Question

Assuming that the critical velocity v_{c} of a viscous liquid flowing through a capillary tube depends only upon the radius r of the tube, density ρ and the coefficient of viscosity η of the liquid, find the expression for critical velocity.

Answer 1

Dear Student,

◆ Answer -

v = k.η/(r.ρ)

◆ Explaination -

First let's write down known dimensions.

v = [L1T-1]

r = [L1]

ρ = [L-3M1]

η = [L-1M1T-1]

Suppose critical velocity of viscous fluid is given by -

v = k.r^x.ρ^y.η^z

In dimensional form,

[v] = [r]^x.[ρ]^y. [η]^z

[L1T-1] = [L1]^x.[L-3M1]^y.[L-1M1T-1]^z

[L1T-1] = [L^(x-3y-z).M^(y+z).T^(-z)]

Comparing two sides,

x-3y-z = 1

y+z = 0

-z = -1

Solving we get,

x = -1

y = -1

z = 1

Substitute this in initial formula -

v = k.r^-1.ρ^-1.η^1

v = k.η/(r.ρ)

Hope this helps you...