Bernoulli's Equation

Bernoulli's equation expresses conservation of energy for an ideal (inviscid), incompressible, steady flow along a streamline:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g z_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g z_2$

The three terms represent pressure energy, kinetic energy, and potential energy per unit volume. Their sum is constant along any streamline.

Solving for Downstream Pressure

Rearranging for the pressure at point 2:

$P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) + \rho g(z_1 - z_2)$

where $g = 9.81$ m/s² is gravitational acceleration, $z$ is elevation (m), and $ho$ is fluid density (kg/m³).

Key insight: speed and pressure trade off. A faster-moving fluid exerts less static pressure — the basis of the Venturi effect and aircraft lift.

Torricelli's Theorem

For a large open tank draining through a small hole at depth $h$ below the surface, we set $v_1 \approx 0$ (large tank, surface barely moves) and $P_1 = P_2 = P_{\text{atm}}$. Bernoulli's equation simplifies to:

$v_2 = \sqrt{2gh}$

This is Torricelli's theorem — the exit velocity equals the free-fall speed from height $h$.

Your Task

Implement:

• bernoulli_pressure(P1, rho, v1, z1, v2, z2) — pressure at point 2 (Pa). Use $g = 9.81$ m/s² inside the function.
• torricelli_velocity(h) — drain velocity (m/s) for a tank with head $h$. Use $g = 9.81$ m/s² inside the function.