Question

Two capillary tubes of bore diameter but of lengths ${l_1}$ and ${l_2}$​ are fitted side by side to the bottom of a vessel containing water. The length of a single tube that can replace the two tubes such that the rate of steady flow through this tube equals the combined rate of flow through the two tubes will be ______

Answer 1

In this solution, we will use Poiseuille’s Law. This law helps us in relating the flow rate in a tube with a pressure difference and resistance and with a resistance $R$ dependent on the viscosity of the liquid in the tube
Formula used: In this solution, we will use the following formula:
Resistance in a tube: $R = \dfrac{{8\eta l}}{{\pi {r^4}}}$ where $\eta$ is the viscosity of the liquid, $l$ is the length of the tube, $r$ is the radius of the tube.

Complete step by step answer:
We’ve been given that two capillary tubes of bore diameter but of lengths ${l_1}$ and ${l_2}$​ are fitted side by side to the bottom of a vessel containing water and we want to find the length of a single tube of the same radius such that the rate of flow remains constant.
The resistance experienced by water when flowing in a tube is calculated by Poiseuille’s Law as
$R = \dfrac{{8\eta l}}{{\pi {r^4}}}$
To experience the same amount of resistance, the net resistance of the two tubes must be equal to the resistance of the new tube. Since the two tubes are placed side by side, the net resistance of the tubes can be calculated as
$\dfrac{1}{{{R_{net}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
$\Rightarrow \dfrac{1}{{{R_{net}}}} = \dfrac{1}{{\dfrac{{8\eta {l_1}}}{{\pi {r^4}}}}} + \dfrac{1}{{\dfrac{{8\eta {l_2}}}{{\pi {r^4}}}}}$
This net resistance must be equal to the resistance of the newer tube and we can calculate it as
${R_{new}} = {R_{net}}$
$\Rightarrow \dfrac{1}{{{R_{new}}}} = \dfrac{1}{{{R_{net}}}}$
So now we can write
$\dfrac{1}{{\dfrac{{8\eta {l_{new}}}}{{\pi {r^4}}}}} = \dfrac{1}{{\dfrac{{8\eta {l_1}}}{{\pi {r^4}}}}} + \dfrac{1}{{\dfrac{{8\eta {l_2}}}{{\pi {r^4}}}}}$
Which can be simplified to
$\dfrac{1}{{{l_{new}}}} = \dfrac{1}{{{l_1}}} + \dfrac{1}{{{l_2}}}$
$\therefore {l_{new}} = \dfrac{{{l_1}{l_2}}}{{{l_1} + {l_2}}}$

Note: Here we have assumed that the pressure difference across the two ends of the two tubes will remain constant in the longer new tube as well. The combination of two tubes can also be remembered as the combination of two electrical resistances in parallel since they have the same relations and the resistance in the flow can be determined using Poiseuille’s Law.