In this illustration we'll analyze the water
pressure in a reservoir.
here the figure shows a large water reservoir
of depth h, and on one side of dam a horizontal
pipe of diameter d is closed by a plug here
we can see in the figure.
the pipe is at a depth h bellow the water
surface.
and we are required to find the friction between
plug and pipe wall and the volume of water
coming out from pipe in time t if the plug
is removed.
here, we can directly write.
the pressure.
acting.
on, plug. will be.
here we can write if this is the plug. then
from the right-side it is experiencing p atmosphere.
and from the left side, in this dam wall,
we can say as here, water is filled.
the pressure will be p and the value of p
will be p atmospheric plus.
h ro g as it is at a depth h bellow the, surface
of water in this situation.
so here we can say.
the net pressure acting.
on the plug will be toward right.
and this will be, equal to, if this p 1. we
right it p 1. this will be p 1 minus p atmospheric
that will be the net outward pressure acting
on the plug. this we can write as net, outward.
pressure.
so, when we substitute p 1 this is simply
h ro g. and if we calculate, the force of,
friction. on plug. this will be equal to,
the force due to.
pressure of water.
as friction is keeping this plug as at rest
so water is the excess pressure exerted by
water on it.
that will be balanced by friction.
so this can be simply given as h ro g multiplied
by its cross sectional area we can write as
pie d square by 4.
so, the value here will be pie.
h ro, g. d square, divided by 4, that is 1
result of the problem.
and, we are required to, find the volume of
water coming out from pipe in time t if plug
is removed.
so we can say if we consider, the area of,
this narrow tube is negligible. in that case
negligible compare to the, total area of the
dam so here we can say.
after.
plug is removed.
velocity of efflux. will be, this we already
studied this is written as root 2 gee h. by
torricelli's theorem we already calculated
it.
so, this implies here we can get the flow
rate, of water at which it'll come out will
be equal to, velocity multiplied by cross
sectional area.
which is pie d square by 4 multiplied by,
root 2 gee h. and this implies, if we calculate
volume, flowing out.
in time t is.
this volume we can write, as, this is the
flow rate multiplied by time.
so, this can be simply written as pie, d square
by 4 multiplied by root 2 gee h multiplied
by t, that is the result of this problem.
