Name: Introduction to Fluid Mechanics
Uploaded: 2019-07-19
Duration: 1 min 36 s
Description: A fluid can be at rest like water in a glass or in motion like water flowing in a pipe. This video explains fluid mechanics dealing with the behaviour of the...

Question

Derive Poiseuille&rsquo;s formula for the volume of a liquid flowing per second through a pipe under streamlined flow

Accepted Answer

Consider a liquid flowing steadily through a horizontal capillary tube. Let $v = (\frac{v}{T})$ be the volume of the liquid flowing out per second through a capillary tube. It depends on coefficient of viscosity (η) of the liquid, radius of the tube (r), and the pressure gradient $\frac{P}{l}$ . Then
$v α η^{a} r^{b} {(\frac{P}{l})}^{c} v = k η^{a} r^{b} {(\frac{P}{l})}^{c} . . . . . (i)$
Where k is a dimensionless constant.
There fore $[v] = \frac{V o l u m e}{T i m e} = [L^{3} T^{- 1}] : [\frac{d P}{d x}] = \frac{P r e s s u r e}{D i s \tan c e} = [M L^{- 2} T^{- 2}] [η] = [M L^{- 1} T^{- 1}] a n d [r] = [L]$
Substituting in equation (i)
$[L^{3} T^{- 1}] = [M L^{- 1} T^{- 1}]^{a} [L]^{b} [M L^{- 2} T^{- 2}]^{c} M^{0} L^{3} T^{- 1} = M^{q + c} L^{- a + b - 2 c} T^{- a - 2 c}$
So, equating the powers of M, L and T on both sides, we get
$a + c = 0$ , $- a + b - 2 c = 3$ , and $- a - 2 c = - 1$
We have three unknowns a, b and c. We have three equations, on solving, we get $a = - 1, b = 4 a n d c = 1$
Therefore, equation (i) becomes,
$v = k η^{- 1} r^{4} {(\frac{P}{l})}^{1}$
Experimentally, the value of k is shown to be $\frac{π}{8}$ , we have
$v = \frac{π r^{4} P}{8 η l}$
The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity (vC).

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