Parallel plate capacitor

Born2bwire

Obviously vk6kro can answer this more clearly but in essence he is talking about a two "plate" capacitor. One plate is a metal sleeve surrounding the pipe, the second is a strip of conductor that runs along on the inside of the pipe. The pipe itself will separate the two "plates" and provide a small amount of dielectric.

I was thinking of a similar idea, running a metal sleeve and then having a central rod, but that was before I realized you wanted to run the pipe horizontally, not vertically.

So in your diagram, the plates outside would be shorted together and curved to fit the exterior. The inner plate stays the same. You can connect the inner plate to ground, which as v6kro mentions, probably will set the water to ground as well since it will probably be contaminated enough to be conductive.

I don't know of a closed form for doing this configuration but you can do a numerical simulation easily enough. The main question would be how you estimate the water but giving it a slight conductivity is probably realistic.

vk6kro

The curve would have to follow the shape of the pipe.

We are assuming the water is conductive, so it fills the pipe to some height making contact with the inside of the pipe.
Then we want just the thickness of the plastic pipe to be the dielectric. So the outer conductor has to be as close to the outside surface of the pipe as possible.

This way, the capacitance should be quite large and measuring it should be easier too.

You can get very thin copper sheet from craft stores.

You could have an oscillator like one of the following
555 and schmitt Osc.JPG
which would generate a square wave whose frequency depended inversely on the depth of water in the pipe. These oscillators have one side of the capacitor grounded, which may be necessary in this case if the water is grounded.

mathew086

Sorry for my lack of knowledge.. I didnt understand it completely.
If two copper plates are curved to fit on the outer surface of the pipe and a stainless steel plate is placed inside in the middle of the pipe,how is the capacitance value calculated? i mean capacitance without the presence of water? When water comes, it changes the overall capacitance of the oscillator, right?

Pipe has a thickness of 4 mm; DImension of copper plates = 130 X 65 (L X B). DIlectric constant of pipe = around 4; height of water varies from 1 mm to 100 mm. lets say 50mm for instance.

Born2bwire

No easy way. If we had two full cylinders within a cylinder then you could do it using closed form equations. Essentually you are just doing a Poisson equation, though with the inhomogeneity of the dielectric pipe adds further complications. You would set the inner strip to be ground and assume a constant potential across the surface of the outer cylinder. Then it is simply finding the charge distribution on the surfaces. Integrate the charge distribution across the surface of a plate to find the total charge, divide by the voltage you chose to set the plates at, and Bob's your uncle (or other suitable relative) and you have capacitance.

The dielectric pipe adds a bit of trouble because it will be polarized by the fields and adds additional boundary conditions. Really, I think the best way to tackle this would be to do a finite element simulation. You could easily then model the dielectric of the pipe and you can also solve for increasing water levels too. You may be able to solve for it in closed form without any water, it would be rather difficult though. If there wasn't a dielectric pipe, then I would suggest a moment method computational solver. That would not be too difficult to write up. Fortunately, you could just assume that you have an infinite pipe so it is a 2D problem and you would get the capacitance per unit length.

Or you could just build it and take measurements with different levels of water and different levels of contaminants in the water to change the conductivity. But where's the fun in that?

Oh, what a fun little problem.

EDIT: Don't forget to insulate the outer plates on the outside of the pipe. If the pipe is buried and you do not insulate it then the outer plates get pulled to earth ground which may be the same as the ground you pull down on the inner conductor.

vk6kro

When there is 50 mm of water in the pipe, the water is occupying half of the circumference of the pipe on the inside.
If the diameter of the pipe is 100 mm, the circumference = pi * 100. Half this is pi * 50

Suppose you have 200 mm length of metal plate close to the surface on the outside.
The area that is directly opposite the water inside would be
pi * 50 * 200 sq mm or 314 sq cm.
The plastic pipe is 4 mm 0r 0.4 cm thick.
K= 4

C= (0.0885 * 4 * 314 /.4) or about 278 pF

When there is no water in the pipe, the capacitance would depend on the dimensions of the stainless steel electrode at the bottom of the pipe, on the inside.

However, all this is just speculation. Some experimentation would be required to see if this would really work.

Born2bwire

I think that would work as a good first order approximation, but I think it should diverge as the pipe fills up with water. As the water nears the top of the pipe, you should get a growing capacitance from the top surface of the water and the outer pipe above it.

vk6kro

This is just a design estimation. It looks like we could get over 400 pF capacitance, so simple oscillators become possible.

The top would be problematic anyway because the little bit of water at the top of the pipe would add a lot of capacitance. However, this would just cause non linearity, not errors and it would be possible to calibrate the pipe to get reasonable readouts.
Micros can do lookup charts, so it is possible.

mathew086

But is the calculation method for capacitance by vk6kro really correct? i mean is that the correct method to find capacitance produced by curved parallel plates?

Or should i need to find the charge distribution over the plates and find the capacitance as mentiond by Born2bwire.


Charge distribution on a surface = dQ/dA . what is it actually in this situation? and total charge = integrating this charge distribution over the complete area ( 175 mm *65mm = 0.011375 m2)

Do we need to findthe charge distribution for both 2 plates and sum up them??

mathew086

Hey vk6kro

I have a doubt with your calculation. Suppose the plates have a dimension of 200 X75 mm (L X B). so when we place it on the side of the pipe, 200 mm is length of the plate upwards and 75 mm is sidewards.
Lets say 50 mm of water is there.
Then thecircumference of the pipe that is covered by water = pi * 50mm.
Area of plates exposed = pi * 50 * (75mm OR 200mm) ?? Is it 75 mm the breadth of the plate or 200 mm the lenght of the plate?? I think 75 mm , breadth is the right one or am I wrong???

