Dimensional Analysis: Buckingham pie Method: problems
Problem-01:
Deduce the general form of the expression for drag force, Fd on a solid sphere of
diameter D which is held fixed in a fluid stream that moves at uniform speed U.
Begin with the assumption that Fd is some function of U, D,μ, and ρ. Where μ and
ρ are fluid viscosity and density respectively. Using this assumption, show that Fd
can be expressed in the form:
FD= ρU2D2 ϕ(Re)
Re= Reynold's number=
Solution:
Lets solve this problem using Buckingham pie method.
FD=f(U,D,μ,ρ)
=> f1(FD,U,D,μ,ρ)=0
----------------------Eqn(1)
Total number of variables, n =5
Dimensions:
FD = [M1L1T-2]
U= [L1T-1]
D= [L1]
μ= [M1L-1T-1]
ρ =[M1L-3]
Number of fundamental dimensions, m= 3
Number of π terms= n-m=5-3=2
f1(π1,π2)=0
Choose D, U, ρ as repeating variables.
Therefore;
π1=Da1Ub1ρc1FD ----------------------Eqn(2)
Substituting dimensions on both sides:
[M0 L0 T0 ]=[ L1 ]a1 [ L1 T-1 ]b1 [M1 L-3 ]c1 [M1 L1 T-2 ]
Equating the powers of M,L,T on both sides:
Power of M:
0= c1+1
=> c1=-1
Power of T:
0=-b1-2
=> b1=-2
Power of L:
0= a1+b1-3c1+1
=>0=a1-2-3(-1)+1= a1+2
=> a1=-2
Therefore Eqn(2) becomes:
π1=D-2U-2ρ-1FD
Rearranging ..
𝐹
𝜋 =
𝜌𝑈 𝐷
π2=Da2Ub2ρc2 μ ----------------------Eqn(3)
Substituting dimensions on both sides:
[M0 L0 T0 ]=[ L1 ]a1 [ L1 T-1 ]b1 [M1 L-3 ]c1 [M1 L-1 T-1 ]
Equating the powers of M,L,T on both sides:
Power of M:
0= c2+1
=> c2=-1
Power of T:
0=-b2-1
=> b2=-1
Power of L:
0= a2+b2-3c2-1
Put c2=-1 and b2=-1, we get:
=>0= a2+1
=> a2=-1
Therefore Eqn(3) becomes:
π2=D-1U-1ρ-1μ
Rearranging ..
𝜇
1
𝜋2 =
=
𝜌𝑈 𝐷 𝑅
f1(𝜋 , 𝜋 )=0
f 1(
𝐹𝐷
𝜌𝑈2 𝐷2
,
)=0
Or
𝐹
f1( 2𝐷 2 , 𝑅 )=0
𝜌𝑈 𝐷
𝐹
= 𝜙(𝑹𝐞)
𝜌𝑈 𝐷
=> FD= ρU2D2 ϕ(Re) Hence Proved!
Any Doubts? Please watch the video lecture on this from below link:
https://youtu.be/peQj9mXvJ0w
Still have doubts? Please comment your doubt in the video
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