# In a holding tank is the air pressure equal to the water pressure?

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**In a holding tank is the air pressure equal to the water pressure?**

I wanted to start by saying I know using the schrader valve to check the air pressure with water in the tank would be inaccurate. That's not what I did.

So I wanted to lower my water pressure from 40/60 (water gauge said 42/64) to 30/50 (by removing 10 psi so actually my plan was 32/54). I did not trust the gauge I had to measure air in the tank, but I do trust my water gauge and the current charge that was in the tank.

What I did was I used the water gauge instead by first shutting water off to the house using the shutoff, then I let air out of the schrader valve until the water gauge read 10 psi less. I then adjusted the the pressure switch so the cut in pressure would be 10 psi less than it was, so 32 psi, my end result is 32/52, cut off pressure was 2 psi different than I was expecting it to be somehow but whatever.

So I guess the real question is what the title is... Is the air pressure in the holding tank going to be the same as the water pressure? Or is there somehow a difference making this method for adjusting air inaccurate? I don't think the pressures would be any different, but I got to thinking about it and maybe they are somehow. Thanks.

So I wanted to lower my water pressure from 40/60 (water gauge said 42/64) to 30/50 (by removing 10 psi so actually my plan was 32/54). I did not trust the gauge I had to measure air in the tank, but I do trust my water gauge and the current charge that was in the tank.

What I did was I used the water gauge instead by first shutting water off to the house using the shutoff, then I let air out of the schrader valve until the water gauge read 10 psi less. I then adjusted the the pressure switch so the cut in pressure would be 10 psi less than it was, so 32 psi, my end result is 32/52, cut off pressure was 2 psi different than I was expecting it to be somehow but whatever.

So I guess the real question is what the title is... Is the air pressure in the holding tank going to be the same as the water pressure? Or is there somehow a difference making this method for adjusting air inaccurate? I don't think the pressures would be any different, but I got to thinking about it and maybe they are somehow. Thanks.

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That's what I was thinking, but with that rubber diaphragm acting as a barrier I thought maybe there was something I didn't understand and the two pressures could differ from one another. Air being compressible and water not, I wasn't positive so I thought I should ask.

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**6**
With your home water pressure tank the only time you will see the air and water pressure different is when the water pressure is less than the air pressure in the tank. Any time the water pressure is equal or greater than your initial air pressure setting the water will compress the air to be equal to the water pressure. That is why the only time you can check the air pressure in the tank is when the water system is totally bled down to zero.

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Yes, water can be turned into a solid (ice) at non-freezing (for Earth) temperatures. A quick search turn up pressures around 300'000 psi. So, if our oceans were deeper than 65'000 feet the water at the bottom could be solid.

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Yes, water can be turned into a solid (ice) at non-freezing (for Earth) temperatures. A quick search turn up pressures around 300'000 psi. So, if our oceans were deeper than 65'000 feet the water at the bottom could be solid.

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**7**
Agree with all of the notes above, one clarification point though-

A "Holding Tank" is for the sewage system, literally a big tank that gets pumped out ~weekly, usually used temporarily when the drain field fails.

A "Pressure Tank" is for the drinking water system, when you have a well.

Only mention because, reading the first part of the subject line, I immediately thought "of course they're the same, it's a big tank of waste water vented to the air. Then I read the 2nd part and thought; "air valve? A pressurized sewage system?! Something's very wrong here..."

A "Holding Tank" is for the sewage system, literally a big tank that gets pumped out ~weekly, usually used temporarily when the drain field fails.

A "Pressure Tank" is for the drinking water system, when you have a well.

Only mention because, reading the first part of the subject line, I immediately thought "of course they're the same, it's a big tank of waste water vented to the air. Then I read the 2nd part and thought; "air valve? A pressurized sewage system?! Something's very wrong here..."

*Last edited by Hal_S; 10-23-20 at 09:43 AM.*

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**8**
Given a typical well water system with bladder pressure tank, the air pressure in the tank will be greater than or equal to the water pressure in the rest of the system.

When the water pressure is less than the preset air pressure in the tank then the tank diaphragm will bottom out and the pressure tank will no longer be supporting the pressure for the system.

