Flux Through Circular Pipe Cross-Sections

Flow Through a Circular Cross-Section

A pipe’s volumetric flow rate measures how much volume passes through a cross-section per unit time. When the velocity points along the pipe, the calculation is a surface-flux integral over a disk. The rate may change from one cross-section to another if the axial velocity varies along the pipe.

We will use a circular pipe with its axis along the \(z\)-axis. Let \(r\) be the distance from that axis, let \(R\) be the pipe radius, and let \(w(r,z)\) be the axial velocity. The velocity field is \(\mathbf{u}=w(r,z)\mathbf{k}\), where \(\mathbf{k}\) is the unit vector in the positive \(z\)-direction. Thus the flow is purely axial: it has no radial or circumferential velocity component.

Definition: The signed volumetric flow rate through a surface \(S\), with chosen unit normal \(\mathbf{n}\), is \(Q=\iint_S \mathbf{u}\cdot\mathbf{n}\,dA\). For a pipe cross-section, positive flow is in the positive \(z\)-direction when the chosen normal is \(\mathbf{k}\). The units of \(Q\) are volume per time, such as \(\text{m}^3/\text{s}\).

The word signed matters. If the velocity points opposite the chosen normal, the flux is negative. For a disk at fixed \(z\), its normal is either \(\mathbf{k}\) or \(-\mathbf{k}\), so the sign depends on which side of a boundary is being considered.

Integrating an Axial Velocity Profile

On a disk perpendicular to the pipe axis, the area element in polar coordinates is \(dA=r\,dr\,d\theta\). With normal \(\mathbf{k}\), the dot product \(\mathbf{u}\cdot\mathbf{k}\) is \(w(r,z)\). If the velocity profile is axisymmetric, so it depends on the disk position only through \(r\), the flow rate at position \(z\) is

Q(z)=\int_0^{2\pi}\int_0^R w(r,z)\,r\,dr\,d\theta =2\pi\int_0^R w(r,z)\,r\,dr.

The factor \(r\) in the integral is essential: a thin ring at radius \(r\) has area approximately \(2\pi r\,dr\), not \(dr\). For a uniform axial speed \(w_0\), the formula reduces to \(Q=w_0\pi R^2\). For a nonuniform profile, the integral accounts for the different speeds at different radii.

Worked Example: A Parabolic Velocity Profile

Suppose a pipe of radius \(R=0.20\text{ m}\) has axial velocity \(w(r)=3(1-r^2/R^2)\text{ m/s}\). Find the flow rate in the positive \(z\)-direction.

Use the disk integral and substitute the profile:

\begin{aligned} Q &=2\pi\int_0^R 3\left(1-\frac{r^2}{R^2}\right)r\,dr\\ &=6\pi\left[\frac{r^2}{2}-\frac{r^4}{4R^2}\right]_0^R\\ &=6\pi\left(\frac{R^2}{2}-\frac{R^2}{4}\right) =\frac{3}{2}\pi R^2. \end{aligned}

With \(R^2=(0.20\text{ m})^2=0.040\text{ m}^2\), this gives \(Q=0.060\pi\text{ m}^3/\text{s}\), or approximately \(0.1885\text{ m}^3/\text{s}\) to four significant digits. The units follow from velocity times area: \((\text{m/s})(\text{m}^2)=\text{m}^3/\text{s}\).

As a check, the average speed over the disk is \(Q/(\pi R^2)=1.5\text{ m/s}\), half the centerline speed \(3\text{ m/s}\). That is consistent with a profile that decreases from its maximum at \(r=0\) to zero at \(r=R\).

Comparing End-Disk Flow with Closed-Surface Flux

A finite pipe segment can be enclosed by a cylinder with end disks at \(z=a\) and \(z=b\), where \(a<b\). Its closed boundary consists of the two disks and the curved lateral wall. On the wall, the outward normal is perpendicular to the \(z\)-axis, while the velocity is parallel to that axis. Therefore \(\mathbf{u}\cdot\mathbf{n}=0\) on the wall: a purely axial velocity contributes no flux through it.

