# Venturi Effect

When a fluid flows through a constricted section of a pipe, the pressure decreases. This can be explained by invoking the equation of continuity and Bernoulli's theorem. The Bernoulli equation describes the relationship between velocity and pressure: \begin{align} p_{1}-p_{2}={\frac {\rho }{2}}\left(v_{2}^{2}-v_{1}^{2}\right) \end{align} where $p$ is pressure, $\rho$ is the fluid density and $v$ is velocity. This equation shows that, in the event of a pressure drop, the velocity increases.

## Venturi meter

A Venturi meter is a device used to measure the flow rate of fluid in a pipe. The flow rate is the volume rate of fluid flow i.e., the volume of fluid that flows in unit time.

A Venturi meter works on the principle of differential pressure, where a constricted section in the pipe creates a drop in pressure, which is proportional to the velocity of the fluid flow. This drop in pressure is measured by pressure taps and used to calculate the flow rate.

The continuity equation gives the flow rate $Q$, \begin{align} Q&=v_{1}A_{1}=v_{2}A_{2} \end{align} and the Bernoulli's theorem gives \begin{align} p_{1}-p_{2}&={\frac {\rho }{2}}\left(v_{2}^{2}-v_{1}^{2}\right) \end{align}

Simplify to get the flow rate \begin{align} Q&=A_{1}{\sqrt {{\frac {2}{\rho }}\cdot {\frac {p_{1}-p_{2}}{\left({\frac {A_{1}}{A_{2}}}\right)^{2}-1}}}} \\ &=A_{2}{\sqrt {{\frac {2}{\rho }}\cdot {\frac {p_{1}-p_{2}}{1-\left({\frac {A_{2}}{A_{1}}}\right)^{2}}}}} \end{align}

## Problems from IIT JEE

Problem (JEE Mains 2022): A liquid of density 750 kg/m3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from $A_1=1.2\times{10}^{-2}$ m2 to $A_2=A_1/2$. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is_________$\times10^{-3}\;\mathrm{m^3/s}$.

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