Question

The rate of the steady volume flow of water through a capillary tube of length ' l ' and radius ' r ' under a pressure difference of P is V. This tube is connected with another tube of the same length but half the radius in series. Then find the rate of steady volume flow through them. (The pressure difference across the combination is P)
A. \[\dfrac{V}{{16}} \\ \]
B. \[\dfrac{V}{{17}} \\ \]
C. \[\dfrac{{16V}}{{17}} \\ \]
D. \[\dfrac{{17V}}{{16}}\]

Answer 1

Hint:Poiseuille's theorem states that the velocity of a liquid flowing through a capillary is directly proportional to the pressure of the liquid and the fourth power of the radius of the capillary and is inversely proportional to the viscosity of the liquid and the length of the capillary.

Formula Used:
The poiseuille's formula is,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Where, \[\eta \] is dynamic viscosity, P is pressure, l is length of the pipe and r is radius of the pipe.

Complete step by step solution:
The rate of fluid flow through a pipe of radius r and length l. If the pipe is in contact with another pipe of a radius half the earlier radius. Then we need to find the rate of fluid flow when the pressure difference remains the same. We will consider the fluid resistance concept here. First of all, we know that,
\[V = IR\]
\[ \Rightarrow R = \dfrac{V}{I}\]
If we consider fluid resistance,
\[P = QR\]
That means, if there is a pressure difference then only the rate of fluid flow occurs.

We can write,
\[ \Rightarrow R = \dfrac{P}{Q}\]…. (1)
Here, Q is the poiseuille's formula, that is,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Substitute the value of Q in equation (1), we obtain,
\[R = \dfrac{P}{{\dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}}} \\ \]
\[ \Rightarrow R = \dfrac{{8\eta l}}{{\pi {r^4}}}\]

In the first case,
\[{R_1} = \dfrac{{8\eta l}}{{\pi {r^4}}} \\ \]
\[\Rightarrow {R_1} = R\]
In the second case,
\[{R_2} = \dfrac{{8\eta l}}{{\pi {{\left( {\dfrac{r}{2}} \right)}^4}}} \\ \]
\[\Rightarrow {R_2} = 16 \times \left( {\dfrac{{8\eta l}}{{\pi {r^4}}}} \right) \\ \]
\[\Rightarrow {R_2} = 16R\]
Since the two pipes are in series combination,
\[QR = {Q^1}\left( {{R_1} + {R_2}} \right)\]

Substitute the value of \[{R_1}\] and \[{R_2}\] in above equation, we obtain,
\[QR = {Q^1}\left( {R + 16R} \right) \\ \]
\[\Rightarrow QR = {Q^1}\left( {17R} \right) \\ \]
\[\Rightarrow Q = {Q^1}\left( {17} \right) \\ \]
\[ \Rightarrow {Q^1} = \dfrac{Q}{{17}}\]
If we write this in terms of V then,
\[ \therefore {V^1} = \dfrac{V}{{17}}\]
Therefore, the rate of steady volume flow through them is \[\dfrac{V}{{17}}\].

Hence, option B is the correct answer.

Note: Poiseuille number is a non-dimensional number which characterizes steady, fully-developed, laminar flow of a constant-property fluid through a duct of arbitrary, but constant, cross section.