Bernoulli's pitot tube jet question

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SUMMARY

The discussion centers on calculating the flow rate and angle θ of a jet using Bernoulli's principle and a Pitot tube. The participant correctly identifies that the horizontal component of velocity remains constant and applies the equation v=sqrt(2gz) to relate velocity to height change in the Pitot tube. The fluid in question is water with a density of 1 g/cm³, which is noted to be irrelevant for the calculations of horizontal and vertical velocities. The consensus confirms the accuracy of the participant's approach and emphasizes the constancy of horizontal velocity in fluid dynamics.

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  • Understanding of Bernoulli's principle
  • Knowledge of fluid dynamics and Pitot tube functionality
  • Familiarity with basic physics equations involving velocity and height
  • Concept of fluid density and its implications in calculations
NEXT STEPS
  • Study the application of Bernoulli's equation in various fluid flow scenarios
  • Learn about the design and calibration of Pitot tubes in aerodynamics
  • Explore the relationship between fluid density and pressure in fluid mechanics
  • Investigate the effects of different fluids on flow rate calculations
USEFUL FOR

Students in physics or engineering disciplines, particularly those focusing on fluid dynamics, as well as professionals working with fluid measurement and analysis using Pitot tubes.

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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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