The outward normal on the upper disk is \(\mathbf{k}\), and on the lower disk it is \(-\mathbf{k}\). Consequently, the total outward flux through the closed cylinder is \(Q(b)-Q(a)\). If the field is continuously differentiable throughout the pipe segment, the Divergence Theorem equates this boundary flux to the volume integral of the divergence. For \(\mathbf{u}=w(r,z)\mathbf{k}\), the divergence is \(\partial w/\partial z\), so

Q(b)-Q(a) = \iiint_V \nabla\cdot\mathbf{u}\,dV = \iiint_V \frac{\partial w}{\partial z}\,dV.

This identity explains how the two calculation strategies agree. The direct disk integral gives the flow at each end. The closed-surface calculation gives the difference between those end flows from the divergence inside the pipe. In particular, if \(w\) does not depend on \(z\), then \(\partial w/\partial z=0\), and the two end-disk flow rates are equal.

Why the Balance Identity Holds

For an axisymmetric profile that is differentiable in \(z\), write the disk flow as \(Q(z)=2\pi\int_0^R w(r,z)r\,dr\). Differentiating with respect to \(z\) gives \(Q'(z)=2\pi\int_0^R (\partial w/\partial z)r\,dr\). This is the cross-sectional integral of the divergence at \(z\). Integrating that expression from \(a\) to \(b\) and applying the Fundamental Theorem of Calculus yields the balance.

\begin{aligned} \iiint_V \frac{\partial w}{\partial z}\,dV &=\int_a^b\int_0^{2\pi}\int_0^R \frac{\partial w}{\partial z}\,r\,dr\,d\theta\,dz\\ &=\int_a^b Q'(z)\,dz\\ &=Q(b)-Q(a). \end{aligned}

This calculation makes the signs explicit: the upper cap contributes positive flow \(Q(b)\), while the lower cap contributes negative flow \(-Q(a)\), because the outward normal there points in the negative \(z\)-direction. The identity requires a sufficiently smooth field throughout the cylinder; a singularity inside the volume would invalidate this direct use of the Divergence Theorem.

Worked Example: A Flow Rate That Increases Along the Pipe

Consider a cylindrical segment \(0\le z\le 2\text{ m}\) of radius \(R=0.10\text{ m}\), with velocity

\mathbf{u}=w(r,z)\mathbf{k}, \qquad w(r,z)= \left(2\frac{\text{m}}{\text{s}}\right) \left(1+\frac{z}{2\text{ m}}\right) \left(1-\frac{r^2}{R^2}\right).

First compute the disk flow at an arbitrary position \(z\):

\begin{aligned} Q(z) &=2\pi\int_0^R \left(2\frac{\text{m}}{\text{s}}\right) \left(1+\frac{z}{2\text{ m}}\right) \left(1-\frac{r^2}{R^2}\right)r\,dr\\ &=\left(2\frac{\text{m}}{\text{s}}\right) \left(1+\frac{z}{2\text{ m}}\right) \frac{\pi R^2}{2}\\ &=\pi R^2\frac{\text{m}}{\text{s}} \left(1+\frac{z}{2\text{ m}}\right). \end{aligned}

Since \(R^2=0.010\text{ m}^2\), the lower and upper end flows are \(Q(0)=0.010\pi\text{ m}^3/\text{s}\) and \(Q(2\text{ m})=0.020\pi\text{ m}^3/\text{s}\). Their difference is \(0.010\pi\text{ m}^3/\text{s}\), approximately \(0.03142\text{ m}^3/\text{s}\) to four significant digits.

Now check by closed-surface flux balance. Differentiating the stated velocity with respect to \(z\) gives

\frac{\partial w}{\partial z} = \left(2\frac{\text{m}}{\text{s}}\right) \left(\frac{1}{2\text{ m}}\right) \left(1-\frac{r^2}{R^2}\right) = 1\text{ s}^{-1}\left(1-\frac{r^2}{R^2}\right).

