The equation of continuity describes conservation of flow in a pipe or duct.
For steady incompressible flow, the same volume of fluid must pass every cross-section each second.
1. Volume flow rate
The volume flow rate is
\[
\begin{aligned}
Q &= Av.
\end{aligned}
\]
Here \(A\) is cross-sectional area and \(v\) is average flow speed.
The unit of \(Q\) is usually \(\mathrm{m^3/s}\).
2. Continuity equation
If the fluid is incompressible and the flow is steady, then
\[
\begin{aligned}
Q_1 &= Q_2.
\end{aligned}
\]
Since \(Q=Av\), this becomes
\[
\begin{aligned}
A_1v_1 &= A_2v_2.
\end{aligned}
\]
This is the equation of continuity.
3. Solving for outlet speed
To find \(v_2\), rearrange:
\[
\begin{aligned}
A_1v_1 &= A_2v_2.
\end{aligned}
\]
\[
\begin{aligned}
v_2 &= \frac{A_1v_1}{A_2}.
\end{aligned}
\]
If \(A_2v_1\). A narrower pipe makes the fluid move faster.
4. Solving for inlet speed
To find \(v_1\), rearrange the same equation:
\[
\begin{aligned}
v_1 &= \frac{A_2v_2}{A_1}.
\end{aligned}
\]
5. Circular pipe area
For a circular pipe of diameter \(D\),
\[
\begin{aligned}
A &= \frac{\pi D^2}{4}.
\end{aligned}
\]
If radius \(R\) is used instead,
\[
\begin{aligned}
A &= \pi R^2.
\end{aligned}
\]
Because area depends on diameter squared, cutting the diameter in half reduces the area by a factor of four.
6. Diameter change and speed change
For two circular pipe sections,
\[
\begin{aligned}
\frac{A_1}{A_2}
&= \frac{\pi D_1^2/4}{\pi D_2^2/4} \\
&= \left(\frac{D_1}{D_2}\right)^2.
\end{aligned}
\]
Therefore,
\[
\begin{aligned}
\frac{v_2}{v_1}
&= \frac{A_1}{A_2} \\
&= \left(\frac{D_1}{D_2}\right)^2.
\end{aligned}
\]
If the diameter changes from \(10\ \mathrm{cm}\) to \(5\ \mathrm{cm}\), the diameter ratio is \(2\), so the speed ratio is \(2^2=4\).
7. Mass flow rate
The mass flow rate is
\[
\begin{aligned}
\dot m &= \rho Q.
\end{aligned}
\]
For an incompressible fluid with constant density, conservation of volume flow rate also means conservation of mass flow rate:
\[
\begin{aligned}
\dot m_1 &= \dot m_2.
\end{aligned}
\]
8. Worked example
A pipe narrows from diameter
\[
\begin{aligned}
D_1 &= 10\ \mathrm{cm}=0.10\ \mathrm{m}
\end{aligned}
\]
to
\[
\begin{aligned}
D_2 &= 5\ \mathrm{cm}=0.05\ \mathrm{m}.
\end{aligned}
\]
The inlet speed is
\[
\begin{aligned}
v_1 &= 1.5\ \mathrm{m/s}.
\end{aligned}
\]
First calculate the areas:
\[
\begin{aligned}
A_1
&= \frac{\pi D_1^2}{4} \\
&= \frac{\pi(0.10)^2}{4} \\
&= 0.00785\ \mathrm{m^2}.
\end{aligned}
\]
\[
\begin{aligned}
A_2
&= \frac{\pi D_2^2}{4} \\
&= \frac{\pi(0.05)^2}{4} \\
&= 0.00196\ \mathrm{m^2}.
\end{aligned}
\]
Then use continuity:
\[
\begin{aligned}
v_2
&= \frac{A_1v_1}{A_2} \\
&= \frac{(0.00785)(1.5)}{0.00196} \\
&\approx 6.0\ \mathrm{m/s}.
\end{aligned}
\]
The volume flow rate is
\[
\begin{aligned}
Q
&= A_1v_1 \\
&= (0.00785)(1.5) \\
&= 0.0118\ \mathrm{m^3/s}.
\end{aligned}
\]
9. Formula summary
| Goal |
Formula |
Use |
| Volume flow rate |
\(Q=Av\) |
Volume of fluid passing per second |
| Continuity equation |
\(A_1v_1=A_2v_2\) |
Steady incompressible flow |
| Outlet speed |
\(v_2=A_1v_1/A_2\) |
Find speed after area change |
| Inlet speed |
\(v_1=A_2v_2/A_1\) |
Find speed before area change |
| Circular area from diameter |
\(A=\pi D^2/4\) |
Pipe diameter problems |
| Mass flow rate |
\(\dot m=\rho Q\) |
Mass passing per second |
10. Assumptions
- The flow is steady.
- The fluid is incompressible.
- The average speed is used at each cross-section.
- There are no leaks or side branches between the two sections.
- Pipe friction and pressure losses are not included.
- For very high speeds or gases, compressibility may need to be considered.
Key idea: when pipe area decreases, flow speed increases so that \(Q=Av\) remains constant.
