Measuring the water volume in a small tank with a manometer

[wacNTN]

This is what I thought was a simple task, but it's causing me headaches.
I have a conical steel tank with approximately 14 liters of fluid capacity. I need to determine the volume of water in the tank when in use, and we are currently planning to measure the pressure at the bottom of the tank with a pressure sensor and relate that mathematically to the volume. (Please accept this now rather than suggest alternatives like weighing the tank, etc.)

The tank is open to the atmosphere.

I have a homemade U-tube manometer that is open on one end, and the other has tubing attached to a rigid tube that is in the tank to the bottom. As we add water to the tank, the fluid in the manometer is displaced as expected. We read the fluid displacement in one tube on a scale on the manometer and then doubled it to get the total displacement (the fluid in the other tube goes down due to the pressure exerted by the water in the tank.) However, the displacement does not equal the water level in the tank. The displacement is approximately 1/2 of the water level in the tank. It appears to be consistent for various depths of water in the tank.

What am I missing? (I'm embarrassed to ask because I have engineering degrees but haven't worked in this hands-on in years.)

Does the ID of the connecting tubes and rigid tube in the tank make a difference?
Does the fact that the tank is a cone vs. a cylinder matter? I don't think so.
Atmospheric pressure should be an issue as the tank and manometer are open to the atmosphere.

 

[Dullard]

It sounds as if your approach is unnecessarily complicated. You have:
Clear tubing
A 'tap' at the bottom of the tank
Geometry/dims for the tank

Rig the tubing to indicate tank level (fluid from the tank will rise in the tubing to be 'level' with the fluid in the tank). Do the math.

Tubing size and tank shape should not be a factor either way.

 

[wacNTN]

Thanks for the fast reply. Sorry, but I left out that, ultimately, this will be connected to a pressure sensor and a microcontroller, which will control draining and filling the tank periodically as part of a more extensive process.

I started with the pressure sensor connected generally as you described, but I was not getting the readings I expected from it. Once I confirmed that it was working correctly, I then connected the manometer and compared the manometer readings to the pressure sensor readings. They match very well.

So the question became why isn't the pressure at the port what we expected from the water level in the tank.

You have confirmed what I believed - tank and tubing shapes and size should not be a factor.

We put the tube in a 2-liter bottle and filled it, and the manometer is still reading about 1/2 the depth of the water. The reading we are making is the difference between the water level in the two halves of the manometer u-tube, so that's not the issue.

It has to be something simple because this is not that complicated!!! I'm sure we will feel like idiots when we realize what it is.

 

[erobz]

Give us a diagram, and the calculations you used.

 

[wacNTN]

There aren't any calculations involved in this. Here's a sketch.

An example of what I have described is that when h = 7", D is approximately 3 1/2" And the ratio between h and D seems to stay relatively constant at about 2.

 

[erobz]

There aren't any calculations involved in this. Here's a sketch.

An example of what I have described is that when h = 7", D is approximately 3 1/2" And the ratio between h and D seems to stay relatively constant at about 2.
View attachment 348583

I don't think its going to work like that. The change in pressure of the air due to the reduction in volume is not linear in depth. I think the air in the tube will be something like this:

 

[wacNTN]

Hmm...I'm not sure what you mean by "change in pressure of the air due to the reduction in volume is not linear in depth" If you mean it won't work because the tank is conical, that should not be a factor. We've tested this with a cylinder as well and have the same issue.

And here's a Youtube video that demonstrates this approach, although he uses a port instead of the drop in tube. It works for him.

 

[erobz]

Hmm...I'm not sure what you mean by "change in pressure of the air due to the reduction in volume is not linear in depth" If you mean it won't work because the tank is conical, that should not be a factor. We've tested this with a cylinder as well and have the same issue.

And here's a Youtube video that demonstrates this approach, although he uses a port instead of the drop in tube. It works for him.

As you submerge the tube the pressure at the interface between the gas and water is not is not $\rho g h$ its $\rho g ( h - \delta)$.

once you figure out $\delta$ ( via ideal gas law ) then you use the manometer equation $$ P_2 = P_1 + \sum_{down} \gamma h_i - \sum_{up} \gamma_i h_i $$

Maybe try oil in the joining section to get around the complexity of the compressing gas. It could be a small effect until the pressures are high... it's hard to say without computation. Just a thought.

 

[Arjan82]

I don't think air compression can explain 31/2'' difference. You need way more pressure than that.

I think the force balance is different than you think.

h is only going to be equal to D if the water level in the part of the tube that is in the tank is indeed equal to the level in the tank outside of the tube. This is just a geometrical consideration.

