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  1. #121
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    Quote Originally Posted by SeymourDumore View Post
    Guys, I am just f'n pissed off!!!

    I have been looking ( and wasting a shitton of time ) at this shit.
    Could someone with a serious fluid dynamics background please chime-in!!!
    Here is a simple and universally used equation:
    P = r * g * d
    Please, fill in the blanks:
    P = pressure gage above atmospheric pressure = ???
    r = rho = density of the fluid. ( OK, so water = 1.0 is OK??? )
    g = gravitational accelerational constant ( is 9.806 OK? )
    d = Depth of fluid ( what is the freakin' unit of measure!!! meters? inches? British pounds or DJT's hair color???)

    For dummies
    https://www.instructables.com/id/Sol...-Dam-by-Water/

    Skip the Width

  2. #122
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    Quote Originally Posted by Bill D View Post
    What if I take a garden hose a squirt the top of a four foot tall pane of glass. Not squirting hard but with a good flow so that there is a 1/8" thick stream of water running down the glass in a narrow river. What is the pressure on the bottom of the window? What if it is a rainstorm with just enough wind to cause the same thing across the width of the window.
    What if it is the domed roof of a stadium with say 50 feet from the top to the eaves. For sure they are designed to take the weight of the water or snow/ice but pressure?
    Downspouts and sewer pipes are pretty much self cleaning. If there is a blockage water builds up above it until the pressure moves the blockage and flushes it down. I have seen blocked downspouts squirting water out of joints eight feet up from the bottom where it clogged in the bend at the foot.
    Bill D
    Bill

    If we add to your imagery a volcano and a scoop of mint chip ice cream, we might put Salvador Dali to shame.

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    Quote Originally Posted by SeymourDumore View Post
    Guys, I am just f'n pissed off!!!

    I have been looking ( and wasting a shitton of time ) at this shit.
    Could someone with a serious fluid dynamics background please chime-in!!!


    Here is a simple and universally used equation:

    P = r * g * d

    Please, fill in the blanks:

    P = pressure gage above atmospheric pressure = ???
    r = rho = density of the fluid. ( OK, so for water = is 1 OK??? )
    g = gravitational accelerational constant ( is 9.806 OK? )
    d = Depth of fluid ( what is the freakin' unit of measure!!! meters? inches? British pounds or DJT's hair color???)
    MKS

    density in kilograms per meter cubed
    acceleration in meters per second
    depth in meters
    pressure in Newtons

    For dummies
    https://www.instructables.com/id/Sol...-Dam-by-Water/

    Skip the Width

  4. #124
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    Quote Originally Posted by 9100 View Post
    Some never accepted quantum mechanics, including Albert Einstein, leading to the famous Copenhagen debate with Niels Bohr. When Einstein said that God didn't roll dice, a frustrated Bohr said "Albert, stop telling God what to do."
    It's possible that ol' Albert was smarter than he looked ....

    The Rebel Physicist on the Hunt for a Better Story Than Quantum Mechanics

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    not quite correct on the units.

    m/s^2 metres per second / per second accell

    pressure has units of Pascals KPa is Kilo pascals (1000 x ) a conversion is 101.3 kpa is 14.7 psi or 1 bar. don't mix units must be SI units. so 1 Pa is a very small pressure.

    Force has units of Newtons and a Newton is kg x 9.81 (x accel due to gravity)

    as i said earlier its been a while since i looked at it so look it up in a good book is best especially if its critical to safety or life.

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    Originally Posted by JohnMartin
    In that case, I guess you’d have to believe that a hydraulic cylinder develops more force the further out the piston is extended, and when it is retracted all the way there is no force at all. Intuition does not always serve us well.




    How is that relevant here??


    Well, if you believe that the pressure on a glass aquarium wall or on a dam face is somehow related to the amount of water pushing on it in the horizontal direction, then you would also have to believe that to be the case with the hydraulic cylinder as well. The more it extends the more fluid is behind it, so the force must increase as well. And, when the piston is fully retracted with very little fluid behind it, there would be very little force exerted. Does a bottle jack get easier to pump the more the ram is extended?

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    Quote Originally Posted by 9100 View Post
    Some never accepted quantum mechanics, including Albert Einstein, leading to the famous Copenhagen debate with Niels Bohr. When Einstein said that God didn't roll dice, a frustrated Bohr said "Albert, stop telling God what to do."
    It’s incorrect to say Einstein never accepted quantum mechanics as he was, in fact, one of the founders of quantum physics. His debate with Bohr was about the interpretation of quantum mechanics.