When the water pressure is greater than the pressure tank preset pressure than the water pressure and air pressure will be approximately equal, slight differences occurring if water is being used or if the pump is running.

The purpose of presetting the pressure tank (with a faucet open and the pump off) to say 2 PSI less than pump start pressure is so the pump will start before the tank diaphragm bottoms out, preventing a momentary stoppage of water during the middle of your shower.

When the water pressure is less than the preset air pressure in the tank then the tank diaphragm will bottom out and the pressure tank will no longer be supporting the pressure for the system.

When the water pressure is greater than the pressure tank preset pressure than the water pressure and air pressure will be approximately equal, slight differences occurring if water is being used or if the pump is running.

The purpose of presetting the pressure tank (with a faucet open and the pump off) to say 2 PSI less than pump start pressure is so the pump will start before the tank diaphragm bottoms out, preventing a momentary stoppage of water during the middle of your shower.

*Last edited by AllanJ; 10-23-20 at 10:58 AM.*

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**9**
A "Holding Tank" is for the sewage system

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So if the two pressures do equal out the method I used ought to work assuming you know/trust the current pressure in your pressure tank. I don't see why you couldn't also use this method to add air then if you wanted to. Draining the tank will be necessary if you do not know/trust the current air pressure in your tank.

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**11**
Originally Posted by

**Marq1**[img]carlift.gif (64×104) (doityourself.com)[/img]
Well, I have a holding tank as part of my water system, with chlorine injection you have a holding tank to allow the iron to oxidize and fall out of solution!

Originally Posted by

**HalS**]A "Holding Tank" is for the sewage system

But that's why in my area, you make it a point to call the chlorine tank a "mixing tank"...

Interesting how you get the "soda vs pop" and "hoagie vs sub" thing with regional terms..

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**12**
<<< Draining the tank will be necessary if you do not know/trust the current air pressure in your tank. >>>

Draining the tank will be desirable if you do not know/trust the current air volume in your tank,

With "not enough" air in the pressure tank, the pump will still cycle on and off at the currently set cut in an cut out pressures on the pressure switch. But the air cushion in the pressure tank will be much smaller, and pump cycles will be shorter and more frequent which is much less than best for the pump.

You can use trial and error to get the right amount of air in the pressure tank but it will probably take several tries and the final result will likely not be quite as accurate compared with draining the tank and presetting its pressure in one try. If you are experiencing pump short cycling you can arbitrarily and summarily add more air to the pressure tank up to 5 PSI over pump cut off pressure. Then see if the pump cycle has lengthened out to seem normal. Add more air if desired.

Should you experience a momentary total loss of pressure when the pump turns on then bleed a little air out of the pressure tank.

I do not know how iron oxide and chlorine holding tanks work, like how much of the tank volume should be occupied by gaseous chlorine, or how much should be occupied by air if there is no gaseous chlorine.

Draining the tank will be desirable if you do not know/trust the current air volume in your tank,

With "not enough" air in the pressure tank, the pump will still cycle on and off at the currently set cut in an cut out pressures on the pressure switch. But the air cushion in the pressure tank will be much smaller, and pump cycles will be shorter and more frequent which is much less than best for the pump.

You can use trial and error to get the right amount of air in the pressure tank but it will probably take several tries and the final result will likely not be quite as accurate compared with draining the tank and presetting its pressure in one try. If you are experiencing pump short cycling you can arbitrarily and summarily add more air to the pressure tank up to 5 PSI over pump cut off pressure. Then see if the pump cycle has lengthened out to seem normal. Add more air if desired.

Should you experience a momentary total loss of pressure when the pump turns on then bleed a little air out of the pressure tank.

I do not know how iron oxide and chlorine holding tanks work, like how much of the tank volume should be occupied by gaseous chlorine, or how much should be occupied by air if there is no gaseous chlorine.

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**13**
hi Et9000–

If I understand how you did it, and I’m not sure that I do, then I think there is a problem with that method. If I understand I think AllanJ has already pointed out the problem.

There is relationship between Pressure and Volume:

Pressure x Volume = Constant, that is, P x V = C

That relationship is a curve.

