Respuesta :
Answer:
[tex]\frac{E}{pc^{2} } =f\frac{h}{D}[/tex]
Explanation:
Using the Buckingham pi theorem and following the series of steps below;
STEP 1
List all the variables involved:
Velocity = c
Artery diameter = D
Wall thickness = h
Density = ρ
Modulus of elasticity = E
(c, D, h, ρ, E) = 5 variables
STEP 2
Express each of the above variables in their basic physical dimension (MLT)
c = L[tex]T^{-1}[/tex]
D = L
h = L
ρ = M[tex]L^{-3}[/tex]
E = M[tex]L^{-1} T^{-2}[/tex]
STEP 3
Determine the required number of pi (π) terms using the below formula;
π = n - m
where n = number of dimensional variables = 5
m = number of dimensions the variables are expressed in = 3 (M,L,T)
Therefore, π = 5-3 = 2; π = 2
STEP 4
Select a number of repeating variables, where the number required is equal to the number of dimensions used i.e. 3. The remaining two variables are the non repeating variables.
Selected repeating variables are c, D, ρ
non repeating variables are h and E
STEP 5
Form a pi term by multiplying one non repeating variable by the product of the repeating variables, each (repeating variables) raised to an exponent that will make the expression dimensionless.
π (1) = h [tex]D^{a} c^{b} p^{c}[/tex] ...............................................(i)
express this pi terms in their fundamental dimension and equate it to [tex]M^{0} L^{0} T^{0}[/tex]
[tex]M^{0} L^{0} T^{0}[/tex] = [tex]L (L)^{a} (LT^{-1} )^{b} (ML^{-3} )^{c}[/tex]
[tex]M^{0} L^{0} T^{0} = M^{c} L^{1+a+b-3c} T^{-b}[/tex]
Equate the powers
[tex]c=0\\b=0\\1+a+b-3c =0\\a = -1[/tex]
substitute c and b into the third expression to find a
a = -1, b = 0, c = 0
substitute the values for a, b and c in (i)
π (1) = [tex]hD^{-1}[/tex]
π (1) = [tex]\frac{h}{D}[/tex]
STEP 6
Repeat step 5 for the second non repeating variable E
π (2) = [tex]ED^{a} c^{b} p^{c}[/tex] .......................................................... (ii)
express in fundamental dimensions
[tex]M^{0} L^{0} T^{0} =(ML^{-1}T^{-2}) (L)^{a}(LT^{-1})^{b}(ML^{-3} )^{c}[/tex]
[tex]M^{0} L^{0} T^{0} = M^{1+c} L^{-1+a+b-3c} T^{-2-b}[/tex]
compare the powers
[tex]1+c=0\\-1+a+b-3c=0\\-2-b=0[/tex]
using elimination method,
a = 0, b = -2, c = -1
substituting the values of a, b and c in (ii)
π (2) = [tex]ED^{0} c^{-2} p^{-1}[/tex] = [tex]\frac{E}{pc^{2} }[/tex]
π (2) = [tex]\frac{E}{pc^{2} }[/tex]
STEP 7
Check if the pi terms are dimensionless
π (1) = [tex]\frac{h}{D}=\frac{L}{L}=M^{0} L^{0} T^{0}[/tex]
π (2) = [tex]\frac{E}{pc^{2} } =\frac{ML^{-1} T^{-2} }{ML^{-3} (LT^{-1} )^{2} } =M^{0} L^{0} T^{0}[/tex]
STEP 8
Express the final equation as a relationship between the pi terms in the form,
π([tex]1[/tex]) = [tex]f[/tex]π([tex]2[/tex])
[tex]\frac{h}{D}=f\frac{E}{pc^{2} }\\ or\\\frac{E}{pc^{2} }=f\frac{h}{D}[/tex]
