First slide

Viscous flow/Laminar flow

Question

Two capillary tubes of the same length but different radii r₁ and r₂ are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is same as before

Moderate

A

$r_{1} + r_{2}$
B

$r_{1}^{2} + r_{2}^{2}$
C

$r_{1}^{4} + r_{2}^{4}$
D

${(r_{1}^{4} + r_{2}^{4})}^{1 / 4}$

Solution

$\begin{array}{l} V = V_{1} + V_{2} \\ \Rightarrow \frac{{πPr}^{4}}{8 ηl} = \frac{{πPr}_{1}^{4}}{8 ηl} + \frac{{πPr}_{2}^{4}}{8 ηl} \Rightarrow r^{4} = r_{1}^{4} + r_{2}^{4} \\ ∴ r = {(r_{1}^{4} + r_{2}^{4})}^{1 / 4} \end{array}$

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