in this video we're given a large tube filled with water and we're using a density of 1000 kilograms
per cubic meter for that and we've placed a metal cylinder on top of the water and this is sealed
with o-rings so the water can't leak past it and the pressure at the top of the water column has
to be sufficient to hold up that metal cylinder so this makes the pressure greater in the water
than it would be otherwise i'm going to call that point number one at the top of the water column
now we're going to be working in gauge pressure through this entire problem and so i could label
the pressure up here at the top of the metal cylinder as atmospheric pressure but in the
gauge pressure units that's zero and in part a we're interested in computing the hydrostatic
gauge pressure at the bottom of the column and i'll call that point number two so that's going
to be the pressure at the top of the water column plus the additional rho gh component from going
deeper into the fluid now that's why the valve is closed at the bottom of this tube and in part b we
open that valve and we find the speed of the water coming out of the tube i'm going to call that 0.3
and just while i'm labeling things let's make a note of the fact that p3 is equal to atmospheric
pressure which again in gauge pressure units is 0. so we start working on part a we want the
hydrostatic pressure at point two at the bottom of the water column and the first thing we have
to do is get the pressure at point one which is sufficient to balance that metal cylinder so this
requires a force analysis on the metal cylinder and the force of gravity is pulling down on that
with a magnitude of mg and then the pressure at point 1 is exerting a force upward on that
and i'll call that f1 now just a quick reminder here if we want to relate pressure to the force
exerted by the pressure we would go back to the definition pressure is force per unit area
which means you can always compute the force exerted by a fluid by taking the pressure and
multiplying by the area that it's exposed to and this means f1 is going to be equal to p1
times the cross-sectional area of that cylinder so let's go ahead and express the fact the forces
balance on this metal cylinder in other words mg pointing down it's got to be equal to p1
times a pointing up and we're interested in solving for p1 here so p1 is going to be
mg over a where a is the cross sectional area of that cylinder so we plug in the
mass of the cylinder that's 0.825 kilograms just making a quick unit conversion to s high units
g is 9.8 meters per second squared and then the cross-sectional area of this thing well it's a
tube with a diameter of 5.2 centimeters so we can use one-quarter pi times diameter squared
that's equivalent to pi times radius squared it just doesn't require you to convert to radius
first and the diameter in si units is 0.052 meters now when we run the numbers we get 3807 pascals
just keeping a little extra precision and again that's the pressure at point one at the top of the
water column now we have to find the pressure at point two and that's given by the pressure at the
top of the water column that's p one plus rho g d where d is the depth of the fluid so we get three
thousand eight hundred seven pascals at the top of the water column plus a 1000 kilograms per cubic
meter for the approximate density of water times 9.8 meters per second squared for little g and
then the depth in the fluid is 0.25 meters when we run the numbers on this we get 6257 pascals
at 0.2 and i traditionally round final answers to 3 sig figs so this is 6260 pascals
now in part b we have a fluid dynamics problem and the idea here is to apply bernoulli's equation in
order to find v3 the speed of the fluid coming out of this little valve we're told we can ignore all
effects of friction and we can ignore the kinetic energy density in the tube itself because it's so
large compared to the size of the valve so that water's not going to be dropping very fast and
there's two different ways we could go here one way we could go is to say that all right let's
look at the energy density at point number two provided we're far enough away from that valve
the velocity of the fluid there is still going to be negligible and then we could compare that
energy density to the energy density at the opening in the valve the other way is we could
compare point number one to point number three and either way we're going to get the same answer
so what i'm going to do is illustrate what happens if you compare point number one to point number
three and then we'll show it's equivalent to comparing point number two to point number three
so our first approach is the point number one to point number three approach and we
apply bernoulli's equation and so there it is and i'll post a link to where i originally discussed
bernoulli's equation but what we're doing here is saying the energy density in the fluid is the same
at these two locations and there's three types of energy density pressure gravitational potential
energy density and kinetic energy density and so at point one we've already calculated the pressure
we'll be able to get the gravitational potential energy density with the y-coordinate there
i'm going to use the simplest coordinate system i can think of which would be putting y equals 0
at the bottom of the tube and y equals 0.25 meters at the top of the tube and then the
kinetic energy density at the top of the tube we were already told we could approximate this away
so the tube is large compared to the valve at the bottom so the fluid velocity is so small that we
don't have to worry about the kinetic energy part now what about p3 that's atmospheric pressure
which engage pressure units gives me zero so that one's gone and then the gravitational potential
energy density component that's gone too because i set up my coordinate system so that y is zero
at that location so now we just have to solve for v three so i multiply by two on both sides
divide by rho and take the square root so there's our general expression for v3 and we just sub our
numbers into this and p1 was 3807 plus a rho g y1 so 1000 times 9.8 times 0.25 all divided by the
density of the fluid approximately one thousand into three sig figs i get three point five
four meters per second now just a quick note on the other way to do it we could have compared
point number two to point number three provided we're far enough away from that valve where we
don't have to think about the actual kinetic energy density in the water as it transitions
from practically motionless over on the left hand side to moving at v3 over on the right hand side
and what i notice in the expression that i already wrote down is that this piece in parentheses
is really just p2 that's p1 plus rho gh we already computed this thing so if i write down bernoulli's
equation here i have p2 i computed that in part a my gravitational potential energy density is zero
at point number two my kinetic energy density is also zero provided i stay away from that valve
p3 is still zero because that's atmospheric pressure and we're using gage pressure
units and then the height at point three is also zero solving for v3
i just get 2p2 divided by rho square rooted and again p2 is really just all of this stuff
right here the pressure at point one plus rho gh symbolically it's this stuff right here so
i know that number is the same when i plug it in and this means i arrive at the same speed at
point three either way i do the problem if you enjoyed this video or at least found it useful
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