Physics Help Forum Static Pressure

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 Jun 5th 2009, 03:55 AM #1 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 Static Pressure First, is it correct that $\displaystyle P+\rho gh$ in Bernoulli Principle the static pressure? For the problem, the answer is 2400Pa By applying Bernoulli Principle,$\displaystyle P_1+\rho gh_1+\rho v_1^2/2=P_2+\rho gh_2+\rho v_2^2/2$ v1=4ms^-1, v2=2ms^-1 If the static pressure is as mentioned above, the difference in static pressure should be simply the change in dynamic pressure 0.5(1200)(4^2-2^2)=7200Pa But the answer of difference in static pressure is 2400Pa which is calculated from P2-P1 =7200-$\displaystyle \rho gh$ =7200-1200(10)(0.4) Why is the answer P2-P1 but not simply the difference in dynamic pressure? Is the answer wrong? Attached Thumbnails
Jun 5th 2009, 06:24 PM   #2
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Join Date: Apr 2008
Posts: 815
Hey werehk!
This is a good problem, I wasn't aware of the difference of total pressure, static pressure and dynamic pressure.
I also get your answer (that is, if I consider that the streamline of A is the same than the one of B. I know I can't deduce this, but well...). From wikipedia (Static pressure - Wikipedia, the free encyclopedia), we can read
 Static pressure and dynamic pressure are likely to vary significantly throughout the fluid but total pressure is constant along each streamline.
so I equaled $\displaystyle P_A+\rho \frac{v_1 ^2}{2}$ with $\displaystyle P_B + \rho \frac{v_2^2}{2}$. From it I got that $\displaystyle P_B-P_A=7200 Pa$.

On the other hand, if I use Bernouilli's equation : $\displaystyle P_A+\rho g h_A + \frac{\rho v_A^2}{2}=P_B+\rho g h_B + \frac{\rho v_B^2}{2} \Leftrightarrow P_B-P_A=\rho (gh_A+\frac{v_A^2}{2}$$\displaystyle -gh_B-\frac{v_B^2}{2})=\rho (gh_A+8-gh_B-2)=\rho(6+g(h_A-h_B))=1200(6-4)=2400$. (I didn't carry the units because it would have been too long to type)
I hope I didn't make any error at the end. So I better wait to see if anyone has the answer.
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 Oct 4th 2009, 02:40 AM #3 Banned   Join Date: Aug 2009 Location: UK Posts: 240 Guys, P=absolute pressure at a point Po= atmospheric pressure at a point pv=velocity pressure at a point Ps= static pressure at a point Pt = total pressure at a point Ps = P-Po Pv = (1/2)pv^2 (p) = density Pt = Ps + Pv I think i would side with Werehk and possibly suggest the answer was wrong. I think Bernoulli is/was a very clever person, but i think other people get mis-conceptions of his principle, another concept to conclude is the compressibility factor. Water is regarded as incompressible when in fact in can compress by some 20-25%. If you don't believe this then get a tube with base/glass, fill it half way with water, get a beer mat or cardboard etc on the top then turn it over, your see it compresses by 20-25%, this will in turn have an effect on the related pressures, (1/2)pv^2, hence static pressure will change.If you dropped a steel tank into the deepest part of the ocean, with the valves open to let the water in then turned the valves off when it hit the bottom, then brought the tank to the surface, the water would expand until it blew either the tank or the valves open, unless the tank or valves could take serious pressure. I realise what your question will be! what has this got to do with Bernoulli principle? well think about it! I hope this helps. Last edited by Paul46; Oct 4th 2009 at 12:26 PM.
 Oct 5th 2009, 12:09 AM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 Paul, Water cannot be compressed 20 -25 % . If it were so all the hydraulic systems which are used universally wouldn't work. Please google and check some reliable source. In fact the compressibility is so low, it can be safely treated as incompressible for practical purposes. Let me quote For each atmosphere increase in pressure, the volume of water would decrease 46.4 parts per million. I remembering you mentioning a video on you tube i think which shows the claim of 20 to 25 %. I havent seen it myself, but in all likelihood it could be a trick or twisted interpretation. Dont go by what floats on you tube unless it is in a lecture demo by some known institute like MIT, Stanford etc. There is circuitry floating around with tall claims which destroy laptops if connected to them!
 Oct 5th 2009, 10:40 AM #5 Banned   Join Date: Aug 2009 Location: UK Posts: 240 physicsquest, Have you tried the trick/test? honestly water does compress alot going by that test, i'm only guessing 20-25% i'll have to do an accurate reading some day. In no way am i trying to disagree with you physicsquest, there must be some reason for it, but i urge you please to try the test out for yourself then please give me an explanation for it? Last edited by Paul46; Oct 5th 2009 at 10:50 AM.
 Oct 5th 2009, 11:35 PM #6 Physics Team   Join Date: Feb 2009 Posts: 1,425 Please give me the link again. Will try to see it when i manage to access you tube. And could you explain the experiment a little more clearly with reference to the procedure and terms? If you don't believe this then get a tube with base/glass, fill it half way with water, get a beer mat or cardboard etc on the top then turn it over, your see it compresses by 20-25%

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