Question

Two capillary of length $L$ and $2\;L$ and of radius $R$ and $2\;R$ are connected in series. The net rate of flow of fluid through them will be (given rate of the flow through single capillary $X = \pi p{R^4}/8\eta L$)
(A) $\dfrac{8}{9}X$
(B) $\dfrac{9}{8}X$
(C) $\dfrac{5}{7}X$
(D) $\dfrac{7}{5}X$

Answer 1

We will first calculate the net resistance of both the capillary tubes. The resistance of any single capillary is given by $Y = \dfrac{{8\eta l}}{{\pi {r^4}}}$, and since in the problem, both the capillary tubes are in series, the individual resistances will be added. Next, we shall divide this total resistance from the total pressure difference to get the required rate of flow of fluid through them using the formula $Q = \dfrac{{\Delta P}}{{{Y_{net}}}}$.

Formula used:
$Y = \dfrac{{8\eta l}}{{\pi {r^4}}}$ and $Q = \dfrac{{\Delta P}}{{{Y_{net}}}}$

Complete step by step solution:
Here we are required to calculate the volume flow rate of the fluid through the entire capillary.
The quantity $Y = \dfrac{{8\eta l}}{{\pi {r^4}}}$ is the equivalent resistance of the capillary. Here $\eta$ is the coefficient of viscosity, which will be the same here since the fluid flowing through both the capillary tubes is the same. $r$ is the radius of the capillary and $l$ is the length of the capillary tube. The rest are constants.
Here since there are two capillary tubes involved, we will have two values of this resistance.
Now for the first capillary tube of length $L$ and of radius $R$, we will calculate the equivalent resistance ${Y_1}$ from the above formula.
$\therefore {Y_1} = \dfrac{{8\eta L}}{{\pi {R^4}}}$
Now for the first capillary tube of length $2\;L$ and of radius $2\;R$, we will calculate the equivalent resistance ${Y_2}$ from the above formula.
$\therefore {Y_2} = \dfrac{{8\eta (2L)}}{{\pi {{(2R)}^4}}}$
$\Rightarrow {Y_2} = \dfrac{{\eta L}}{{\pi {R^4}}}$
Here we will use the equivalent electrical analogy i.e. the volume flow rate $Q$ is equal to $\dfrac{{\Delta P}}{{{Y_{net}}}}$, where $\Delta P$ is the pressure difference between the two sides of the entire capillary and ${Y_{net}}$ is the net resistance. This equation is similar to the equation $I = \dfrac{{\Delta V}}{R}$, where $I$ is current $\Delta V$ is the voltage difference and $R$ is resistance. Since the two capillary tubes are in series, we will add these two calculated resistances just like any electrical resistances in series.
Therefore ${Y_{net}} = {Y_1} + {Y_2}$.
$\Rightarrow {Y_{net}} = \dfrac{{8\eta L}}{{\pi {R^4}}} + \dfrac{{\eta L}}{{\pi {R^4}}}$
$\Rightarrow {Y_{net}} = \dfrac{{9\eta L}}{{\pi {R^4}}}$
Now the rate of flow is
$Q = \dfrac{{\Delta P}}{{{Y_{net}}}}$.
Here according to the question, the pressure difference between the two ends of the entire capillary setup is $p$. Thus
$Q = \dfrac{p}{{{Y_{net}}}}$.
$\Rightarrow Q = \dfrac{{\pi p{R^4}}}{{9\eta L}}$
Now in terms of
$X = \pi p{R^4}/8\eta L$,
we see that $Q$ becomes $\dfrac{8}{9} \times \dfrac{{\pi p{R^4}}}{{8\eta L}}$.
$\Rightarrow Q = \dfrac{8}{9} \times \dfrac{{\pi p{R^4}}}{{8\eta L}}$
$\Rightarrow Q = \dfrac{8}{9}X$
Thus we see that option (a) is the correct answer.

Note:
The equivalent electrical analogy is required for solving such kinds of problems where capillary tubes of various diameters and lengths are joined together. Here the voltage difference is equivalent to the pressure difference; the current is equivalent to the rate of fluid flow and the electrical resistance $(R = \rho \dfrac{l}{A})$, is equivalent to the quantity $Y = \dfrac{{8\eta l}}{{\pi {r^4}}}$. We then solve the problem as an electrical circuit.