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.