Also i did not quitely understood the diagram in your post #18. Is it the same as you said in post #3?? Using an Oscillator and connectings its output in series with a R and the test capacitor in parallel.???

Also in post #15 u showed the mounting of the plates. Is it ok that if we mount the outside plates as in figure 2( below). This would be easier to keep the plates very close to the pipe. i.e taking a cylindrical plate and cutting a small portion out from the top portion.

Or else can one use screw and nut to mount the plates to the pipe inorder to have it as in case 1 in figure 2.

Attached Images

Figure 2.jpg (33.5 KB, 2 views)

vk6kro

In the calculation, there is 200 mm of outer copper along the pipe and assuming it encloses the pipe completely around its circumference.
The only part of the pipe which has significant capacitance is where there is water on the inside. This has an area of 200 mm along the pipe * half the circumference (157 mm).

water level 4.PNG
This is to show the calculation figures. The copper plate is curved and covers the whole of the pipe for 200 mm of its length, but only the bit with water opposite it counts for this calculation. (The dark blue bit is the top of the water. The artist got carried away a bit :) )

The two copper plates got replaced by one plate on the outside of the plastic pipe which covers the entire outside surface.

Also in post #15 u showed the mounting of the plates. Is it ok that if we mount the outside plates as in figure 2( below). This would be easier to keep the plates very close to the pipe. i.e taking a cylindrical plate and cutting a small portion out from the top portion.
Yes, just like that.

Would it be useful to calibrate the readings for equal area steps rather than height? This would give you a scale that showed equal increments in water volume in the pipe.

Born2bwire

It will be a good approximation. If we have a parallel plate capacitor of plates with area A, then the capacitance is

where d is the distance between the plates and \epsilon is the permittivity of the dielectric between the plates. All v6kro has done is roll these plates up into cylinders. This has a few problems (not just from rolling them up though). First, the equation ignores fringing fields, it assumes that the fields are perfectly perpendicular to the plates which is not true. However, you get good results for large plate area to width ratios since the fringing fields are only over a small part of the capacitor. In addition, since the plates are now curved, the inner plate will have a smaller area than the outer plate. Plates with large radii and small separations will minimize this error. Finally, because the outer plate is a full cylinder and we are assuming that the water will behave as a conductor (this should be a verification done in experiment, perhaps vary the salinity of the water, *shrug*) and so the inner "plate" is now going to be a semi-cylindrical volume with a flat surface on top. There will be capacitance contributing from this upper surface, which will become more prominent as the water gets higher.

So, if you want a rough calculation, use the simple parallel plate capacitor equation. This may give you a decent curve to fit against measurement. But if you want it to be exact, you will probably need to do a numerical method. Ideally, what you will do is calculate the capacitance for different heights of the water. You can then store these values in a lookup table and use a simple interpolating polynomial (also calculated ahead of time and stored in memory perhaps) to find the capacitance as a function of water height.

vk6kro

I calculated a graph of capacitance vs height for this setup (but without any allowance for fringing etc).
pF vs height.PNG
It is surprisingly linear away from the very top and bottom. Quite adequate for this purpose.

I think I would be worried about the time the top of the inside of the pipe stays wet when the water level falls. If it stayed wet, the pipe could be empty and appear full.
I tested some new PVC pipe and it did seem to shed the water quickly after being dunked in water. Some plastic pipe has a greasy feel and this might be more suitable.

mathew086

Hello guys...

I started to mount the plates and design an oscillator. I mounted a 65mm long copper plate around the pipe and connected it to + 9VDC. Then I placed a small plate of staineless steel inside the pipe in the middle and connected it to 0V. After a few minutes i connected th eplates to a LCR meter to check the capacitance( Without water inside the pipe).
The value showed as around 10pF. Then i poured a few drops of water in to the pipe and the value changed to around 25pF. ( At the moment i dont have a large container to fit in the pipe and check for the full level of water.) When some small particles like paper pieces were placed in the pipe, the value changed very negligibly. 0.X pF change.

What happend next is interesting.. The few drops in the pipe started flowing outside and some of them went in to the gap between the pipe and the outer copper plate. Suddnely the reading was out of range. The copper plate is 0.5mm thick. I treid the very maximum to fix it completely over the surface. Now i need to find a solution for it.

I am designing an astable osciliator that produces around 100kHz frequency and try the output.

I hope there will be considerable change detection.

I also calculated a therotical approch with the mehtod you mentioned. The graph was also linear ... like in the figure below.

Attached Images

graph.jpg (19.2 KB, 2 views)

vk6kro

Sounds like the water was conductive enough to upset your LCR meter on its C range.
That might be good news.
Could you put a plastic bag of water in the tube with an electrode inside the bag?

I did a calculation for the 555 oscillator in #18 above.
Using 100 K as the resistor, the frequency curve was hyperbolic, but the period curve followed the same shape as the Capacitance vs Height curve above. ie it was quite linear.

So, that raises some interesting possibilities. How would you like a digital readout with 1 mm accuracy? Probabaly an overkill, but it would look impressive.

Anyway, let's see if the thing even works first.