Filters for Electromagnetic Compatibility Applications
UNIT-III FMM. DIMENSIONAL ANALYSIS
1. UNIT – III DIMENSIONALANALYSIS
FLUID MECHANICS AND
MACHINERY
2. UNIT – III DIMENSIONALANALYSIS
Need for dimensional analysis – methods of
dimensional analysis – Similitude –types of
similitude - Dimensionless parameters- application
of dimensionless parameters – Model analysis.
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 m2 L2
Volume V m3 L3
Angle 𝜃 Deg. Or Rad M0 L0 T0
Velocity V m/s LT-1
Angular Velocity 𝜔 Rad/s T-1
Speed N rpm T-1
Acceleration a m/s2 LT-2
Gravitational Acceleration g m/s2 LT-2
Discharge Q m3 /s L3T-1
Mass Density 𝜌 Kg/m3 M L-3
Sp. Weight or Unit Weight w N/m3 ML-2 T-2
5. Quantities Symbol Unit Dimension
Force or Weight F or W N MLT-2
Dynamic Viscosity µ N-s/m2 M L−1 T−1
Kinematic viscosity 𝛾 m2 /s L2 T-1
Shear stress 𝜏 N/m2 ML-1T-2
Surface Tension 𝜎 N/m MT-2
Pressure or Pressure intensity p N/m2 or Pa ML-1T-2
Modulus of Elasticity E N/m2 or Pa ML-1T-2
Bulk Modulus K N/m2 or Pa ML-1T-2
Work done or Energy W or E N-m ML2T-2
Torque T N-m ML2T-2
Power P N-m/s or J/s or Watt ML2T-3
6. Problem 1: Determine the dimensions of the quantities given below: (i) velocity, (ii)
density, (iii) angular velocity, (iv) angular acceleration, (v) discharge, (vi) kinematic
viscosity, (vii) force, (viii) specific weight and (ix) dynamic viscosity.
(i) Velocity =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
=
𝐿
𝑇
= 𝐿𝑇 -1
(ii) Density =
𝑀𝑎𝑠𝑠
𝑉𝑜𝑙𝑢𝑚𝑒
=
𝑀
𝐿3 = ML-3
(iii) Angular velocity =
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒
=
1
𝑇
= 𝑇 −1
(iv) Angular Acceleration =
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒
=
1
𝑇
𝑥
1
𝑇
= 𝑇 −2
(v) Discharge, Q = A* V = m2 *m/s = m3 /s = L3 T −1
7. (vi) Dynamic viscosity, 𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
=> 𝜇 =
𝜏
𝑑𝑢
𝑑𝑦
=
M L−1 T −2
𝑇 −1 = M L−1 T −1
where, 𝜏 =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
=
M L T −2
𝐿2 = M L−1 T −2
𝑑𝑢
𝑑𝑦
=
𝐿𝑇 −1
𝐿
= 𝑇 -1
(vii) Force = mass x acceleration = kg x m/s2 = M L T−2
(viii) Kinematic viscosity =
Dynamic viscosity
𝐷𝑒𝑛𝑠𝑖𝑡𝑦
=
M L−1 T −1
ML−3 = L2 T −1
(ix) Specific Weight, w =
weight
𝑣𝑜𝑙𝑢𝑚𝑒
=
N
𝑚3 =
M L T −2
L3 = M L−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 = 2𝑔𝐻
Units: V – m/s => LT-1
G – m/s2 => LT-2 => L/T2
H – m => L
LT-1 = L/T2 ∗ 𝐿
LT-1 = L2/T2
LT-1 = L/T
LT-1 = LT-1 ---> This equation is Dimensionally Homogeneous.
9. Methods of dimensional Analysis:
• Rayleigh method,
• Buckinghum π method
Rayleigh method:
This method is used for determining the expression for a variable which depends upon
maximum three or four variables only. If the number of independent variables becomes
more than four, then it is very difficult to find the expression for the dependent variable.
Let X is a variable, which depends on X1, X2 and X3 variables. Then according to
Rayleigh’s method, X is function of X1, X2 and X3 and mathematically it is written as
X = f [X1, X2, X3].
This can also be written as X = K X1
a, X2
b and X3
c
Where, K is constant and a, b and c are arbitrary powers.
