Question

A capillary tube is attached horizontally to a constant pressure head arrangement. If the radius of the capillary tube is increased by $10\%,$ then the rate of flow of the liquid shall change nearly by :

• A. $+10\%$
• B. $46\%$
• C. $-10\%$
• D. $-40\%$

Answer 1

The correct option is B $46\%$

A constant pressure head cannot be maintained in a horizontal tube unless we consider viscosity of the fluid.
Poiseuille's equation can be stated also as $\Delta P=\frac{8\mu LQ}{\pi r^4}$
where:
$\Delta$ P is the pressure loss;
$L$ is the length of pipe
$\mu$ is the dynamic viscosity
$Q$ is the volumetric flow rate
$r$ is the radius
$\Delta$ P , $\mu ,L$ remain constant so $Q\propto$ $R^4$
$⟹Q=K{R}^4$
$R^{\prime }=R+0.1R=1.1R$
$Q^{\prime }=K(1.1R{)}^4$
$\frac{Q^{\prime }-Q}{Q}=\frac{(1.1R{)}^4-R^4}{R^4}=0.4641$
Percentage increase in the rate of flow of the liquid} $=46.41\%$