Derivation of Torricelli's Law From Bernoulli's Equation

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I am reading through my fluid mechanics book and there is a derivation of Torricelli's theorem i.e. [itex]V = \sqrt{2gh}[/itex].

The author's pick the datum line at the middle of the jet and show that:
[itex]h = \dfrac{p}{\gamma} + \dfrac{V^2}{2g}[/itex]
where [itex]h[/itex] is the distance from the jet to the surface of the liquid.

The author's then assume that the streamlines are straight and parallel and say that the acceleration of the fluid is due only to the pressure from above and below the infinitesimal particle. Why is this so? Why doesn't the horizontal pressure have any effect on the acceleration?
 

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I am reading through my fluid mechanics book and there is a derivation of Torricelli's theorem i.e. [itex]V = \sqrt{2gh}[/itex].

The author's pick the datum line at the middle of the jet and show that:
[itex]h = \dfrac{p}{\gamma} + \dfrac{V^2}{2g}[/itex]
where [itex]h[/itex] is the distance from the jet to the surface of the liquid.

The author's then assume that the streamlines are straight and parallel and say that the acceleration of the fluid is due only to the pressure from above and below the infinitesimal particle. Why is this so? Why doesn't the horizontal pressure have any effect on the acceleration?
So you need derivation , that's all huh ? I am giving that right away. :wink:

Let P be the atmospheric pressure , ρ be the density of liquid and v be the velocity of fluid coming out of orifice of a container. Depth of orifice from free surface is h and total depth till bottom is H. Pressure of liquid at free surface and orifice is atmospheric and has nothing to do with horizontal velocity coming out of orifice. Liquid coming out has both kinetic energy and potential energy while liquid at free surface has only potential energy.

According to Bernoulli's theorem , sum of pressure and total energy per unit volume at surface and every point of orifice is constant.

p+0+ρgH = P+ρv2/2 + ρg(H-h)
Since potential energy of free surface has to convert to kinetic energy of motion.

ρgH = ρv2/2

Now solve for v.
 
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