The volume integral is therefore

\begin{aligned} \iiint_V \frac{\partial w}{\partial z}\,dV &=\left(1\text{ s}^{-1}\right)(2\text{ m}) \left(2\pi\int_0^R\left(1-\frac{r^2}{R^2}\right)r\,dr\right)\\ &=\left(1\text{ s}^{-1}\right)(2\text{ m}) \left(\frac{\pi R^2}{2}\right)\\ &=\pi R^2\frac{\text{m}}{\text{s}} =0.010\pi\text{ m}^3/\text{s}. \end{aligned}

This matches \(Q(2\text{ m})-Q(0)\). The lateral-wall flux is zero because the velocity is axial and the wall normal is perpendicular to the pipe axis.

A Constant Profile with a Different Radial Shape

The same method applies beyond parabolic profiles. The next example uses a profile that decreases more sharply near the pipe wall. Because it has no dependence on \(z\), the flow rate is the same at both ends of a pipe segment, even though the speed varies with radius.

Worked Example: A Squared Radial Profile

Let \(R=0.15\text{ m}\) and \(w(r)=6(1-r^2/R^2)^2\text{ m/s}\). Integrate over a disk with normal \(\mathbf{k}\):

\begin{aligned} Q &=12\pi\frac{\text{m}}{\text{s}}\int_0^R \left(1-\frac{2r^2}{R^2}+\frac{r^4}{R^4}\right)r\,dr\\ &=12\pi\frac{\text{m}}{\text{s}} \left[\frac{r^2}{2}-\frac{r^4}{2R^2}+\frac{r^6}{6R^4}\right]_0^R\\ &=12\pi\frac{\text{m}}{\text{s}}\left(\frac{R^2}{6}\right) =2\pi R^2\frac{\text{m}}{\text{s}}. \end{aligned}

Using \(R^2=0.0225\text{ m}^2\), the result is \(Q=0.045\pi\text{ m}^3/\text{s}\), approximately \(0.1414\text{ m}^3/\text{s}\) to four significant digits. Since \(w\) is independent of \(z\), \(\partial w/\partial z=0\). The closed-cylinder balance therefore gives zero net outward flux: the upper cap contributes \(+Q\), the lower cap contributes \(-Q\), and the wall contributes zero. This verifies the equal flow rates at the two ends.

Common Mistake: Replacing a Profile with Its Maximum

A frequent error is to multiply the centerline speed by the full disk area even when the speed is not uniform. For the parabolic profile \(w(r)=3(1-r^2/R^2)\text{ m/s}\), this incorrect shortcut gives \(Q_{\text{wrong}}=(3\text{ m/s})\pi R^2\). It treats every point in the disk as if it had the maximum speed.

Q_{\text{wrong}}=3\pi R^2\frac{\text{m}}{\text{s}}, \qquad Q_{\text{correct}}=\frac{3}{2}\pi R^2\frac{\text{m}}{\text{s}}.

The correct integral averages the velocity over the whole disk and gives half the centerline speed as the mean speed. Another sign error is to add the two outward cap contributions: the lower cap has outward normal \(-\mathbf{k}\), so its contribution is \(-Q(a)\), not \(+Q(a)\). A useful final check is to confirm both the direction of each cap normal and the units of every term.

Interpreting the Result

For an axial velocity field, direct cross-sectional integration answers how much volume crosses a particular disk per unit time. A closed-cylinder flux calculation answers how much that rate changes between the two ends, provided the field is smooth throughout the segment. If the divergence is positive across the volume, the net outward flux is positive, so the upper-end flow exceeds the lower-end flow in this axial setting. If the divergence vanishes, the signed end flows agree.

These statements concern volumetric flow, measured in volume per time. They do not by themselves calculate mass flow, which would also require density. Keeping the surface orientation, velocity profile, and units visible in the calculation makes the flux result both interpretable and checkable.

Check Your Understanding

Use the disk integral and the closed-cylinder orientation consistently when answering.

For an axial profile \(w(r,z)\), what polar-coordinate area element is used on a circular cross-section?
Why does a purely axial velocity have zero flux through the curved side of a cylindrical pipe segment?
For a pipe segment from \(z=a\) to \(z=b\), why is the net outward flux through the two end disks \(Q(b)-Q(a)\)?
If \(w\) is independent of \(z\), what does the Divergence Theorem imply about the end-disk flow rates?
What physical or mathematical error is made by multiplying the centerline speed by the full cross-sectional area for a nonuniform profile?