>> This is exactly the kind of
contrived example in fluid statics
that you might see on a quiz or an exam.
I can't picture this kind of
box existing in real life,
but it gives you a good chance
test out your skills.
So, there's three different
air pockets inside this box.
Down here, we've got some water
in this lower section down here,
and there's the interface
between the air and the water.
This is an interface between
the water and some oil.
And then an interface back
up here to the air again.
So this piece here, it's a solid wall.
Likewise, this piece here is a solid
and it's containing a pool of mercury,
something that would be a horrible safety
hazard if we were actually doing it outside
of an example out in the real world.
Then finally, there's another
surface here for the mercury.
So this surface 6 is a little lower
than the surface 5 for the mercury,
so there must be a higher pressure
of air pushing down on here to raise
that mercury surface up a little bit.
So our intuitive understanding of
what's going on physically helps us out.
And if we assume that P1 is 1 atmosphere
because this is open out to atmosphere,
what is the pressure going to be at
location 7 measured on this pressure gauge?
And we need to know some elevation information.
So let's say that location 1 is at a height 20
metres, location 2 is at a height of 12 metres,
location 3 is at a height of 7 metres,
location 4 at 15 metres, location 5 at 8 metres
above datum, location 6, 7
metres and location 7, 13 metres.
And these are all relative
to some datum line down here,
and it really doesn't matter what we're going
to use as our datum elevation because all
of our pressure differences are going
to depend on the change in elevation,
the delta h. We also need to
know something about the fluids.
Now that air, anyway, is
at atmospheric pressure.
These air pockets may be a little
different from atmospheric pressure,
but we don't know what they're at yet.
So let's just assume that for all of our cases,
the air density is 1.2 kilograms
per cubic metre.
And we'll come back and check on that later on.
Density of water, for the sake of round numbers,
let's make it 1000 kilograms per cubic metre.
It's a little bit cooler than room temperature.
The density of our oil is given
as 900 kilograms per cubic metre.
That makes sense.
It's a little bit lighter than the
water, it floats on top of the water.
And finally, the density of mercury is enormous,
about 13,550 kilograms per cubic metre,
making it vastly denser than any other
liquids, about 13 times the density of water,
and the water is about 800
times the density of air.
So I'm suspecting that maybe
the density of air isn't going
to matter in our calculation very much.
So, we want to find out what is
the difference between P7and P1.
Well, we could start by getting
P2 in terms of P1.
P2 will be equal to P1 plus the density of
the air because we're going from location 1
to location 2, positive because we're
going down, times g times h1 minus h2.
So I've set all of these as positive
and I'm accounting for the sign,
the fact that we're going down in an increasing
pressure by making sure that's positive there.
I can get P3 in terms of P2.
And P3, well, location 3 is lower,
we're going on a path through the water,
continuous path around under the
bottom here and back up to location 3.
So we'll have an increase in
pressure, so positive times the density
of the water times g times h2 minus
h3, again, making all of those positive
and accounting for the sign here.
Now the pressure at location 4 is going to
be lower than the pressure at location 3
because we're going up through the oil.
So the pressure at location 4 will be
equal to the pressure at location 3 minus,
to account for the fact that we're going up,
the density of the oil times g times h4
minus h3, it gives us a positive number.
So we're not tracking the signs, we're just
making sure that we catch everything here
in this sign to elaborate
the fact that it's going up.
Now, over the top here through a
continuous path through the air and down
to 5, the pressure will increase.
So P5 equal to P4, it's increasing so positive,
times the density of the air because we're going
through air times g times h4 minus h5,
so again, positive number h4 minus h5.
Then we can do P6 the same way we're going
from 5 to 6, we're going through the mercury,
a continuous path around underneath and
we've got an elevation change from 5 to 6.
P6 equal P5 and it's going to be an
increase in pressure, so positive,
times the density of the mercury times g times
h5 minus h6, again, keeping everything positive.
And finally, I can get P7 because I'm going
from 6 through the air up to location 7
at 13 metres off the ground, a little higher up.
So, that pressure is going to go down.
So P7 will be equal to P6 minus the
density of the air, because we're going
through air, times g times h7 minus h6.
So that's the elevation difference there.
And again, we've kept everything
positive so it all woks out.
Now, that means that I could take all of
this together if I summed all of these up
and subtracted the sum of all
of those intermediate pressures,
P2 through 6 from both sides, I'd wind up with
just P1-- or sorry, P7 equal to P1 plus, plus,
minus, plus, minus, minus, so
I can write P7 directly based
on all the delta Ps along the way.
So it'll be equal P1 plus rho
air g, h1 minus h2 plus rho h2o,
density of water times g times h2 minus h3,
minus density of the oil
times g times h4 minus h3,
plus for density of the air
times g times h4 minus h5,
plus the density of the mercury
times g times h5 minus h6,
minus the density of the air
times g times h7 minus h6.
Now, can we simplify it all here?
If we make the assumption that the density of
the air is much, much less than the density
of the others, then that term is going to
go away and that term is going to go away
and that term is going to go away.
And that's going to give
us half as much work to do.
We've only got to account for the
three changes through the three liquids
that have got much higher
density than the air density.
So, I wind with P7 equal to P1 plus about
1000, that's the density of the water,
times g times h2 minus h3, so
h2 is 12, H3 is 7, so that's 5.
Then I'll have negative for
the oil so minus density
of the oil is 900 times g times the
elevation difference between 3 and 4,
I'm going from 3 to 4, 3 is at 7 and 4 is at 15.
So that's 8 metres.
And then the last one I've
got is plus for the hg,
so plus 13,550 times g times the
elevation change between 5 and 6.
And 5 is at 8 and 6 is at 7, so that's 1 metre.
So you see, with each of these elevations,
I was able to take just the positive value
of the elevation and not have to worry
about the sign because I was keeping track
of it right here by whether I added or
subtracted, by knowing whether I was going
down in elevation or going up with in elevation.
And if I plug all of that in, I wind up
with P7 minus P1, the difference being equal
to 111,344 pascals or 111
kilopascal or about 1.1 atmospheres.
So, given that my range in pressure
variations went over about 1 atmosphere,
I might see a change by a factor of 2 in
the density of air between 1.2 and 2.4.
But I'm still in a situation
where this density of air is much,
much smaller than the density of the liquids.
So, I think and I'm perfectly valid here in
neglecting these terms for the density of air,
the pressure change from 1 to 2, pressure
change from 4 to 5, pressure change from 6 to 7.
And in most practical mechanical
engineering situations,
you can get away with making that assumption.
So here's a contrived example we've done.
This is the kind of thing you're likely
to see on a quiz or in an assignment.
It'll test out your ability to make
sure that you get the signs right here,
that you've got this understanding of
being to able to chain around and just walk
through from one end to the other to find
out what the pressure is in fluid statics.
And if you get to this answer at the end,
following the right path
will give you full marks.
So you should practice some questions like this.
