The Venturi Meter

Figure 1. Venturimeter

A Venturi meter, named after the Italian physicist Giovanni Battista Venturi (1746 – 1822), is a differential pressure flow meter that generates a flow measurement by measuring the pressure difference at two different locations in a pipe. The device demostrates the Venturi effect, which states that the velocity of the flowing fluid increases due to a pressure drop created by constricting the diameter of the pipe.

Problem

Consider an ideal fluid of density $\rho$ flowing through a horizontal pipe of variable cross-sectional area (Figure 1). The fluid at the portion of the pipe with cross-sectional area ${A}_{1}$ is moving at ${v}_{1}$ and has a measured pressure of ${P}_{1}$. The fluid at the portion of the pipe with cross-sectional area ${A}_{2}$ is moving at ${v}_{2}$ and has a measured pressure of ${P}_{2}$.

Part 1: Determine ${v}_{1}$ in terms of $\rho, A_1, A_2, P_1$ and $P_2$.

Part 2: Two different fluids of density ${\rho}_{1}$ (fluid in the horizontal pipe) and ${\rho}_{2}$ (fluid in the u-shaped manometer), where ${\rho}_{1} < {\rho}_{2}$. The height difference of the fluid in the manometer is $h$. It helps to let $h = h_1 - h_2$. Determine ${v}_{1}$ in terms of ${\rho}_{1}, {\rho}_{2}, A_1, A_2$ and $h$.

Hints

There are two important equations for this problem:

(1) The continuity equation

$$\rho A_1 v_1 = \rho A_2 v_2$$

(2) Bernoulli’s equation

$$P_1 + \rho gh_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2$$

Solution

Part 1

From the first equation we find

$$v_2 = \frac{A_1 v_1}{A_2}$$

Since there is no change in the altitude of the pipe, we can deduce that

$$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho \left(\frac{A_1 v_1}{A_2} \right)^2$$

Isolate for $v_1$

$$P_1 - P_2 = \frac{1}{2} \rho v_1^2 \left[\left(\frac{A_1}{A_2} \right)^2-1\right]$$

$$v_1 = \sqrt{\frac{2(P_1-P_2)}{\rho \left[\left(\frac{A_1}{A_2}\right)^2-1 \right]}}$$

Part 2

Treat the manometer as two static column problems.

$$\begin{align} P_1 + {\rho}_{1} g h_1 &= P_2 + {\rho}_{1} g h_2 + {\rho}_{2} g (h_1 - h_2) \\[1.5 pt] P_1 - P_2 &= -{\rho}_{1} g h_1 + {\rho}_{1} g h_2 + {\rho}_{2} g (h_1 - h_2) \\[1.5 pt] P_1 - P_2 &= ({\rho}_{2} - {\rho}_{1}) g (h_1 - h_2) \\[1.5 pt] P_1 - P_2 &= ({\rho}_{2} - {\rho}_{1}) gh \end{align}$$

Substitute the difference of pressure into the result from part 1 to get

$$v_1 = \sqrt{\frac{2({\rho}_{2} - {\rho}_{1})gh}{{\rho}_{1} \left[{\left(\frac{A_1} {A_2}\right)}^{2} - 1 \right]}}$$