Manometer: Calculating Height of Water

[Richard Spiteri]

Consider a manometer as shown above with different widths W and w.

If we take a mass (red) that is frictionless and does not allow water to leak of mass M, then I would like to calculate the height of water H in the narrow tube of width w. I arrive at a non sensical answer of H = 0 in one instance (and H = ΔabW/w) but I cannot see my mistake.

 

[Ibix]

Your first expression, from the pressures, is almost correct. You should be using areas instead of widths, unless your tubes are rectangular and have the same depth into the page as each other.

Your second equation isn't correct. You should use the fact that the water volume cannot change.

 

[Richard Spiteri]

Your first expression, from the pressures, is almost correct. You should be using areas instead of widths, unless your tubes are rectangular and have the same depth into the page as each other.

Your second equation isn't correct. You should use the fact that the water volume cannot change.

@Ibix, you are right, I should have been more specific in stating this is a 2D (or a 2½D) model. To keep things simple assume the "volume" is always (area)*(unit depth).

In that case the "volume" of water displaced must give you H = Δab(W/w) as shown in my figure.

I cannot, however, see the problem with my logic using equal pressures at level b on the left- and right-hand sides. Should the correct equation be:

Mg/W + P₀ = (H+Δab)ρg + P₀ (as per my original post) or
Mg/W + P₀ = (H+Δab)ρg/w + P₀ ?​

Once I see what I am doing wrong, I will follow up with another question.

 

[Ibix]

I cannot see the problem with my logic using equal pressures at level b on the left- and right-hand sides.

Apart from the area thing (which you've resolved) there's nothing wrong with your first expression, which is the one that equates pressures. It's the second one that isn't helping because the red mass is sitting on top of the water not floating in it. Instead, equate the volume that the left hand side decreased by and the volume the right hand side increased by and it should work out.

 

[Richard Spiteri]

It's the second one that isn't helping because the red mass is sitting on top of the water not floating in it.

This is good for me to hear. So, how should I treat the red mass if not floating?

If it is sitting on the water and we assume no friction and no leaks between the manometer wall and the red mass, is that different from floating? Does it still displace its own weight or is the physics different?

Should I treat it as a piston exerting a downwards force of Mg? In that case, must the pressures at level b (just below the piston) be equal on the left and right?

 

[Chestermiller]

This is good for me to hear. So, how should I treat the red mass if not floating?

If it is sitting on the water and we assume no friction and no leaks between the manometer wall and the red mass, is that different from floating? Does it still displace its own weight or is the physics different?

Should I treat it as a piston exerting a downwards force of Mg? In that case, must the pressures at level b (just below the piston) be equal on the left and right?

Your first equation is a correct equilibrium force balance on the mass. $$[p_0+\rho g (H+\Delta ab)]W-p_0W=Mg$$The first term is the upward pressure force on the mass, and the 2nd term is the downward pressure force on the mass. This is equal to the weight of the mass.

 

[Richard Spiteri]

Your first equation is a correct equilibrium force balance on the mass. $$[p_0+\rho g (H+\Delta ab)]W-p_0W=Mg$$The first term is the upward pressure force on the mass, and the 2nd term is the downward pressure force on the mass. This is equal to the weight of the mass.

Great, so my equation

is correct. It was my assumption (thanks, @Ibix ) that the mass M displaces its own weight that was incorrect. Here is my follow up question:

 

[Ibix]

Does it still displace its own weight or is the physics different?

If you drop a ship into a lake, the water level goes up slightly. The displacement is the volume of the ship below the new waterline, not the old one. Your problem here is working out where the new waterline is, since there's no water round the side of your mass to look at and the only actual water level you have isn't helpful. Furthermore, I suspect that if you do the calculation correctly you will find the result reproduces your first equation because it's the same force balance thing.

Instead, do what I suggested in the last sentence of #4. Combined with your first equation it should give you a sensible answer.

 

[Richard Spiteri]

This was my follow up question but I think I have enough feedback to try and figure it out for myself. I would say that:

[p₀ + ρg(H - x + Δac)]W - p₀W = Mg​

and then solve for Δac .

 

[Chestermiller]

View attachment 285944

This was my follow up question but I think I have enough feedback to try and figure it out for myself. I would say that:

[p₀ + ρg(H - x + Δac)]W - p₀W = Mg​

and then solve for Δac .

This figure is not consistent with your original figure. Your definition of $\Delta ab$ has changed.

I don't know what you did here, but there is only one force balance, and you got that correct in post #7 (for the definition of $\Delta ab$ in your original figure). There is no other force balance required. As @Ibix points out, the only other equation needed is the conservation of mass equation: $$W(\Delta ab)=wH$$

 

[Richard Spiteri]

Please look at my thread #9 for the diagram.

I cut a bit off the top (equal to x) and then defined the extra depth that the mass sinks as Δbc for a total of (Δab + Δbc) = Δac. It should be clear in the figure in #9.

 

[Chestermiller]

Please look at my thread #9 for the diagram.

I cut a bit off the top (equal to x) and then defined the extra depth that the mass sinks as Δbc for a total of (Δab + Δbc) = Δac. It should be clear in the figure in #9.

The layout in the original figure is all that is needed to solve this. So let's, for the time being, stick with that. You have the force balance equation correct in post #7, and you have the mass balance equation I wrote in post #10. So you have 2 equations and 2 unknowns.

