Corrections To Poiseuille’s Equation:

Let us consider a horizontal tube of radius R and length l, If P is the pressure difference between the two ends of the tube then the total volume that passes through the tube in unit time is given by,

\( \displaystyle{V=\frac{\pi{P}R^4}{8\eta{l}}} \)

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Where \( \eta \) is the coefficient of viscosity of the liquid.

In deducing Poiseuille’s equation the following two factors have not been considered and therefore two corrections are necessary.

(i) Length:

As the pressure at the inlet end of the tube is high, so when the liquid enters the tube its motion is accelerated at the inlet end and as a result, the velocity distribution becomes uniform and streamline only when the liquid has travelled an appreciable length of the flow tube. To correct this, a factor \( 1.64R \) should be added to the length \( l \) of the tube. So the effective length of the flow tube should be \( (l+1.64R) \), where \( R \) is the radius of the tube.

(ii) Pressure:

When the liquid leaves the tube it has some velocity, hence the entire pressure difference \( P \) is not used up to overcome the viscous force. A part of it is spent in imparting kinetic energy of the liquid. Let \( p \) be the effective pressure difference that overcomes viscosity.

Now the work done against the viscous force in one second is \( pV \), the kinetic energy given to the liquid of density \( \rho \) per second is

\( \displaystyle\int_{0}^{R}\frac{1}{2}\rho(2\pi{r}\,dr{v})v^2 =\displaystyle\int_{0}^{R}\rho\pi{v}^3r\,dr \\=\displaystyle\int_0^{R}\rho\pi{\left(\frac{p}{4l\eta}\right)}^3(R^2-r^2)r\,dr \\= \displaystyle{\rho\pi{\left(\frac{p}{4l\eta}\right)}^3\frac{R^8}{8}} \)

Where \( v=\frac{p}{4\eta{l}}\left(R^2-r^2\right) \)

\( \displaystyle\int_0^{R}{\left(R^2-r^2\right)}^3\,dr \\= \displaystyle\int_0^{R}\left(R^6{r}-3R^4{r^3}+3R^2{r}^5-R^7\right)\,dr \\=R^6\frac{R^2}{2}-3R^4\frac{R^4}{4}+3R^2\frac{R^6}{6}-\frac{R^8}{8} \\=R^8\left(\frac{1}{2}-\frac{3}{4}+\frac{1}{2}-\frac{1}{8}\right) \\=\frac{R^8}{8} \)

Therefore the total loss of energy per second is,

\( pV+ \rho\pi{\left(\frac{p}{4l\eta}\right)}^3\frac{R^8}{8} \)

which is equal to \( PV \).

\( pV+ \rho\pi{\left(\frac{p}{4l\eta}\right)}^3\frac{R^8}{8} = PV \)

or, \( pV+\frac{\rho}{{\pi}^2{R}^4}{\left(\frac{\pi{p}R^4}{8\eta{l}}\right)}^3=PV \)

[where, \( V= \left(\frac{\pi{p}R^4}{8\eta{l}}\right) \)]

or, \( pV+\frac{\rho}{{\pi}^2{R}^4}V^3=PV \)

or, \( p=P-\frac{\rho{V}^2}{{\pi}^2R^4} \)

It is however only approximately true and the correction should be,

\( p=P-\frac{k\rho{V}^2}{{\pi}^2R^4} \),

where \( k \) is the constant whose value depends upon the form of the apparatus.

Now,

\( \displaystyle{\eta=\frac{\pi{R^4}\left( P-\frac{k\rho{V}^2}{{\pi}^2R^4}\right)}{8V\left(l+1.64R\right)}} \)