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Posers and Puzzles

Posers and Puzzles

  1. 23 Jun '10 08:41
    This is a nice and easy poser.................

    If it takes 15 minutes to fill a bath with water, (plug in and 2 taps on full), and it takes 20 minutes for the bath to empty, how long would it take to fill the bath with taps on full (and plug out, of course!)?

    answers please, with your explanations!
  2. 23 Jun '10 10:21
    one hour
  3. 23 Jun '10 10:32
    If we can assume the draining rate is constant.
  4. 23 Jun '10 21:21
    yes, drainage rate is constant. This is not a trick question
  5. Standard member forkedknight
    Defend the Universe
    23 Jun '10 22:51
    Originally posted by michael liddle
    yes, drainage rate is constant. This is not a trick question
    I think his point was that drainage in a real bathtub is not constant. Water pressure out the drain is proportional to how much water is in the tub.

    The actual answer would be much more complicated and have several more variables.
  6. 24 Jun '10 18:38
    Originally posted by forkedknight
    I think his point was that drainage in a real bathtub is not constant. Water pressure out the drain is proportional to how much water is in the tub.
    Thank you . Yes, that was my point.
  7. 25 Jun '10 07:06
    assume the fill and empty rates are constant.
  8. 25 Jun '10 07:59
    Originally posted by michael liddle
    This is a nice and easy poser.................

    If it takes 15 minutes to fill a bath with water, (plug in and 2 taps on full), and it takes 20 minutes for the bath to empty, how long would it take to fill the bath with taps on full (and plug out, of course!)?

    answers please, with your explanations!
    Assuming fill and empty rates constant:

    Fill Rate = 1/15 baths per minute
    Empty Rate = -1/20 baths per minute
    Net rate = 1/15 - 1/20 = 1/60 baths per minute
    Therefore 1 bath will fill in 60 minutes.

    It's much harder if the emptying rate is a function of the water height, for instance if the emptying rate is kH
  9. Standard member PBE6
    Bananarama
    25 Jun '10 18:46
    As noted above, the fill behaviour is better modeled using Bernoulli's equation. For this problem, we choose point 1 to be at the surface of the water, and point 2 to be at the drain. Bernoulli's equation for this situation is as follows:

    P1/(rho*g) + v1^2/(2g) + h1 = P2/rho*g + v2^2/(2g) + h2

    where:

    P1, P2 = pressures
    rho = density of water
    g = acceleration due to gravity
    v1, v2 = velocities
    h1, h2 = heights

    To simply the above equation, we make a few assumptions:

    (1) The drain is a free jet (i.e. the water surface at the drainage point is open to the atmosphere), so P1 = P2.
    (2) The velocity of the water at the surface of the tub is negligible compared to the drainage velocity, so v1 = 0.
    (3) We arbitrarily set h2 = 0.

    Using these simplifying assumptions, we are left with:

    h1 = v2^2/(2g)

    Solving for v2, we have:

    v2 = SQRT(2g*h1)

    We make one more simplifying assumption here, that the cross-sectional area of the tub B is constant all the way up. In that case, V = B*h1. Subbing in, we have:

    v2 = SQRT(2g*B*V)

    Now, the product of the drainage velocity and the cross-sectional area of the drain Ac gives us the rate of change in volume due to drainage D. Carrying out this operation, we have:

    D = Ac*v2 = Ac*SQRT(2g*B*V)

    Lumping all the constants into one constant k, we are left with:

    D = k*SQRT(V)

    We know it takes 20 minutes to drain a full tub, so using a mass balance we have:

    dV/dt = -D = -k*SQRT(V)

    dV/(-k*SQRT(V)) = dt

    Integrating the above expression, with the limits of integration being V = 1...0 (as the tub is being drained) and t = 0...20, we get:

    (2/k)*(SQRT(1)-SQRT(0)) = 20-0

    2/k = 20

    k = 1/10

    Now, we know from iamatiger's post that the fill rate F is 1/15. Therefore, the overall balance for the tub is as follows:

    dV/dt = F - D = 1/15 - (1/10)*SQRT(V)

    This equation is surprisingly involved to tackle, but luckily the Wolfram Alpha site eats equations like this for breakfast! Here is a link to a clear version of the final answer:

    http://www.wolframalpha.com/input/?i=dV/dt+%3D+1/15+-+(1/10)*SQRT(V)

    However, a simple asymptote test will gives us some valuable information. Setting dV/dt = 0, we have:

    0 = 1/15 - (1/10)*SQRT(V)

    SQRT(V) = 2/3

    V = 4/9

    Therefore the graph of the water volume over time becomes asymptotic as V approaches 4/9, which means that the water will never fill the tub beyond 4/9 of the total volume.