 

[Richard Spiteri]

Apart from the area thing (which you've resolved) there's nothing wrong with your first expression, which is the one that equates pressures. It's the second one that isn't helping because the red mass is sitting on top of the water not floating in it. Instead, equate the volume that the left hand side decreased by and the volume the right hand side increased by and it should work out.

@Ibix , my equation which equates pressures (that you say is correct) is:

H = M/(ρW) - Δab​

This is still confusing to me as there is no dependency on lower case w (the width of the right hand tube). It says that H does not change regardless of lower case w ...unless w = f(Δab)

Your first equation is a correct equilibrium force balance on the mass. $$[p_0+\rho g (H+\Delta ab)]W-p_0W=Mg$$The first term is the upward pressure force on the mass, and the 2nd term is the downward pressure force on the mass. This is equal to the weight of the mass.

I pressed the wrong tag earlier. Sorry about that.

 

[Chestermiller]

@Ibix , my equation which equates pressures (that you say is correct) is:

H = M/(ρW) - Δab​

This is still confusing to me as there is no dependency on lower case w (the width of the right hand tube). It says that H does not change regardless of lower case w ...unless w = f(Δab)

The equation for the variation of pressure with spatial location in the vertical direction is $$\frac{\partial p}{\partial z }=-\rho g$$ where z is the elevation in the upward direction. The variation of pressure with respect to the two horizontal coordinates is zero. So no distance parameters in the horizontal direction ever come into play.

 

[Richard Spiteri]

Thank you @Chestermiller and @Ibix for the threads above. I have worked it out using simpler and clearer variables and definitions and have come up with the following which should all be correct.

Figure i above defines the variables and there is no question.

Figure ii is what I postulated (using different terms and variables) and which was confirmed as correct.

Figure iii above takes what I gathered from @Chestermiller and formulates the question using a variation of my previous lab set-up in order to help me understand the fundamentals better. My answers a) - d) attempt to answer my own questions but feel free (anyone) to comment.

 

[Chestermiller]

Is the 2nd tube present initially and then just cut off, or is the 2nd tube connected (after the system in ii equilibrates), allowed to fill, and overflow?

 

[Richard Spiteri]

Is the 2nd tube present initially and then just cut off, or is the 2nd tube connected (after the system in ii equilibrates), allowed to fill, and overflow?

In my mind, it is there initially at the same height as the one in the middle, the system equilibrates and then it is cut off by Δy.

Is the 2nd tube present initially and then just cut off, or is the 2nd tube connected (after the system in ii equilibrates), allowed to fill, and overflow?

In my mind, it is there initially at the same height as the one in the middle, the system equilibrates and then it is cut off by Δy.

 

[Chestermiller]

In my mind, it is there initially at the same height as the one in the middle, the system equilibrates and then it is cut off by Δy.

Then the initial values of H and $\Delta x$ are different for this two-tube case. What are the initial values here?

 

[Richard Spiteri]

Then the initial values of H and $\Delta x$ are different for this two-tube case. What are the initial values here?

I think what you are getting at is the conservation of mass that affects my H and Δx in Figure iii when compared to the results shown in Figure ii. If it makes it easier to remain consistent so we can compare, then let's change the areas a in each of the right-hand side tubes in Figure iii to (a/2). That works for my purposes.

Alternatively, we can leave the right-hand side tube as a and consider the middle tube as infinitesimally narrow and ignore any capillary and viscous effects. It is really the idea that the level in the middle tube drops while the right-hand level spills that I am trying to understand.

Please be clear which one you choose so it is easy for me (everyone!) to follow.

 

[Chestermiller]

Let's go with a/2. In that case, the values of $\Delta x$ and H are the same as in ii. In iii, after the right tube is cut, does the vertical distance between the upper fluid surfaces on the right- and left sides change, or does it remain the same as before the right tube was cut?

 

[Richard Spiteri]

Let's go with a/2. In that case, the values of $\Delta x$ and H are the same as in ii. In iii, after the right tube is cut, does the vertical distance between the upper fluid surfaces on the right- and left sides change, or does it remain the same as before the right tube was cut?

Figure ii cf Figure iii: I get the same result if we go with a/2. The height H and value Δx must be the same in both figures.

Figure iii: In my solution I work out that if I cut either one of the two tubes on the right by Δy , then the whole system experiences a bulk shift of Δy resulting in the piston of mass M sinking by that further distance and the uncut tube also drops it's level H by the same amount.

 

[Chestermiller]

Figure ii cf Figure iii: I get the same result if we go with a/2. The height H and value Δx must be the same in both figures.

Figure iii: In my solution I work out that if I cut either one of the two tubes on the right by Δy , then the whole system experiences a bulk shift of Δy resulting in the piston of mass M sinking by that further distance and the uncut tube also drops it's level H by the same amount.

This is correct. So, based on that, how much water leaks out?

 

[Richard Spiteri]

This is correct. So, based on that, how much water leaks out?

(Δy.A)leaks out the right hand side tube
Δy is the drop in level in the uncut tube

 

[Chestermiller]

(Δy.A)

$$\Delta y(A+a)$$

 

[sophiecentaur]

In the simplest example, it seems to me that the height of the right hand column above the bottom of the metal plug is independent of the area of the right hand tube at equilibrium. The weight of metal is balanced by the water pressure times its area.
Cutting that extra tube short will put the plug at the same level difference and the longer tube just adds its volume to the amount displaced by the fallen plug.