UNIT-III FMM. DIMENSIONAL ANALYSIS

III DIMENSIONAL ANALYSIS

1. UNIT –III DIMENSIONAL ANALYSIS
FLUID MECHANICS AND
MACHINERY

2. UNIT –III DIMENSIONAL ANALYSIS
Needfordimensionalanalysis–methodsof
dimensionalanalysis–Similitude–typesof
similitude-Dimensionlessparameters-application
ofdimensionlessparameters–Modelanalysis.

3. DIMENSIONAL ANALYSIS:
It is a mathematical technique used in research work for design and conducting model tests. It
deals with the dimensions of the physical quantities involved in the phenomenon.
➢It helps in testing the dimensional homogeneity of any equation of fluid motion.
➢It helps in deriving equations expressed in terms of non-dimensional parameters.
➢It helps in planning model tests and presenting experimental results in a systematic manner.
Primary or Fundamental quantities:
The various physical quantities used to describe a given phenomenon can be described by a set
of quantities which are independent of each other. These quantities are known as fundamental
quantities or primary quantities.
Mass (M), Length (L), Time (T) and Temperature (θ) are the fundamental quantities.
Secondary or Derived quantities:
All other quantities such as area, volume, velocity, acceleration, energy, power, etc. are termed as
derived quantities or secondary quantities because they can be expressed by primary quantities.

4. Quantities Symbol Unit Dimension
Area A m
2
L
2
Volume V m
3
L
3
Angle ?????? Deg. Or Rad M
0
L
0
T
0
Velocity V m/s LT
-1
Angular Velocity ?????? Rad/s T
-1
Speed N rpm T
-1
Acceleration a m/s
2
LT
-2
Gravitational Acceleration g
m/s
2
LT
-2
Discharge Q
m
3
/s L
3
T
-1
Mass Density � Kg/m
3
M L
-3
Sp. Weight or Unit Weight w
N/m
3
ML
-2
T
-2

5. Quantities Symbol Unit Dimension
Force or Weight F or W
N MLT
-2
Dynamic Viscosity µ
N-s/m
2 ML
−1
T
−1
Kinematic viscosity ?????? m
2
/s L
2
T
-1
Shear stress ?????? N/m
2
ML
-1
T
-2
Surface Tension ?????? N/m MT
-2
Pressure or Pressure intensity p
N/m
2
or Pa ML
-1
T
-2
Modulus of Elasticity E
N/m
2
or Pa ML
-1
T
-2
Bulk Modulus K
N/m
2
or Pa ML
-1
T
-2
Work done or Energy W or E
N-m ML
2
T
-2
Torque T
N-m ML
2
T
-2
Power P
N-m/s or J/s or Watt ML
2
T
-3

6. Problem1:Determinethedimensionsofthequantitiesgivenbelow:(i)velocity,(ii)
density,(iii)angularvelocity,(iv)angularacceleration,(v)discharge,(vi)kinematic
viscosity,(vii)force,(viii)specificweightand(ix)dynamicviscosity.
(i)Velocity =
????????????��??????���
�??????��
=
�
??????
=????????????
-1
(ii)Density =
�??????��
??????�����
=
�
�
3=ML
-3
(iii)Angular velocity =
??????�??????��??????�????????????���??????������
�??????��
=
1
??????
=??????
−1
(iv)Angular Acceleration =
??????�??????��??????�??????����??????�??????
�??????��
=
1
??????
??????
1
??????
=??????
−2
(v)Discharge, Q = A* V = m
2
*m/s = m
3
/s = L
3
T
−1

7. (vi) Dynamic viscosity, ??????=??????
��
�??????
=> ??????=
??????
????????????
????????????
=
ML
−1
T
−2
??????
−1 = ML
−1
T
−1
where, ??????=
??????����
??????��??????
=
MLT
−2
�
2=ML
−1
T
−2
��
�??????
=
�??????
−1
�
= ??????
-1
(vii) Force = mass x acceleration = kg x m/s
2
= M L T
−2
(viii) Kinematic viscosity =
Dynamicviscosity
??????���??????�??????
=
ML
−1
T
−1
ML
−3=L
2
T
−1
(ix) Specific Weight, w =
weight
������
=
N
�
3=
MLT
−2
L
3= ML
−2
T
−2

