1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Water pressure

  1. Feb 12, 2010 #1
    Can anybody help?

    I am not a physics or science expert. So be patient and laymens terms please.

    I am trying to calculate the pressure of water on the side walls of a water barrel that I make to store rain water collected from roofs.

    The barrels have an inside bladder that is clad with wood.

    Is the pressure greater at the bottom? What is the pressure per square metre. Is there a formula that i can use to the calculate the water pressure according to the height, length and width of my barrel.
     
  2. jcsd
  3. Feb 12, 2010 #2

    Lok

    User Avatar

    Yes, there is it is called Archimedes law.

    P=Density of medium(water 1000kg/sq m)*g(gravitational acceleration)*h(height of barrel)

    A water barrel has about 1 m in height so the P=9810 pascals that is about 0.0981 atm (that is normal atmospheric pressure)

    The width and length do not matter at all just the height.

    In order to feel the pressure on your own take a plastic bag put it over your hand and insert it into the barrel, because the pressure is not evenly distributed you will feel a lot of force.
     
  4. Apr 21, 2010 #3
    Hello,

    I have a follow-up question related to water pressure that has been plagueing me for some time:

    Water pressure is calculated by: P = Density of fluid * acceleration of gravity * height of fluid above.

    I once saw a professor say he knew of an expierment where he filled a wood barrel with water, and it was okay...he then took a small (in diamater) buy LONG tube and put it in the barrel, got on his roof, and proceeded to fill that tube with water...his claim was the pressure at the bottom of the barrel increased dramatically, as the Height of fluid above climbed...and eventually, (presumabely with a 20 foot section of tube (just a guess here)) the barrel exploded since it was subject to pressures it could not contain.

    My question is: does the science here work out? Woud I feel the same pressure if I were in the bottom of a barrel, and somebody had a 50 foot section of garden hose coming out of the water (vertically) and filled it with water? Would this be the same pressure I'd feel 50 feet under water?

    I'm very new to physics, so please bear with my ignorance here, I'm trying to learn, and that's why I'm asking.

    Thanks for your time and thought on this.
     
  5. Apr 21, 2010 #4

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Yes - but only over the area under the bottom of the hose, whereas if you were underwater you would have that same pressure everywhere.
     
  6. Apr 21, 2010 #5
    If the barrel had a lid, and the hose was in a hole in the lid...wouldn't the entire pressure of the barrel increase? (assuming the lid is sealed well, and the junction of the hose to lid was also sealed).

    Would it be possible to over-pressurize a barrel this way?

    Assuming the above (sealed barrel, sealed lid) is there any difference between having a garden hose full of water 50 feet high vs. a 12 inch diameter pipe filled with water 50 feet high?

    My common sense tells me there would be a difference...but the formula for calculating fluid pressure tells me there should be no difference.
     
  7. Apr 21, 2010 #6

    SpectraCat

    User Avatar
    Science Advisor

    Think of it in terms of mass ... the mass of water in a 12" diameter pipe will be a heck of a lot bigger than in your garden hose. Under static conditions, the water in the barrel is effectively supporting all of the water in the tube ... so in short, yes, there will always be a point where you add enough water for the barrel to reach its breaking point. You will have to go much higher to reach that point for a [STRIKE]thin[/STRIKE] narrow tube than for a [STRIKE]thick[/STRIKE] wide tube, because the only thing that matters is the mass being supported (assuming that the tube itself is strong enough that to support the column of liquid).
     
    Last edited: Apr 21, 2010
  8. Apr 21, 2010 #7

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Yes it would, you would have to keep pouring water into the tube until the barrel was almost full (the small amount of air in the top would be compressed) then the pressure would be the head of water in the tube.

    Since the air inside the barrel is compressed to the same pressure - it is pushing down on the rest of the surface with the same pressure the water in the bottom of the hose is pressing down with.

    It's like a tire pump, the pressure in the pump is only across the end of the hose. But you keep pushing more air in so that the pressure everywhere in the tire is the same as this pressure.
    The water filled hose is just another form of pump.
     
  9. Apr 21, 2010 #8
    This is where I keep getting lost! In the formula to calculate pressure, Mass isn't a factor, just density of the liquid, gravity, and depth...so, please bear with me, but a follow-up question....if we're factoring in the mass of the liquid above (as makes sense to me)...then, does that mean the pressure experienced one meter under water in a swimming pool is less than the pressure experienced one meter under water in the ocean...because, the mass of the liquid above me (one meter of the entire ocean vs. one meter of a swimming pool) must be more; right?

    Am I just over thinking this?

    Thanks for your help!
     
  10. Apr 21, 2010 #9
    So, am I correct is saying the pressure in a barrel with 50 feet of garden hose full of water above would be the same pressure in a barrel with 50 feet of 12 inch pipe full of water above? What about 50 feet of drinking straw full of water above? That's what I can't wrap my head around.

    Thanks for your help.
     
  11. Apr 21, 2010 #10

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    It is really mass = density * volume = density * area * height

    No because pressure just depends on the weight of water above you, some water 1000mi away doesn't gave any effect.
    Although because seawater is denser than fresh water (the extra salt) the pressure a certain distance under the ocean is higher than in a swimming pool ( by about 2%)

    Yes - because that smaller weight is concentrated on a smaller area.

    If you really want a confusing one think about floating.
    How much water does it take to float a battleship?

    You learned that a floating object displaces it's own weight in water right?
    So if you put a boat in a dry dock you only need a bit of water around and underneath it.
    Now move the walls in toward the hull, you need less water.
    Now imagine a perfectly designed dry dock the same shape as and a fraction of a mm larger than the boat - you would only need a teaspoon of water to float it.
     