However, if the waterlevel in the tube that is in the tank is indeed equal to the water level in the tank itself, the pressure at this waterlevel inside and outside of the tube should also be equal, i.e. 1 atm. But this can never be the case since that would mean that D has to be zero.

In other words, you need to take into account the height of the water column inside the tube that is in the tank. Now, I haven't done the analysis, but it wouldn't surprise me that this is guaranteed to be equal to D.

 

[erobz]

I don't think air compression can explain 31/2'' difference. You need way more pressure than that.

I think the force balance is different than you think.

h is only going to be equal to D if the water level in the part of the tube that is in the tank is indeed equal to the level in the tank outside of the tube. This is just a geometrical consideration.

However, if the waterlevel in the tube that is in the tank is indeed equal to the water level in the tank itself, the pressure at this waterlevel inside and outside of the tube should also be equal, i.e. 1 atm. But this can never be the case since that would mean that D has to be zero.

In other words, you need to take into account the height of the water column inside the tube that is in the tank. Now, I haven't done the analysis, but it wouldn't surprise me that this is guaranteed to be equal to D.

Seems to me like we are saying the same thing?

 

[Arjan82]

Well, the 'air level' (rather watersurface I would say) in the tube is indeed not going to be equal to the water surface of the tank, that much we agree on.

It has however nothing to do with compression of air, I think you can assume the air to be incompressible at these pressure levels (and otherwise you need a tube with a smaller diameter to minimize this effect)

Just to be clear, by compression I mean a difference of the volume of the air part. This air part is of coarse trapped in the tube.

 

[wacNTN]

I'm getting a bit confused so let's set the manometer aside for a moment and just look at the tank.

The fluid in the tank is tap water and the tank top is open to atmosphere.
Is the pressure at the bottom of the tank, caused by the water in it, equal to h inches of water?

If so, how would we measure that pressure with a manometer or pressure sensor?
It seems like we'd to measurement device at the bottom of the tank. I can do that as shown, with a tube in the water that puts the measurement port at the bottom OR put a port at the bottom of the tank and connect my measurement device to it. The manometer becomes our measurement device.

If I pull the tube out of the water, the water in the manometer tubes stabilizes to the same level in both sides of the u-tube.

 

[Arjan82]

I'm getting a bit confused so let's set the manometer aside for a moment and just look at the tank.

The fluid in the tank is tap water and the tank top is open to atmosphere.
Is the pressure at the bottom of the tank, caused by the water in it, equal to h inches of water?

yes, it can be computed with $\rho g h$

If so, how would we measure that pressure with a manometer or pressure sensor?
It seems like we'd to measurement device at the bottom of the tank. I can do that as shown, with a tube in the water that puts the measurement port at the bottom OR put a port at the bottom of the tank and connect my measurement device to it. The manometer becomes our measurement device.

The pressure sensor should give the pressure equal to $\rho g h$, assuming it is equal to 0 if in the open air.

If I pull the tube out of the water, the water in the manometer tubes stabilizes to the same level in both sides of the u-tube.

I would hope so indeed ;) But what is your point here?

 

[wacNTN]

Well, the 'air level' (rather watersurface I would say) in the tube is indeed not going to be equal to the water surface of the tank, that much we agree on.

It has however nothing to do with compression of air, I think you can assume the air to be incompressible at these pressure levels (and otherwise you need a tube with a smaller diameter to minimize this effect)

Just to be clear, by compression I mean a difference of the volume of the air part. This air part is of coarse trapped in the tube.

I understand what you are saying now.

 

[erobz]

Yeah, I think the that $D = h - \text{water column}$

Could be asleep at the wheel though.

 

[wacNTN]

yes, it can be computed with $\rho g h$



The pressure sensor should give the pressure equal to $\rho g h$, assuming it is equal to 0 if in the open air.



I would hope so indeed ;) But what is your point here?

In retrospect, no particular point.

I think you are saying there is a difference between the water level in the tank and the water level in the measuring tube because the air in the tube is being compressed. And the compression is being done by the water in the tank, but the air is only compressed 'so far' and then water goes up into the measuring tube.

That does make sense. And the smaller the tube, the less air in it so the less of the pressure from the tank's water is used to compress the air.

This seems to be reasonable since the manometer is measuring less than the level of the water in the tank.

 

[Arjan82]

If so, how would we measure that pressure with a manometer or pressure sensor?
It seems like we'd to measurement device at the bottom of the tank. I can do that as shown, with a tube in the water that puts the measurement port at the bottom OR put a port at the bottom of the tank and connect my measurement device to it. The manometer becomes our measurement device.