The values of a, b and c are obtained by comparing the powers of the fundamental
dimension on both sides. Thus the expression is obtained for dependent variable.
10. Problem 2: Find the expression for the power p, developed by a pump when p
depends upon the head h, the discharge q and specific weight w of the fluid.
Solution:
P = f [H, Q, w]
P = K Ha Qb wc
Substitute the fundamental units,
ML2T-3 = K (L)a (L3T-1)b (ML-2 T-2)c
Equate the Powers of M, L, T
Power of M => 1 = c
11. Power of 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
Subtituting the values of a, b & c in power equation,
P = K H1 Q1 w1
or
P = K H Q w
12. Problem 3: Find an expression for the drag force on smooth sphere of diameter d,
moving with uniform velocity v in a fluid of density 𝜌 and dynamic viscosity 𝜇.
Solution:
F = f [D, v, 𝜌, 𝜇]
F = K Da Vb 𝜌c 𝜇d
Substitute the fundamental units,
MLT-2 = K (L)a (LT-1)b (M L-3)c (M L−1 T −1)d
Equate the Powers of M, L, T
Power of M => 1 = c + d
=> c = 1 - d
13. Power of 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) 𝜌(1 – d) 𝜇d => F = K D2 D-d V2 V- d 𝜌1 𝜌 - d 𝜇d
F = K 𝜌1 D2 V2 𝜇
𝐷 𝑉𝜌
𝑑
F = K ∅ 𝜌1 D2 V2 𝜇
𝐷 𝑉𝜌
14. Problem 4: The resisting force r of a supersonic plane during flight can be considered as dependent
upon the length of the aircraft l, velocity v, air viscosity 𝜇, air density 𝜌 and bulk modulus of air k.
express the functional relationship between these variables and the resisting force.
Solution:
R = f [l, V, 𝜇, 𝜌, K]
R = A la Vb 𝜇𝑐 𝜌𝑑 Ke
Substitute the fundamental units,
MLT-2 = A (L)a (LT-1)b (M L−1 T −1)c (M L-3)d (M L−1 T −2)e
Equate the Powers of M, L, T
Power of M => 1 = c + d + e
Power of 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 (𝜇 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 (𝜇)𝑐 𝜌 1 – c – e (K)e
= A 𝜌 1 l2 V2 (l- c V– c 𝜇𝑐 𝜌– c) (V– 2e 𝜌– e Ke)
= A 𝜌 l2 V2 𝜇
𝐿 𝑉𝜌
𝑐 𝐾
𝜌𝑉2
𝑒
R = A 𝜌 l2 V2 ∅
𝜇
𝐿 𝑉𝜌
𝐾
𝜌𝑉2
16. Buckingham Π-theorem Method:
If there are n variables (dependent and independent variables) in a dimensionally
homogeneous equation and if these variables contain m fundamental dimensions
(such as M, L, T, etc.) then the variables are arranged into (n-m) dimensionless
terms. These dimensionless terms are called π -terms.
Mathematically, if any variable X1, depends on independent variables, X2, X3,
X4, ....Xn; the functional equation may be written as: X1 = f (X2, X3,…Xn)
Eqn. can also be written as: f1(X1, X2, X3, ...Xn) = 0
It is a dimensionally homogeneous equation and contains n variables. If there are
m fundamental dimensions, then according to Buckingham’s π-theorem, it can be
written in terms of number of π -terms (dimensionless groups) in which number
of π -terms is equal to (n - m).
Hence, eqn. becomes as: f1 (π1, π2, π3 ...πn – m) = 0
17. Contd…
Each π -term 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.
Let in the above case X2, X3 and X4 are repeating variables if the
fundamental dimension m(M, L, T) = 3. Then each π –term is written as
π1 = X2
a1, X3
b1, X4
c1 . X1
π2 = X2
a1, X3
b1, X4
c1. X5
:
:
πn-m = X2
an-m, X3
bn-m, X4
cn-m. Xn
18. Contd…
Each equation is solved by considering dimensional homogeneity and values
of a1, b1, c1; a2, b2, c2 etc. are obtained. These values are substituted in eqn. and
values of 𝜋1 ,𝜋2 ,𝜋3….,𝜋n-m are obtained.