  12. Apr 21, 2010 #11
    I had never considered that before...but it makes perfect sense...that is too crazy!

    Thank you for your patience in answering my questions...it seems common sense was in fact right here, I think I must not fully understand the factors that go into calculating pressure. Thanks again!
     
  13. Apr 21, 2010 #12
    Yes, but very little.
    Assuming all other factors equal (an ocean side pool perhaps), the difference in pressure between the ocean and the pool will only be based on the difference in the density of the two fluids.
    From a quick search, ocean water is about 2.7% more dense than pure water (1000 kg/m^3 for pure water, 1027 kg/m^3 for ocean water due to the salt).

    Too late (mgb_phys beat me too it).
     
    Last edited: Apr 21, 2010
  14. Apr 21, 2010 #13

    SpectraCat

    User Avatar
    Science Advisor

    Well, mass doesn't appear in the formula explicitly, but it is certainly there implicitly in the density. You are correct that the pressure at the bottom of the tube is the same for any diameter tube, and depends only on the height of the column. However, what you seem to be missing is that the force exerted by the column of water on the barrel below it does depend on the total mass as I said. Since the barrel is the same in all cases, we can just focus on the column. The pressure from the column on the water in the barrel is applied over the cross sectional area of the column ... therefore the total applied force is F=PxA. That force is then added to the force from the water already in the barrel, and distributed over the walls of the barrel.

    So, hopefully this helps you to see now why the pressure in the barrel depends on the total mass of water in the column.
     
  15. Apr 21, 2010 #14

    russ_watters

    User Avatar

    Staff: Mentor

    No. This is completely wrong. The pressure in the barrel is completely independent of the horizontal geometry of the tube. It depends on the height of the tube only. That's the entire point of the concept of hydrostatic pressure and the equation given above (which includes only m, g, and h - nothing about the area/geometry).
     
  16. Apr 21, 2010 #15

    russ_watters

    User Avatar

    Staff: Mentor

    Am I reading this correctly? You seem to be saying if you're under the hose you'd feel one pressure but if you're not under the hose - but at the same depth - you'd feel less pressure? That's not correct. The pressure is a function of the height of the hose and depth in the barrel only.

    Heck, the tube doesn't even have to be attached to the top of the barrel: A tube coming out the bottom of the barrel, then bending upwards to the same height will cause the same pressure at the same depth.

    Make sure you guys are clear on what these diagrams are saying: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html [Broken]
     
    Last edited by a moderator: May 4, 2017
  17. Apr 21, 2010 #16

    SpectraCat

    User Avatar
    Science Advisor

    Yes, I am aware of the concept of hydrostatic pressure, but I don't think it applies in this case the way you are using it. You may notice that I already stated that the hydrostatic pressure at the bottom of the tube is the same, but I think the situation inside the barrel is different. Consider the following illustration:

    Imagine a membrane at the top of the barrel, now take a weight equivalent to the mass of the column of water and place it on top of the membrane ... the force from that mass will be transferred through the membrane, increasing the hydrostatic pressure in the barrel. Clearly, if you make the mass large enough, the barrel will fail. I believe this analogy is correct in this case, because if the height is the same, the mass from the water in a wider tube will clearly be greater than that from a narrower column, and all of this mass ultimately has to be supported by the walls of the barrel.

    I cannot see how the above analysis is wrong ... however I am perfectly willing to accept that it is if you can show me where I have made a logic error. Actually, I think I may try this experiment, using a balloon to illustrate whether or not there is really a pressure increase from the wider water column.

    (*Ahem* I am pretty sure I already did a related experiment in college ... more than once ... using an apparatus consisting of a funnel and a long tube suspended from a dorm window ... however for some reason :rolleyes: my memories of it are a little fuzzy, but I am pretty sure I would not have gotten the same effect if the tube were as narrow as say, a soda straw :wink:).
     
  18. Apr 21, 2010 #17

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Yes you would (ignoring effects of surface tension/friction etc)

    Think of a compressed airline (or a bike pump) if you have 100psi then you have the same pressure whether it's a 1/4" bike pump hose or a gas pipeline. More force but spread over more area = same pressure.
     
  19. Apr 21, 2010 #18

    SpectraCat

    User Avatar
    Science Advisor

    Right, but it is the total force that matters here .. not the pressure, that is the point I have been trying to make. Like I said, I am not 100% sure my analysis here is correct, and I understand why it seems to go against what one would normally predict from the hydrostatic pressure equation.

    But, c'mon .. you are trying to get me to believe that the pressure in the barrel is the same if you have a 10 m column of water in a soda straw pressing down on it, as opposed to having a 10 m column of water in a 30-cm diameter tube. I simply cannot see how that can be correct .. what am I missing?
     
  20. Apr 21, 2010 #19

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Actually the breaking of the barrel depends on pressure, but the total force is the pressure * the area.

    Imagine a steel rod instead of water.
    A 10m long nail with a small tip would have a very high pressure and would go through a wooden plank with it's own weight.
    A truck driving over the plank with it's much larger weight spread over a larger area wouldn't have as much pressure and wouldn't go through the wood
     
  21. Apr 22, 2010 #20
    Hi all,

    There is some great discussion here; thanks! This is exactly where my friends and I got stuck. I'm glad to see smarter people than I are working though it!

    For reference sake: This idea / problem came from an old (80's?) video that played on public television; it was a video of a physics class in Hawaii..the physics professor was really excited about science (his students, so-so) but he was fun to watch, and he always had great ideas. He was the one who said he had a friend who made a barrel fail using the garden-hose and standing on a roof trick...(increasing the pressure simply by increasing the depth, independent of the volume of water, just the height).

    Thanks again for the continued discussion!

    Russ_waters, thank you for the link to that hydrostatic pressure page; that was a great lesson on what I'm trying to figure out!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook