Dimensional analysis offers a method for reducing complex physical problems to the simplest (that is, most economical) form prior to obtaining a quantitative answer.

1. Dimensions
&
Dimensional Analysis
Peter Huruma Mammba
Department of General Studies
DODOMA POLYTECHNIC OF ENERGY AND EARTH RESOURCES MANAGEMENT (MADINI INSTITUTE) –
DODOMA

2. What is dimension?
 Is a physical property which describes a
way any physical quantity is related to
fundamental physical quantities
 E.g. dimension of velocity is 1 in length
and -1 in time.

3. What is dimensional analysis?
 Is the analysis of the relationships
between different physical quantities by
identifying their fundamental dimensions
such as length, mass, time.

4. Physical quantity
 Is a physical property of a phenomenon, body
or substances that can be quantified by
measurement.
 Can be expressed as the combination of
a number and a unit or combination of units.

5. Types of physical quantity
 There are two main types of physical
quantity which are:-
1. Fundamental (primary/ basic) physical
quantities
2. Derived (secondary) physical quantities

6. Fundamental Physical Quantity
 Fundamental quantities are the quantities
which cannot be expressed in terms of any
other physical quantity.
 E.g. Mass, length, Time, Temperature,
Intensity of light, Electric Current,
etc.

8. Derived physical quantities
 These are quantities whose definitions are
based on other physical quantities (base
quantities).
 E.g. Pressure ( 𝑲𝒈𝒎−𝟏
𝒔−𝟐
), speed
(𝒎𝒔−𝟏
),Young modulus (𝑲𝒈𝒎−𝟏
𝒔−𝟐
).

9. Other examples of Derived quantity

10. Unit
 Unit is the reference used as the standard
measurement of a physical quantity.
 The unit in which the fundamental quantities
are measured are called fundamental unit and
the units used to measure derived quantities
are called derived units.

11. System of unit
 G.S. system – It is the centimeter gram
second system which are units of length,
mass and time.
 P.S. system – it is foot pound second
system. Britishers used this system.

12. System of unit…
 K.S. System – it is metre kilogram second
system. European countries like France use
this system.
 S.I. System – It is the international system of
unit. It is universally accepted and has seven
fundamental units.

13. Law of dimensional analysis
(principle of homogeneity)
 State that “The equation is dimensionally
correct if the dimensions on the left hand
side of the equation are equal to the
dimensions on the right hand side of the
equation, if not the equation is not
dimensionally correct”.

14. Checking a Result
 Terms do not match  Terms match, this could be a
valid formula.
2
2
1
ghv 
2
2
L
T
L
T
L

2
3
T
L
T
L

ghv 
L
T
L
T
L
2

T
L
T
L


15. Dimensional formulas of physical quantities

16. Four category of physical quantities
I. Dimensional variables
II. Dimensionless variables
III. Dimensional constant
IV. Dimensionless constant

17. Dimensional variables


18. DIMENSIONLESS VARIABLES
Those physical quantities which have
neither dimensions nor fixed value.
E.g.. Specific gravity, strain, angle,
etc.

19. DIMENSIONAL CONSTANT
Those physical quantities which
possess dimensions and have fixed
value.
E.g. Gravitational contact, planks
constant, velocity of light, etc.

20. Dimensionless constant
Those quantities which do not
possess dimensions but have fixed
value.
E.g. 1, 2, 3, 4, 5, π, e, etc.

21. Uses of dimension equations
 To check the correctness of the physical
relation.
 To recapitulate a forgotten formula.
 To derive the relationship between
different physical quantities.

22. Uses of dimension equations…
 To convert one system of unit to
another.
 To find the dimensions of constant in a
given relation.

23. Limitations
 Dimensional analysis only checks the
units.
 Numeric factors have no units and can’t be
tested.
is also valid. is not valid.
3
gh
v  4 ghv

24. Limitations...
 Dimension analysis cannot be used to
derive the exact form of a physical
relation if the physical quantity depends
upon more than three physical quantities
(M,L & T).

25. Limitations...
 Dimension analysis can not be used to derive
the relation involving trigonometrical and
exponential function.
 Dimensional analysis does not indicate
whether a physical quantity is scalar or
vector.