If you put a port [edit] at the bottom of the tank, such that the tube is attached from the bottom[/edit] the water column will still have an influence but then negative (D larger than h). The column of water inside the tube that is at the side of the tank needs to be taken into account. It has mass and therefore influences the pressure.

 

[jack action]

I think I found the problem.

First, a geometric constraint. If the water from the tube in the tank goes up by $H$, then the level at the entrance of the U-tube must drop by $H$ also. It follows that the level at the exit of the U-tube rises by $H$ as well. Thus, $D$ must be equal to $2H$.

If $D= 2H$ then the level in the tube $H$ cannot be equal to the level in the tank $h$. As the level rises in the U-tube, it pushes back on the column of water in the tube from the tank.

• The pressure at the bottom of the U-tube of initial depth $Y_0$ is $P_0 + \rho g \left(Y_0 + \frac{D}{2}\right)$;
• The pressure at the entrance of the U-tube is $P_0 + \rho g \left(Y_0 + \frac{D}{2} - \left(Y_0 - \frac{D}{2} \right)\right)$ or $P_0 + \rho g D$;
• The pressure above the column of water in the tube in the tank is the same: $P_0 + \rho g D$;
• The pressure at the bottom of the tube is then $P_0 + \rho g (D + H)$ but it also must be $P_0 + \rho g h$;

Therefore, $h= D + H$.

But we have already established that $D=2H$, so:
$$h= \frac{3}{2}D \ \ \ \ \ \ \ \text{or}\ \ \ \ \ \ \ H = \frac{h}{3}$$
Note that I have assumed the same density for all liquids.

 

[russ_watters]

This isn't how manometers work. Consider if you let the tube fill up completely inside the tank (ya know, like a straw in a drink). What pressure do you have?

 

[russ_watters]

It sounds as if your approach is unnecessarily complicated. You have:
Clear tubing
A 'tap' at the bottom of the tank
Geometry/dims for the tank

Rig the tubing to indicate tank level (fluid from the tank will rise in the tubing to be 'level' with the fluid in the tank). Do the math.

Aka, a u-tube manometer(or a sight glass) OP has invented a W-tube manometer.

[Hint: the tank is the first "leg"]

 

[berkeman]

I have a conical steel tank with approximately 14 liters of fluid capacity. I need to determine the volume of water in the tank when in use, and we are currently planning to measure the pressure at the bottom of the tank with a pressure sensor and relate that mathematically to the volume. (Please accept this now rather than suggest alternatives like weighing the tank, etc.)

Okay, this silliness has gone on long enough, IMO. Please explain why you are standing on your head (and wasting our time) to do this simple task in such a poorly-engineered way. Is this a schoolwork assignment from a demented professor? Or a Tik-Tok challenge?

Please justify all of the effort that you are asking from our technical helpers for this project/assignment, or the thread will be closed. Thank you.

 

[Arjan82]

I think I found the problem.

First, a geometric constraint. If the water from the tube i the tank goes up by $H$, then the level at the entrance of the U-tube must drop by $H$ also. It follows that the level at the exit of the U-tube rises by $H$ as well. Thus, $D$ must be equal to $2H$.

This is the key point that I overlooked indeed.

If $D= 2H$ then the level in the tube $H$ cannot be equal to the level in the tank $h$. As the level rises in the U-tube, it pushes back on the column of water in the tube from the tank.

• The pressure at the bottom of the U-tube of initial depth $Y_0$ is $P_0 + \rho g \left(Y_0 + \frac{D}{2}\right)$;
• The pressure at the entrance of the U-tube is $P_0 + \rho g \left(Y_0 + \frac{D}{2} - \left(Y_0 - \frac{D}{2} \right)\right)$ or $P_0 + \rho g D$;
• The pressure above the column of water in the tube in the tank is the same: $P_0 + \rho g D$;
• The pressure at the bottom of the tube is then $P_0 + \rho g (D + H)$ but it also must be $P_0 + \rho g h$;

Therefore, $h= D + H$.

But we have already established that $D=2H$, so:
$$h= \frac{3}{2}D \ \ \ \ \ \ \ \text{or}\ \ \ \ \ \ \ H = \frac{h}{3}$$
Note that I have assumed the same density for all liquids.

Although I find your reasoning a bit convoluted and I didn't get what $Y_0$ is, I did also find the $h= \frac{3}{2}D$ part in my own way. So I think that is correct indeed. (I think your first bullet and the first part of your second bullet are unnecessary).

So, the OP found

$D = \frac{1}{2} h$ (1)

but theory states that $D = \frac{2}{3} h$. So either the OP is a bit sloppy, or maybe the compressibility of air is a factor after all. So let's nerd this out...