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    If the thick glass bows as much as you say there could be another problem. Touching the outside with a diamond ring could cause it to break and make a huge mess. Best to cover the glass with one of those clear protective films used on cars. Expel is a good one, and not easy to scratch as it has some bit of self healing to it. If applied properly you will not know its there.

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    2020-06-30-08_40_27-window.jpg

    ................

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    Quote Originally Posted by SeymourDumore View Post
    Guys, I am just f'n pissed off!!!

    I have been looking ( and wasting a shitton of time ) at this shit.
    Could someone with a serious fluid dynamics background please chime-in!!!


    Here is a simple and universally used equation:

    P = r * g * d

    Please, fill in the blanks:

    P = pressure gage above atmospheric pressure = ???
    r = rho = density of the fluid. ( OK, so for water = is 1 OK??? )
    g = gravitational accelerational constant ( is 9.806 OK? )
    d = Depth of fluid ( what is the freakin' unit of measure!!! meters? inches? British pounds or DJT's hair color???)
    Your concern with d is noted BUT wht didn't you also "wonder" a bit about P ??? The identical concern should have been there??? :-) They are related to what "system" you're working in.
    ...lewie...

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    I'm still waiting for someone to expalin, in plain English, why the weight of the medium inside the tank does not matter.

    Why does the tank have a ton of pressure with water (which is heavy at about 8lbs per gallon heavy) but hardly any with a liquid that weighs .005 lbs per gallon.

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    Quote Originally Posted by GregSY View Post
    I'm still waiting for someone to expalin, in plain English, why the weight of the medium inside the tank does not matter.

    Why does the tank have a ton of pressure with water (which is heavy at about 8lbs per gallon heavy) but hardly any with a liquid that weighs .005 lbs per gallon.

    Weight doesn't matter, Density does.

    Empty, the tank is filled with AIR a very "un dense" liquid. It pushes against the inside tank walls at 14.7 pounds per square inch . Just like the "fluid air" on the outside of the tank walls.
    At suitable temperatures, you could fill the tank with liquid air. That would be "more dense" than gaseous air and would press against the glass with more pressure under the influence of GRAVITY.

    Water is higher density yet, and will press against the tank glass harder yet. Also due to gravity attraction.

    Take the tank and it's contents sufficiently distant from any single gravity source (like the earth) and all the pressure will disappear regardless of the density of the contents.

    Like MAGIC

    Gravity turns density into weight. Weight is a result, it is not a characteristic of any material. We just are accustomed to weight, because we seldom leave earths gravity with anything in our pockets. Check with the folks up in the international space station for comments..

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    Archimedes and Blaise Pascal would turn in their graves if they saw this thread.

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    Quote Originally Posted by MattiJ View Post
    Archimedes and Blaise Pascal would turn in their graves if they saw this thread.
    Truer words never spoken.

    I used to do a lot of scuba diving. Every 33 feet equals one atmosphere. So the pressure on the surface, 14.7 psi. At 33 feet, twice that. 66 feet, 3 atmospheres, etc.

    If the pressure was many thousands of psi at say 100 feet deep, you would not be able to take a breath. Steel tanks are only pressurized to 2250 psi. The second stage regulator drops that down to equal the water pressure however deep you are, so your diaphragm can overcome the pressure.

    80 ft^3 of air, pressurized at 3000 psi fits into a backpack size package will last about one hour at 60 feet with reserve. If the pressure was many thousands of psi, you would consume the entire tank of air in just a few breaths.

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    Quote Originally Posted by Scruffy887 View Post
    If the thick glass bows as much as you say there could be another problem. Touching the outside with a diamond ring could cause it to break and make a huge mess. Best to cover the glass with one of those clear protective films used on cars. Expel is a good one, and not easy to scratch as it has some bit of self healing to it. If applied properly you will not know its there.
    When I was putting this tank together I did research on glass, not as basic as metals. It seems every piece of glass is it's own entity. A flawless piece would be exceptionally strong but a piece with a sharp 'V' scratch however small would lead the a stress riser that would weaken it considerably. The figure for the strength spread in glass from strongest to weakest due to imperfections "could" be from 100% to 10%, or so I vaguely remember reading. When my tank was filled I remember tiptoeing around the house whispering as I talked. I was thinking the tank could be at 99.99% of it's breaking strength. If someone slammed a door I would flip out. Now it has lived through terrible thunder storms and all sorts of threats. It would be nice to know, in percentage, how close it is to letting go. My living room is 14 feet by 28 feet and if the tank let go there would be 4" of water to mop(?) up. I must be confident now as I have slept like a baby on the couch beneath it several times.

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    This discussion reminds me of one here a few years back when someone mentioned that Young's Modulus for steel doesn't (on any practical level) change with heat treatment.

