Poiseuille’s formula for the volume ‘V’ of a liquid of density ‘ρ’ flowing in time ‘t’ through a capillary tube of length ‘L’ and radius ‘r’ under a pressure difference ‘P’ between the ends of the tube is

V = (πPr^{4}t)/ (8Lη) where ‘η’ is the coefficient of viscosity of the liquid. [You should note that this formula holds good only if the flow is slow and steady (stream-lined)].

The rate of flow (volume flowing per second) ‘Q’ is given by

Q = (πPr^{4})/ (8Lη)

Here P = hρg where ‘h’ is the height of the liquid column which produces the pressure difference so that

Q = (π hρg r^{4})/ (8Lη)

You will often find questions based on Poiseuille’s formula in entrance tests for admitting students to various courses. Consider the following question:

**A capillary tube of length ‘L’ and radius ‘r’ is joined to another capillary tube of length L/4 and radius r/2 A liquid flows through this series combination. If the pressure difference between the ends of the first tube is P, that between the ends of the second tube is**

**(a) P (b) 4P (c) 8P (d) 16P (e) P/4 **

Since the tubes are in series, the rates of flow through the tubes are equal so that from Poiseuille’s equation

πPr^{4}/8Lη = πP'(r/2)^{4}/[8(L/2)η], where P’ is the pressure difference between the ends of the second tube.

From this P' = 16P/2 = 8P.

Now, consider the following MCQ:

**Under a constant pressure head, the volume of a liquid flowing per second through a capillary tube of radius 1 mm and length 16 cm is 4 cm ^{3}. If another tube of radius 0.5 mm and length 8 cm is connected in series with it and the same pressure head is applied across the combination, the volume of liquid flowing per second will be ( in cm^{3})**

**(a) 5/3 (b) 5/6 (c) 4/3 (d) 4/5 (e) 4/9**

When tubes are connected in series, the net rate of flow (Q_{net}) under a given pressure head is given by the reciprocal relation,

1/Q_{net} = 1/Q_{1} + 1/Q_{2} + 1/Q_{3} +…..etc. where Q_{1}, Q_{2}, Q_{3}.....etc. are the individual rates of flow when the tubes are connected separately to the same pressure heaed.

In the present case, since there are two tubes only, 1/Q_{net} = 1/Q_{1} + 1/Q_{2} so that

Q_{net} = Q_{1}Q_{2}/(Q_{1}+Q_{2}). Here Q_{1} = 4 cm^{3}.

Since the rate of flow through a tube is given by Q = (πPr^{4})/ (8Lη), we have Q α r^{4}/L.

Since the radius of the second tube is half that of the first, the rate of flow is *reduced* to (½)^{4} = 1/16 of that through the first tube on account of this. Since the length of the second tube is half that of the first, the rate of flow is* increased* to twice that through the first tube on this account. Therefore, rate of flow through the second tube is given by

Q_{2 }= Q_{1}×(2/16) = Q_{1}/8 = 4/8 cm^{3} = 0.5 cm^{3}.

The net rate of flow is therefore given by Q_{net} = (4×0.5)/(4+0.5) = 4/9 cm^{3} [Option (e)].

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