8. Dimensionally Homogeneous:
An equation is said to be dimensionally homogeneous if the dimensions of the
terms on its LHS are same as the dimensions of the terms on its RHS.
Example:Let us consider, V = 2????????????
Units:V –m/s => LT
-1
G –m/s
2
=> LT
-2
=> L/T
2
H –m => L
LT
-1
= L/T
2
∗??????
LT
-1
= L
2
/T
2
LT
-1
=L/T
LT
-1
= LT
-1
---> This equation is Dimensionally Homogeneous.

9. Methods of dimensional Analysis:
•Rayleigh method,
•Buckinghumπ method
Rayleigh method:
Thismethodisusedfordeterminingtheexpressionforavariablewhichdependsupon
maximumthreeorfourvariablesonly.Ifthenumberofindependentvariablesbecomes
morethanfour,thenitisverydifficulttofindtheexpressionforthedependentvariable.
LetXisavariable,whichdependsonX
1,X
2andX
3variables.Thenaccordingto
Rayleigh’smethod,XisfunctionofX
1,X
2andX
3andmathematicallyitiswrittenas
X=f[X
1,X
2,X
3].
ThiscanalsobewrittenasX=KX
1
a
,X
2
b
andX
3
c
Where,Kisconstantanda,bandcarearbitrarypowers.
Thevaluesofa,bandcareobtainedbycomparingthepowersofthefundamental
dimensiononbothsides.Thustheexpressionisobtainedfordependentvariable.

10. Problem2:Findtheexpressionforthepowerp,developedbyapumpwhenp
dependsupontheheadh,thedischargeqandspecificweightwofthefluid.
Solution:
P = f [H, Q, w]
P = K H
a
Q
b
w
c
Substitute the fundamental units,
ML
2
T
-3
= K (L)
a
(L
3
T
-1
)
b
(ML
-2
T
-2
)
c
Equate the Powers of M, L, T
Power of M => 1 = c

11. Powerof L => 2 = a + 3 b –2c
=> 2 = a +3 -2
=> 2 = a +1
=> a = 1
Power of T => -3 = -b –2c
=>-3 = -b -2
=> -b = -3 + 2
=> -b = -1 or b = 1
Subtitutingthe values of a, b & c in power equation,
P = K H
1
Q
1
w
1
or
P = K H Q w

12. Problem3:Findanexpressionforthedragforceonsmoothsphereofdiameterd,
movingwithuniformvelocityvinafluidofdensity�anddynamicviscosity??????.
Solution:
F = f [D, v, �,??????]
F = K D
a
V
b
�
c
??????
d
Substitute the fundamental units,
MLT
-2
= K (L)
a
(LT
-1
)
b
(M L
-3
)
c
(ML
−1
T
−1
)
d
Equate the Powers of M, L, T
Power of M => 1 = c + d
=> c = 1 -d

13. Powerof L => 1 = a + b -3c –d
=> 1 = a + ( 2 -d) –3(1 –d) –d => 1 = a + 2 –d -3 + 3d –d
=> 1 = a –1 + d
=> a = 2 –d
Power of T=> –2 = –b –d
=> 2 = b + d
=> b = 2 –d
F = K D
(2 –d)
V
(2 –d)
�
(1 –d)
??????
d
=> F = K D
2
D
-d
V
2
V
-d
�
1
�
-d
??????
d
F = K �
1
D
2
V
2
??????
??????????????????
�
F = K ∅�
1
D
2
V
2
??????
??????????????????

