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The velocity of fluid of viscosity $\eta$ through a capillary of radius $r$ and length $l$ at a distance $x$ from the center of the capillary is given by; $v=\frac{P}{4l \eta }(r^2-x^2)$ (where $P$ is the pressure difference at the two ends of capillary). With the help of this I can find the rate of flow of fluid out of the capillary equal to $\frac {dV_{out}}{dt} = \frac{\pi Pr^4}{8l \eta }$.

But what happens when the capillaries are in series with different radius and length?

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up vote 2 down vote accepted

Assuming the fluids are incompressible, the flow through each capillary must be the same. Also the sum of the pressures across each capillary must equal the total pressure. Therefore, you have the equations:

$P_1+P_2 = P$

$V_1 = \frac{\pi P_1 r_1^4}{8 l_1\eta} = V_2 = \frac{\pi P_2 r_2^4}{8 l_2\eta}$

Solve this system for $P_1$ and $P_2$ then plug back in to find the flow rate in terms of $P, r_1, l_1, r_2, l_2$.

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To add to that: a system of pipes with laminar incompressible flow is (at least mathematically) extremely similar to an electrical circuit. You have a potential (pressure), you have a current (flow) and you have a resistance (hydrodynamic resistance). So all the stuff you learned about resistances in parallel and in series, also holds here and can make your life easy – Michiel Feb 13 '13 at 21:31

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