So the air pocket in the tank would be the V in the above, i.e., the volume in the above equation. When your system operates that air volume expands and contracts as water is used from the tank and pumped back in. If you set the air pressure with no water pressure on the air pocket, that is, no water pressure on the air volume, as you are supposed to do, then without water in the tank that air volume will be at some starting volume V. Let’s say V = 32 gallons and you set the air pressure to 28 ( 2 below the cut-in pressure of 30).

So P x V = 28 x 32 = 896 (i.e. P x V =C so for this curve the constant C = 896 ).

When your pump runs and pushes water in the tank, the air is compressed and the volume shrinks and the pressure goes up. There will be some point at which the pressure reaches 50 - the cut-out pressure. That point has to be where:

P x V= 896 that is where 50 x V= 896 ---- therefore V = 17.92 gallons when the pressure hits 50.

In other words, after you set the cut-in to 28 and the system runs, the volume of air goes from 32 gallons down to 17.92 gallons and back again in a cycle. So the amount of water you are getting between cut-in pressure(28) and cut-out pressure(50) is 32 – 17.92 = 14.08 gallons. That’s how much water you would draw from your tank between cut-out (50) back down to cut-in (28) pressure.

But if instead you start the above adjustment procedure where the air in the tank is already compressed from water pressure because the tank was not emptied, say for example, where the air volume is already compressed down to 24 gallons, then:

P x V = 28 x 24 = 672 (so this new curve has this new constant C= 672).

In other words, now you are starting with a different curve. But the product here still has to follow the rule P x V = C a constant,

- so for the cut-out pressure 50 on this new curve: 50 x V = 672 and so V = 13.44 in this case.

That is, the air volume on this new curve will be down to 13.44 gallons when the cut-out pressure of 50 is hit. So now the operating range on this new curve will be 24- 13.44 = 10.56 gallons.

In other words, because you set the air pressure with water already in the tank, you changed the curve and you reduced the number of gallons between cut-in/cut-out pressure (the drawdown) from 14.08 down to 10.56. I don’t think that’s what you want to do.

The above is a long way of saying you need to empty the tank of water when you set cut-in/cut-out pressure.

So I wanted to lower my water pressure from 40/60 (water gauge said 42/64) to 30/50 (by removing 10 psi so actually my plan was 32/54). I did not trust the gauge I had to measure air in the tank, but I do trust my water gauge and the current charge that was in the tank.

What I did was I used the water gauge instead by first shutting water off to the house using the shutoff, then I let air out of the schrader valve until the water gauge read 10 psi less. I then adjusted the the pressure switch so the cut in pressure would be 10 psi less than it was, so 32 psi, my end result is 32/52, cut off pressure was 2 psi different than I was expecting it to be somehow but whatever.

What I did was I used the water gauge instead by first shutting water off to the house using the shutoff, then I let air out of the schrader valve until the water gauge read 10 psi less. I then adjusted the the pressure switch so the cut in pressure would be 10 psi less than it was, so 32 psi, my end result is 32/52, cut off pressure was 2 psi different than I was expecting it to be somehow but whatever.

There is relationship between Pressure and Volume:

Pressure x Volume = Constant, that is, P x V = C

That relationship is a curve.

So the air pocket in the tank would be the V in the above, i.e., the volume in the above equation. When your system operates that air volume expands and contracts as water is used from the tank and pumped back in. If you set the air pressure with no water pressure on the air pocket, that is, no water pressure on the air volume, as you are supposed to do, then without water in the tank that air volume will be at some starting volume V. Let’s say V = 32 gallons and you set the air pressure to 28 ( 2 below the cut-in pressure of 30).

So P x V = 28 x 32 = 896 (i.e. P x V =C so for this curve the constant C = 896 ).

When your pump runs and pushes water in the tank, the air is compressed and the volume shrinks and the pressure goes up. There will be some point at which the pressure reaches 50 - the cut-out pressure. That point has to be where:

P x V= 896 that is where 50 x V= 896 ---- therefore V = 17.92 gallons when the pressure hits 50.

In other words, after you set the cut-in to 28 and the system runs, the volume of air goes from 32 gallons down to 17.92 gallons and back again in a cycle. So the amount of water you are getting between cut-in pressure(28) and cut-out pressure(50) is 32 – 17.92 = 14.08 gallons. That’s how much water you would draw from your tank between cut-out (50) back down to cut-in (28) pressure.

