Bernoulli equation of fluid mechanics is a powerful equation used in the Lord of engineering applications. Which involves fluids or liquids. This article explains fundamental Bernoulli’s equation derivation and its application in very basic terms.
Bernoulli equation is actually an energy conservation equation in fluid mechanics. It’s a one-dimensional equation derived by Swiss Mathematician “Euler” was an associate of “Daniel Bernoulli”.
I recommend you to read the article on the derivation of the Euler equation of motion in which I have explained the derivation of the Euler equation from the very basic physics.
This is the Euler equation from which the Bernoulli’s equation is derived. Now let’s go ahead and integrate the Euler equation as it flows as soon to be incomprehensible. The density remains constant. Hence the equation can be write as follows.
and dividing all the downs by ‘g’ You’ll be left with this expression.
In this Bernoulli’s equation, The first time is the energy in the form of pressure. The second time is the energy in the form of velocity and the last time is the energy in potential or Datum form. You can observe a specialty in this equation. Yes, all the energies in this equation are express in a common unit called a head. That will be mention in a meter of water or meter of mercury.
Let’s see a very basic application of Bernoulli’s equation in fluid mechanics:
Here we have a converging pipeline with the inlet area of 0.2 meters square and the outlet area of 0.1 meters square. The inlet velocity is 40 meters per second. We need to find out the velocity at the outlet. Then the inlet pressure is one need to tend to the power of seven newtons per meter square.
We need to find out the pressure at the outlet. So two things we need to calculate the outlet velocity and the outlet pressure. The outlet velocity can be calculated by the continuity equation. See the velocity is getting increased as the cross-section area decreases. Let us use Bernoulli’s equation to find out the outlet pressure.
In this case, both the inlet and outlet are in the same height from the datum. Hence Z1 and Z2 can be canceled out.
Re-write the equation to find the pressure at the outlet because that is needed.
Substitute the known inlet pressure velocities and the density of water to get the outlet pressure. As the cross-sectional area of the pipeline gets reduced. The velocity gets increase and the pressure decreases.
It shows that a part of pressure energy available at the inlet is converted into velocity energy at the outlet. In other words, the velocity Got increased at the cost of a decrease in pressure.
Thank you. Hope it helps. Don’t forget to comment.