Misconception in fluid (HELP!)

Atheist

2010-12-08 3:28 am

Please help, my teacher is a 誤人子弟 so that I can hardly learn anything from him. I really, really need a help, so please..

There are two questions I want to ask.

Consider a pitot tube measuring a flow of water, P(s) and P(t) denote the static pressure and the total pressure of two points at the same level that below two open tubes.
1) Since static pressure P(s) is acted by fluid at rest, why static pressure P(s) can also be measured from moving fluid? Now the static pressure P(s) in this case is then higher than that when the fluid is at rest about pgh. Why?

2) Consider the tube that measuring the static pressure of water. Whats wrong if I put two points which are at the same axis but have different positions of a tube (say, at the highest level of the water and that just at the mouth of the tube) into the Bernouli's equation ?
If I do so, I can obtain an equation: (Po+pgh) + (1/2)pv^2 = Po + pgh, where Po, v and p are atmospheric pressure, velocity of the flowing water and density of the water respectively.

更新1:

I know that there are some restrictions in using the Bernouli's equation but I can not thoroughly realize. Please tell what the misconception is.

回答 (1)

天同

2010-12-19 9:03 pm

✔ 最佳答案

1. Honestly, your question is not clear to me. The three terms in Bernouli's Equation denotes three different pressures for a fluid, namely the static pressure P(s), the dyamic pressure P(d) and the hydrostatic pressure P(dy). The term (pgh) is the hydrostatic pressure, which is given by the specific weight (weight per unit volume) of the fluid. It has no relation to the static pressure P(s) (which, according to the kinetic theory, is due to random bombardment of fluid molecules) of the fluid. I think the follwoing web-page enables you to have a better understanding of the Pitot Tube in measureing pressures in a moving fluid:

http://www.engineeringtoolbox.com/pitot-tubes-d_612.html

2. You said that the two points are at the same axis, there is no difference in elevation of the fluid (I suppose your axis is a horizontal one). The term (pgh) is the same on both sides. Then your equation is clearly not correct, as you would have,.

Po + (1/2)pv^2 = Po

There is no difference in pressure to cause a fluid flow.Basically, the Bernouli's Principle is dervied from the Law of Conservation of Energy for an imcompressible fluid (hence, the continuity equation can apply). The underlying physical principle is that a pressure difference casues the fluid to flow.

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