A Venturi tube, or Venturi meter, measures flow rate by using a narrow throat in a pipe.
As the fluid enters the throat, the cross-sectional area decreases, the speed increases, and the static pressure decreases.
The pressure drop is then used to infer the flow rate.
1. Continuity equation
For steady incompressible flow, the volume flow rate is constant:
\[
\begin{aligned}
Q &= Av.
\end{aligned}
\]
Between the inlet and the throat:
\[
\begin{aligned}
A_1v_1 &= A_2v_2.
\end{aligned}
\]
Therefore:
\[
\begin{aligned}
v_1 &= \frac{Q}{A_1},\\
v_2 &= \frac{Q}{A_2}.
\end{aligned}
\]
Since the throat area \(A_2\) is smaller than the inlet area \(A_1\), the throat speed \(v_2\) is greater than the inlet speed \(v_1\).
2. Bernoulli equation in a horizontal Venturi tube
In a horizontal ideal Venturi tube, the inlet and throat are at the same height, so \(z_1=z_2\).
Bernoulli’s equation becomes:
\[
\begin{aligned}
P_1+\frac{1}{2}\rho v_1^2
&=
P_2+\frac{1}{2}\rho v_2^2.
\end{aligned}
\]
Rearranging gives the pressure drop:
\[
\begin{aligned}
\Delta P
&= P_1-P_2\\
&= \frac{1}{2}\rho(v_2^2-v_1^2).
\end{aligned}
\]
This shows the Venturi effect: where the speed is higher, the static pressure is lower.
3. Flow rate from pressure drop
Substitute \(v_1=Q/A_1\) and \(v_2=Q/A_2\) into the pressure-drop equation:
\[
\begin{aligned}
\Delta P
&= \frac{1}{2}\rho\left[
\left(\frac{Q}{A_2}\right)^2
-
\left(\frac{Q}{A_1}\right)^2
\right].
\end{aligned}
\]
Factor out \(Q^2\):
\[
\begin{aligned}
\Delta P
&= \frac{1}{2}\rho Q^2
\left(\frac{1}{A_2^2}-\frac{1}{A_1^2}\right).
\end{aligned}
\]
Solve for the ideal flow rate:
\[
\begin{aligned}
Q_{\mathrm{ideal}}
&=
\sqrt{
\frac{2\Delta P/\rho}
{\frac{1}{A_2^2}-\frac{1}{A_1^2}}
}.
\end{aligned}
\]
4. Discharge coefficient
Real Venturi meters are not perfectly ideal. Friction, turbulence, boundary-layer effects, and nonuniform velocity profiles reduce the actual flow compared with the ideal value.
This is represented by the discharge coefficient \(C_d\):
\[
\begin{aligned}
Q_{\mathrm{actual}} &= C_dQ_{\mathrm{ideal}}.
\end{aligned}
\]
For an ideal Venturi tube, \(C_d=1\). For real meters, \(C_d\) is usually slightly below \(1\).
5. Pressure drop from known flow rate
If the actual flow rate is known, first convert it to the ideal-equivalent flow rate:
\[
\begin{aligned}
Q_{\mathrm{ideal}} &= \frac{Q_{\mathrm{actual}}}{C_d}.
\end{aligned}
\]
Then:
\[
\begin{aligned}
\Delta P
&= \frac{1}{2}\rho Q_{\mathrm{ideal}}^2
\left(\frac{1}{A_2^2}-\frac{1}{A_1^2}\right).
\end{aligned}
\]
6. Throat pressure
Once the pressure drop is known, the throat pressure is:
\[
\begin{aligned}
P_2 &= P_1-\Delta P.
\end{aligned}
\]
The throat pressure is lower than the inlet pressure because part of the pressure energy has been converted into kinetic energy.
7. Areas from diameter
For a circular pipe:
\[
\begin{aligned}
A &= \frac{\pi D^2}{4}.
