Poiseuille's Relation and its Applications

Poiseuille's relation for the rate of liquid flow

Poseuille made the following assumptions to deduce the rate of flow of liquid through a tube:

- The liquid flow is steady with the streamlines parallel to the axis of the tube.

- The liquid layer in contact with the wall of the tube is stationary and there is no radial flow of the liquid.

- The tube is narrow and horizontal.

Consider a liquid having coefficient of viscosity \(\eta\) be flowing steadily through a narrow horizontal tube of radius \(r\) and length \(l\) such that the liquid pressure \(P\) be uniform throughout the tube. The cylindrical volume of the liquid can be considered to be composed of a large number of co-axial elementary hollow cylinders. Let us take any one of such cylinders. Let us take any one of such cylinders of radius \(x\) and thickness \(dx\). Thus, the viscous drag force acting on this particular layer is \(F_{v}=-\eta A \frac{dv}{dx}---1\).

Here, A is the surface area of the liquid layer of interest, \(\frac{dv}{dx}\) is the velocity gradient. We know, \(A=2\pi xl\). So, \(F_{v}=-\eta 2\pi xl \frac{dv}{dx}---2\).

Let, P be the pressure which acts on the whole cross section of the liquid. Then, the driving force on the elementary hollow cylinder is, \begin{align*} F_{d}=Pa=P\pi x^2---3 \end{align*} But for the steady flow, viscous drag must be equal to the driving force, ie. \begin{align*} -\eta2\pi xl \frac{dv}{dx}=P\pi x^2 \end{align*} \begin{align*} or, dv=\frac{-P}{2\eta l}xdx \end{align*} integrating, we get,\begin{align*} v=\frac{-P}{2\eta l}\frac{x^2}{2}+c---4 \end{align*} To find c, let us use the boundary condition that v=0 when x=r. Then, from 1,\begin{align*} c=\frac{P}{4\eta l}r^2 \end{align*} Using c in 4, we get,\begin{align*} v=\frac{-P}{4\eta l}x^2+\frac{Pr^2}{4\eta l} \end{align*}\begin{align*} \therefore v=\frac{P}{4\eta l}(r^2-x^2)---5 \end{align*} Equatoin 5 gives the velocity of any liquid layer in the tube. Equation 5 also suggests that the liquid in tube is parabolic as v is directly proportional to square of x.

Rate of flow of liquid

Let, \(dQ\) be the rate of liquid flow through the elementary cylinder, then,\begin{align*} dQ=\frac{volume}{fluid}=\frac{2\pi x l dx}{t}=2\pi x v dx \end{align*} where, \(v=\frac{l}{t}\) is the velocity of the liquid layer. Using \(v\) from equation 5, we get,\begin{align*} dQ=2\pi x \frac{P}{4\eta l}(r^2-x^2)dx \end{align*}\begin{align*} or, dQ=\frac{\pi P}{2\eta l}x(r^2-x^2)dx \end{align*} Since, the elementary cylinders vary from x=0 to x=r, the rate of liquid flow is given by,\begin{align*} Q=\int dQ \end{align*}\begin{align*} or, Q=\int_{0}^{r} \frac{\pi P}{2\eta l}x(r^2-x^2)dx\end{align*}\begin{align*} or, Q=\frac{\pi Pr^4}{8\eta l} \end{align*} This is the required Poiseuille's relation.

Application of Poiseuille's formula

Rate of flow of fluid when the capillaries are in series

We have two capillaries AB and BC having radius \(r_{1}\) and \(r_{2}\) such that the pressure difference across the combination be \(P_{1}\) and \(P_{2}\) respectively. Thus, the pressure difference across the combination is \(P=P_{1}+P_{2}\). For an incompressible fluid the rate of flow will be the same across both the tubes. From Poiseuille's realtion,\begin{align*} Q=\frac{\pi PR^4}{8\eta l}=\frac{P}{\frac{8\eta l}{\pi R^4}}=\frac{P}{r} \end{align*} where, \(r=\frac{8\eta l}{\pi R^4}\) is the viscous resistance. So, \(P=Qr\). For the tube AB of length \(l_{1}\), \(P_{1}=Qr_{1}\) and for the tube BC of length \(l_{2}\), \(P_{2}=Qr_{2}\). Then, we can write, \(Qr=Qr_{1}+Qr_{2}\) and \(r=r_{1}+r_{2}\). Thus, effective viscous resistance is the sum of individual viscous ressitance. Also, the rate of flow through the combination is,\begin{align*} Q=\frac{P}{r}=\frac{P}{r_{1}+r_{2}} \end{align*}\begin{align*} or, Q=\frac{\pi P}{8 \pi}[\frac{l_{1}}{R_{1}^4}+\frac{l_{2}}{R_{2}^4}]^-1 \end{align*}

Rate of flow of fluid when the tubes are in parallel

Here, \(Q=Q_{1}+Q_{2}\) and \(P=P_{1}=P_{2}\) So, \(\frac{P}{r}=\frac{P}{r_{1}}+\frac{P}{r_{2}}\) or, \(\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r_{2}}\). So, the effective viscous resistance reduces when the tubes are connected in parallel. Also, the rate of flow is,\begin{align*} Q=P[\frac{1}{r_{1}}+\frac{1}{r_{2}}] \end{align*}\begin{align*} Q=\frac{\pi P}{8 \eta}[\frac{R_{1}^4}{l_{1}}+\frac{R_{2}^4}{l_{2}}] \end{align*}

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.