- #1

I attempted to follow the outline linked below, but I seem to be running afoul somewhere. Thanks for any help you are willing to provide.

https://www.physicsforums.com/threads/lateral-force-of-water-on-the-walls-of-a-tank.581539/

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- Thread starter David In Kentucky
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- #1

I attempted to follow the outline linked below, but I seem to be running afoul somewhere. Thanks for any help you are willing to provide.

https://www.physicsforums.com/threads/lateral-force-of-water-on-the-walls-of-a-tank.581539/

- #2

Some of the math above is wrong, but the formula I am using is:

where my integral is from 0 to 560.

- #3

SteamKing

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Do you want to calculate the pressure acting on the wall, or the total force, which is what F is?Some of the math above is wrong, but the formula I am using is:

where my integral is from 0 to 560.

According to Pascal's Law, P = ρ ⋅ g ⋅ h, where ρ ⋅ g = 64.336 lbf / ft

At the top of your wall, P = 0 lbf / ft

The total force acting on a strip of wall 1 foot wide is 64.336 (560 y - y

For a wall 20 statute miles long, the force F would be 10,087,885 lbs / ft. * 105,600 ft. = 1.065 × 10

- #4

I want to calculate the pressure acting on the wall; it is 250 PSI at the base? What would the average PSI be? Thanks again for the help.Do you want to calculate the pressure acting on the wall, or the total force, which is what F is?

According to Pascal's Law, P = ρ ⋅ g ⋅ h, where ρ ⋅ g = 64.336 lbf / ft^{3}

At the top of your wall, P = 0 lbf / ft^{2}, while at a depth of 560 ft, P = 560 ⋅ 64.336 = 36,028 lbf / ft^{2}= 250 p.s.i.

The total force acting on a strip of wall 1 foot wide is 64.336 (560 y - y^{2}/2), evaluated for 0 ≤ y ≤ 560 ft, or F = 10,087,885 lbf / ft. width

For a wall 20 statute miles long, the force F would be 10,087,885 lbs / ft. * 105,600 ft. = 1.065 × 10^{12}lbf = 532,640,628 tons (1 ton = 2000 lbf)

- #5

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0 at the top, 250 psi at the bottom, 125 psi average.I want to calculate the pressure acting on the wall; it is 250 PSI at the base? What would the average PSI be? Thanks again for the help.

- #6

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then the psi on the wall at any given point along the wall at a depth of 280 ft

would be 1/2 of 250psi

250psi/2=125psi

the psi of a standing column of water can be found by multiplying the weight of

1 cu in of the water x the height of the column of water in inches.

you set the psi at the base to 250psi

and the height of the column of water to 560 ft

(250psi/560ft)/12 inches = .037202 lbs/cu in

.037202 lbs/cu in * ((560ft / 2)*12)=124.9987psi

- #7

SteamKing

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Or the weight of 1 cubic foot of water by the depth of water in feet. Same difference.

then the psi on the wall at any given point along the wall at a depth of 280 ft

would be 1/2 of 250psi

250psi/2=125psi

the psi of a standing column of water can be found by multiplying the weight of

1 cu in of the water x the height of the column of water in inches.

you set the psi at the base to 250psi

and the height of the column of water to 560 ft

(250psi/560ft)/12 inches = .037202 lbs/cu in

.037202 lbs/cu in * ((560ft / 2)*12)=124.9987psi

- #8

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actually theres a really big difference.Or the weight of 1 cubic foot of water by the depth of water in feet. Same difference.

that would deliver an answer of 17999.82 lbs vertical pressure per horizontal square ft.

there is no way that he could use that number unless he divided it by the 144 sq inches

in the horizontal square foot.

which results in 124.998 psi

he was wanting the average pressure that acts horizontally against the wall.

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- #9

SteamKing

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It's not clear what you are talking about.actually theres a really big difference.

that would deliver an answer of 8035.632 lbs vertical pressure per horizontal square ft.

there is no way that he could use that number.

he was wanting the average pressure that acts horizontally against the wall.

Pascal's Law works the same for cubic feet as it does for cubic inches.

If the depth of water is 560 feet and the weight density of seawater is 64.336 lbf / ft

P = 560 ⋅ 64.336 = 36,028 lbf/ft

Since there are 144 in

The hydrostatic pressure at the surface is P = 0 p.s.i. by definition, therefore the average hydrostatic pressure = 250 p.s.i. / 2 = 125 p.s.i.

The pressure profile is triangular w.r.t. depth, like so:

But let's check the numbers anyway.

1 cu. ft = 1728 cu. in.

1 cu.ft. of seawater weighs 64.336 lbf, so 1 cu.in. seawater weighs 64.336 / 1728 = 0.0372 lbf

At a depth of 560 feet = 6720 inches, the hydrostatic pressure P = 0.0372 ⋅ 6720 = 250 lbf / in

Looks like the same numbers to me.

- #10

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1 cu ft of sea water by the depth of the sea water @ 280 ft to find the average pressure

of the sea water that acts against the wall.

and that he would need to divide the result of your suggested math by 144 to find the average psi.