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    Jancolic, The basis of the question here, if I can remember it through the noise level, is that a small hydrostatic head may only have a small pressure but acting on a lot of square inches can generate a lot of force. The actual question is whether it will still generate the same head when the column becomes very narrow. The answer is that it will until the column becomes so narrow that other forces dominate. These forces are always there but for most spacings they are so small that they can be ignored. There is no magic dimension where there is a sudden change.

    Bill

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    Quote Originally Posted by SeymourDumore View Post
    Guys, I am just f'n pissed off!!!

    I have been looking ( and wasting a shitton of time ) at this shit.
    Could someone with a serious fluid dynamics background please chime-in!!!

    Here is a simple and universally used equation:

    P = r * g * d

    Please, fill in the blanks:

    P = pressure gage above atmospheric pressure = ???
    r = rho = density of the fluid. ( OK, so for water = is 1 OK??? )
    g = gravitational accelerational constant ( is 9.806 OK? )
    d = Depth of fluid ( what is the freakin' unit of measure!!! meters? inches? British pounds or DJT's hair color???)
    So, first of all, calm down and get less pissed off: your equation is correct. That said, this is a fluid STATICS equation. Nothing moving. Fluid dynamics would involve moving fluid.

    Units is a stumbling block for many. I'll give you a simple factor below, but r * g * d has units

    mass/volume * length/(time^2) * length

    When you collect terms you get mass * length/(time^2) * length/volume, or mass * length/(time^2) /(length^2), or

    mass * length/(time^2) /area

    the mass*length/(time^2) is force, so this checks out: Pressure = force/area.

    If P = r * g * d, then using metric (meters, seconds, grams) we have

    P = g/(m^3) * m/(s^2) * m, or
    P = g *m /(s^2) /(m^2)

    The g * m/(s^2) unit is called a gram force unit and is analogous to the kg force unit. But both of these are non-standards. 1 kg force = 9.80 N. One g force = 9.80 mN.

    If you want simple, in psi, P = 0.036 * d, where d is inches assuming water at 4° C (1.000 g/cm^2).

    in Bar, you get P = 0.098 * d, where d is in meters, again assuming that water is at 4° C

    If you want to work out the equations yourself, I suggest writing out the equations with the units but with variables and value, like this:

    P [ g *m /(s^2) /(m^2) ] = r [g/m^3] * 9.80 m/(s^2) * d [m]

    Then remember that x/y is 1 if x = y and y is not zero. Why is this important? Because it allows you to calculate conversion factors. Suppose you want to get pressure in the same units but using a depth in inches. A conversion factor from d in meters, to d'' in inches. We know that 100cm = 1m. So we can multiply any factor in the equation by 1m/100cm. And by 2.54 cm/inch. So

    P [ g *m /(s^2) /(m^2) ] = r [g/m^3] * 9.80 m/(s^2) * d'' [inches] * [m/100cm] * [2.54 cm/inch]

    P [ g *m /(s^2) /(m^2) ] = r [g/m^3] * 9.80 m/(s^2) * d'' (2.54/100) [m]

    The point is that d'' in inches * 2.54 /100 gives you d in meters.

    Clear as mud?

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    Quote Originally Posted by 9100 View Post
    Jancolic, The basis of the question here, if I can remember it through the noise level, is that a small hydrostatic head may only have a small pressure but acting on a lot of square inches can generate a lot of force.
    The force can never exceed the weight of the water, and it's just the weight divided by the square inches or whatever unit you want to use.

    A column of water one square inch and 33 feet high weighs about 13.7 lbs. Pressure at the base is 13.7 psi.

    If you spread that same volume of water in a column 1 square foot in cross section, you get 13.7/144 or 1.3 psi.

    OP's tank has a base cross section of 48 ft^2 and holds 9445 lbs of water. That's 196lbs/ft^2 or 1.36 psi.

    The pressure on the sides is equally divided on all sides. So the average pressure is .68 psi on the sides and the total force is about 8680 lbs. (average force .68 psi times total side area of 12,768 in^2). This is lower than the 9445 lbs of water, so we're okay.

    For our 1" thick aquarium, the weight of the water is only 131 lbs. you get the same 1.36 psi at the base. The total side area is 7372 in^2, so multiply the .68 psi times 7372 in^2 (the total side area) and you get 5012 lbs as the total force.

    This is a nonsense number, you cannot generate 5000 lbs of force with 131 lbs. of water.

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    Jancollc, you seem usually pretty able &smart in technical matters so take a deep breath and think this over once more.

    People on this thread are mixing up weight, pressure and force in amazing amounts. 😬

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