Speed of water and height involving fluid mechanics

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Homework Help Overview

The problem involves fluid mechanics, specifically analyzing the speed of water exiting a nozzle from a tank and the height it reaches after leaving the nozzle. The scenario includes parameters such as absolute pressure, depth of the nozzle, and assumptions regarding air resistance and viscous effects.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using energy conservation principles to determine the velocity of water exiting the nozzle and the subsequent height it reaches. There are questions about the application of Bernoulli's equation and the implications of nozzle size on the analysis.

Discussion Status

The discussion is active, with participants providing various approaches to the problem, including energy conservation and potential concerns regarding the assumptions made about the nozzle size. Some participants seek clarification on the equations involved and how they relate to the problem setup.

Contextual Notes

There is mention of assumptions such as ignoring air resistance and viscous effects, as well as the implications of the nozzle being small versus not small, which may affect the application of fluid dynamics principles.

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Homework Statement



In a very large closed tank, the absolute pressure of the air above the water is 6.01x10^5 Pa. The water leaves the bottom of the tank through a nozzle that is directed straight upward. The opening of the nozzle is 4.00m below the surface of the water. a) Find the speed at which the water leaves the nozzle. b) Ignoring air resistance and viscous effects, determine the height to which the water rises.

Thanks!

Homework Equations



P+rhov^2/2+rhogh= constant

P2=P1+rhogh

kinematics?


The Attempt at a Solution



I know that I can set the initial height to zero, but that's about it. I really don't know how to treat this problem.
 
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Use energy conservation to get velocity(u) of water just outside nozzle.You will get u=sqrt(2gs). Assuming nozzle is very small.Here s is distance of nozzle from topmost layer of water(4.00m). After getting velocity use kinematic equation (v^2)=(u^2)-2gh.
For highest point v=0.Thus getting h=u^2/2g.
Which is same as s(4.00m) as it should be from energy conservation(similar to ball thrown in air and returning back to Earth and again reaching same height h after colliding from ground(assuming it to be elastic collision).
 
1994Bhaskar said:
Use energy conservation to get velocity(u) of water just outside nozzle.You will get u=sqrt(2gs).

Can you show me what the entire (energy conservation) equation is before you simplified it?
Is it Bernoulli's equation?

thanks
 
Just take a small element of water of mass dm at a height s above the nozzle.
Initial energy=Potential Energy=dm*g*h
Final energy after just coming out from nozzle(assuming it to be at height 0)=0.5*dm*u^2
Equating final and initial energy you can get u=sqrt(2gs).
Here however we assume that nozzle is very small and let's only point particles out.Only then we can do mechanical energy conservation.
 
Can you tell me that what would you do if nozzle is not very small, i.e. it has a small area 'a' and let's finite amount of water out(not point sized particles)??You also know area of topmost layer of water as 'A'.
Think it yourself.The answer becomes more interesting.
Hint:In that case don't apply energy conservation.You will apply only equations of fluids-->
Bernoulli and Equation of continuity.
 

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