Question

If the rate of flow of liquid through horizontal pipe of length l & radius R is Q;What is rate of flow of liquid if length & radius of tube is doubled?

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Answer 1

Answer:

• 8 times the original value

Steps:

This is a question based on Poiseuille's theorem.

According to him, the rate of flow of a particular volume is given by the formula:

$\boxed{ {Q} = \dfrac{(P - P_o) \pi r^4}{8 \mu h}}$

where,

'Q' refers to the rate of flow of volume, (P - P₀) refers to Pressure difference,

'μ' refers to viscosity of the liquid and 'h' refers to the height (or) length of the pipe.

Let us first calculate 'Q' with the initial parameters.

$\implies Q = \dfrac{\Delta P \pi R^4}{8 \mu l} \hspace{20} ...(i)$

Now let us apply the new parameters in (i). Hence we get:

$\implies Q' = \dfrac{ \Delta P \pi (2R)^4}{8\mu (2l)}\\\\\\\implies Q' = \dfrac{16\:(\Delta P \pi R^4)}{ 2 (8 \mu l )}$

On simplifying we get:

$\implies Q' = 8\:(\dfrac{\Delta P \pi R^4}{8 \mu l })\\\\\\\implies \boxed{\textbf{Q' = 8 (Q)}}$

Hence on doubling the radius and the length of the tube, the rate of flow of liquid becomes 8 times the original value.