Physics Bootcamp

Subsection 18.1.1 Flow in a Pipe

Turn on the faucet and collect water in a bucket. Suppose it takes \(15\text{ sec}\) to fill a \(4\text{-liter}\) bucket. Dividing the volume by the time, i.e., \(16\text{ liter/min} \text{,}\) gives us an important measure of flow, called the volume rate of flow. This is denoted by letter \(Q\text{.}\)

We can also speak of the speed of flow, \(v\text{,}\) by looking inside the pipe. Observe that all the fluid that was within a distance of \(v\Delta t\) would have gone to the bucket in time \(\Delta t\text{.}\) That is, if the cross-section of the pipe is \(A \text{,}\) then, the volume flowed is

\begin{equation} \Delta V = A v\Delta t.\label{eq-volume-of-flow-pipe}\tag{18.1.2} \end{equation}

Using this in Eq. (18.1.1) we find the relation between the two measures of flow, viz., volume rate of flow and the speed of flow.

On multiplying \(Q \) by the density of the fluid, \(\rho\text{,}\) we will get the mass rate of flow. The mass rate of flow is also called current. It is usually denoted by letter \(I\text{.}\)

When we divide current by the area of cross-section \(A\text{,}\) we get current density, denoted by \(J\text{.}\)

Subsection 18.1.2 Incompressibility and Steady Flow

When the volume flow rate \(Q = Av\) is same throughout and not changing with time, we say that the fluid flow is steady. An object is incompressible if volume change is negligible when you compress the object. Below, we derive an important relation for an incompressible fluid flow.

To study analytically, let us consider a steadily flowing incompressible fluid in a pipe whose cross-section varies as shown in Figure 18.1.1.

Let \(A_1\) and \(A_2\) be the cross-sectional areas of the entrance and exit respectively, and \(v_1\) and \(v_2\) be the corresponding speeds. In time \(\Delta t \text{,}\) all fluid within a volume \(A_1 v_1 \Delta t \) would have entered the space between 1 and 2, and during this time all fluid within a volume \(A_2 v_2\Delta t \) would have exited the space. Equating the two, and canceling out \(\Delta t\) gives the following important result, which is sometimes called the equation of continuity.

This says that the fluid will be faster (higher \(v \)) in the narrower part (ie., where \(A\) is smaller) and vice-versa.

Subsection 18.1.3 (Calculus) Current Density

Current per unit area of cross-section is called the current density. We defined current above by \(I = \rho A v\text{.}\) Dividing this by \(A \) we get the current density, which will be denoted by letter \(J\text{.}\)

If the flow in a tube is not uniform, for instance, flow may be faster in the middle of the tube than near the edge, then current density will vary over the cross-section. Current density will also vary if density of the material varies. We then speak of current density as local current density, and can think of \(J\) as a function of position \((x, y, z) \text{.}\)

Current density is a vector with magnitude given here and the direction in the direction of the flow.

If current density does not vary over the cross-section, then current is simply the product of current density and the area of cross-seciotn.

otherwise, we need to integrate current density over the cross-section area. This is written as a conceptual integral.

In a particular situation, you will need to decide how to write \(dA\text{.}\) In a rectangular cross-section, you might say \(dA=dxdy\) and in a circular cross-section, polar coordinates will be more appropriate with \(dA = rdrd\theta\text{.}\) If we have a circular symmetry, you might use \(dA=2\pi r dr\text{.}\)

Subsection 18.1.4 (Calculus) Laminar Flow in a Cylindrical Tube

It will not surprise you that fluid flows faster in the center than near the wall - this is what happens in a river where water flows faster in the middle and hardly any near the bank.

A more advanced treatment of fluid flow shows that at low speeds of flow, speed of flow at radial distance \(r\) from the center of a tube of radius \(R\) varies quadratically as shown diagrmatically in the figure on the right.

\begin{equation} v = v_0 - \dfrac{r^2}{R^2} v_0,\label{eq-velocity-profile-cylindrical-pipe}\tag{18.1.10} \end{equation}

where \(v_0\) is the speed at the center.

To find current through the tube, we look at flow through a ring of infinitesimal area \(dA = 2\pi rdr\text{,}\) which is between radius \(r\) and \(r+dr\text{.}\) Let \(\rho\) be uniform density.

\begin{align*} I \amp = \int J\, dA = \int \rho v 2\pi rdr \\ \amp = 2\pi \rho v_0 \int_0^R \left( 1 - \dfrac{r^2}{R^2}\right) r dr \\ \amp = \dfrac{1}{2}\rho v_0 \pi R^2 = \dfrac{1}{2}\rho v_0 A. \end{align*}

Note that if the speed was \(v_0\) at all points of the cross-section, we would get \(2 \times \) as much current, but that is not what actually happens. You get only half as much flow as dictated by the speed at the center since fluid is flowing at different speeds across the cross-section.