14. Problem4:Theresistingforcerofasupersonicplaneduringflightcanbeconsideredasdependent
uponthelengthoftheaircraftl,velocityv,airviscosity??????,airdensity�andbulkmodulusofairk.
expressthefunctionalrelationshipbetweenthesevariablesandtheresistingforce.
Solution:
R = f [l, V, ??????,�,K]
R = A l
a
V
b
??????
�
�
�
K
e
Substitute the fundamental units,
MLT
-2
= A (L)
a
(LT
-1
)
b
(ML
−1
T
−1
)
c
(M L
-3
)
d
(ML
−1
T
−2
)
e
Equate the Powers of M, L, T
Power of M => 1 = c + d + e
Powerof L => 1 = a + b –c –3d –e
Power of T=> –2 = –b –c –2e
There are five unknowns but equations are only three. Expressing the three unknowns in terms of two
unknowns (??????and K).

15. Express the values of a, b and d in terms of c and e.
d = 1–c –e
b = 2 –c –2e
a = 1 –b + c + 3d + e = 1 –(2 –c –2e) + c + 3 (1 –c –e) + e
= 1 –2 + c + 2e + c + 3 –3c –3e + e
a = 2 –c
Substituting these values in eqn, we get
R = A (l)
2 -c
(V)
2 –c –2e
(??????)
�
�
1 –c –e
(K)
e
= A �
1
l
2
V
2
(l
-c
V
–c
??????
�
�
–c
)(V
–2e
�
–e
K
e
)
=A �l
2
V
2
??????
�????????????
�
�
????????????
2
�
R = A �l
2
V
2
∅
??????
�????????????
�
????????????
2

16. Buckingham Π-theorem Method:
Iftherearenvariables(dependentandindependentvariables)inadimensionally
homogeneousequationandifthesevariablescontainmfundamentaldimensions
(suchasM,L,T,etc.)thenthevariablesarearrangedinto(n-m)dimensionless
terms.Thesedimensionlesstermsarecalledπ-terms.
Mathematically,ifanyvariableX1,dependsonindependentvariables,X2,X3,
X4,....Xn;thefunctionalequationmaybewrittenas:X1=f(X2,X3,…Xn)
Eqn.canalsobewrittenas:f
1(X
1,X
2,X
3,...X
n)=0
Itisadimensionallyhomogeneousequationandcontainsnvariables.Ifthereare
mfundamentaldimensions,thenaccordingtoBuckingham’sπ-theorem,itcanbe
writtenintermsofnumberofπ-terms(dimensionlessgroups)inwhichnumber
ofπ-termsisequalto(n-m).
Hence,eqn.becomesas:f
1(π
1,π
2,π
3...π
n–m)=0

17. Contd…
Eachπ-termisdimensionlessandisindependentofthesystem.Divisionor
multiplicationbyaconstantdoesnotchangethecharacteroftheπ–term.Each
π-termcontains(m+1)variables,wheremisthenumberoffundamental
dimensionsandisalsocalledrepeatingvariables.
LetintheabovecaseX
2,X
3andX
4arerepeatingvariablesifthe
fundamentaldimensionm(M,L,T)=3.Theneachπ–termiswrittenas
π
1=X
2
a1
,X
3
b1
,X
4
c1
.X
1
π
2=X
2
a1
,X
3
b1
,X
4
c1
.X
5
:
:
π
n-m=X
2
an-m
,X
3
bn-m
,X
4
cn-m
.X
n

18. Contd…
Each equation is solved by considering dimensional homogeneity and values
of a
1, b
1, c
1; a
2, b
2, c
2etc. are obtained. These values are substituted in eqn. and
values of �
1 ,�
2 ,�
3….,�
n-mare obtained.
These values of �’s are substituted in eqn. The final general equation for the
phenomenon is obtained by expressing any one of the π-terms as a function of
the other as
π
1= ∅(π
2, π
3,π
4...π
n –m)
π
2= ∅(π
1, π
3,π
4...π
n –m)
Method of Selecting Repeating Variables:
Thenumberofrepeatingvariablesareequaltothenumberoffundamental
dimensionsoftheproblem.Thechoiceofrepeatingvariablesisgovernedbythe
followingconsiderations:

