Pitot tube, 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 1/2 (points are at same height) is d. At point 1 the fluid velocity is V1 and at point 2 it is zero (stagnation point). Calculate the water velocity V1. (Figure below for help)

First I calculate the stagnation pressure Ps, by using Bernoulli from 1 to 2. This yields:

(1/2)*V1^2 + P1/rho = 0 + P2/rho

Ps = rho*(1/2)*V1^2 + P1

Then I calculate the pressure through the tube, where we have hydrostatic conditions. P0 is the atmospheric pressure.:

Ps = P0 + rhogl + rhogd.

My question is:

under which conditions can we assume that P1 = rhogd, i.e. under which 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 ONLY horizontal?

Chet Miller · Answer

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

$$ufrac{partial u}{partial x}+vfrac{partial u}{partial y}+wfrac{partial u}{partial z}=-frac{1}{rho}frac{partial p}{partial x}$$

$$ufrac{partial v}{partial x}+vfrac{partial v}{partial y}+wfrac{partial v}{partial z}=-frac{1}{rho}frac{partial p}{partial y}$$

$$ufrac{partial w}{partial x}+vfrac{partial w}{partial y}+wfrac{partial w}{partial z}=-frac{1}{rho}frac{partial p}{partial z}-g$$
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?

Pitot tube, assumption of hydrostatic pressure distribution

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