But if instead you start the above adjustment procedure where the air in the tank is already compressed from water pressure because the tank was not emptied, say for example, where the air volume is already compressed down to 24 gallons, then:

P x V = 28 x 24 = 672 (so this new curve has this new constant C= 672).

In other words, now you are starting with a different curve. But the product here still has to follow the rule P x V = C a constant,

- so for the cut-out pressure 50 on this new curve: 50 x V = 672 and so V = 13.44 in this case.

That is, the air volume on this new curve will be down to 13.44 gallons when the cut-out pressure of 50 is hit. So now the operating range on this new curve will be 24- 13.44 = 10.56 gallons.

In other words, because you set the air pressure with water already in the tank, you changed the curve and you reduced the number of gallons between cut-in/cut-out pressure (the drawdown) from 14.08 down to 10.56. I don’t think that’s what you want to do.

The above is a long way of saying you need to empty the tank of water when you set cut-in/cut-out pressure.

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**14**
I do not know how iron oxide and chlorine holding tanks work

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**15**
Pressure x Volume = Constant, that is, P x V = C

That relationship is a curve.

That relationship is a curve.

P x V = C determines the slope of a line.

The available volume of the tank determines where that sloped line crosses the axis.

The net area between P-low and P-high (quadrilateral) determines how well the pressure tank works.

Pressure tanks are simply "air springs". To get the most spring, you fill them to the minimum pressure with the tank empty.

This is because, to get the largest useful volume, you pressurize the tank at empty. If the tank is 1/2 full of water, then you are loosing '1/2 of the possible volume of the "air spring" available if you used an empty tank.

To get the most work, you want pressure to be applied through the ENTIRE change of height of the water in the tank. So, you drain the tank, pressurize the tank to the minimum/cut in pressure.

That ensures that the ENTIRE tank volume is available to generate the "air spring" effect that pushes water through the pipes.

*Last edited by Hal_S; 10-26-20 at 06:40 PM.*

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**16**
Um, actually, it's a straight line.

P x V = C determines the slope of a line.

P x V = C determines the slope of a line.

P x V = C is in fact a curve. In fact, it is a very famous curve which describes a very famous law – Boyle’s Law. It is in fact expressed as a curve (a parabola). For example, this graph above is from Wikipedia – and from Boyle’s original data. You’ll see this relationship expressed that way everywhere.

If you mean you can plot V = C/P or P = C/V and get a straight line, as some prefer, that doesn’t change a thing. That just means some people like to look at straight lines instead of curves. It’s just an alternate, but less straight forward way of expressing the same relationship.

If on the Y axis you plot the variable “P”, and on the X axis you plot the variable “C/V” - you get a straight line . That just falls out of the equation P x V = C. It doesn’t add any new knowledge.

I think you will be extremely hard pressed to find a place where the relationship is first described as a straight line V=C/P or P=C/V. I doubt that place exists.

The reason I mentioned the curve in the first place is because I think the OP was confused on Pressure Volume relationship and I thought numerical examples would help. The fact that the constant was just made up (psi x gallons) I don’t think makes any difference in demonstrating the point – at least as far as I can see.

The available volume of the tank determines where that sloped line crosses the axis.

The net area between P-low and P-high (quadrilateral) determines how well the pressure tank works.

The net area between P-low and P-high (quadrilateral) determines how well the pressure tank works.

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**17**
What I never remember is if one needs to account for atmospheric pressure when estimating results. In other words. A 30 psi water pressure is 100% more then a 15 psi water pressure, but that fails to account for atmospheric pressure. When atmospheric pressure is added, those numbers are actually 45psi compared to 30 psi, which is only a 50% increase in pressure when one adds about 15 psi for atmospheric pressure to both of them.

So if I want to know the change in volume of air, created by those pressures above, which pressure do I use. Gauge pressure or absolute pressure?

So if I want to know the change in volume of air, created by those pressures above, which pressure do I use. Gauge pressure or absolute pressure?

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**18**
Originally Posted by

**Zoesdad**Can you explain that a little further. What does the axis represent?

At 29.99 psi air pressure, the volume of air in the tank would effectively be 0%.

