Pitot tube and the assumption of hydrostatic pressure distribution

Question

We have water flowing in an open channel. A small tube is placed in the channel, and the water raises to a height $l$ above the water surface. The distance from the water surface to point $1$ OR $2$ (points are at same height) is $d$. At point $1$ the fluid velocity is $v_1$ and at point $2$ it is zero (stagnation point). Calculate the water velocity $v_1$.

Figure above depicts the flow of interest.

First, I calculate the stagnation pressure $P_s$, by using Bernoulli's application for the flow from $1$ to $2$. This yields

$$\frac{1}{2}v_1^2 + \frac{P_1}{\rho} = 0 + \frac{P_2}{\rho},$$

$$P_s = \rho\frac{1}{2}v_1^2 + P_1.$$

Then, I calculate the pressure through the tube, where we have hydro-static conditions as follows, where $P_0$ denotes the atmospheric pressure.

$$P_s = P_0 + \rho gl + \rho gd.$$

My questions are as follows. Under which conditions can we assume that $P_1 = \rho gd$, that is, under what conditions can we assume that the pressure at point $1$ is independent of the fluid flow at that point? Is it only when the fluid flow is purely horizontal?

Answer 1

Here are the Euler (differential force balance) equations for steady, incompressible flow of an inviscid fluid:

$$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}=-\frac{1}{\rho}\frac{\partial p}{\partial x}$$

$$u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}=-\frac{1}{\rho}\frac{\partial p}{\partial y}$$

$$u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}=-\frac{1}{\rho}\frac{\partial p}{\partial z}-g$$ where u is the (horizontal) velocity component in the x direction, v is the (horizontal) velocity component in the y direction, and w is the (vertical) velocity component in the z direction. What do these equations tell you about the answers to your questions?