Question

In a fuel injection system small droplets are formed as the liquid jet breaks up. Assume the droplet diameter, d, is a function of the liquid density, ρ, viscosity, µ, surface tension, ϒ, the jet velocity, V, and the jet diameter D. Derive this relationship in dimensionless form.

Answer 1

droplet diameter in terms of function

d = f (ρ, µ, ϒ, V, D)

From The Buckingham π Theorem

f (d, ρ, µ, ϒ, V, D) = 0

Number of variables n = 6

Number of fundamental dimensions m = 3

Number of π functions = n - m = 6 - 3 = 3

Repeating variables are D, V, ρ

Remaining variables are d, µ, ϒ

π1 = ρ^x1V^y1D^z1d

π2 = ρ^x2V^y2D^z2μ

π3 = ρ^x3V^y3D^​z3ϒ

For π1 by using fundamental dimensions

M⁰L⁰T⁰ = (ML^-3)^x1 (LT^-1)^y1 (L)^z1 (L)

Solve for x1, y1 and z1

M : x1 = 0

L : - 3x1 + y1 + z1 + 1 = 0

z1 = - 1

T : - y1 = 0

π1 = ρ⁰V⁰D^-1d = d/D

Similarly for π2

M⁰L⁰T⁰ = (ML^-3)^x2 (LT^-1)^y2 (L)^z2 (ML^-1T^-1)

Solve for x2, y2 and z2

M : x2 + 1 = 0

x2 = - 1

L : - 3x2 + y2 + z2 - 1 = 0

z2 = - 1

T : - y2 - 1= 0

y2 = - 1

π2 = ρ^-1V^-1D^-1μ = μ/(ρVD)

Similarly for π3

M⁰L⁰T⁰ = (ML^-3)^x3 (LT^-1)^y3 (L)^z3 (MT^-2)

Solve for x3, y3 and z3

M : x3 + 1 = 0

x3 = - 1

L : - 3x3 + y3 + z3 = 0

z3 = - 1

T : - y3 - 2 = 0

y3 = -2

π3 = ρ^-1V^-2D^​-1ϒ = ϒ/(ρV²D^​)

f (π1, π2, π3) = 0

f [d/D, μ/(ρVD), ϒ/(ρV²D^​)] = 0

d/D = μ/(ρVD), ϒ/(ρV²D^​)

d = D f[μ/(ρVD), ϒ/(ρV²D^​)]