# Water pressure dilema

1. Jan 3, 2009

### bluechipx

I have a problem that 'seems' to defy physics. I've been told that when your mind and physics don't mesh, go with physics, so I will, but I'm not taking it well. Okay, here's the problem. I was building a 1000+ gal aquarium in my home. Eight feet long, six feet from front to back and 38 in. deep. I was worried about the 3/4" glass not being able to withstand the pressure, so I found comparable tanks that were working. The comparable tanks were way less than my six foot dim. from front to back, but I knew that only depth should matter for pressure on the glass, not front to back dim. At this point I'm sure we all agree that if the back of the tank were only a foot from the glass or a mile, the pressure on the glass would be the same, right? The glass had an area of 3648 sq. in. (38x96) Water pressure is constant at any given depth, so I calculated the total pressure on the glass at approx 2500 lbs. I was fine with the fact that 1138 gallons of water could easily generate the 2500 lb number. In fact as I filled the tank the pressure began to bow the glass somewhat and was evident with a straightedge. Now here's the problem, if you were to move the back of the tank (6 feet) toward the glass until there was about 1/32" from front to back, and pour in one glass of water, I really can't envision a 2500 lb load on the glass, and the glass bulging from pressure. So what's up here?

2. Jan 3, 2009

### flatmaster

You are correct that preasure is a function of depth only, however you have this pressure acting on a large area.

P = F/A
F = PA
Where F is force, P is pressure, and A is area.

Your tank is 8 feet by 6 feet by 6 feet. I will find the total force on one wall of glass. Because pressure is linear with depth, I can find the pressure half-way down the glass and multiply by the area of the glass.

I go to MKS units. 2.66m X 2m X 2m

p = dgh
where d is density, g is acceleration due to gravity, and h is the depth.

p = 9,800 Pa
A pascal is a rather small unit. Now to find the net force.
F=PA
F=52,000 N

Let's take that to pounds

F = 11,700 Pounds

So you have a total force of over 5 tons pushing on one wall. Again, you are correct that water pressure is a function of depth only. However, you need to think of the total force acting on the wall. As shall pressure acting over a large area creates a large force.

3. Jan 3, 2009

### bluechipx

Flatmaster stated; "Your tank is 8 feet by 6 feet by 6 feet. I will find the total force on one wall of glass. Because pressure is linear with depth, I can find the pressure half-way down the glass and multiply by the area of the glass".

The tank is 6' x 8' x 38", not six feet tall. But your method of figuring force by taking the half-depth and multiplying that figure would be correct. I got a figure of approx 2500# with a depth of 38" and you arrived at twice that number with a depth of twice as much. So we are both doing our math correctly. For a moment let's say the distance from the glass to the back of tank was .001" and a thimble full of water was needed to fill the tank, would the thimble of water exert 2500 lbs of force on the glass?

4. Jan 3, 2009

### pallidin

Ok, maybe someone is getting their equations wrong. I wish I knew math better, but I do know that a thimble amount of water, from its own negligible weight and from negligible static atmospheric pressure conditions, can not result in that amount of force.
Just my opinion.

On the other hand, it depends on the configuration.
A hydraulic jack can exert tremendous side pressures with relatively low downward pressure.

Now I think my brain is fried.

5. Jan 3, 2009

### Staff: Mentor

It would be tough to get that much pressure with a thimblefull of water since the sides would have to be so close together surface tension would take over. But the concept is still valid: the sides of the tank do not have to resist pressure, they have to resist force. So the wider the sides, the higher the force, and the stronger it needs to be.

6. Jan 4, 2009

### gmax137

well dont confuse force and pressure. The average pressure is only about 0.7 psi, you get a big force only because you have alot of square inches (8 * 12 * 38 = 3648). This may be where your intuition is causing you to doubt your calculation

7. Jan 4, 2009

### Dr.D

You are demonstrating here the principle that makes hydraulic power and hydraulic jacks work so effectively. A relatively low pressure applied over a very large area develops a very large force so that area becomes a tremendously effective multiplier for the applied pressure.

8. Jan 4, 2009

### technutz

Also note that salt water is heavier than fresh water. If you decide to use salt water plan your construction for that:

Roughly, salt water weighs 64 lbs/cu. ft., as opposed to fresh water, which is about 62.2 lbs/cu. ft. There are 7.4805 gallons in a cubic foot. So a gallon of salt water weighs 8.556 lbs.

I hope that helps too.