26. Example 1


27. Solution
 𝑣 ∝ 𝑟𝜌𝜂
 𝒗 = 𝒌𝒓 𝒂
𝝆 𝒃
𝜼 𝒄
……(i)
 But;
 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
=
𝐿
𝑇
= 𝑳𝑻−𝟏 …..(ii)
 𝑅𝑎𝑑𝑖𝑢𝑠 = 𝑳 ….(iii)
 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑀
𝐿3
= 𝑴𝑳−𝟑….(iv)
 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡
𝐹𝑜𝑟𝑐𝑒 = 𝑀𝐿𝑇−2
𝐴𝑟𝑒𝑎 = 𝐿2
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝐿𝑒𝑛𝑔ℎ𝑡
𝐿𝑇−1
𝐿
= 𝑇−1
𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 =
𝑀𝐿𝑇−2
𝐿2 𝑇−1
= 𝑴𝑳−𝟏 𝑻−𝟏….(v)

28. Solution…
 Substitute equation ii, iii, iv & v into
equation i;
 𝑳𝑻−𝟏=k 𝑳 𝑎 𝑴𝑳−𝟑 𝑏
𝑴𝑳−𝟏 𝑻−𝟏 𝑐
𝑇−1 = 𝑇−1𝑐
∴ c = 1……vi
𝑀0 = 𝑀 𝑏 𝑀 𝑐
𝑀0 = 𝑀 𝑏+𝑐
0= b + c; but c = 1
∴ b = -1 ….vii
𝐿 = 𝐿 𝑎 𝐿−3𝑏 𝐿−1𝑐
1 = 𝑎 − 3𝑏 − 1𝑐
But; c = 1 & b = -1
1 = 𝑎 + 3 − 1
𝒂 = −𝟏….viii

29. Solution…
 Substitute equation vi, vii & viii into
equation i;
𝑣 = 𝑘𝑟−1
𝜌−1
𝜂1
∴ 𝑣 = 𝑘
𝜂
𝑟𝜌

30. Example 2
 A gas bubble from an explosion under water
is found to oscillate with a period T, which is
proportional to 𝑷 𝒙
, 𝝆 𝒚
and 𝝐 𝒛
where P is
pressure, 𝝆 is the density and 𝝐 is the energy
of explosion. Find the units of the constant of
proportionality.

31. Solution
 𝑇 ∝ 𝑃 𝑥 𝜌 𝑦 𝜖 𝑧
 𝑻 = 𝒌𝑷 𝒙 𝝆 𝒚 𝝐 𝒛….(i)
But;
 𝑇𝑖𝑚𝑒 = 𝑻…..(ii)
 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑀
𝐿3
= 𝑴𝑳−𝟑
….(iii)
 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑚𝑎𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 2
= 𝑴𝑳 𝟐 𝑻−𝟐….(iv)
 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
=
𝑀𝐿𝑇−2
𝐿2
= 𝑴𝑳−𝟏 𝑻−𝟐 ….(v)
 Substitute equation ii, iii, vi &
v into equation i;
 𝑻 = 𝒌 𝑴𝑳−𝟏
𝑻−𝟐 𝒙
𝑴𝑳−𝟑 𝒚
𝑴𝑳 𝟐
𝑻−𝟐 𝒛
..vi
 𝑇1 = 𝑇−2𝑥 𝑇0𝑦 𝑇−2𝑧
𝟏 = −𝟐𝒙 + 𝟎𝒚 − 𝟐𝒛 … a

32. Solution…
 𝑀0 = 𝑀1𝑥 𝑀1𝑦 𝑀1𝑧
𝑀0 = 𝑀1𝑥+1𝑦+1𝑧
0= 𝟏𝒙 + 𝟏𝒚 + 𝟏𝒛…b
 𝐿0 = 𝐿−1𝑥 𝐿−3𝑦 𝐿2𝑧
𝐿0 = 𝐿−1𝑥−3𝑦+2𝑧
𝟎 = −𝟏𝒙 − 𝟑𝒚 + 𝟐𝒛 …c
 From quadratic equations x,
y & z;
−𝟐𝒙 + 𝟎𝒚 − 𝟐𝒙 = 𝟏
𝟏𝒙 + 𝟏𝒚 + 𝟏𝒛 = 𝟎
−𝟏𝒙 − 𝟑𝒚 + 𝟐𝒛 = 𝟎
𝑥 = − 5
6 = −0.83333
𝑦 = 1
2 = 0.5
𝑧 = 1
3 = 0.3333333

33. Solution…
 Type equation here.Substitute the value of x,
y and z into equation vi;
𝑻 = 𝒌 𝑴𝑳−𝟏
𝑻−𝟐 −5
6
𝑴𝑳−𝟑
1
2
𝑴𝑳 𝟐
𝑻−𝟐
1
3
𝑘 =
𝑇1
𝑀 −5
6+1
2+1
3 𝐿
5
6−3
2+5
3 𝑇
5
3−2
3
∴ 𝒌 = 𝟏 no unit (hence shown)

34. Example 3
 If the viscous force F is defined by 𝐹 = 𝜂𝐴
Δ𝑣
Δ𝐿
where 𝜂 is the coefficient of viscosity, A is the
cross sectional area and
Δ𝑣
Δ𝐿
is the velocity
gradient. Find the dimensions and units of 𝜼.

