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Pressure in a tank.

  1. Jun 1, 2013 #1
    Hello people,
    This is my first forum post, so i hope I'm posting my problem in the right area.
    I have been looking trough books, internet and forum posts but i could not find anything helpful.

    1. The problem statement, all variables and given/known data
    A circular tank holds a liquid with a density of 13000 kg/m3.
    The tank has the following dimension:
    radius: 14m
    height: 12.15m

    The problem is that the tank broke and i have figure out why the tank broke.
    The tank probably broke at a hight of 10m.
    I do have a tensile test result at 10m which is 200 to 300 N/mm2.

    3. The attempt at a solution

    Volume =¼* π *diameter2*h
    Volume = ¼* π *14^2*2,15
    Volume = 331,0 m3
    331,0 m3=331000 liter

    p=m/v
    1,3 ton/m3=m/331,0 m3
    m=430300 kg

    I hope one of you guys can point me in the right direction
     
  2. jcsd
  3. Jun 1, 2013 #2
    check

    can you please check the density of your the liquid.For density 13000kg/m3 the material cannot be liquid.the density of iron is 7870 kg/m3.
     
  4. Jun 1, 2013 #3

    SteamKing

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    Ever heard of mercury? It's known to be a dense liquid.
     
  5. Jun 1, 2013 #4

    haruspex

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    Probably? Is this part of the given information or a guess on your part?
    Do you mean it broke at a point 10m above the base of the tank, or that it broke when the liquid filled it to that height?
    What does that mean? Are you saying that the vessel is known to withstand that pressure? If so, where does the 10m come into it?
    I would have thought a relevant quantity would be the pressure near the base of the tank.
     
  6. Jun 2, 2013 #5

    SteamKing

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    It's gonna be hard to figure out the stress in the tank since no thickness information is given.
     
  7. Jun 2, 2013 #6
    Thank you for all your replies.

    The liquid in the tank is biological compound which makes the density that high.
    A short time before the tank broke the tank filled to its maximum capacity.

    After the tank broke, four pieces were sent in for further investigation. A tensile test was performed on only one of the plates.
    Different pieces where cut out of the plate for a tensile test a few over a weld and some are not. All pieces have a yield from 200 to 300 N/mm2
    The pieces that wend over a weld had a tensile strength of 250 to 300 N/mm2 and one has a tensile strength of 150 N/mm2.
    The pieces that did not go over the weld have a tensile strength of 300-400 N/mm2
    There is alot of corrosion on the plate which can explain the weak weld.

    This plate was located at approximately at a height of 10m, this makes me think the tank broke at that height.

    I do have results from a thickness measurement with ultrasonic all four plates:
    The thickness on the plate where the tensile test was performed on varies from 3,0mm to 5,6mm on a surface 4665 cm2

    I do have some more information but i don't know if it is relevant.
    Again thank you guys for your time and effort.
     
    Last edited: Jun 2, 2013
  8. Jun 2, 2013 #7

    SteamKing

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    As a first cut, I would calculate the hoop stress in the side of the tank.

    Hoop stress = P*r / t, where P is the pressure at the location, r is the radius of the tank, and t is the wall thickness of the side. P is equal to rho*g*h, where rho is the density of the fluid in kg/m^3, g is 9.81 m/s^2, and h is the depth of fluid in meters at the location.

    If the hoop stress > tensile test, then you have your answer
     
  9. Jun 2, 2013 #8
    Ok so if I understand you correctly.
    The result of the tensile test are from a height of 10 meter so that leaves 2,15 meter
    p=rho*g*h
    p=1,3 kg/m3*9.81m/s2*2.15m= 27.41 (unit=kg/m/s?)

    My earlier post of the radius is wrong the diameter=14m and the radius=7m
    lets take a thickness of 5mm for the stress calculation

    stress = P*r / t
    stress = 27.41*7/5
    stress = 38.39 (unit=N/mm2?)

    but when one takes a smaller number for the thickness for example 3mm,
    stress = P*r / t
    stress = 27.41*7/3
    stress = 63.97 (unit=N/mm2?)

    This means that a thinner wall has more stress and is there for more likely to break.
    Can you check if I used the right units?
     
    Last edited: Jun 2, 2013
  10. Jun 2, 2013 #9

    SteamKing

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    The pressure P will have units of N/m^2, or Pascals.

    Remember, rho is in units of kg/m^3. Your OP indicates rho = 13000 kg/m^3, but you have used 1,3 in your calculations. Confirm which density is correct.
     
  11. Jun 2, 2013 #10

    SteamKing

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    In your hoop stress calculations, the thickness must be measured in m rather than mm. The stress will have units of Pascals. For your tensile test data, 1 N/mm^2 = 1 MPa = 10^6 Pa.
     
  12. Jun 2, 2013 #11
    You are right,
    I made a big mistake, i typed one 0 too much. The correct density is 1300kg/m^3

    P=rho*g*h
    P=1300*9.81*2.15 = 27419 N/m^2

    (for the stress test again 5mm wall thickness)
    stress= P*r / t
    stress= 27419 N/m^2 * 7 meter/ (5*10^-3) meter = 38 386 600 Pa

    Tensile test = 250 N/mm^2 = 250 000 000 Pa

    Thank you very much.
    I'm happy i found this forum. People are really helpful towards each other and there are some interesting forum posts and discussions.
     
    Last edited: Jun 2, 2013
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