Bernoulli's Equation

The mathematical representation of Bernoulli's principle is known as Bernoulli's equation. As mentioned in previous post (Bernoulli's Principle), the equation is just a restatement of the conservation of energy.

We start out with a "piece" of a fluid in phase A, where its velocity is $v_1$, its pressure $P_1$ and its height is at $h_1$. At a later point in time, the same "piece" of fluid is at phase B, where its velocity is $v_2$, its pressure is $P_2$ and its height is at $h_2$.

Due to the conservation of energy, we know that the amount of energy should be the same for that "piece" of fluid at phase A and B. The energy here is split into 3 parts: the kinetic energy $\left(\frac{1}{2}mv^2\right)$, the gravitational potential energy $\left(mgh\right)$ and the potential energy stored in the form of pressure $\left(PV\right).$

By conservation of energy we get: $$$$ $$P_1 V + \frac{1}{2}mv_1^2 + mgh_1 = P_2 V + \frac{1}{2}mv_2^2 + mgh_2$$ $$$$ However, we can express $m$ as $\rho V$, and since the fluid is non-compressible, we know that $V$ did not change.

Therefore we can divide through by $V$: $$\boxed{P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2}$$

And that is Bernoulli's Equation.

[Note: Bernoulli's equation is ONLY applicable for incompressible and nonviscous fluids i.e. for ideal fluids ]

In other words we can say - for an ideal fluid: $$$$ $$P + \frac{1}{2}\rho v^2 + \rho gh = Constant$$ $$$$ Or: $$$$ $$\Delta \left( P + \frac{1}{2}\rho v^2 + \rho gh \right) =0$$