Hydrostatic Pressure Calculation (Pressure on tank walls)

nolapete

Jack Dempsey
MFK Member
Jun 1, 2007
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I know that the pressure is greater at the bottom than the top, but as I said with everything it was approximate.

I understand the link page quite well. I got the same results as the hydrostatic paradox page did, so I guess they don't understand it either.

Regardless of the overall thrust in PSI across front/back/side, the 310 lbs./^foot @ 5' depth is accurate and works on any of those.

I challenge you to simplify your mathematical genius to a formula that the rest of us can use.

FuzzyDuck, for the total capacity of my tanks, I always calculate 10 lbs. per gallon to allow for decorations/gravel, so that's where the approx. 37000 lbs. number was coming from. 3665 * 10 = 36650 rounded to approx 37000.
 

Jgray152

Feeder Fish
MFK Member
Dec 23, 2006
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I challenge you to simplify your mathematical genius to a formula that the rest of us can use.
How more simple can this get? I even broke it down for you. If your way works for you, using 62lbs/ft^2, great.
JGray152 said:
To get pressure at depth
5' / 2.31' = 2.164 PSI (Pounds per sqaure inch)

2.31 = 1PSI - Just FYI.

You tank bottom is 144" x 98" which equals out to be 14112 sq/in

To get weight at same depth
14112 * 2.164 = 30,538 lbs. This is at the bottom of the tank. So there is roughly 311.61 lbs per sqaure foot at the bottom of your tank.

So if you wanted to, you could say you have 30,538lbs of horixontal thrust at the base of your tank.

To get pounds per sqaure foot
14112 / 144 = 98 sq/ft
30538 / 98 = 311.61 lbs/ft^2

You can not use the PSI rating or the lbs/foot^2 at 5' for your side walls because at depth of 1 ft, there is only .432 PSI which equals out to be 62.20 lbs per sqaure foot.

To get weight at different depth and pounds per sq/ft
.432 * 14112 = 6096.38 lbs
6096.38 / 98 = 62.20 lbs/ft^2
I know that the pressure is greater at the bottom than the top, but as I said with everything it was approximate.
Approximate? Approximate you have a few values that are close to one another.
I don't think 311lbs/ft^2 at the bottom of your tank and 62.20 lbs/ft^2 which is only 1ft down from the surface, are both Approximate numbers to each other.....do you? They are approx at the specific depths. You can not get an approx for the entire surface of the glass.

I bet if you try to an an approx. pounds for the entire glass surface using the 1ft mark, you will get a number that has miles between the number you got when using 311lb/ft^2 :grinno:

The bottom of the glass has way more pressure than the top.
ap·prox·i·mate (
-pr
k
s
-m
t)
adj. 1. Almost exact or correct: the approximate time of the accident.
2. Very similar; closely resembling: sketched an approximate likeness of the suspect.
3. Botany Close together but not united


I understand the link page quite well. I got the same results as the hydrostatic paradox page did, so I guess they don't understand it either.
Then why are you not applying your knowledge that you say you have, to what you are figuring out? BTW, that pages says NOTHING about figuring pressure for the side of the tank, so what do you exactly understand on that page?

Regardless of the overall thrust in PSI across front/back/side, the 310 lbs./^foot @ 5' depth is accurate and works on any of those.
I know its accurate, close enough anyways. Its actually 311lbs/ft^2. That number ONLY works at the 5 foot specific depth. But if you are using that figure how to you come up with below? Because 36650 is 373.97lbs/ft^2, not 311lbs/ft^2

FuzzyDuck, for the total capacity of my tanks, I always calculate 10 lbs. per gallon to allow for decorations/gravel, so that's where the approx. 37000 lbs. number was coming from. 3665 * 10 = 36650 rounded to approx 37000.
Do you know that when you do this, you are adding an extra 6,112 Pounds to the tank weight? Do you really think with sand and decorations you will accumulate over 6000 lbs in your sized tank?

I'll keep watch on this thread :popcorn:
 

nolapete

Jack Dempsey
MFK Member
Jun 1, 2007
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New Orleans, LA
How does the pressure change on a side? You have 4 sides to a rectangular or square tank don't you? How does the water know the difference? That makes absolutely no sense.

