Venturi Tube Problem

I'm trying to calculate the volume flow. I've tried setting up Bernoulli and using the hydrostatic pressure and the conservation of mass, as one does to solve a horizontal tube, but I keep hitting up against needing to know the heights z1 and z2, which are not given.

I can't figure out a way to remove it from my equations.

Density is constant.

The system is in equilibrium.

No turbulence.

No friction.

Pressures are constant across the diameters.


1 Answer 1


If I understand correctly, the $\Delta p$ is the net pressure drop shown by the gauge. The pressure indicator will always show the 'net effect', i.e. it doesn't understand the physical reasons contributing towards the reading. Thus, it includes the hydrostatic pressure drop as well as dynamic pressure drop. Assuming there is the pressure drop along the direction of flow, and assuming that the pressure gauge shows absolute value of the pressure drop (i.e. magnitude), we can write the Bernoulli's equation (which we can safely do since the flow is irrotational, incompressible and inviscid) along the streamline as,

\begin{equation} p_1 + \frac{1}{2} \rho v_1^2 + \rho g z_1 = p_2 + \frac{1}{2} \rho v_2^2 + \rho g z_2 \end{equation} i.e. \begin{equation} \frac{1}{2} \rho \left( v_1^2 - v_2^2 \right) = \underbrace{\overbrace{\left(p_2 - p_1 \right)}^{-ve} + \overbrace{\rho g \left( z_2 - z_1 \right)}^{+ve}}_{-\Delta p} \end{equation}

We can see that, tilting of the Venturi exerts back-pressure at section $1$. i.e. it resists the flow.

According to the continuity equation,

\begin{equation}\label{cont} v_1 = \frac{A_2}{A_1} v_2 \end{equation}

Substituting above equation in Bernoulli's equation,

\begin{equation} \frac{1}{2} \rho \left( \left(\frac{A_2}{A_1}\right)^2 - 1\right)v_2^2 = {-\Delta p} \end{equation}

Which gives,

\begin{equation} v_2 = \sqrt{ \left(\frac{-2 \Delta p}{\rho}\right) \left( \frac{A_1^2}{A_2^2 - A_1^2}\right) } = 0.1326 \frac{m}{s} \end{equation}

and hence, $v_1 = 0.0398 \frac{m}{s}$

The required flow rate is then; $v_1 A_1 = v_2 A_2 = 0.004 \frac{m^3}{s}$.

  • $\begingroup$ Wow thanks! The exam's been and gone but this was still bugging me! $\endgroup$
    – Andy Grey
    Dec 13, 2017 at 11:52

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