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Fluid Mechanics: Bernoullis Theorem Demonstration? 
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#1
Oct2710, 08:59 AM

P: 1

Having trouble using the bernoulli equation to work out the dynamic head for this experiment.
the equation I am using is as follows: P1/ρg+(v1^2)/2g+z1=P2/ρg+(v2^2)/2g+z2 Where: P= static pressure v = fluid velocity z = vertical elevation of fluid so z1 = z2 for horizontal tube so equation is now: P1/ρg+(v1^2)/2g=P2/ρg+(v2^2)/2g now the measurements taken for static presure head (h), in meter which is related to P using this realtionship: h = P/ρg which allows the bernoulli equation to be rewritten as: h1+(v1^2)/2g=h2+(v2^2)/2g now i am having trouble rearanging this part of the equation to work out the the dynamic head for the experiment 


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