Explain Poiseuille's Equation - QS Study
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Poiseuillie investigated the steady flow of a liquid through a capillary tube. He derived an expression for the volume of the liquid flowing per second through the tube.

Consider a liquid of co-efficient of viscosity η flowing, steadily through a horizontal capillary tube of length / and radius r. If P is the pressure difference across the ends of the tube, then the volume V of the liquid flowing per second through the tube depends on n. r and  the pressure gradient (P/∫).

But, V ∞ ηx ry (P/l)z

So, V = K ηx ry (P/l)z …. …. (1)

Where k is a constant of proportionality. Rewriting the  equation (1) in terms of dimensions,

[L3T-1] = [ML-1T-1]x

Equating the powers of L. M and T on both sides we get x = -1, y = 4 and z= 1

Substituting in equation (1),

V = K η-1 r4 (P/l)1

V = kPr4/η∫

Experimentally k is found to be equal to π/8

So, V = πPr4 / 8η∫

It is known as Poiseuille’s equation.