These values of 𝜋’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
π1 = ∅ (π2, π3, π4 ...πn – m)
π2 = ∅ (π1, π3, π4 ...πn – m)
Method of Selecting Repeating Variables:
The number of repeating variables are equal to the number of fundamental
dimensions of the problem. The choice of repeating variables is governed by the
following considerations:
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. Problem 5: The resistance r experienced by a partially submerged body depends
upon the velocity v, length of the body l, viscosity of the fluid µ, density of the
fluid ρ and gravitational acceleration g. obtain a dimensionless expression for r.
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 𝜋1 ,𝜋2 ,𝜋3 are formed.
The equation may be written as: f1 (𝜋1 ,𝜋2 ,𝜋3) = 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
π1 = la1, Vb1, ρc1 R
π2 = la2, Vb2, ρc2 µ
π3 = la3, Vb3, ρc3 g
22. Step 4: Each π-term is solved by the principle of dimensional homogeneity, as follows:
π1 – term:
M0L0T0 = La1, (LT-1)b1, (ML-3)c1 (MLT-2)
Equating the exponents of M, L and T respectively, we get:
For M: 0 = c1 + 1
For L: 0 = a1 + b1 – 3 c1 + 1
For T: 0 = – b1 – 2
∴ c1 = -1; b1 = -2; a1 = -2
Substituting the a1, b1 & c1 values in π1 equation, we get
π1 = l-2, V-2, ρ-1 R
π1 =
𝑅
𝑙2
𝑉2
𝜌
23. π2 – term:
M0L0T0 = La2, (LT-1)b2, (ML-3)c2 (ML-1T-1)
Equating the exponents of M, L and T respectively, we get:
For M: 0 = c2 + 1
For L: 0 = a2 + b2 – 3 c2 - 1
For T: 0 = – b1 – 1
∴ c2 = -1; b2 = -1; a2 = -1
Substituting the values of a2, b2, c2 in π2, we get,
Π2 =
µ
𝑙 𝑉 𝜌
24. π3 – term:
M0L0T0 = La3, (LT-1)b3, (ML-3)c3 (LT-2)
Equating the exponents of M, L and T respectively, we get:
For M: 0 = c3
For L: 0 = a3 + b3 – 3 c3 + 1
For T: 0 = – b3 – 2
∴ c3 = 0; b2 = -2; a3 = 1
Substituting the values of a3, b3, c3 in π3, we get,
Π3 =
𝑙 g
𝑉2
26. Problem 6: Using buckingham’s π-theorem, show that the velocity through a circular orifice is
given by v = 𝟐𝒈𝑯∅
𝑫
𝑯
µ
𝜌𝑽𝑯
where, h = head causing flow, d = diameter of the orifice,
µ = co-efficient of viscosity, ρ = mass density, and g = acceleration due to gravity
Solution:
V = f (g, H, D, µ, 𝜌)
f1 (V, g, H, D, µ, 𝜌) = 0 ----------------- 1
n - no of variables = 6; m – fundamental units = 3
π = (n − m) = (6 - 3) = 3
Equation 1 can also be written as
f1 (π1, π2, π3) = 0 ----------------- 2
Selecting Repeating variables
Geometric property - H, Flow Property - g, Fluid property - 𝜌
27. 𝛑 – term: π1 = Ha1, gb1, 𝜌c1 D
π2 = Ha2, gb2, 𝜌c2 V
π3 = Ha3, gb3, 𝜌c3 µ
𝛑1 – term:
π1 = Ha1, gb1, 𝜌c1 D
M0L0T0 = (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
π1 = H-1, g0, 𝜌0 D
𝛑1 =
𝑫
𝑯
28. 𝛑2 – term:
π2 = Ha2, gb2, 𝜌c2V
M0L0T0 = (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 = -2b2 -1
b2 = -1/2 ; c2 = 0; a2 = -1/2
π2 = H-1/2, g-1/2, 𝜌0 V
π2 =
𝑉
𝑔𝐻
30. Multiply and divide the π3 - term with V
π3 =
µ V
𝜌𝐻𝑉 𝑔𝐻
=
µ
𝜌𝐻𝑉
π2
f1 (π1, π2, π3 ) = 0
f1 (
𝐷
𝐻
,
𝑉
𝑔𝐻,
µ
𝜌𝐻 𝑔𝐻
) = 0
𝑉
𝑔𝐻
= f1 (
𝐷
𝐻
,
µ
𝜌𝐻𝑉
π2)
V = 2𝑔𝐻 ∅ (
𝐷
𝐻
,
µ
𝜌𝐻𝑉
π2)
Multiply by a constant does not change the character of π - terms.
