Dimensional analysis - Part 1

1. Dimensional Analysis definition necessity
2. Fundamental quantities- Primary Dimensions
3. Non Primary Dimensions
4. Secondary derived Dimensions
5. Dimensionless quantities
6. Necessity and objectives of Dimensional Analysis
7. Dimensional Homogeneity
8.Methods of Dimensional Analysis
9. Rayleigh's Method of dimensional analysis
10. Numerical Problem on Rayleigh's Method of dimensional analysis

1. DIMENSIONAL ANALYSIS
PART-I
Dr.RAMESH B R
[email protected]

2. Dimensional Analysis
•It is a pure mathematical technique to establish a relationship
between physical quantities involved in a fluid phenomenon by
considering their dimensions.
•In dimensional analysis, from a general understanding of fluid
phenomena, we first predict the physical parameters that will
influence the flow
•then we group these parameters into dimensionless combinations which
enable a better understanding of the flow phenomena.
•Dimensional analysis is particularly helpful in experimental work because it
provides a guide to those things that significantly influence the phenomena
•thus it indicates the direction in which experimental work should go.

3. Fundamental quantities-(Primary dimensions)
Thephysicalquantitieswhichcanbetreatedasindependentofotherphysical
quantitiesandarenotusuallydefinedintermsofotherphysicalquantities,are
calledfundamentalquantities.(primarydimensions.Primary(sometimes
calledbasic)dimensionsaredefinedasindependentorfundamentaldimensions,
fromwhichotherdimensionscanbeobtained.
Ex.:Mass,Length,Timeetc.,
•Unitsarethestandardelementsweusetoquantifythesedimensions.
Ex.:Kg,Metre,Secondsetc.,
•Forexample,lengthisadimensionthatismeasuredinunitssuchasmicrons(m),
feet(ft),centimeters(cm),meters(m),kilometers(km),etc.Thereareseven
primarydimensions(alsocalledfundamentalorbasicdimensions.Theyare
mass,length,time,temperature,electriccurrent,amountoflight,and
amountofmatter

4. Non primary dimensions
•Allnonprimarydimensionscanbeformedbysomecombinationof
thesevenprimarydimensions
Forexample,forcehasthesamedimensionsasmasstimesacceleration(by
Newton’ssecondlaw).Thus,intermsofprimarydimensions,
Dimensionsofforce:

5. EXAMPLE :
Primary Dimensions of Surface Tension
SOLUTION
Theprimarydimensionsofsurfacetensionaretobedetermined.
Analysis
Forcehasdimensionsofmasstimesacceleration,or{mL/t
2
}.Thus,

6. Geometric Units Dimensions
Area �
2
�
2
Volume �
3
�
3
Kinetic
Velocity m/s L/T (L�
−1
)
Acceleration m/�
2
L�
2
(L�
−2
)
Discharge m
3
/s L
3
/T(L
3
T
-1
)
Dynamic
Force N ML/T (ML�
−1
)
Density g/�
3
M/�
3
(M�
−3
)
Thephysicalquantitieswhosedefiningoperationsarebasedonotherphysical
quantities,arecalledderivedquantities.
Allphysicalquantitiesotherthanthesevenbasequantitiesarederived
quantities
Swcondarydimensionsarethosequantitieswhichpossesmorethanone
fundamentaldimensions
Secondary or Derived Dimensions:

7. Dimensionless quantity
Physicalquantitieswhichdonotpossessdimensionsarecalled
dimensionlessquantities.
Example:Angle,specificgravity,strain.
Ingeneral,physicalquantitywhichisaratiooftwoquantitiesofsame
dimensionwillbedimensionless.
Example:
Reynold’snumber�
�=
���
??????
isadimensionlessnumber
ThereforetheDimensionsofnumerator??????��=(M�
−3
)(L�
−1
)(L)
Simplifyingweget��
−1
�
−1
Dimensionsofdenominator??????=��
−1
�
−1
Therefore�
�=
���
??????
=
��
−1
�
−1
��
−1
�
−1
hencedimensionless.

