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Posers and Puzzles

Posers and Puzzles

  1. Standard member PBE6
    Bananarama
    17 May '05 18:37
    Here's a problem I tried recently in my quest to further my understanding of spill protection.

    Suppose you have a tank with a height of 10 metres and a radius of 1 metre. It's siting on the ground and filled with an ultralow viscosity fluid. You also have a 1-metre high spill barrier of negligible thinkness, long enough to make a complete ring around the tank wherever it is placed. How far back from the tank would you put the barrier to ensure all fluid would be captured while minimizing the spill protection area?

    Please make an a*s out of you and me by stating your assumptions in your answer.
  2. 17 May '05 18:54 / 1 edit
    Originally posted by PBE6
    Here's a problem I tried recently in my quest to further my understanding of spill protection.

    Suppose you have a tank with a height of 10 metres and a radius of 1 metre. It's siting on the ground and filled with an ultralow viscosity fl ...[text shortened]... a*s out of you and me by stating your assumptions in your answer.
    A concentric circle at 2.16 m (radius = sqrt(10)) from the tank.
  3. Standard member PBE6
    Bananarama
    17 May '05 19:09
    Originally posted by Mephisto2
    A concentric circle at 2.16 m (radius = sqrt(10)) from the tank.
    AHEM! Assumptions please?
  4. 17 May '05 19:24
    Originally posted by PBE6
    AHEM! Assumptions please?
    I did not calculate it, but assumed that if the tank was damaged at any height between 1m and 10m, the pressure (also depends on the surface and form of the damage) would not give the fluid a horizontal velecoty that would send it over the protection. Perhaps it can even be demonstarted that it is impossible, regardless.

    No other assumptions.
  5. 17 May '05 19:26
    Originally posted by Mephisto2
    A concentric circle at 2.16 m (radius = sqrt(10)) from the tank.
    I will say 10.5 meters from the center of the tank.
    While it is true that the volume of the tank is much less than that of my recommended spill barrier, I will assume that the tank is rigid enough to maintain it's shape while falling over.
  6. 17 May '05 19:31
    Originally posted by Gremlinzzz
    I will say 10.5 meters from the center of the tank.
    While it is true that the volume of the tank is much less than that of my recommended spill barrier, I will assume that the tank is rigid enough to maintain it's shape while falling over.
    nice. what if the tank jumps (e.g. after an explosion) over the barrier
  7. Standard member PBE6
    Bananarama
    17 May '05 20:02
    Originally posted by Mephisto2
    nice. what if the tank jumps (e.g. after an explosion) over the barrier
    These kinds of catastrophic failures would not included as part of a regular spill protection plan designed to handle leaks. They would be part of an emergency response plan.

    Gremlinzzz was pretty close to my answer, but it's still a little too small.
  8. Standard member Daemon Sin
    I'm A Mighty Pirateā„¢
    18 May '05 14:14
    Oh..... I thought the title of the thread was 'Water sports'
    I was expecting some logical problem about wakeboarding or something!

    I'm all disapointed now =(
  9. Standard member PBE6
    Bananarama
    18 May '05 15:06
    Originally posted by Daemon Sin
    Oh..... I thought the title of the thread was 'Water sports'
    I was expecting some logical problem about wakeboarding or something!

    I'm all disapointed now =(
    Usually you have to pay to see "water sports" on the internet. There's no free lunch here!


    Yeah, I wanted to change the title of the thread as soon as I wrote it, but I couldn't figure out how to do that.
  10. Standard member PBE6
    Bananarama
    18 May '05 17:11 / 1 edit
    Really what you need to worry about for spill protection are the following:

    1) If the tank springs a leak, how far will the fluid go when it spurts out?

    2) If the tank leaks all of its contents, will the volume be contained within the berm?

    There are some other environmental issues too, like nearby watercourses, sewers, proper spill response and clean-up procedures, and the like. But for this question, figuring out the maximum distance of the fluid jet and the volume will be enough.
  11. 18 May '05 17:16 / 1 edit
    Originally posted by PBE6
    Really what you need to worry about for spill protection are the following:

    1) If the tank springs a leak, how far will the fluid go when it spurts out?

    2) If the tank leaks all of its contents, will the volume be contained within the be ...[text shortened]... e maximum distance of the fluid jet and the volume will be enough.
    OK, I understand. Now explain why you need 10+ m.
  12. Standard member PBE6
    Bananarama
    18 May '05 18:15 / 2 edits
    OK. I'll state my assumptions in bold as a I go along.

    Assume that the fluid is ultralow viscosity (no head losses due to friction), and that it is incompressible (so we can use Bernoulli's equation).

    Now we take two points along the streamline and use Bernoulli's equation. Point 1 will be at the top of the fluid inside the tank, and point 2 will be right at the top of the jet, just outside the leak. So we have:

    P1/gamma + v1^2/(2*g) + z1 = P2/gamma + v2^2/(2*g) + z2

    Now we can simplifiy. Assuming that P1 = ambient atmospheric pressure, v1 is negligible (no movement in the fluid so far from the leak), and P2 = P1 = ambient atmospheric pressure (free jet, no change in air pressure with height), we get:

    z1 = v2^2/(2*g) + z2

    Solving for v2, the horizontal velocity of the free jet is:

    v2 = (2*g*(z1-z2))^0.5

    The jet will only continue to move horizontally through the air until it hits the ground, so we have to calculate the time for the fluid to drop. Assuming no air resistance, we have:

    z3 = z2 - (g/2)*t^2

    and solving for t we get:

    t = (2*(z2-z3)/g)^0.5

    So the distance the jet will travel is just:

    d = v2*t = [(2*g*(z1-z2))^0.5]*[(2*(z2-z3)/g)^0.5]

    which simplifies to:

    d = 2*((z2-z3)*(z1-z2))^0.5

    Taking the derivative of d with respect to z2, we can find out what leak height gives the furthest distance:

    dd/dz2 = (z1-2*z2+z3)/((z2-z3)*(z1-z2))^0.5)

