# How does changing the speed of a fluid affect its pressure?

Apr 21, 2015

Hi, sorry for having kept you waiting,
A fluid is either a gas or liquid, right?

For a liquid in a tube( like crude oil in a pipeline, or so)
The relationship between Pressure of the liquid and the velocity of flow of its volume, can be written using Bernoulli's equation; $\frac{P}{\rho} + {v}^{2} / 2 + g h = \text{constant}$
where,
$P$ is the Pressure exerted by the flowing liquid
$\rho$ is the density of the liquid
$v$ is the velocity with which the liquid's volume is flowing
$g$ is the acceleration of free fall
$h$ is the vertical height of the liquid above the ground

Bernoulli's equation is got from the assumption that, " total energy per init mass of a liquid flowing in a pipe or a tube is constant "

That is, $\text{Pressure energy" + "Kinetic energy" + "Potential energy" = "constant}$
(and i'll spare us of the details... unless requested)

From the above relation, if $g$ and $h$ are kept constant,
we have, $\frac{P}{\rho} + {v}^{2} / 2 = \text{constant}$
This also means, ${P}_{1} / \rho + {\left({v}_{1}\right)}^{2} / 2 = {P}_{2} / \rho + {\left({v}_{2}\right)}^{2} / 2$

This implies that if ${v}^{2} / 2$ increases then $\frac{\text{P}}{\rho}$ must decrease, in order for the total of the two terms to remain the same as before.

Finally, this means that there is like an inverse proportionality relationship between Pressure $P$ and velocity $v$

For a gas(an ideal gas inside a closed container for the matter)
You can simply use the formula $P = \frac{1}{3} \rho {c}^{2}$
$P$ is the pressure exerted by the gas
$\rho$ is the density of the gas
${c}^{2}$ is the root mean square speed(rms)

So increasing the speed of the molecules(by heating for example) will increase the rms and hence the Pressure P.

An also, Bernoulli's equation concerns gases too.

It predicts that, for a moving fluid(such as air) the higher the speed of the gas, the smaller the pressure it exerts." P " alpha  $v$