At 50.01 psi air pressure, , the amount of air in the tank would effectively be 100%.

Given the pressures the OP is talking about, the volume & height of water in the tank scales linearly to pressure

In practicable terms, the volume of water in the tank scales linearly, given the stated limits of 30 PSI and 50 PSI. Assuming the air bladder is 20" tall, then every 1 PSI over 30 will roughly correspond to a 1" difference in the level of water in the tank.

Originally Posted by

**Zoesdad**Neither pressure or volume can go to zero.

What is the volume of gaseous basalt at 72 degrees Fahrenheit?

Don't make me dig out my physical chemistry notes and get into entropy, enthalpy and integrating the value of Gibbs free energy during an endothermic gaseous reaction...

*Last edited by Hal_S; 10-28-20 at 09:45 AM.*

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**20**
"Don't make me dig out my physical chemistry notes and get into entropy, enthalpy and integrating the value of Gibbs free energy during an endothermic gaseous reaction..."

Yes. Let's not make him do that. lol

Yes. Let's not make him do that. lol

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**21****You use the word
**

*In practicable terms, the volume of water in the tank scales linearly, given the stated limits of 30 PSI and 50 PSI. Assuming the air bladder is 20" tall, then every 1 PSI over 30 will roughly correspond to a 1" difference in the level of water in the tank.*

**. I think the whole point of looking at this as a curve is that it tells you that as you apply pressure to the volume and the volume shrinks, it takes more and more pressure to shrink the volume by the same amount. It is not linear. I think that for some reason you don’t want to look at this the straight forward way that is expressed in textbooks and articles all over the place. Why confuse a simple concept that was introduced into the discussion for, what I think at least, was a legitimate purpose – to understand what is happening as your pressure and volume cycle in your tank.**

*roughly*I haven’t done calculus for many years and at 76 many of my neurons have retired – but I think the above graph has it right – I think. Maybe it will shed some light, or someone who has better math skills will point out any error(s).

**
**

*What is the vapor pressure of basalt at 72 degrees Fahrenheit?*

What is the volume of gaseous basalt at 72 degrees Fahrenheit?

Don't make me dig out my physical chemistry notes and get into entropy, enthalpy and integrating the value of Gibbs free energy during an endothermic gaseous reaction...

What is the volume of gaseous basalt at 72 degrees Fahrenheit?

Don't make me dig out my physical chemistry notes and get into entropy, enthalpy and integrating the value of Gibbs free energy during an endothermic gaseous reaction...

I think you are really misunderstanding what we are talking about here. We are not trying to understand other physical and chemical processes – other than Boyles’ very simple observation: Pressure and Volume of a gas vary inversely proportional to each other in a “closed volume” at a constant temperature. We are talking about the air in a “closed system” here, i.e., the closed area in a pressure tank which is designed to hold air via a diaphragm or bladder which can expand and contract. We are not talking about other gases and other environments.

Unless the tank is broken that pressure will not go to zero nor will the volume go to zero. It’s designed to operate in a range of varying pressure (e.g., 30-50) and volume – but neither goes to zero. We are not talking about the physics and chemistry of gases in other environments. It is much simpler than that. You will be wasting your time getting out your chemistry notes.

The reason I posted the Boyle’s curve was because I thought that the OP seemed to think (I certainly could be wrong) that somehow if you lower the pressure by 10 psi while you were at some unknown volume, that somehow everything would self-adjust and you would still have the same volume operating range that you had before the adjustment. I thought Boyle’s curve and example would help clarify that.

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**22**
I can't fix everything, or everyone. Gave up on THAT decades ago.

Nobody cares WHY PV=nRT.

Twice the pressure, half the volume. Graph it right, and you get a nice straight line that people understand.

Nobody cares WHY PV=nRT.

Twice the pressure, half the volume. Graph it right, and you get a nice straight line that people understand.

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**23**
SandburRanch: <<< Atmospheric is always added when calculating compression ratios. >>>

No wonder someone got a smaller drawdown before pump restart than the calculations I did for him three weeks ago,

Warning: What follows is long.

Warning: What follows contains gluten, er, peanuts, er, math.

