In this illustration, we'll be calculating
the force on a tube ejecting a liquid. here
the figure shows a tank of which side wall
is having an attached narrowing tube of which
cross sectional area, decreases from s 1 to
s 2. and we are required to find the horizontal
force acting on the tube due to ejection of
the liquid. so in this situation, first thing
we can directly write down as the tube is
at a depth h bellow the open, surface of the
liquid. here we can directly write down that,
the ejection velocity, of liquid is. this
ejection velocity we can directly write down
this will be root 2 gee h. and if we talk
about this tube here we can redraw. the enlarged
view of this tube. where we can consider,
if this cross sectional area is s 1 this is
s 2. here the radius, is b and here we consider
radius to be, ay, and here liquid is coming
out with the velocity v which is given as
root 2 gee h. and this can be written as v
2, which is coming out over here from, the
cross sectional area, s 2. now if, at a distance
x inside the tube we consider small elemental
section. which is of width d x. and its radius
if we consider as r then we can write. the
radius of, elemental. ring or we can also
call it strip. considered, as shown, is given
as. this radius can be given as, r, as it
is varying linearly from b to ay, we can write
it b plus, b minus ay by l times, x. as we
have got the velocity. we have got the radius
of this section we can find out the velocity
of fluid at the section. so we can write velocity
of, liquid. at elemental section, is, this
we can calculate by continuity equation, applied
between this point and this point, here we
can write v pie r square should be equal to
v 2 pie b square, so this will give us the
value of v. as, pie b square multiplied by
root 2 gee h divided by pie, r square, r we
already calculated. now, using. bernoulli's
theorem. at element. and s 2, we get. if at
element in s 2 we apply bernoulli's theorem,
here we can see at s 2. pressure is say p.
plus, kinetic energy of fluid is half ro v
square, must be equal to at s 2 pressure is
p not, plus kinetic energy is half ro, multiplied
by ro 2 gee h whole square. so from this relation
we'll get the value of pressure difference,
p minus p not. which is equal to, ro g h.
plus, ro g h here we can substitute, when
we substitute the value of v over here this
gives us b to power 4 by, r to power, 4. so,
this is the pressure difference between, this
element pressure and the atmospheric pressure.
now we can see the pressure over here will
exert a force d f, on this elemental strip
in normal direction. of which vertical component,
will get canceled out and horizontal component
will be added up. so if this angle is theta
we can consider this angle is theta. and here
we can write the net force acting on this
tube will be d f sine theta integration, so
we can calculate. the, total, horizontal force.
on tube is, this force we can write as integration
of d f sine theta. and if we calculate the
value of d f it'll be pressure difference
multiplied by the area of, this element, so
this p minus p not, multiplied by if its radius
is r we can consider it 2 pie, small r multiplied
by the width, as its, along the axis width
is d x this will be taken as d x by coz theta.
so this will be d x by coz theta, multiplied
by sine theta. and, here x we integrate from,
zero to the total length. of the tube, but
rather then integrating it in d x you can
see, the value of d x tan theta this sine
theta by coz theta is tan theta. so d x tan
theta will be the elemental increase in radius
we are getting, so we can write this value
as integration of p minus p not. multiplied
by. 2 pie, r, and d x tan theta we can write
as d r, and the limits of radius from 1 point
to another we can change from, ay to b. so
if we calculate this value of force, this
you can see p minus p not will substitute
from here, which is ro g, h. 1 minus, b to
power 4 by, r to power 4. multiplied by 2
pie r d-r, and we integrate this term from,
ay to, b. we have, less space but, i'll try
to integrate and give the result and try to
evaluate it carefully here. so, this gives
us, after integration. the first term will
be getting here is, ro g h pie also can be
taken as a constant and integration of r will
be r square by 2. so after substituting limits
this becomes b square. minus ay square. and,
the next term we are getting is b to power
4 by this is r cube when we integrate this
minus 1 by 2 r square. so this becomes plus.
when we integrate this this becomes plus.
and this can be given as. ro g h. pie, b to
power 4 is also a constant. and when we substitute
the limits, this will be 1 by, ay square,
minus 1 by, b square because integration of
1 by r cube will be minus 1 by 2 r square
1 minus will be changing this sign to plus,
and substituting the limits from ay to b gives
us this value. and in this situation when
we further simplify this result. and substitute
here s 1 as pie, ay square and s 2 as pie
b square. on simplifying this result. the
final result you will be getting is ro g h
multiplied by, this will give us, s 1 minus
s 2 whole square, divided by, s 1. i am leaving
it as an exercise for you to evaluate and
verify the result by substituting s 1 as pie
square and s 2 as, pie b square the final
result we'll be this which is the final answer
for this problem.