The amount of approximation is relative to the tank size. Sure, I agree I'm not going to have 6000 lbs. of decorations in the big tank. You could probably agree that in my 210 gallon it would be closer with 150 - 200 lbs of gravel/pebbles,rock. I know that there's the overlap because the water is displaced by the decorations, but a cubic foot of gravel/pebbles/rock weighs more than a cubic foot of water.

How far off would it be to say that?: At the 5' of depth the pressure would be 5 times what it is at 1'.

So you'd have to do:

pressure/^ft @ 1' x length
+
pressure/^ft @ 2' x length
+
pressure/^ft @ 3' x length
+
pressure/^ft @ 4' x length
+
pressure/^ft @ 5' x length
=
total pressure on any given tank surface

Would that work?

I really don't care who is right. I just want to have a reliable, easy-to-use formula for this puzzling measurement that is fairly critical in building tanks. I think it's a huge factor in correctly determining what thickness of glass/acrylic to use and how much reinforcement you need.
 

Jgray152

Feeder Fish
MFK Member
Dec 23, 2006
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How does the pressure change on a side? You have 4 sides to a rectangular or square tank don't you? How does the water know the difference? That makes absolutely no sense.
As you know, the pressure changes with depth and not with horizontal measurments. At a specific depth, the pressure is the exact same in all 360 degrees.

The pressure, at a specific depth, say 2', is the same pressure on every piece of glass no matter the size. But its only at THAT depth, where the pressure, in this case, .865 PSI, is the same. At 2.5', you have a pressure of 1.082 PSI and at 1.5' there is a pressure of .649 PSI. This tells you that trying to get an "approx." pressure on the sides is impossible because the different in pressure can be drastically different then another depth.

So you could find the FORCE on the glass its self at that specific depth of 2' (.865psi). If you have a side glass that is 48" long, then considering we will be using PSI which is pounds per sqaure inch. We will need to convert 48 linear inches into sqaure inches. Simple enough, 48" * 1" = 48 sq/in. lol. So you can perform the following, 48 sq/in * .865psi = 41.52 pounds of force on all side glass pieces that are 48" long.

Obveously you want to figure out what the maximum pressure will be so you want to take a measurment at the bottom of the tank to figure out what that pressure is.

You can think of it this way, you "could" use glass that was thicker at the bottom and got thinner to the top only because the higher you get, the less force is applied to the glass.

How far off would it be to say that?: At the 5' of depth the pressure would be 5 times what it is at 1'.

So you'd have to do:

pressure/^ft @ 1' x length
+
pressure/^ft @ 2' x length
+
pressure/^ft @ 3' x length
+
pressure/^ft @ 4' x length
+
pressure/^ft @ 5' x length
=
total pressure on any given tank surface

Would that work?
Ah, close, except for the last part in bold. Yes, 5' deep would be 5 times greater than at 1'. Adding all together would just give you the pressure at 5' and not the overall pressure on the glass since you can't figure that out.

I think you know what you are talking about but possible wording it wrong. You should say something along the lines of, "= total pressure on all sides at the specific depth". Something like that anyways. Because pressure/ft^2 at 5' is drastically different than pressure/ft^2 at 1ft.

SO you really buying glass that can support the maximum pressure at the deepest part of the tank. There is a "safty factor" too, don't know how they come up with it, but lets throw a number out there saying 3 is good for the bottom. Well the safty factor is like 15 at 1ft depth which is out of this world good (atleast in this post).

I really don't care who is right. I just want to have a reliable, easy-to-use formula for this puzzling measurement that is fairly critical in building tanks. I think it's a huge factor in correctly determining what thickness of glass/acrylic to use and how much reinforcement you need.
I do believe you are making this harder than it should be considering you are trying to figure something out that cannot be done.

Just do Pressure per ft^2 * Length Like you were doing. At specific depths.

I like being more accurate and using these forumlas.

Head Height (ft) / 2.31 = PSI
PSI * SQ/IN AREA = POUNDS IN FORCE

I think I said what I wanted but I have to go do something so i'll be back later.
 

nolapete

Jack Dempsey
MFK Member
Jun 1, 2007
2,726
9
38
New Orleans, LA
I see where we're clashing and you're right. It's impossible to use a total or even an average here since at each 1' increment it's different. Probably the better way of saying what I want to say is:

On (insert variable) depth, the psi ranges from minimum (insert result a) to maximum (insert result b).

