Amount of pressure at the bottom of a tank from water

[DumpsterKing]

I am not sure how to calculate any of this or if i am even doing it correctly. I pulled these formulas from different sites and not sure if I applied them correctly. Or even if these were the right formulas to apply.

**I also did all calculations acting as if the bladder was square and had not round corners.**

1. The problem and given/known data. ( the attached PDF shows the data and shape of the bladder)

Calculate the total force, on the bottom of the bladder.

2.Equations used

Pressure= force/area
Force= length x Width x Height x water density
Area of bottom in square inches= width x length x 144

3. My attempt at a solution

First I solved for force

Force= length x Width x Height x water density
15.166 x 7.166 x 6 x 62.5
f= 40,754.831 lbs

Next I solved for area of the bottom

Area of bottom in square inches= width x length x 144
A= 7.166 x 15.166 x 144
A= 15649.855 sq in

Then I solved for pressure

Pressure= force/area
p= 40,754.831 lbs/15,649.855 sq in
p=2.604 lbs/sq.in.

I understand that pressure increases as the depth increases... so pressure isn't the same in all levels (right?)

But does any of this make sense or seem right. Any help or insight on this problem is greatly appreciated.

 

[sophiecentaur]

Is this a practical problem? What is in the bladder? Air?
If it is air filled then it will distort due to the differences in pressure. The shape it will distort to will correspond to the lowest potential (that is a fundamental) - what is keeping it at the bottom and how rigid is it? The details matter in this problem - as ever.

 

[DumpsterKing]

Yes this is practical. In reality it can be anything but for this problem assume it is filled with water.

This is a stationary holding tank. It is inside a steel frame and is rigid.

 

[sophiecentaur]

Yes this is practical. In reality it can be anything but for this problem assume it is filled with water.

This is a stationary holding tank. It is inside a steel frame and is rigid.

[Edit -nearly a complete re-write of what I first wrote]

I really didn't get what you were describing but I think I have it now. (I thought we were talking 'submarines' at first lol) Yes, the pressure will be greatest on the bottom but will it be resting on a 'floor'? The pressure on the sides, near the bottom, will be high and will cause the sides to 'bulge' - but not much, if your tank is steel. I think a more detailed description is needed about what you have in your head - and your reasons for choosing this particular construction. Wouldn't a simple steel box do the job?

 

[DumpsterKing]

Nope no subs here lol.. I work for a roll off container manufacturer. This item is called a poly tank. This bladder is actually made out of high weight molecular polyethylene. I attached an image to help you get a idea of what I'm working with. The black bladder is the high weight molecular polyethylene. We use poly because our customers store all sorts of chemicals and liquids in the tank. We are just trying to figure out the pressure on at the bottom of the tank on the sides of the tank. Whether this be in lbs\sq ft or psi. Hope this helps and sorry for not being detailed to start with.

 

[sophiecentaur]

Right. No need to apologise at all. When you are not heavily into Physics it's not always easy to ask the question in the most understandable way. Yes, the pressure at the bottom of the sides is the same as the pressure on the floor and it goes to zero at the top. (Dams are always made fatter at the bottom than at the top because of this)
I notice that the bottom of the cage is supported with a solid wall so someone has presumably taken this into account. I realize that your question is not about designing one - just healthy curiosity.
The easiest way to calculate pressure is the formula
Pressure = height * density * g which is much shorter than your sum.
It doesn't matter what shape the container is - just the height from surface to the depth of interest. Because you are using those beastly Imperial units, you need to particularly careful about sticking to the units you choose (feet or inches) and use the appropriate density units.

A quick mental calculation /estimate would tell me that, if the depth is 6', that's 1/5 of 30'. 33' of water corresponds to one atmosphere (14 pounds / sq ft) so the pressure would be about 14/5 = 2.8 lbs/sq ft. Your sums are more accurate than mine and give 2.604 lbs/sq ft but we 'agree' within the same ball park.

 

[jbriggs444]

A quick mental calculation /estimate would tell me that, if the depth is 6', that's 1/5 of 30'. 33' of water corresponds to one atmosphere (14 pounds / sq ft) so the pressure would be about 14/5 = 2.8 lbs/sq ft. Your sums are more accurate than mine and give 2.604 lbs/sq ft but we 'agree' within the same ball park.

Those beastly imperial units indeed. Atmospheric pressure is about 14.7 pounds per square inch. Dividing by 5 and rounding differently gives me about 3 psi.

 

[DumpsterKing]

Thank you all for these answers. Yes, this is all in healthy curiosity. Yes imperial units are a pain to use and when it comes down to it confusing lol. Yes i just wasn't sure of my math or if i applied anything that was even right.

Now a second question. Say i wanted to pick this tank up at one end at a set angle say 40° would i be able to figure out the pressure at the back bottom corner of the bladder. With some ease??

 

[jbriggs444]

Thank you all for these answers. Yes, this is all in healthy curiosity. Yes imperial units are a pain to use and when it comes down to it confusing lol. Yes i just wasn't sure of my math or if i applied anything that was even right.

Now a second question. Say i wanted to pick this tank up at one end at a set angle say 40° would i be able to figure out the pressure at the back bottom corner of the bladder. With some ease??

That turns it into a trigonometry problem. How high does the fluid rise when the shape changes from a rectangle to a triangle and you still need to hold all the fluid? Does it just spill out over the downhill end?

Based on the picture, I'd say that the "spill out" result applies.

 

[sophiecentaur]

If the water can't spill out then the maximum head of water will be the diagonal distance from highs to lowest point - assuming the tank is full.

DumpsterKing: your maths looks right enough- you just used a bit of a round about way of getting there.

 

[DumpsterKing]

Okay, Say that the tank is only half filled. Which would be roughly 5,000 gallons of water. The water cannot spill out i can assure you of this.

sophiecentaur: If the water can't spill out then the maximum head of water will be the diagonal distance from highs to lowest point - assuming the tank is full.

Does this mean I would use the same formulas to go about this problem and just plug in different numbers?

 

[sophiecentaur]

Just treat the problem as if it were pipe work. In pipework, the (static) pressure is just due the head of water. The route is not relevant.
Pressure = density times g times head of liquid (always, whatever the shape is involved)
Finding the head is just geometry.

 

[DumpsterKing]

I might be asking a dumb question but your description of the formula confused me.

Pressure = denisty (assuming this is density of the liquid?) times g (gravity?) times head of liquid (height of the liquid) what do you mean by shape involved?

 

[sophiecentaur]

I might be asking a dumb question but your description of the formula confused me.

Pressure = denisty (assuming this is density of the liquid?) times g (gravity?) times head of liquid (height of the liquid) what do you mean by shape involved?

I am amazed that we can get this far (and it happens so often on PF) and yet I get the impression that you have not looked any further than this PF thread. I wrote the formula in words, expecting that you would possibly not be confident enough to use the formulae that you can find in about a thousand easily found physics sites. Have you actually looked anywhere else about this? I thought that the "pipework" reference would make it more approachable.
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Pressure/HydroStatic.html and come back with any particular problems you find. You will need to read what is written there and not just dismiss it as 'too hard' but it says all you need to know about pressure in liquids. PF expects people to put in a bit of their own work, in addition to getting answers.