Flow Through Orifice Homework: Bernoulli's & Motion Eqs

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samjohnny
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Homework Statement



Please see attached.

Homework Equations


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Bernoulli equation written as p1/(rho g) + v1^2/(2g) + z1 = p2/(rho g) + v2^2/(2g) + z2, with pressure, velocity and elevation terms on each side.

v = u + at
s = ut + ½ at²
v² = u² + 2as

The Attempt at a Solution



I've managed to get the first part of the question applying Bernoulli's principle and then the equations of motion above by considering separately the vertical and horizontal.

As for the second part, I'm not sure how to go about it. I believe that I would simply need to plug what's known into a certain formula and then solve for h; I expect that would yield a quadratic equation with two solutions for h. But I'm not sure what I need to be using to get this?
 
Attachments
  • Diagram of a water tank of height H with a horizontal hole at depth h, water jet curving to the floor with horizontal distance x marked from the tank base.
    Flow-through-orifice.JPG
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Suppose you get a distance x if you drill the hole at h=h1. What distance do you get if you drill the hole at h=H-h1?

Chet
 
Thank you for the replies! I've managed to work through and get the answer; it was so simple but I just couldn't see it. For the last part of the question though I'm not too sure how to proceed. Essentially I'm trying to maximise the distance x, but I'm not sure how to do that. I'm assuming I could either differentiate an equation for x and set it to zero and solve, or make an assumption to maximise x. However I'm not sure on an equation to use, or an appropriate assumption to make.
 
You already have an equation for x as a function of the relevant variable. I suggest you do what you just suggested.
 
Thanks for the reply. Ok so I got the depth at which the maximum distance occurs as being at a height H/2 below the surface. Is that correct?
 
samjohnny said:
Thanks for the reply. Ok so I got the depth at which the maximum distance occurs as being at a height H/2 below the surface. Is that correct?
Yes.