What is the static pressure in this Bernouilli experiment?

Question

I have done an experiment in my class for proving the Bernouilli equation. The instrumentation in this video was similar to the one I used in my practice. The water flows through a horizontal Venturi pipe which has manometric tubes connected in a direction tangential to that of the fluid.

I simplified Bernouilli equation $p_1+\frac{1}{2}\rho v_1^2 +\rho gz_1=p_2+\frac{1}{2}\rho v_2^2 +\rho gz_2$ and $ \ z_1=z_2; \ \ p_1=\rho g h_j $ where $p_1$ is the static pressure measured by the manometric tubes, so Bernouilli equation can be simplified to: $h_1+\frac{v_1^2}{2g}=h_2+\frac{v_2^2}{2g}$. So the first term will represent the static pressure and the second the dynamic pressure.

My main problem is that I don't understand what is the static pressure in this case and why it can be measured by the manometric tubes. If I open the water inlet tap more, the flow rate increases, so it is logical that the dynamic pressure increases because the velocity increases, but why the static pressure increases too? And why If I close the water outlet faucet more, the static pressure increases as well?

Answer 1

Perhaps this might help. Just brainstorming here, since I also have the same question. My interpretation is the following. If you graph pressure versus height, you will get a linear correlation. The graph line will be of the form Y = mx + b or Pressure = m(Height) + b. Therefore, we can make the observation that static pressure would represent the offset of this graph, meaning b = Static Pressure. An interpretation would be that static pressure is the initial pressure at height zero or the very surface of the liquid. Making it, in most cases, the atmospheric pressure or "atmospheric weight" on the liquid.