\end{aligned}
\]
Therefore:
\[
\begin{aligned}
A_1 &= \frac{\pi D_1^2}{4},\\
A_2 &= \frac{\pi D_2^2}{4}.
\end{aligned}
\]
The diameter ratio is:
\[
\begin{aligned}
\beta &= \frac{D_2}{D_1}.
\end{aligned}
\]
The area ratio is:
\[
\begin{aligned}
\frac{A_2}{A_1} &= \beta^2.
\end{aligned}
\]
8. Equivalent pressure head
A pressure drop can also be written as an equivalent liquid head:
\[
\begin{aligned}
h &= \frac{\Delta P}{\rho g}.
\end{aligned}
\]
This is useful when the Venturi meter is connected to a manometer.
9. Worked example
Suppose water flows through a Venturi tube with:
\[
\begin{aligned}
D_1 &= 8\ \mathrm{cm}=0.08\ \mathrm{m},\\
D_2 &= 4\ \mathrm{cm}=0.04\ \mathrm{m},\\
\rho &= 1000\ \mathrm{kg/m^3},\\
P_1 &= 200\ \mathrm{kPa},\\
\Delta P &= 30\ \mathrm{kPa},\\
C_d &= 0.98.
\end{aligned}
\]
First calculate the areas:
\[
\begin{aligned}
A_1
&= \frac{\pi(0.08)^2}{4}
= 0.00503\ \mathrm{m^2},\\
A_2
&= \frac{\pi(0.04)^2}{4}
= 0.00126\ \mathrm{m^2}.
\end{aligned}
\]
The ideal flow rate is:
\[
\begin{aligned}
Q_{\mathrm{ideal}}
&=
\sqrt{
\frac{2(30000)/1000}
{\frac{1}{(0.00126)^2}-\frac{1}{(0.00503)^2}}
}\\
&\approx 0.0100\ \mathrm{m^3/s}.
\end{aligned}
\]
With \(C_d=0.98\):
\[
\begin{aligned}
Q_{\mathrm{actual}}
&= 0.98(0.0100)\\
&\approx 0.0098\ \mathrm{m^3/s}.
\end{aligned}
\]
The throat pressure is:
\[
\begin{aligned}
P_2 &= P_1-\Delta P\\
&= 200\ \mathrm{kPa}-30\ \mathrm{kPa}\\
&= 170\ \mathrm{kPa}.
\end{aligned}
\]
10. Formula summary
| Goal |
Formula |
Use |
| Continuity |
\(A_1v_1=A_2v_2\) |
Relate inlet speed and throat speed |
| Pressure drop |
\(\Delta P=\frac{1}{2}\rho(v_2^2-v_1^2)\) |
Find pressure decrease caused by speed increase |
| Ideal flow rate |
\(Q_{\mathrm{ideal}}=\sqrt{\frac{2\Delta P/\rho}{1/A_2^2-1/A_1^2}}\) |
Find ideal flow rate from measured pressure drop |
| Actual flow rate |
\(Q_{\mathrm{actual}}=C_dQ_{\mathrm{ideal}}\) |
Correct ideal result for real Venturi behavior |
| Pressure drop from actual flow |
\(\Delta P=\frac{1}{2}\rho(Q/C_d)^2(1/A_2^2-1/A_1^2)\) |
Find required pressure drop for a known actual flow |
| Throat pressure |
\(P_2=P_1-\Delta P\) |
Find pressure in the throat |
| Circular area |
\(A=\pi D^2/4\) |
Convert diameter to cross-sectional area |
11. Assumptions
- The flow is steady.
- The fluid is incompressible.
- The Venturi tube is treated as horizontal.
- The inlet and throat average velocities are used.
- Viscous and turbulence effects are represented only through \(C_d\).
- The throat area must be smaller than the inlet area.
- For high-speed gas flow, compressibility corrections may be required.
Key idea: in a Venturi tube, the narrow throat makes the fluid faster and the static pressure lower, so the pressure drop reveals the flow rate.