19. 1.As far as possible, the dependent variable should not be selected as repeating
variable.
2.The repeating variables should be chosen in such a way that one variable
contains geometric property (e.g. length, l; diameter, d; height, H etc.), other
variable contains flow property (e.g. velocity, V; acceleration, a etc.) and third
variable contains fluid property (e.g. mass density, ρ; weight density, w;
dynamic viscosity, µ etc.).
3.The repeating variables selected should not form the dimensional group.
4.The repeating variables together must have the same number of fundamental
dimensions.
5.No two repeating variables should have the same dimensions.
In most of fluid mechanics problems, the choice of repeating variables, may
be: (i) l, V, ρ (ii) d, V, ρ (iii) l, V, µ (iv) d, V, µ.

20. Problem5:Theresistancerexperiencedbyapartiallysubmergedbodydepends
uponthevelocityv,lengthofthebodyl,viscosityofthefluidµ,densityofthe
fluidρandgravitationalaccelerationg.obtainadimensionlessexpressionforr.
Solution:
Step 1:The resistance R is a function of:
(i) Velocity V, (ii) Length l, (iii) Viscosity µ, (iv) Density ρ & (v) Acceleration g.
Mathematically, R = f (V, l, µ, ρ, g)
(or)
f1(R, V, l, µ, ρ, g) = 0
∴Total number of variables, n = 6
m is obtained by writing dimensions of each variable as: R = MLT
-2
, V = LT
-1
, µ = ML
-1
T
-1
,
ρ = ML
-3
, g = LT
-2
.Thus the fundamental dimensions in the problem are M, L, T and hence m = 3.

21. Number of dimensionless π-terms = n –m = 6 –3 = 3
Thus three π-terms say �1 ,�2 ,�3are formed.
The equation may be written as: f1(�1 ,�2 ,�3) = 0
Step 2: Selection of repeating variables:
Out of six variables R, V, l, µ, ρ, g three variables (as m = 3)are to be selected as repeating variables. R is
a dependent variable and should not be selected as a repeating variable.
Out of the remaining five variables one variable should have geometric property, second should have
flow property and third one should have fluid property; these requirements are met by selecting l, V and ρ as
repeating variables.
Step 3:Each π-term (= m + 1 variables) is written as
π
1= l
a1
, V
b1
, ρ
c1
R
π
2= l
a2
, V
b2
, ρ
c2
µ
π
3= l
a3
, V
b3
, ρ
c3
g

22. Step 4:Each π-term is solved by the principle of dimensional homogeneity, as follows:
π
1–term:
M
0
L
0
T
0
= L
a1
, (LT
-1
)
b1
, (ML
-3
)
c1
(MLT
-2
)
Equating the exponents of M, L and T respectively, we get:
For M:0 = c
1+ 1
For L:0 = a
1+ b
1–3 c
1+ 1
For T: 0 = –b
1–2
∴c
1= -1;b
1= -2; a
1= -2
Substituting the a
1, b
1& c
1 values in π
1 equation, we get
π
1= l
-2
, V
-2
, ρ
-1
R
π
1=
??????
�
2
??????
2
??????

23. π
2–term:
M
0
L
0
T
0
= L
a2
, (LT
-1
)
b2
, (ML
-3
)
c2
(ML
-1
T
-1
)
Equating the exponents of M, L and T respectively, we get:
For M:0 = c
2+ 1
For L:0 = a
2+ b
2–3 c
2-1
For T: 0 = –b
1–1
∴c
2= -1;b
2= -1; a
2= -1
Substituting the values of a
2, b
2, c
2in π
2, we get,
Π
2=
µ
�????????????