    Letting dd/dz2 = 0, and solving for z2, we get:

    z1 - 2*z2 + z3 = 0
    z2 = 0.5*(z1+z3)

    So the worst case scenario happens when the leak occurs at a height half the combined height of the tank and the barrier. Subbing this value into our distance formula, we get:

    d = 2*[(0.5*z1 + 0.5*z3 - z3)*(z1 - 0.5*z1 + 0.5*z3)]^0.5
    ...= 2*(0.5*z1-0.5*z3)
    ...= (z1-z3)

    So the fluid will fly horizontally the height of the tank minus the height of the barrier. This is where we must place the barrier. So using the values given, we get:

    d = (10-1) = 9

    So the barrier will be placed in a concentric circle around the tank, at a distance of 9 metres from the tank wall (a total radius of 10 metres).

    Now we do a quick check for the total volume. The volume will be the volume of the system minus the volume of the tank up to a height of the barrier:

    V(barrier) = pi*[(9+1)^2]*(1)
    V(tank in barrier) = pi*p(1)^2]*(1)

    V(barrier) - V(tank in barrier) = pi*(100-1) = 99*pi

    The total volume of the tank is:

    V(tank) = pi*[(1)^2]*(10) = 10*pi

    So the barrier has sufficient containment to hold the entire contents of the tank.

    Oops! It looks like Gremlinzzz's answer of 10 metres from the tank wall (10.5 metres total radius) was pretty close and definitely acceptable, but it didn't minimize the footprint of the system and no assumptions were spelled out. Mephisto2 was a bit off.
  13. Standard member The Plumber
    Leak-Proof
    19 May '05 19:04
    Your calculation doesn't take splashing into account. At the outset, the water jet hits the top of your barrier and splashes back in, but as the stream velocity diminshes with reduced volume in the tank, the stream will eventually hit the ground in front of the barrier, and has every right to splash over the barrier....
  14. Standard member PBE6
    Bananarama
    19 May '05 20:06
    Originally posted by The Plumber
    Your calculation doesn't take splashing into account. At the outset, the water jet hits the top of your barrier and splashes back in, but as the stream velocity diminshes with reduced volume in the tank, the stream will eventually hit the ground in front of the barrier, and has every right to splash over the barrier....
    Leave it to a plumber...

    In practice, splashing does not occur except for a few dribbly-drops here and there. The reason is that the free jet loses a lot of its energy when it hits the ground or a pre-existing puddle due to dispersion of the stream, friction, air resistance and inelastic collisions. Also, the barrier is usually placed a bit further away than required, and the worst-case leak is a very infrequent occurence.

    If we assumed that the jet kept all of its energy, it would keep bouncing back to the original leak height forever, bounding its way down Main St. like a retarded skip rope. We could assume some constant energy loss for each bounce, too. Let's assume the jet only gets back to 18% of its original height (so a 5.5 metre high leak only bounces back up 0.99 metres, then 0.18 metres, then 0.03 metres, etc...until it fizzles out). I think this number is pretty reasonable. If you look at Niagara Falls, I don't think the water bounces back any higher. It's also a pretty convenient number because it will never bounce over the barrier, and saves me a bunch of work! Yay.

    I was surprised to find out that the minimum set back for a spill barrier is not the height of the tank minus the height of the barrier - it's actually half that! The reason is that there are significant energy losses when the fluid exits the tank through an oddly shaped hole (leak). It reduces the velocity of the jet to about 60% of what you'd expect. Viscosity and air resistance dissapate some of the energy too, and when balancing risk vs. cost (big footprints cost big money), a barrier placed half as far away is often the preferred choice.
  15. Standard member The Plumber
    Leak-Proof
    19 May '05 21:43
    Originally posted by PBE6
    Leave it to a plumber...
    At the risk of letting the cat out of the bag....

    I'm actually not a plumber. I'm a licensed professional engineer (mechanical) - the alias comes from the fact that I am in an extended home improvement project of replacing most of the pipes in my 35+ year old home (kitchen, laundry room, and one bathroom down - two bathrooms to go).

    I noticed that 18% was a very convenient assumption. In my experience, when we look at containment (for example, a dike around a diesel fuel tank), we generally aren't thinking in terms of the liquid jetting out but are primarily concerned with making sure the capacity of the dike is sufficient to containe the contents of the tank. If you have an example where you're really concerned about a potential catastrophic leak (e.g., a bullet hole), the solution is to provide a double wall tank. The jet from the interior tank will for the most part lose it's energy when it hits the outer tank, even if the two holes are lined up pretty well. In extreme cases you can go with a ballistic-proof (forget the techincal term) tank which has concrete between the two walls. I saw a pretty impressive demonstration on these guys a couple years back - they'll take quite a bit of abuse without leaking at all.

    I suspect a standard double-wall tank with a typical containment dike would be the most practical solution if you're really concerned about positive leak containment. Your 1 meter high wall would then only need to have a radius of SQRT (10) to contain the contents of the tank - being an engineer I'd had a good "engineering safety factor" of 50%, and go with a radius of about 4.75 meters.