To estimate this, we would need the length of the trapped air part of the tube and the tube inner diameter and a value for h for which (1) holds however.

So, a very ball park estimate: Length of trapped air: 1meter (40''), inner diameter 5mm (0.2''), if then D = 3.5'' then the gauge pressure inside the trapped air part would be 872Pa ($\rho g D$ with $\rho = 1000kg/m^3$, $g = 9.81m/s^2$ and $D=89mm$).

We're going to apply

$$\left( \frac{p_2}{p_1} \right) = \left( \frac{V_1}{V_2} \right)^\gamma$$

With this I compute a difference in length of the trapped air part of about 6mm (0.24'').

For (1) to hold we must have a length difference of air of about 44mm (1.75'') if I'm correct. This is because:

$h = D + \delta H$ (2)

if $\delta$ is taken to be some factor that takes compressibility into account (i.e. the waterlevel in the part of the tube that is in the tank wil rise a bit). Then we still have

$D = 2H$ (3)

Note that this is the other side of the trapped air part, so at the 'manometer' side. This is to say that $\delta H$ change in waterlevel in the tube part will only push the waterlevel at the manometer side a distance $H$ downward (and upward at the open side...). Take (2) and (3) together:

$$ h = D + \delta \frac{D}{2} = \left( 1 + \frac{\delta}{2} \right) D$$
or

$$ D = \frac{1}{1+\frac{\delta}{2}} h$$

For $\delta = 2$ we find the relation $D = \frac{1}{2} h$ meaning (1) holds.

So, after compression of the air $D$ should still be 3.5'' (89mm) since otherwise the pressures aren't correct anymore. This means the watercolumn in the tube that is in the tank is $H=44mm$ before compression. After compression it is $\delta H$ so $(\delta - 1) H = 44mm$ is the length that the trapped air has contracted due to compression, if the relation (1) is to hold.

Since there is quite a difference between 6mm (estimated contraction of trapped air) and 44mm (needed contraction of trapped air for $D = \frac{1}{2} h$ to hold) I think compression is not important here. But I must admit that it was closer than I thought it would be...

So, that was really much more verbose than I wanted it to be... But yeah...

 

[jack action]

Please explain why you are standing on your head (and wasting our time) to do this simple task in such a poorly-engineered way.

Although not stated clearly, I think one constraint is that he cannot drill a hole in his tank.

Even if there was a better solution for this problem, I find it rather challenging (in a fun way) to try to explain why this real-life result actually happens.

 

[jack action]

I didn't get what $Y_0$ is

The height of the fluid in the U-tube at rest (when both "legs" are of equal length).

I think your first bullet and the first part of your second bullet are unnecessary

I was going step-by-step, evaluating pressure at every point from one end to the other, making sure I had it correctly.

 

[russ_watters]

Even if there was a better solution for this problem, I find it rather challenging (in a fun way) to try to explain why this real-life result actually happens.

...it's because as the water goes up the tube in the tank (toward the liquid level), the pressure goes down, not up. You have two manometers, one starting neutral and one starting pressurized, and when you connect them they meet in the middle at half height/pressure.

Now knowing that, he can use the device, if he is constrained to not add any holes in the tank.

 

[sophiecentaur]

Okay, this silliness has gone on long enough, IMO. Please explain why you are standing on your head (and wasting our time) to do this simple task in such a poorly-engineered way. Is this a schoolwork assignment from a demented professor? Or a Tik-Tok challenge?

I'm inclined to agree with you when you say it's been made highly over-complicated.

The pressure inside the enclosed tube will be the same both ends (if you ignore the density of air). So the relative vertical position of the tube and the tank is irrelevant within a very small amount. The internal air pressure and any consequent density change is negligible so you can treat the two halves independently. How is the red line on the diagram of any consequence if the vertical column of air doesn't contribute to pressure?

The internal air pressure will be due to the difference in column heights
P=ΔHρg
It will also be due to the difference between tank water level height (h) and the position of the meniscus in the tube
P=(h-d)ρg
So ΔH = h-d
If you introduce air into the tube until bubbles can be heard then d is zero and H will then be equal to h. Job done? Alternatively, you could detect when the sensing tube is 'just full of air' with a float and contact system.

Needless to say, there has to be a conversion table to relate water level with actual volume if the tank is not a cylinder.

 

[Dullard]

In minimum syllables:
The manometer that you built must be 'read' correctly. The pressure at the bottom of the tank is the difference in levels for the 2 legs of your 'U.' Using only 1 level is acceptably close for very low pressure measurements (like HVAC duct balancing), but will become increasingly incorrect as pressure increases. Note post 25 - the effective level may not be what you think it is.