24. π
3–term:
M
0
L
0
T
0
= L
a3
, (LT
-1
)
b3
, (ML
-3
)
c3
(LT
-2
)
Equating the exponents of M, L and T respectively, we get:
For M:0 = c
3
For L:0 = a
3+ b
3–3 c
3+ 1
For T: 0 = –b
3–2
∴c
3= 0;b
2= -2; a
3= 1
Substituting the values of a
3, b
3, c
3in π
3, we get,
Π
3=
�g
??????
2

25. Step 5:Substitute the values of �
1,�
2,�
3.
The functional relationship becomes:
f
1
??????
�
2
??????
2
??????
,
µ
�????????????
,
�??????
??????
2= 0
??????
�
2
??????
2
??????
= ∅
µ
�????????????
,
�??????
??????
2
R = ??????
2
??????
2
�∅
µ
�????????????
,
�??????
??????
2

26. Problem6:Usingbuckingham’sπ-theorem,showthatthevelocitythroughacircularorificeis
givenbyv=??????????????????∅
??????
??????
µ
??????????????????
where,h=headcausingflow,d=diameteroftheorifice,
µ=co-efficientofviscosity,ρ=massdensity,andg=accelerationduetogravity
Solution:
V = f (g, H, D, µ, �)
f
1 (V, g, H, D, µ, �) = 0 -----------------1
n -no of variables = 6; m –fundamental units = 3
π=(n−m)= (6 -3) = 3
Equation 1 can also be written as
f
1 (π
1, π
2, π
3) = 0 -----------------2
Selecting Repeating variables
Geometric property -H, Flow Property -g, Fluid property -�

27. ??????–term: π
1= H
a1
, g
b1
, �
c1
D
π
2= H
a2
, g
b2
, �
c2
V
π
3= H
a3
, g
b3
, �
c3
µ
??????
1–term:
π
1= H
a1
, g
b1
, �
c1
D
M
0
L
0
T
0
= (L)
a1
, (LT
-2
)
b1
, (ML
-3
)
c1
L
Equate the powers of M: 0 = c1
Equate the powers of L: 0 = a1 + b1 –3c1 + 1
Equate the powers of T: 0 = -2b1
b1 = 0; c1 = 0; a1 = -1
π
1= H
-1
, g
0
, �0D
??????
1=
??????
??????

28. ??????
2–term:
π
2= H
a2
, g
b2
, �
c2
V
M
0
L
0
T
0
= (L)
a2
, (LT
-2
)
b2
, (ML
-3
)
c2
(LT
-1
)
Equate the powers of M: 0 = c2
Equate the powers of L: 0 = a2 + b2 –3c2 + 1
Equate the powers of T: 0 = -2b
2-1
b2 = -1/2 ; c2 = 0; a2 = -1/2
π
2= H
-1/2
, g
-1/2
, �
0
V
π
2=
??????
????????????

29. ??????
3–term:
π
3= H
a3
, g
b3
, �
c3
µ
M
0
L
0
T
0
= (L)
a3
, (LT
-2
)
b3
, (ML
-3
)
c3
(ML
-1
T
-1
)
Equate the powers of M: 0 = c3 + 1 => c3 = -1
Equate the powers of L: 0 = a3 + b3 -3c3 -1
Equate the powers of T: 0 = -2b
3-1
Solving the above equations,
b
3= -1/2 ; c
3= -1 ; a
3= -3/2
π
3= H
-3/2
, g
-1/2
, �
−1
µ
π
3=
µ
????????????????????????

30. Multiply and divide the π
3-term with V
π
3=
µV
??????????????????????????????
=
µ
??????????????????
π
2
f
1 (π
1, π
2, π
3) = 0
f
1 (
??????
??????
,
??????
????????????
,
µ
????????????????????????
) = 0
??????
????????????
= f
1 (
??????
??????
,
µ
??????????????????
π
2)
V = 2????????????∅(
??????
??????
,
µ
??????????????????
π
2)
Multiply by a constant does not change the character of π-terms.
V = ??????????????????∅
??????
??????
µ
??????????????????