35. Solution
 𝐹 = 𝜂𝐴
Δ𝑣
Δ𝐿
 𝜂 =
𝐹Δ𝐿
𝐴Δ𝑣
…(i)
But;
𝐹𝑜𝑟𝑐𝑒 = 𝑴𝑳𝑻−𝟐
... (ii)
𝐴𝑟𝑒𝑎 = 𝑳 𝟐
…(iii)
 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
=
𝐿
𝑇
= 𝑳𝑻−𝟏
…..(iv)
 𝐿𝑒𝑛𝑔ℎ𝑡 = 𝑳 ….(v)

36. Solution…
Substitute equation ii, iii, iv & v into
equation i, we get;
𝜂 =
𝑴𝑳𝑻−𝟐
𝑳
𝑳 𝟐 𝑳𝑻−𝟏
∴ 𝜂 = 𝑴𝑳−𝟏
𝑻−𝟏
(Dimension)
∴ 𝜂 = Kg𝑚−1
𝑠−1
(Unit)

37. Example 4
 While moving through liquid at speed, v a
spherical body experiences a retarding force
given by 𝑭 = 𝒌𝑹 𝒙
𝝆 𝒚
𝒗 𝒛
. Where k is constant,
𝝆 is density of liquid and R is radius of the
body. Determine the numerical values of x, y
and z by the method of dimension.

38. Solution
 𝑭 = 𝒌𝑹 𝒙 𝝆 𝒚 𝒗 𝒛…..(i)
But;
 𝐹𝑜𝑟𝑐𝑒 = 𝑴𝑳𝑻−𝟐
…(ii)
 𝑅𝑎𝑑𝑖𝑢𝑠 = 𝑳 ….(iii)
 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑀
𝐿3
= 𝑴𝑳−𝟑….(iv)
 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
=
𝐿
𝑇
= 𝑳𝑻−𝟏 …..(v)
 Substitute the dimensions of
equation ii, iii, iv & v into
equation i;
 𝑴𝑳𝑻−𝟐 = 𝒌 𝑳 𝒙 𝑴𝑳−𝟑 𝒚
𝑳𝑻−𝟏 𝒛
 𝑀1 = 𝑀 𝑦
∴ 𝑦 = 1
 𝑇−2
= 𝑇−1𝑧
−2 = −1𝑧
∴ 𝑧 = 2
 𝐿1 = 𝐿1𝑥 𝐿−3𝑦 𝐿1𝑧 = 𝐿 1𝑥−3𝑦+1𝑧
1 = 1𝑥 − 3 + 2; ∴ 𝑥 = 2
∴ 𝒙 = 𝟐, 𝒚 = 𝟏 & 𝒛 = 𝟐

39. Question 1
 The equation below is called Bernoulli's
equation which is applied to fluid flow and it
is stated that 𝑷 + 𝝆𝒈𝒉 +
𝟏
𝟐
𝝆𝒗 𝟐
= 𝒌. Where
P = pressure, h = height, 𝝆 = Density, v =
velocity, g = acceleration due to the gravity
and k = constant. Show that K = 1

40. Question 2
 The stress between two planes of molecules
in a moving liquid is given by 𝝈 =
𝜼𝒗
𝒙
. Where
v is the velocity difference between the
planes, x is their distance apart and 𝜼 is a
constant for a liquid. Show that the
dimensions of 𝜼 are 𝑴𝑳−𝟏
𝑻−𝟏

41. Question 3
 The velocity of wave of wavelength 𝜆 on the
surface of a pool of liquid whose surface tension
and density are 𝛾 and 𝜌 respectively is given by
𝑣2 =
𝜆𝑔
2𝜋
+
2𝜋𝛾
𝜆𝜌
. Where g is the acceleration due
to the gravity. Show that the equation is
dimensionally or not dimensionally correct.