8. Necessity of Dimensional analysis
•Insomeofthepracticalrealflowproblemsinfluidmechanicscanbe
solvedbyusingequationsandanalyticalprocedures.
•Solutionsofsomerealflowproblemsdependheavilyonexperimental
data.
•Sometimes,theexperimentalworkinthelaboratoryisnotonlytime
consuming,butalsoexpensive.
•So,themaingoalistoextractmaximuminformationfromfewest
experiments.
•Inthisregard,dimensionalanalysisisanimportanttoolthathelpsin
correlatinganalyticalresultswithexperimentaldataandtopredictthe
prototypebehaviourfromthemeasurementsonthemodel.

9. Objectives of Dimensional Analysis
1) Checking the dimensional homogeneity of any fluid flow equation.
2) Deriving fluid mechanics equations expressed in terms of non-dimensional
parameters to show the relative significance of each parameter.
3) Planning tests and presenting experimental results in a systematic manner.
4) Analysing complex flow phenomena by use of scale models (model similitude).
5) Conversion from one dimensional unit to another
6)Checking units of equations (Dimensional Homogeneity)
7) Defining dimensionless relationship using
a) Rayleigh’s Method
b) Buckingham’s π-Theorem
8) Model Analysis

10. DIMENSIONAL HOMOGENEITY
Principle of dimensional homogeneity states that an equation which
expresses a physical phenomenon of fluid flow must be algebraically correct
and dimensionally homogeneous
Anequationissaidtobedimensionallyhomogeneous,ifthedimensionson
itslefthandsidearesameasthedimensionspfthetermsonthelefthand
side
Example:ConsidertheequationV=2�ℎ
DimensionsofLHSV=L/T=LT
-1
DimensionsofRHS=2�ℎ=
�
�
2
xL=
�
2
�
2
=
�
�
=LT
-1
DimensionsofLHS=DimensionsofRHS
Hencetheequationishomogeneous

11. Example:Considertheequations=ut+
1
2
a�
2
DimensionsofLHSS=distance=m=L
1
DimensionsofRHS=ut=m/sxs=L/TxT=L
1
DimensionsofRHS=½at
2
=m/s
2
xs
2
=L/T
2
xT
2
=L
1
DimensionsofLHS=DimensionsofRHS
Hencetheequationishomogeneous

12. CheckthedimensionalhomogeneityofBernoulli’sequationofenergy
Bernoulli’sequationisP+1/2??????�
2
+??????�ℎ=�
Solution :
Dimension of P =
??????����
??????���
=
����??????
????????????�????????????ℎ
????????????�??????
2
���??????�ℎ
2
=
�
??????
??????
2
�
2
=
�
�
2
�
Dimensions of 1/2 ??????�
2
=
����
�����
���??????�ℎ
�??????��
2
=
�
�
3
�
�
2
=
�
�
2
�
Dimensions of ??????�ℎ=
����
�����
���??????�ℎ
�??????��
2
�����ℎ=
�
�
3
�
�
2
�=
�
�
2
�
Therefore all the additive terms of the Bernoulli’s equation are having the same
dimensions
Therefore from the law of homogeneity the dimensions of the constant shall also
have the dimensions on the left hand side
Therefore the dimension of the right hand side C is
�
�
2
�

13. METHODS OF DIMENSIONAL ANALYSIS
There are two methods of dimensional analysis used.
(i)Rayleigh's method
(ii) Buckingham π Theorem
RAYLEIGH'S METHOD
In this method, the expression is determined for a variable for
maximum three or four variables only.
If the number of independent variables becomes more than four, it is
very difficult to find the expression for the dependent variable

14. Steps involved in Rayleigh's method
1.First, the functional relationship is written with the given data.
Consider X as a variable which depends on X
1, X
2, X
3,… X
n
So, the functional equation is written X=f(X
1, X
2, X
3,…X
n)
2.Thentheequationisexpressedintermsofaconstantwithexponentslike
powersofa,b,c...Therefore,theequationisagainwrittenas:
X=ϕ(X
1
a
,X
2
b
,X
3
c
,...X
n
z
)
Here,(ϕ)=Constanta,b,c,...z=Arbitrarypower
3.Thevaluesofa,b,c,...zaredeterminedwiththehelpofdimensional
homogeneity.Itmeans,thepowersofthefundamentaldimensionsonboth
sidesarecomparedtoobtainthevaluesofexponents.
4.Finally,theseexponents/powervaluesaresubstitutedinthefunctional
equationandsimplifiedtoobtainthesuitableform.