31. Problem7:Usingbuckingham’stheorem,showthatthedischargeqconsumedbyanoilringis
givenbyq=nd
3
∅
??????
??????�
2
�
3
µ
??????��
2,
??????
??????�
2
�
wheredistheinternaldiameterofthering,nisrotational
speed,�isdensity,µisviscosity,??????issurfacetensionandwisthespecificweightofoil.
Given:
d, N, �, µ, ??????, w, Q
Solution:
Q = f(d, N, �, µ, ??????, w)
or
f
1(Q, d, N, �, µ, ??????, w) = 0 --------------------------1
No. of variables, n = 7,
No. of Fundamental dimensions, m = 3
Number of dimensionless π-terms = n –m = 7-3 = 4

32. Equation 1 can also be written as
f
1 (π
1, π
2, π
3, π
4) = 0 ---------------------------2
Selecting Repeating variables:
Geometric property -d,
Flow property -N,
Fluid property -�
π–terms:
π
1= d
a1
, N
b1
, �
c1
??????
π
2= d
a2
, N
b2
, �
c2
Q
π
3= d
a3
, N
b3
, �
c3
w
π
4= d
a4
, N
b4
, �
c4
µ

33. Each π-term is solved by the principle of dimensional homogeneity, as follows:
π
1–term:
π
1= d
a1
, N
b1
, �
c1
??????
M
0
L
0
T
0
= L
a1
, (T
-1
)
b1
, (ML
-3
)
c1
(MT
-2
)
Equating the exponents of M, L and T respectively, we get:
M –0 = c
1+ 1 => c
1= -1
L –0 = a
1-3c
1=> a
1+3 => a
1= -3
T –0 = -b
1-2 => b
1= -2
π
1= d
-3
, N
-2
, �
−1
??????
π
1=
??????
d
3
N
2
??????

34. π
2–term:
π
2= d
a2
, N
b2
, �
c2
Q
M
0
L
0
T
0
= L
a2
, (T
-1
)
b2
, (ML
-3
)
c2
(L
3
T
-1
)
Equating the exponents of M, L and T respectively, we get:
M –0 = c
2
L –0 = a
2-3c
2+3 => a
2= -3
T –0 = -b
2-1 = b
2= -1
π
2= d
-3
, N
-1
, �
0
V
π
2=
Q
d
3
N

35. π
3–term:
π
3= d
a3
, N
b3
, �
c3
w
M
0
L
0
T
0
= L
a3
, (T
-1
)
b3
, (ML
-3
)
c3
(ML
-2
T
-2
)
Equating the exponents of M, L and T respectively, we get:
M –0 = c
3+ 1 = c
3= -1
L –0 = a
3-3c
3-2 => a
3+3 -2 => a
3= -1
T –0 = -b
3-2 = b
3= -2
π
3= d
-1
, N
-2
, �
−1
w
π
3=
w
??????dN
2

36. π
4–term:
π
4= d
a4
, N
b4
, �
c4
µ
M
0
L
0
T
0
= L
a3
, (T
-1
)
b3
, (ML
-3
)
c3
(ML
-1
T
-1
)
Equating the exponents of M, L and T respectively, we get:
M –0 = c
4+ 1 = c
4= -1
L –0 = a
4-3c
4-1 => a
4+3 -1 => a
4= -2
T –0 = -b
4-1 = b
4= -1
π
4= d
-2
, N
-1
, �
−1
µ
π
4=
µ
??????d
2
N

37. f
1 (π
1, π
2, π
3) = 0
Substituting π-values in the above eqn.
f
1 (
??????
d
3
N
2
??????
,
Q
d
3
N
,
w
??????dN
2,
µ
??????d
2
N
) = 0
Q
d
3
N
= ∅(
??????
d
3
N
2
??????
,
w
??????dN
2,
µ
??????d
2
N
)
Q = Nd
3
∅
??????
??????�
2
�
3
µ
??????��
2,
??????
??????�
2
�

38. Thank You