Answer:
[tex]\Delta P / pv^{2} = f ( L/ d , vpd / \mu)[/tex]
Explanation:
Buckingham pi theorem is used to find the relationship between variables in a physical phenomenon.
First, we need to identify the variables involved in the physical problem of pressure loss (ΔP). These are: d, L, p, μ and v
hence, we have
f(ΔP, d, L, p, μ,v) = 0
Counting the number of variables n, (ΔP, d, L, p, μ,v) we have them = 6
The number of fundamental dimensions (Mass, Length, Time), m = 3 (That is, [M], [L], [T])
By Buckingham's pi theorem,
Number of dimensionless [tex](\Pi)[/tex] groups = number of variables - number of fundamental dimensions.
∴No. of dimensionless groups = n -m = 6-3 = 3
To apply the Buckingham Pi theorem, the reoccurring set must contain three variables that cannot themselves be formed into a dimensionless groups.
For this problem we have the following variables chosen as the recurring set : d, v and p.
The others were left out for the following reasons:
By dimensional analysis,
The dimensions of these variables are :
d = [1]
v = [[tex]LT^{-1}[/tex]]
p = [[tex]ML^{-3}[/tex]]
we can form dimensionless groups from each of the remaining variables ΔP, L and μ in turn
Dimensions of ΔP
Dimensions of [tex]\Delta P =[ML ^{-1}T^{-2}][/tex]
Hence we have that [tex]\Delta P M^{-1} L T^{2}[/tex] is a dimensionless quantity
Plugging in the dimensions of each of the variables, we obtain the first Pi group as
[tex]\Pi _{1} = \Delta P (pd^{3})^{-1} (d) (dv^{-l})^{2}[/tex]
[tex]= \Delta P/pv^{2}[/tex]
Dimensions of L
L has the dimensions of [L]
to make it dimensionless, we will have to multiply it by the inverse [tex][L^{-1}][/tex]
[tex]L[L ]^{-1}[/tex] is therefore dimensionless
The second Pi group we have is
[tex]\Pi_{2} = L/d[/tex]
Dimensions of μ
μ has the dimensions of [tex][M L^{-1} L^{-1}][/tex]
[tex]\mu [M^{-1}LT][/tex] is therefore dimensionless
substituting in the dimensions of each variable, we have the third dimensionless group as
[tex]\Pi_{3} = \mu(pd^{3})^{-1} (d) (dv^{-1})^{2}[/tex]
[tex]\Pi_{3}= \mu /dvp[/tex]
Thus we have [tex]f ( \Delta P/pv^{2} , L/d, \mu/vpd)[/tex]
∴ A relationship for the pressure loss is [tex]\Delta P / pv^{2} = f ( L/ d , vpd / \mu)[/tex]
