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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. enter image description here

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?

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  • $\begingroup$ Please see this related post and explicitly stated assumptions in the answer on distinguishing between the hydro-static and hydro-dynamic pressure as well as application of the Pitot tube to compute it. $\endgroup$
    – kbakshi314
    Commented Apr 24, 2021 at 7:27
  • $\begingroup$ Why would you assume that $d$ is of any relevance at all here? Surely, for zero flow the water level in the tube should be level with the water surface, so the flow would only lift the level by $l$. Unless I misunderstand something here. $\endgroup$
    – Thomas
    Commented Apr 24, 2021 at 15:29

1 Answer 1

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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?

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  • $\begingroup$ I guess every term on the left side in those equations turn to zero apart from the first term in the first equation, as we only have velocity in the x-direction? $\endgroup$
    – Jmei
    Commented May 13, 2018 at 7:53
  • $\begingroup$ That doesn't answer your questions. What would these equations reduce to and what would they tell you if (a) w were zero everywhere (b) w were constant everywhere, with u and v were independent of z? $\endgroup$ Commented May 13, 2018 at 12:12

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