Follow one parcel of fluid along its path. If nothing adds energy and nothing takes it away, the three ways it can hold energy must add to a constant. That is Bernoulli's equation:

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

Every symbol, in words. $P$ is the static pressure at a point, in pascals — what a gauge tapped flush into the pipe wall reads. $v$ is the velocity there, in m/s. $z$ is the elevation of that point above whatever datum you chose, in metres — the datum is yours to pick, and only differences in $z$ ever matter. $\rho$ is the fluid density in kg/m³ and $g$ is 9.81 m/s². The subscript convention is the same one continuity used: 1 is the upstream point, 2 is the downstream point, and every term on the left belongs to point 1.

Read it as a budget and the physics becomes obvious. Speed the fluid up and the velocity term grows, so something must shrink — and on a level run the only candidate is the pressure. That is why a venturi throat reads low, why a carburettor draws fuel, and why a fast-moving stream past a branch can pull rather than push. Lift the fluid instead and the elevation term grows, and again the pressure pays.

Now open both ends of the budget to atmosphere: a tank vented at the top with a hole in its side. The two pressures are equal and cancel, the surface is barely moving so $v_1$ is essentially zero, and the whole elevation difference has nowhere to go but into velocity. What survives is Torricelli's law, $v = \sqrt{2gh}$ — v equals root two g h, where $h$ is the height of fluid ABOVE the opening, in metres, and $v$ is the efflux speed in m/s. The density is gone again. A tank of kerosene jets at exactly the speed a tank of water does — and it is the same $\sqrt{2gh}$ a dropped stone arrives at, which is not a coincidence but the same energy trade, told twice.

What Bernoulli leaves out is friction, and over any real length of pipe friction is not small. The next chapter puts it back.