Bernoulli's pitot tube jet question

AI Thread Summary
The discussion revolves around calculating the flow rate and angle θ using a Pitot tube at the top of a jet. The user expresses uncertainty about their calculation of the horizontal velocity component, assuming it remains constant throughout the trajectory. They reference the height change in the Pitot tube being related to this velocity, using the equation v=sqrt(2gz). Other participants confirm that the user's approach is correct, noting that the density of water does not affect the calculations since it cancels out. The conversation emphasizes the constancy of horizontal velocity in fluid dynamics applications.
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Homework Statement



The tip of the Pitot tube is at the top of the jet. Calculate the flowrate and the angle θ .

Homework Equations





The Attempt at a Solution


Please see attached images. 1 is the question and 1 is my attempted solution.

I think I perhaps have done the first part wrong where I calculate the horizontal component of the velocity. I assume the horizontal velocity remains constant throughout the trajectory, and that that the change in height z in the pitot tube is due to this velocity such that:

v=sqrt(2gz)

also, the lecturer has provided information that the fluid is water (density 1g/cm^3) and I haven't used this information hence my post here regarding my method.

Thanks for your time.
 

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I think everything you did was correct. I checked for the functioning of a pitot tube and your formula for vx corresponds to what they state. The density drops out both for vx and vy. Maybe you prof was alluding to the fact that the fluid from the spout and that in the manometer were the same substance. Had they not been, their relative densities would have come into play.

Good job!

BTW it's pretty obvious that vx stays constant, right? No different than throwing a ball in the air at an angle etc.

http://en.wikipedia.org/wiki/Pitot_tube
 
Last edited:
thanks for your time rude man
 
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