As I see it now, you can do this to satisfy your curiosity. Take your pump turn on pressure and turn off pressure and add 15 PSI atmospheric pressure to each. For example 30/50 on the gauge becomes 45/65 absolute. The ratio is 0.69. Ignore the fact that atmospheric pressure is less in Denver compared with in Miami and also ignore for now that you set (should have set) the pressure tank to 28 PSI gauge which corresponds to 43 PSI absolute.

The ratio of minimum air cushion size at pump turn off to maximum air cushion size at pump turn on is 0.69, you can round it to 0.7.

So with the pressure tank properly preset (including depressuring the rest of the system first) the air cushion will be nearly filling the pressure tank at pump turn on and occupying the top 70% of the whole tank after pumping up. So diaphragm goes up and down from near 0% (the bottom) to 30% the way up the tank giving you a drawdown of about 30% of the tank volume.

Now imagine that the pressure tank was partially waterlogged so at pump turn it was half full of water (the bottom half was full of water).. At pump turn off the air cushion will now occupy 70% of the top half leaving 30% of the top half or 15% of the total tank volume for you to draw down before the pump restarts.

Because you cannot see where in the tank the diaphragm is, the only way you can tell that the tank is waterlogged is to observe how much water you can draw before the pump restarts.

You could cheat by adding atmospheric pressure only to the pump turn off pressure for the purpose of computing the air cushion size. So you set the pump turn off to the same 50 PSI on the gauge you used before which corresponds to 65 PSI absolute but you set the pump turn on to your original 30 but as 30 PSI absolute instead of the apples to apples 45 PSI absolute. Yes you will get a larger draw down and less frequent pump starts but you will feel the difference during your shower given the larger spread..

If you are not curious, you can safely depressurize the system and then preset the pressure tank, the way the experts say, say to 30/50 on the gauge, or to 40/60 on the gauge, or whatever your choice is but not exceeding the rating on the pressure tank or well pump. It is not mandatory to learn the math preceding.

Please correct my math in case I made another mistake.

No wonder someone got a smaller drawdown before pump restart than the calculations I did for him three weeks ago,

Warning: What follows is long.

Warning: What follows contains gluten, er, peanuts, er, math.

As I see it now, you can do this to satisfy your curiosity. Take your pump turn on pressure and turn off pressure and add 15 PSI atmospheric pressure to each. For example 30/50 on the gauge becomes 45/65 absolute. The ratio is 0.69. Ignore the fact that atmospheric pressure is less in Denver compared with in Miami and also ignore for now that you set (should have set) the pressure tank to 28 PSI gauge which corresponds to 43 PSI absolute.

The ratio of minimum air cushion size at pump turn off to maximum air cushion size at pump turn on is 0.69, you can round it to 0.7.

So with the pressure tank properly preset (including depressuring the rest of the system first) the air cushion will be nearly filling the pressure tank at pump turn on and occupying the top 70% of the whole tank after pumping up. So diaphragm goes up and down from near 0% (the bottom) to 30% the way up the tank giving you a drawdown of about 30% of the tank volume.

Now imagine that the pressure tank was partially waterlogged so at pump turn it was half full of water (the bottom half was full of water).. At pump turn off the air cushion will now occupy 70% of the top half leaving 30% of the top half or 15% of the total tank volume for you to draw down before the pump restarts.

Because you cannot see where in the tank the diaphragm is, the only way you can tell that the tank is waterlogged is to observe how much water you can draw before the pump restarts.

You could cheat by adding atmospheric pressure only to the pump turn off pressure for the purpose of computing the air cushion size. So you set the pump turn off to the same 50 PSI on the gauge you used before which corresponds to 65 PSI absolute but you set the pump turn on to your original 30 but as 30 PSI absolute instead of the apples to apples 45 PSI absolute. Yes you will get a larger draw down and less frequent pump starts but you will feel the difference during your shower given the larger spread..

If you are not curious, you can safely depressurize the system and then preset the pressure tank, the way the experts say, say to 30/50 on the gauge, or to 40/60 on the gauge, or whatever your choice is but not exceeding the rating on the pressure tank or well pump. It is not mandatory to learn the math preceding.

Please correct my math in case I made another mistake.

*Last edited by AllanJ; 10-29-20 at 08:10 AM.*