In this illustration we'll analyze, fluid
ejecting out from a rotating tube. here the
figure shows a horizontal tube a-b. of length
l, and rotating about a vertical axis o o-dash,
passing through its open end ay at an angular
speed omega. the tube is filled with amount
an ideal fluid. and we are given that due
to rotation fluid comes out from a small orifice
as at end b, of tube, we are required to find
the ejection velocity of the fluid, from the
orifice as a function of column height h which
is filled with a liquid. here, to analyze,
the flux velocity coming out. we can analyze
first the pressure distribution in the liquid
and apply bernoulli's theorem, from open end
to, the closed end, where, liquid is coming
out. or we can also apply bernoulli's theorem
just inside and outside points. of the orifice.
so if we redraw the figure situation is this,
here we are having a tube. and in this situation
if, the liquid is filled upto, a column height
h. and the total tube length is l. then to
analyze the pressure, in the liquid we consider
an elemental section. of width d x, within
the liquid. we are taking this section at
a distance x and the width of this section
we consider as, d x. now in this situation
if we consider pressure on, 1 side, of the,
element is p. then on the other side the pressure
should be p plus, d p. because pressure would
be continuously increasing here. so we can
write, first the pressure difference, across.
the, element. of width. d x. as shown is.
we can write this should be d p and the pressure
difference here is created due to the centrifugal
force acting on it. in the frame of tube as
tube is rotating at omega. and at this point
the centrifugal acceleration will be omega
square x with respect to tube. so, we write
this d p is equal to. d x. ro, and instead
of g we take it as omega square x. and i hope,
you have got how i have calculated this d
p because in vertical column we consider it
d x ro g here instead of g i am taking this
centrifugal acceleration is effective gravity.
on this elemental column of width d x, in
horizontal direction in radially outward direction.
so here if we integrate this d p. this will
give us, ro omega square integration of x
d x. and we can integrated from this point
to where, the value of x is l minus h. to
the final end b which is l. and here pressure
we can write as p atmospheric, which we can
write as p not. here we can consider the atmospheric
pressure. to be equal to, p not, here we can
consider p atmosphere to be p not. and at
end b. the pressure we can consider as p b.
so this gives us the value of p b which is
p b minus p not is equal to. integrating this
will give us, ro omega square. this is x square
by 2. if we apply limits from l minus h to
l. the result will be half. ro omega square
h, multiplied by 2 l minus h, you can subtitute
the limit and simplify. now, once we have
got the value of pressure at point b in terms
of atmospheric pressure, just inside and outside
b we consider 2 points x, and y. here, we
can write now, we apply. bernoulli's theorem.
at points. x and y, which are just inside
and outside the orifice b from which, the
water is ejecting out at a speed v. then we
can write at point x, pressure is p b. and,
we can neglect the velocity of fluid at x
because the area of tube is very large compare
to the area of orifice which is given to be
small. so, on the righthand side at point
y we can write pressure is p atmospheric plus
kinetic energy. per unit volume of the fluid
is half ro, v square. so, in this relation.
we'll get the value of flux velocity to be
equal to under the root. twice of, p b minus
p not, divided by, ro. and here we can substitute
the value of p b minus p not which we already
calculated. so this will give us the value
of v is equal to. when you substitute this
value over here, this will give us omega h.
root of, 2 l by h. minus, 1. that will be
the result of this problem.