I appreciate you working this through with me. I never had a clue on how to figure it out before now.

That said, we really only need to know what the PSI is at the max depth and overbuild accordingly to handle that. That makes it to where we're safe at any depth less than the max.
 

Jgray152

Feeder Fish
MFK Member
Dec 23, 2006
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NH
I appreciate you working this through with me. I never had a clue on how to figure it out before now.
I think I just met someone that has a mind set like me. Untill someone has some facts to proove me wrong or make me look like an idiot (which happends more than I would like), or atleast words something in a way to make me understand differently, I will think.... I am correct and they are wrong. haha.

You are welcome though. ;)

That said, we really only need to know what the PSI is at the max depth and overbuild accordingly to handle that. That makes it to where we're safe at any depth less than the max.
Now your getting it :headbang2

You know how 1 US gallon weighs approx 8.34 lbs? You can pour 1 gallon into a reservoir of choice and you can get the PSI at the lowest depth, multiply that by the area of the bottom of the reservoir and you will get around 8.34lbs. Just a little tid bit I wanted to share.

On (insert variable) depth, the psi ranges from minimum (insert result a) to maximum (insert result b).
Welp, if you calculate the pressure for a sqaure foot, you would have a minimum and maximum, but if you calculated down to 1 sqin, there min and max would be so close it may not be worth saying.
 

Fuzzy Duck

Feeder Fish
MFK Member
Mar 7, 2009
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Hi again Nolapete and Jgrey152. I would like to know wat you two came up with about my 8x2x2 tank as regards to total weight and total pressure on the side been a different result.
BUT ON A LIGHTER NOTE Don't all our tanks hold water? Oh happy days:)
 

nolapete

Jack Dempsey
MFK Member
Jun 1, 2007
2,726
9
38
New Orleans, LA
The pressure is the same at the same depth on the sides as it is on the front. The pressure is based on depth, not the area of a surface. It doesn't matter that the front of the tank is 8' and the side is 2'. They both have 2' depth. It's not how much water contacts the surface of the glass, but the depth of the entire body of water contained.
 

Jgray152

Feeder Fish
MFK Member
Dec 23, 2006
1,659
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NH
Hi again Nolapete and Jgrey152. I would like to know wat you two came up with about my 8x2x2 tank as regards to total weight
To get the total weight, there are a few ways of doing it. You will be close to each other either way.

First way is to get the total volume of the tank and multiply it by the weight of water in one gallon
Weight (Lbs) = (Length * Width * Height / 231) * (Water Weight in one Gallon)

(Length * Width * Height / 231)= 96 * 24 * 24 / 231 = 239.37 Gallons
(Water Gallon Weight 8.33)= 239.37 * 8.32487 = 1992.72 lbs.

2nd way is to find the pressure at the lowest depth and multiply it by the sq/in area surface of the bottom.
Weight (Lbs) = (Height 'ft' / 2.31) * (Length 'in' * Width 'in')

(Height 'ft' / 2.31) =2 / 2.31 = .865 PSI
(Length 'in' * Width 'in') = 96 * 24 = 2304 sq/in area
Weight (Lbs) = 2304 * .865 = 1992.96


The third way is to use the weight of a cubic foot of water. Which you are really only getting the weight of the "cube" 12 inches down from the top of the cube.

Water Weight Cubed = (Height '1ft' / 2.31) * (Length * Width)
(Height '1ft' / 2.31)= 1 / 2.31 = .4329 PSI
(Length * Width)= 12" * 12" = 144 sq/in Area
Water Weight Cubed= 144 * .4329 = 62.337 lbs per ft^2


3rd way continued.. Now find the area of the total cubic foot volume of your tank, multiply it by the lbs per ft^2 above.
Cubic Foot Volume = (Length 'ft' * Width 'ft' * Height 'ft') * (Pounds per ft^2)
(Length 'ft' * Width 'ft' * Height 'ft')= 8 * 2 * 2 = 32 sq/ft
(Pounds per ft^2)= 32 * 62.337 = 1994.78 lbs

total pressure on the side been a different result.
You can only know the total pressure AT a SPECIFIC DEPTH. Please go back and read my posts.

Example I have shown before, Water weight in pounds per ft^2 @ 1ft depth = 62.3376 lbs. Water weight in pounds per ft^2 @ 5ft depth is 311 lbs.
 
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