Welcome to this screencast about the operation of manometers and hydrostatic head.
So the idea behind hydrostatic head is that is that if we have this column of fluid,
then that mass of fluid accelerated by gravity
will produce some force on whatever's sitting below here,
and that force divided by the cross-sectional area here will be equal to a pressure.
So if we want to look at that formally, we have this equation on the right here:
the pressure is equal to the force exerted by that fluid
divided by this area--which is the area of a circle, if it's a regular column of fluid.
So we can write this force as the density of fluid times the volume of fluid
which is the height of the cylinder times that area of the circle to make up the cylinder.
And then we have the acceleration of gravity--
and we optionally should probably go ahead and put in "gc" here,
in case we're dealing with English units. Then we can divide through by the area.
Of course the areas cancel, so then we have this
"rho", "h", times the gravimetric constant you see in the textbook.
So, again, this pressure is just caused by the weight of the fluid divided by the area.
And then potentially, if we just have vacuum above the fluid--
so no additional pressure--then this is all the pressure that's exerted at this point.
But we could have some kind of atmospheric or other pressure, like other fluid pressure
on top of this column of fluid-- and we would just add that force divided by an area,
which again would be a pressure,
to our total pressure term, associated with the column of fluid.
So now, let's consider a manometer and how it works.
It really just uses this principle.
So, again, the idea is that the height of fluid
is somewhat equivalent to a pressure, because we can relate
that height of a fluid by a force exerted by that amount of fluid.
So, in a manometer, if we have both ends of the manometer tube
exposed to atmosphere, then of course we expect the manometer
fluid to equilibrate out at the same level.
The pressures here and here should be equal, and should be equal to the atmospheric pressure.
But, now if we take one end of the tube, and hook it up to a gas source at another pressure,
then if that source is at a higher pressure, then we expect it to push down on the fluid--
so again, this is hooked up to pressure "P", while this is the atmospheric pressure--
so now, if "P" is greater than the atmospheric pressure,
then it's going to push down on the fluid and cause the fluid to rise up
against the atmospheric pressure. So the difference
between atmospheric pressure and this pressure
is equal to that height of fluid that was pushed up:
in order to lift that weight of fluid, you have to have a force differential. Or, in other words,
the pressure differential between this source with pressure "P" and this atmospheric pressure.
So we can just write the "P" minus "P atmosphere" is equal to "rho*g*h", where "h" is here.
This is how a manometer basically works.
Now we can have added layers of complexity. So as I mentioned,
this is for both sources being in the gas phase.
Let's suppose that we had a liquid-phase measurement we were going to do.
If we're going to do a liquid-phase measurement, then we would actually
have to worry about the height of fluid here as well, because that fluid could exert a force at this point.
So in addition to the pressure that you're feeling up here at this point,
this height of fluid is adding to that pressure, which would help to push this down. So we can write
that as the pressure plus, we'll call this distance, I think it may be listed as "d2" in the textbook, I'll go ahead and put that there.
So that pressure plus "rho*g*d2", where "rho" is the density of the fluid in the tank, or in the pipe that
you're measuring. So this is the total force that is being exerted at that point;
it's equal to the external atmospheric pressure on the other side of the manometer tube,
plus the extra weight of fluid that it's feeling at this point.
So, plus "rho*g*h", where this "rho" again is the density of the manometer fluid.
So the idea is that you should be able to draw a line straight across the manometer.
And if the pressures are equal, then you would expect the level to be the same on both sides.
That is, if the force being felt at both surfaces of the manometer fluid were the same,
then of course, the levels would be equal.
So the height of the manometer fluid-- and thus, the total volume of that liquid
and thus, the total weight of that liquid-- gives you a way of measuring the force difference
between this point and this point.