BUCKINGHAM'S PI THEOREM

BUCKINGHAM'S PI THEOREM 

• The Buckingham π theorem is a key theorem in dimensional analysis.

Buckingham ' s Pi theorem states that:  

•  If there are n variables in a problem and these variables contain m primary dimensions (for example M, L, T) the equation relating all the variables will have (n-m) dimensionless groups.  
• Buckingham referred to these groups as π groups.  
• The final equation obtained is in the form of :  

                                                           π_l = f(π₂, π₃ ,….. π_n-m )  

• The π groups must be independent of each other and no one group should be formed by multiplying together powers of other groups.  
• This method offers the advantage of being more simple than the method of solving simultaneous equations for obtaining the values of the indices (the exponent values of the variables).  In this method of solving the equation, there are 2 conditions:  

1. Each of the fundamental dimensions must appear in at least one of the m variables  
2. It must not be possible to fom1 a dimensionless group from one of the variables within a recurring set. A recurring set is a group of variables forming a dimensionless group. 

EXAMPLE

The relationship of the effect on efficiency ɳ of the variables D, ω, 𝞺, µ and Q

STEP 1;

ɳ is a function of D, ω, 𝞺, µ and Q

i.e ɳ = f( D, ω, 𝞺, µ, Q)

STEP 2;

finding dependent and independent variables

HERE 

ɳ is independent variable
D, ω, 𝞺, µ and Q are dependent variables

the functional relations between dependent and independent variables is given as

f(ɳ ,D, ω, 𝞺, µ , Q) = 0

STEP 3;

total no.of variables n=6

STEP 4;

dimensions of variables

QUANTITY	UNITS	DIMENSIONS
ɳ	No units	M0 L0T0
D	m	M0 L1T0
ω	Rad/s	M0 L0T-1
𝞺	Kg/m3	M1 L-3 T0
µ	n.s/m2=kg/s.m	M1 L-1T-1
Q	m3/s	M0 L3T-1

STEP 5;

M,L,T are the three primary dimensions used

i.e no.of primary dimentions  m=3

STEP 6;

no.of π groups = n-m = 6-3 =3

the resulting function is 

f(π_{l ,}π₂, π₃ )=0

π_l is geometry property  

π₂ is  flow property

π₃  is fluid property

note π_{l ,}π₂, π₃ are dimensionless

STEP 7;

finding π groups

π_l = D^a ω^b 𝞺^c ɳ

π₂ = D^p ω^q 𝞺^r µ

π₃ = D^x ω^y 𝞺^z Q

                     π_l = D^a ω^b 𝞺^c ɳ 

 using dimensional analysis

                           M⁰ L⁰T⁰= (M⁰ L¹T⁰)^a   (M⁰ L⁰T^-1 )^b  (M¹ L^-3 T⁰ )^c    M⁰ L⁰T⁰

                           M⁰ L⁰T⁰= M^c L^a-3c T^-b

equating power we will get  a=b=c=0

 so                π_l = ɳ 

similarly 

π₂ =  µ / (D² ω 𝞺)

π₃ = Q / (D³ ω)

the resulting function

 f(π_{l ,}π₂, π₃ )=f(  ɳ ,  µ / (D² ω 𝞺) , Q / (D³ ω)  ) = 0

 the efficiency can be expressed in terms of dimensionless groups as 

ɳ =f(  µ / (D² ω 𝞺) , Q / (D³ ω) )