Question

Two capillary tubes of the same length but different radii \[{r_1}\] and \[{r_2}\] 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 the same as before?
A. \[{r_1} + {r_2}\]
B. \[r_1^2 + r_2^2\]
C. \[r_1^4 + r_2^4\]
D. \[{\left( {r_1^4 + r_2^4} \right)^{\dfrac{1}{4}}}\]

Answer 1

Hint:The single tube can be chosen such that the rate of flow is same as the configuration of two different tubes connected in parallel keeping the pressure head constant in both the cases as well as the length of the tubes.

Formula used:
Hagen-Poiseuille equation is,
\[V = \dfrac{{\pi p{r^4}}}{{8\eta l}}\]
where V is the rate of flow of fluid whose coefficient of viscosity \[\eta \], p is the pressure difference across the end of the tube with radius r and length l.

Complete step by step solution:
For the initial case, the radius of the first tube is \[{r_1}\] and the length of the tube is l. If the viscosity of the fluid is \[\eta \], then the rate of flow through first tube is,
\[{V_1} = \dfrac{{\pi \Pr _1^4}}{{8\eta l}}\]

The radius of the second tube is \[{r_2}\]. As both the tubes are connected in parallel across the same pressure head, so the length of the second tube will be equal to the first tube, i.e. l.
If the viscosity of the fluid is \[\eta \], then the rate of flow through second tube is,
\[{V_2} = \dfrac{{\pi \Pr _2^4}}{{8\eta l}}\]

Then the total amount of fluid flowing through the two tubes per unit time, i.e. the total flow rate is;
\[V = {V_1} + {V_2}\]
\[\Rightarrow V = \dfrac{{\pi \Pr _1^4}}{{8\eta l}} + \dfrac{{\pi \Pr _2^4}}{{8\eta l}}\]
\[\Rightarrow V = \dfrac{{\pi {\mathop{\rm P}\nolimits} }}{{8\eta l}}\left( {r_1^4 + r_1^4} \right)\]

For the final case, let the required radius of the tube is r so that the flow rate through the length l of the same fluid is same as before.
\[V = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
As the flow rate in final configuration is same as before,
\[\dfrac{{\pi {{\Pr }^4}}}{{8\eta l}} = \dfrac{{\pi {\mathop{\rm P}\nolimits} }}{{8\eta l}}\left( {r_1^4 + r_2^4} \right) \\ \]
\[\Rightarrow {r^4} = \left( {r_1^4 + r_2^4} \right) \\ \]
\[\therefore r = {\left( {r_1^4 + r_2^4} \right)^{\dfrac{1}{4}}}\]
So, to get the flow rate as before, the radius of the single tube should be \[{\left( {r_1^4 + r_2^4} \right)^{\dfrac{1}{4}}}\].

Therefore, the correct option is D.

Note: The tubes connected in parallel are analogous to the resistors connected in parallel connection and the tubes connected in series are analogous to the resistors connected in series. This analogy between the fluid flow rate and electric current flow was given by Hagen-Poiseuille equation