15. Example:
Let us consider the frictional resistance of fluid flow per unit area of the
inside surface of the pipe
A reasonable assumption can be made that the resistance which causes
pressure drop of the fluid (Δp)is a function of diameter of pipe (D),
fluid density (??????) fluid velocity u and fluid viscosity ??????, or
Δp= f[u,D,??????,??????]
Δp= c u
a
D
b
??????
c
??????
d
where C is a dimensionless constant. The dimensional equation of the
above expression in fundamental dimensions M, L and T are
���
−2
�
2
=��
−1�
�
�
��
−3�
��
−1
�
−1�
M�
−1
�
−2
=�
�−�−3�−�
�
−�−�
�
−�+�

16. For the homogeneity of
M : 1 = c + d,
L : − 1 = a + b –3 c –d and,
t : − 2 = − a –d.
On solving these equations
we have
b = − d,
c = 1 –d and
a = 2 –d.
∆??????=��
2−�
�
−�
??????
1−�
??????
�
= C??????�
2
??????
���
�
=C
��
2
??????
??????
??????
Where �
�
�
=
���
??????
= Reynolds number.
The values of constants C and D have to
be determined by experiment

17. Problem:Fluidflowthroughasmallorificedischargingfreelyintoatmosphereunderaconstantheaddependsonthe
parametersdischargeQdiameterd,constantheadH.ρthemassdensityandµthedynamicviscosityofthefluidflowing
throughtheorifice.
Q= f(µ, ρ, d, H, g)
Q = C(μ
a
,ρ
b
,d
c,
H
d
g
e
) where C is a dimensionless constant.
Substituting the proper dimensions for each variable in this exponential equation in M-L-T system,
(L
3
/T) = (M°L°T°) (M/LT)
a
(M/L
3
)
b
(L)
c
(L)
d
(L/T
2
)
e
For dimensional homogeneity the exponents of each dimension on the both sides of the equation must be identical.
Thus
for M : 0 = a + b
for L : 3 = –a –3b + c + d + e
for T :–1 = –a –2e
Since there are five unknowns in three equations, three of the unknowns must be expressed in terms of the other two
b=-a
e=
1
2
-
�
2
c=
5
2
−
3�
2
-d

18. Q=c??????
�
??????
−�
�
5
2
−
3??????
2
−�
??????
�
�
1
2
−
??????
2
=c�
5
2�
1
2??????
�
??????
−�
�
−3??????
2�
−??????
2,??????
�
�
−�
Q=c
??????
��
3
2??????
1
2
�??????
�
�−
1
2
�
2
??????
1
2�
1
2( multiplying and dividing by
�
4
2)
=
�
??????
4
2
??????
��
3
2
??????
1
2
�
??????
�
�−
1
2�
4
�
2
2�??????
=??????2�??????�
1
??????
��
3
2??????
1
2
,
??????
�
This expression may be written in usual form i. e., Q=�
�a2�??????
Where �
�is the coefficient of discharge of the orifice which can be
expressed as �
�=�
1
??????
��
3
2??????
1
2
,
??????
�
Note: �
�is a non dimensional factor

19. Example :Find the equation for the power developed by a pump if it
depends on head H ;discharge Q and specific weight ??????of the fluid
P = f (H, Q,??????)
P = C H
a
Q
b
??????
c
[P] = [H]
a
[Q]
b
[??????]
c
-----(a)
[L
2
MT
-3
] = [LM
o
T
o
]
a
[L
3
M
0
T
-1
]
b
[L
-2
MT
-2
]
c
for M : 1 = c ---(i)
for L : 2 = a + 3b –2c ---(ii)
for T :–3 = –b –2c ---(iii)
(i)in( iii) gives b=1
substituting b and c value in( ii) we get a=1
Substituting a,band c in (a) we get
[P] = [H]
1
[Q]
1
[??????]
1
When K = 1
P= ??????�??????