Dimensional analysis

Buckinghams II method

Let x1 , x2, x3, x4 ,xn be thenvariables

Let x1 be the dependent variable

And x2, x3, x4,xn are the independent variables

Let the no of variables involved in a problem is m’

Let the variable having ‘n’ no of fundamental dimensions

According to buckinghamII method

Variables are rearranged as (m-n) dimensionless terms

Let the dimensionless terms are II1, II2,II3, …II(m-n)

II(m-n)= f(repeating and non repeating variables)

Examples:

Calculating the frictional pressure drop in a long circular pipe

Variables

Pressuredrop P

Length L

Diameter D

Height of roughness e

Density of fluid ſ

Viscosity u

Velocity v

P f(L,D,e,d’, u’,u)

Total no of variables 7

No of fundamental variables = 3

Dimensionless terms = 7-3 =4

Repeating variables are

Length , velocity density

II 1 = [L]a1[v]b1 [ſ] c1 P ---à1

II 2 = [L]a2[v]b2 [ſ] c2 e ---à2

II 3 = [L]a3[v]b3 [ſ] c3 D ---à3

II 4 = [L]a4[v]b4 [ſ] c4 u ---à4

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s.no

quantity

units

dimensions

1

pressuredrop

Kg/Nsec2

M L -1T-2

2

length

M

L

3

velocity

m/sec

LT-1

4

diameter

M

L

5

density

Kg/m3

M L-3

6

viscosity

Kg/msec

M L -1T-1

7

roughness

m

L

From eq 1

M0 L 0T0 = [L] a1[LT-1]b1 [ML-3]c1 [M L -1T -2]

Equating the powers of M terms on both sides

0 = c1+1 ----à5

C1 = -1

Equating the powers of L terms on both sides

0= a1+b1-3c1-1-à6

Eqauting powers of T terms on both sides

0 = -b1 -2 --à7

;; b1= -2

Substitute c1 and b1 values in eq 6

: a1 -2 -3(-1)-1=0

; a1= 0

Substituting values of a1 b1 and c1 in eq 1

II 1 = [L ] 0 [u ]-2 [ſ ] -1 P

II 1 = P/u2ſ

Solve in the same way for II2 , II3, II4, and analyse your capability

On solving you ‘ll get

II2 = e/L

II3 = D/L

II4 = u/LVſ

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Rayleigh’s method

Let Q1 , Q2 , Q3 – Qn are n no of variables ,

Let Q1 is the dependent variable and remaining are independent variables

According to rayleighs method

Q1 = f(Q2, Q3 , Q4,--Qn)

Q1 = k [Q2]a [Q3]b [Q4]c -----[Qn]z

with the same example

P = f( L, D, v,u , ſ)

P = [D] a[v]b[ ſ]c [u] d[L]e

s.no

quantity

units

dimensions

1

pressuredrop

Kg/Nsec2

M L -1T-2

2

length

M

L

3

velocity

m/sec

LT-1

4

diameter

M

L

5

density

Kg/m3

M L-3

6

viscosity

Kg/msec

M L -1T-1

7

roughness

m

L

M L -1T-2 = [L]a [LT-1]b [ML-3]c [M L-1 T-1]d [L]e

Equating powers of M

1= c+d--à1

Equating powers of L

-1 = a+b-3c-d+e-à2

Equating powers of T

-2 = -b-d-à3

From eq 1

;;C= 1-d -à4

;; b= 2-d-à5

Substitute eq 4 & 5 in eq 2

-1 = a+(2+d)-3(1-d)+e-d

;; a+2-d-3+3d-d+e

;;a +d+e =0

‘’ a= -(d+e)

P = k [D]-(d+e) [v]2-d [ſ]1-d [L]e [u]d

P/ ſv2 = K[ Dvſ/ u]-d [ L/D]e

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