Bernoulli's equation explained
Water passes a pinched Venturi tube. When you squeeze the throat and the water speeds up, what happens to the pressure? Check it yourself on the gauges. If the column falls at the fast throat, the flow gave up pressure as the price of going faster. It is a picture of the same water moving the seat of its energy.
Drag the probe along the tube and read the speed v and the pressure p, and the split is clear. A wide slow place has high pressure and a long column, and the narrow fast throat has low pressure and a short column. Move only the place and the two readings move opposite each other.
Now squeeze the throat and the more you narrow it the more the speed at the throat rises and the pressure falls. Speeding up and a drop in pressure happen at the same time. Loosen the squeeze and the speed comes down and the pressure rises again. The two are tied to one handle.
The sum of three energy terms is constant: the pressure p, the motion ½ρv², and the height. Drag the speed v and the motion term grows by as much as the pressure term shrinks, the price of going faster paid in pressure. The sum of the three keeps one height on the bars, and when one side grows the other gives.
In symbols, p + ½ρv² + ρgh = constant, the Bernoulli equation. Drag the speed v and the pressure p follows down at once. This is the story when the flow is steady, there is no stickiness, and the water is not squeezed, and this sum is one value only on the same streamline.
Look at an atomizer and the principle shows as it is. Blow air fast over a tube and the pressure above falls, so the liquid below is sucked up and sprayed, and when you drag the air speed up that process shows. Fast air empties the pressure and pulls the liquid up.