000 6.72
∴ ∆𝑃𝑝 = ∆𝑃𝑚 ×
Even though we do not have the
diameters, we know the ratio of
the prototype and the model.