###
Trajectory of Jet issued from an orifice at the side of a tank opened to atmosphere

Home ->
Lecture Notes
->
Fluid Mechanics -> Unit-II

At the tip of the opening:

The horizontal component of jet velocity V_{x} = (2gh)^{0.5 } = d*x*/dt

And the vertical component V_{z} = 0

One the jet is left the orifice, it is acted upon by gravitational forces. This makes the vertical component of velocity to equal ‘-gt’.

i.e., V_{z} = -gt = d*z*/dt

The horizontal and vertical distances covered in time ‘t’ are, obtained from integrating the above equations.

*x* = (2gh)^{0.5} t

and *z* = -gt^{2}/2

And elimination of ‘t’ can be done as,

*z* = -g [*x*^{2}/(2gh)] / 2

i.e,

*z* = - *x*^{2}/(4h)

Let us take downward direction as positive *z*. Then

*x* = 2 (hz)^{0.5}

Table of Contents

HOME

Last Modified on: 14-Sep-2014

Chemical Engineering Learning Resources - msubbu

e-mail: msubbu.in[AT]gmail.com

Web: http://www.msubbu.in