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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo
    Strange Egg
    18 Oct '10 22:12 / 1 edit
    A cylindrical tank is being drained by a duct with

    Area duct: A1= 3*10^(-4) m^2
    Velocity of flow through duct: V1 = sqrt(2g*Z) where Z is the height of the liquid in the tank

    initial mass of 2500 kg
    density of water: p=10^3 kg/m^3

    It asks to find time in minutes when the tank contains 900 kg

    so I start with mass rate balance

    dm/dt = p*A1*V1=-p*A1*sqrt( 2g*Z) ....eq1

    from here where do I go?

    can I say the mass sa a function of time is

    m(t) = p*A_base*Z(t);

    where as far as I can tell the Area of the tank base cannot be found or remains variable at this point

    the subbing into eq1

    d(p*A_base*Z)/dt = -p*A1*sqrt( 2g*Z) which becomes

    A_base* dZ/dt = -A1*sqrt(2g*Z)

    separating variables and integrating

    Z(t)= -A1*sqrt(2g)*t + 2*A_base*Z1^(1/2) where Z1 is the initail height

    here is where I lose it

    m(t) = p*A_base* Z(t) does this bear any fruit A_base is unknown as well as Z1

    I think I can solve for Z1 using the initial condition m(0)=2500 kg, but that still leaves A_base

    I tried using some initial conditions in Z but these don't seem to be independent of each other and thus dont work.

    what can i do?
  2. 19 Oct '10 03:54 / 1 edit
    Originally posted by joe shmo
    A cylindrical tank is being drained by a duct with

    Area duct: A1= 3*10^(-4) m^2
    Velocity of flow through duct: V1 = sqrt(2g*Z) where Z is the height of the liquid in the tank

    initial mass of 2500 kg
    density of water: p=10^3 kg/m^3

    Hmm, are you sure they haven't given you the area of the tank?

    Ahh you can relate that to Z_zero (start height), because you know the density

    rearrange the flow velocity to get the units of velocity right, you need delta_height/delta_t, you will need you rexpression for the area of the tank to do this

    you then need to integrate that and it should work
  3. Subscriber joe shmo
    Strange Egg
    19 Oct '10 16:27
    Originally posted by iamatiger
    Hmm, are you sure they haven't given you the area of the tank?

    Ahh you can relate that to Z_zero (start height), because you know the density

    rearrange the flow velocity to get the units of velocity right, you need delta_height/delta_t, you will need you rexpression for the area of the tank to do this

    you then need to integrate that and it should work
    Im not seeing this, can you elaborate?
  4. Standard member Igloo
    Fishing
    19 Oct '10 17:47
    As the question stands, it isn't solvable. You need either the area of the tank, or another given, such as the initial flowrate through the duct.

    Nothing tells you whether the tank is shaped like a pan or like a straw, which will affect the velocity in the duct.

    Not to be rude, but is this a stand alone question? Maybe is Q1.3, where the tank area would be given in Q1.1?
  5. Subscriber joe shmo
    Strange Egg
    19 Oct '10 18:49 / 1 edit
    Originally posted by Igloo
    As the question stands, it isn't solvable. You need either the area of the tank, or another given, such as the initial flowrate through the duct.

    Nothing tells you whether the tank is shaped like a pan or like a straw, which will affect the velocity in the duct.

    Not to be rude, but is this a stand alone question? Maybe is Q1.3, where the tank area would be given in Q1.1?
    It is a stand alone problem. Believe it or not, no such Area for the base nor 2nd initial condition exists in the problem statement or in the problem diagram.
  6. 20 Oct '10 03:58 / 2 edits
    Originally posted by joe shmo
    A cylindrical tank is being drained by a duct with

    Area duct: A1= 3*10^(-4) m^2
    Velocity of flow through duct: V1 = sqrt(2g*Z) where Z is the height of the liquid in the tank

    initial mass of 2500 kg
    density of water: p=10^3 kg/m^3

    It asks to find time in minutes when the tank contains 900 kg
    Its a bit simpler than I first thought:

    Velocity = sqrt(2g*Z) m/s

    Volume Rate = Velocity * Area
    Volume_Rate = 3*10^(-4) * sqrt(2g*Z) m^3/s

    Mass_Rate = Volume_rate/Density
    Mass_Rate = 0.3 * sqrt(2g*Z) kg/s

    now integrate that from 0 to t and you should be there.
  7. 20 Oct '10 07:53
    Originally posted by iamatiger
    Its a bit simpler than I first thought:

    Velocity = sqrt(2g*Z) m/s

    Volume Rate = Velocity * Area
    Volume_Rate = 3*10^(-4) * sqrt(2g*Z) m^3/s

    Mass_Rate = Volume_rate/Density
    Mass_Rate = 0.3 * sqrt(2g*Z) kg/s

    now integrate that from 0 to t and you should be there.
    Hmm, its a bit harder than I thought second.
  8. Standard member Igloo
    Fishing
    20 Oct '10 15:59
    I tutor these types of questions to our first year engineers. I've yet to see one where their isn't some sort of information about the tank.

    Unless iamatiger can work some magic, I think you should call this question a dead loss.
  9. Subscriber joe shmo
    Strange Egg
    21 Oct '10 14:23
    Originally posted by Igloo
    I tutor these types of questions to our first year engineers. I've yet to see one where their isn't some sort of information about the tank.

    Unless iamatiger can work some magic, I think you should call this question a dead loss.
    I will now assess this to be true also, thanks for the reassurance.
  10. 21 Oct '10 22:55 / 3 edits
    Originally posted by Igloo
    I tutor these types of questions to our first year engineers. I've yet to see one where their isn't some sort of information about the tank.

    Unless iamatiger can work some magic, I think you should call this question a dead loss.
    Yes, most probably impossible.

    Imagine a very thin tall tank, the flow will be fast and the tank will empty quite quickly. Now imagine a very wide short tank, as the width increases the height of water deceases and the flow gets slower and slower, tending to zero flow rate with an infinitely wide, zero height tank. all flow rates between these two are possible given the information provided, and so I think there is no unique solution.