# Inverted U manometer

I have some doubts on inverted U manometer.

In the picture there is one U inverted manometer, the space above the liquid in the manometer is filled with air which can be admitted or expelled through the tap on the top, in order to adjust the level of the liquid in the manometer.

In this link the inverted manometer is described considering that the pressures at surface passing through $X$ and $X'$ is the same. For the situation I want to describe I added a tube at the bottom with fluid which is moving. I have two main questions.

$1.$ Why the isobaric surface chosen is the one passing through $X$ and $X'$ and not the one passing through $Y$ and $Y'$?

As far as I can see there is no reason why I cannot say that the pressure is the same on the surface passing through the red line ($Y$ and $Y'$). Is this possibly correct?

It should not because it leads to wrong conclusions (as $p_2>p_1$) but still $Y$ and $Y'$ are at the same height and the fluid considered is the same. I do not see anything different from a normal U tube manometer (instead of reversed) like the one in the picture, where the surface considered is the one passing through 1 and 2. $2.$ The valve in the picture is open, what happens if I close that valve on the top? Does something change?

In particular I was thinking that maybe if I let the valve open I do not have to take in account the weight of air in the manometer, while if I close the valve then I have to. But in the link I posted the valve is open and the weight of air is still taken into account, so probably that's not the point.

Any help is really appreciated.

• why should the pressure be the same at y and y', is it your believe that presuure depends only on height? – user83548 Jun 5 '16 at 19:16
• en.wikipedia.org/wiki/Bernoulli%27s_principle – user83548 Jun 5 '16 at 19:17

1. Why the isobaric surface chosen is the one passing through $X$ and $X′$ and not the one passing through $Y$ and $Y′$?

Equating the pressure at the level XX'(pressure at the same level in a continuous body of static fluid is equal)

In your example you would not regard a static fluid since there is a flow underneath the tube. Therefore the assumption that the pressure in $Y$ and $Y'$ is equal does not hold.

So you have to look at your manometric fluid and find $X$ and $X′$ as suiteable points to state that they have the same pressure.

To repeat and clarify, the difference between this setup and your example is again that the example regards static fluid bodies.

2. The valve in the picture is open, what happens if I close that valve on the top? Does something change?

First of all the weight of the air does not vanish simply because the container is open towards the environment.

Someone might correct me or add to it now but from what I see I'd say the valve is for calibration purposes only. You can shift the fluid level in the limbs via this valve. For the pressure difference itself it will have no effect.

There is a difference between the two images shown, and the 1st is analysed wrong.

The fluids are reversed, in 1st case air is at the top and liquid is at the bottom, but then they switch position for the 2nd image. In air, pressure is constant across the U manifold because it's very light (the weight of air shouldn't really be taken into account). In the liquid, the pressure varies with height because it is much more dense.

The equations in the mediums are different, so they yield different results: $1-2$ is isobaric, but $X-X'$ isn't.

1. Why the isobaric surface chosen is the one passing through XX and X′X′ and not the one passing through YY and Y′Y′?

The pressure is the same in all connected air in the tube, so pressure is the same in $X$, $X'$ (which is useless) and $Y'$.

This makes sense because you get that $P_X>P_Y$ and you know how much. This means speed is higher in the moving fluid at (1) (slow fluid) than at (2) (fast fluid). This nets you the speed difference.

1. The valve in the picture is open, what happens if I close that valve on the top? Does something change?

Nothing will change. If the valve is open, then air in the tube is at the same pressure as the air outside. Note that there is no flow, nor speed nor anything dynamic happening at the the valve. Therefore if you close it, the air in the tube is still at the same ambient pressure.

This analysis may change if the main fluid is at standstill before the experiment, or before opening-closing the valve, but I don't think that was your question.