    FINAL ANSWER

    The tub will never fill with the taps on and the drain open. Neat!
  10. Standard member uzless
    The So Fist
    28 Jul '10 07:27
    So, according to your calculation, one of the timeframes given in the OP must change in order for the bathtub to never be filled.

    Does the water coming out of the tap slow down, or does the drainage rate increase?
  11. Subscriber Ponderableonline
    chemist
    28 Jul '10 10:53
    Originally posted by uzless
    So, according to your calculation, one of the timeframes given in the OP must change in order for the bathtub to never be filled.

    Does the water coming out of the tap slow down, or does the drainage rate increase?
    evidently the tab volume flows have been kept constant (which is a well enough approximation for bath armatures.

    Drainage increases with height of level.

    If we however factor in the increase in pressuer drop in the drain due to the higher mass flow the result becomes still more complicate.

    Thank you PB6 for the nicely done calculation.
  12. Subscriber sonhouse
    Fast and Curious
    28 Jul '10 22:00
    Originally posted by PBE6
    As noted above, the fill behaviour is better modeled using Bernoulli's equation. For this problem, we choose point 1 to be at the surface of the water, and point 2 to be at the drain. Bernoulli's equation for this situation is as follows:

    P1/(rho*g) + v1^2/(2g) + h1 = P2/rho*g + v2^2/(2g) + h2

    where:

    P1, P2 = pressures
    rho = density of water
    g = ac ...[text shortened]...
    [b]FINAL ANSWER


    The tub will never fill with the taps on and the drain open. Neat![/b]
    So you could have a feedback system where the drain goes into the inlet and thus keep the level at 4/9 forever

    But if the drain takes 20 minutes, why does the intuitive answer, it drains 3 times in one hour and the fill rate is 4 times per hour, that sounds like what the one answer was 1 hour. Why is that wrong?
  13. Standard member PBE6
    Bananarama
    29 Jul '10 02:08 / 1 edit
    Originally posted by sonhouse
    So you could have a feedback system where the drain goes into the inlet and thus keep the level at 4/9 forever

    But if the drain takes 20 minutes, why does the intuitive answer, it drains 3 times in one hour and the fill rate is 4 times per hour, that sounds like what the one answer was 1 hour. Why is that wrong?
    I don't think the drainage could ever be fed back into the tub without a pump. Even if the outlet of the drain could be lowered progressively to keep level with the top of the liquid in the tub, losses in the drain would cause the tub liquid to feed back into the drain instead of having the drain feed into the tub. A submerged drain (just below the tub water surface) would run into the same problem.

    The reason the intuitive answer doesn't make physical sense is because the drainage rate is not constant. In the simplified scenario I posited, the drainage rate is proportional to the square root of the tub water level, i.e. the drainage rate decreases as the drainage progresses. As Ponderable mentioned, a more complicated model can be constructed in which the back pressure from the liquid in the drain creates an additional back-pressure on the tub liquid, thereby lowering the drainage rate, but the results would be similar - the tub level would reach a minimum based on the steady-state relation between the fill rate and the drainage rate (which would be a function of the tub level and the drainage velocity). Regardless, the most physical answer is that the tub will never drain completely.
  14. Subscriber joe shmo On Vacation
    Strange Egg
    29 Jul '10 02:11 / 1 edit
    Originally posted by sonhouse
    So you could have a feedback system where the drain goes into the inlet and thus keep the level at 4/9 forever

    But if the drain takes 20 minutes, why does the intuitive answer, it drains 3 times in one hour and the fill rate is 4 times per hour, that sounds like what the one answer was 1 hour. Why is that wrong?
    Its not wrong given the original parameter (constant rates of flow). The other guys added more constraints to make the problem more physically stringent (in essence they took a math problem and turned it into a physics problem).

    Edit: I should probably not say that the original problem was a "math" problem, rather just a less rigourous physics problem.
  15. Subscriber sonhouse
    Fast and Curious
    29 Jul '10 03:24 / 1 edit
    Originally posted by PBE6
    I don't think the drainage could ever be fed back into the tub without a pump. Even if the outlet of the drain could be lowered progressively to keep level with the top of the liquid in the tub, losses in the drain would cause the tub liquid to feed back into the drain instead of having the drain feed into the tub. A submerged drain (just below the tub water velocity). Regardless, the most physical answer is that the tub will never drain completely.
    Interesting topic. That feedback comment was really a joke. You would of course have to have a pump, consuming energy when all you need to maintain the level is to turn off the drain and inlet valves and it stays at the same level with no energy needed.

    I like the example of a bunch of Peltier elements, where one side gets hotter and the other gets colder when you input electrical energy. So picture a string of them clamped together with a slight wedge (heat conductive of course) and enough of them to form a torous shape, then apply power, follow the heat flow! Of course they are all physically connected hot side to adjacent cold side. It would be a thermal-merry go round