V = 𝟐𝒈𝑯 ∅
𝑫
𝑯
µ
𝜌𝑽𝑯
31. Problem 7: Using buckingham’s theorem, show that the discharge q consumed by an oil ring is
given by q = nd3∅
𝝈
𝜌𝑁2
𝑑3
µ
𝜌𝑁𝑑2 ,
𝒘
𝜌𝑁2
𝑑
where d is the internal diameter of the ring, n is rotational
speed, 𝜌 is density, µ is viscosity, 𝝈 is surface tension and w is the specific weight of oil.
Given:
d, N, 𝜌, µ, 𝝈, w, Q
Solution:
Q = f(d, N, 𝜌, µ, 𝝈, w)
or
f1(Q, d, N, 𝜌, µ, 𝝈, 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
f1 (π1, π2, π3, π4) = 0 --------------------------- 2
Selecting Repeating variables:
Geometric property - d,
Flow property - N,
Fluid property - 𝜌
π –terms:
π1 = da1, Nb1, 𝜌c1 𝝈
π2 = da2, Nb2, 𝜌c2 Q
π3 = da3, Nb3, 𝜌c3 w
π4 = da4, Nb4, 𝜌c4 µ
33. Each π-term is solved by the principle of dimensional homogeneity, as follows:
π1 – term:
π1 = da1, Nb1, 𝜌c1 𝝈
M0L0T0 = La1, (T-1)b1, (ML-3)c1 (MT-2)
Equating the exponents of M, L and T respectively, we get:
M – 0 = c1+ 1 => c1 = -1
L – 0 = a1 -3c1 => a1 +3 => a1 = -3
T – 0 = -b1 -2 => b1 = -2
π1 = d-3, N-2, 𝜌−1 𝝈
π1 =
𝝈
d3 N2𝜌
34. π2 – term:
π2 = da2, Nb2, 𝜌c2 Q
M0L0T0 = La2, (T-1)b2, (ML-3)c2 (L3T-1)
Equating the exponents of M, L and T respectively, we get:
M – 0 = c2
L – 0 = a2 -3c2 +3 => a2 = -3
T – 0 = -b2 -1 = b2 = -1
π2 = d-3, N-1, 𝜌0 V
π2 =
Q
d3 N
35. π3 – term:
π3 = da3, Nb3, 𝜌c3 w
M0L0T0 = La3, (T-1)b3, (ML-3)c3 (ML-2T-2)
Equating the exponents of M, L and T respectively, we get:
M – 0 = c3 + 1 = c3 = -1
L – 0 = a3 -3c3 -2 => a3 +3 -2 => a3 = -1
T – 0 = -b3 -2 = b3 = -2
π3 = d-1, N-2, 𝜌−1 w
π3 =
w
𝜌 d N2
36. π4 – term:
π4 = da4, Nb4, 𝜌c4 µ
M0L0T0 = La3, (T-1)b3, (ML-3)c3 (ML-1T -1)
Equating the exponents of M, L and T respectively, we get:
M – 0 = c4 + 1 = c4 = -1
L – 0 = a4 -3c4 -1 => a4 +3 -1 => a4 = -2
T – 0 = -b4 -1 = b4 = -1
π4 = d-2, N-1, 𝜌−1 µ
π4 =
µ
𝜌 d2 N
37. f1 (π1, π2, π3 ) = 0
Substituting π - values in the above eqn.
f1 (
𝝈
d3 N2𝜌
,
Q
d3 N,
w
𝜌 d N2,
µ
𝜌 d2 N
) = 0
Q
d3 N
= ∅ (
𝝈
d3 N2𝜌
,
w
𝜌 d N2,
µ
𝜌 d2 N
)
Q = Nd3∅
𝝈
𝜌𝑁2
𝑑3
µ
𝜌𝑁𝑑2 ,
𝒘
𝜌𝑁2
𝑑
