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Math Help - Question 12

  1. #1
    Grand Panjandrum
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    Question 12

    Water leaves a tap (faucet) in a steady stream at a speed of 20 cm/s
    which is 1 cm wide.

    How wide is the stream 30 cm below the tap (assuming that the stream
    does not break up due to surface tension induced instability).

    RonL
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  2. #2
    Eater of Worlds
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    Hey Cap'n.

    No one bit on this problem, so I will give my farthings worth.

    I think I have seen a problem like this before, a while back.

    I think the width would be proportional to how far it is from the tap.

    If I remember right the formula is R(y)=ky^{\frac{-1}{4}}

    Where R is the radius at some point y down the stream and y is the distance down the stream form the tap.

    We could probably stack an infinite number of itty-bitty disks of thickness h up the stream with volume {\pi}(R(y))^{2}h

    Because of gravity and all that, the rate this disk is passing y is:

    F={\pi}(R(y))^{2}h\sqrt{19.6y}

    Where F is the rate the water is leaving the tap.

    Now, solve for R(y).


    Now, maybe we could integrate. Maybe construct a DE. That's all I have for now.

    This does look like an interesting little problem, though.
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    Grand Panjandrum
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    Quote Originally Posted by galactus View Post
    Hey Cap'n.

    No one bit on this problem, so I will give my farthings worth.

    I think I have seen a problem like this before, a while back.

    I think the width would be proportional to how far it is from the tap.

    If I remember right the formula is R(y)=ky^{\frac{-1}{4}}

    Where R is the radius at some point y down the stream and y is the distance down the stream form the tap.

    We could probably stack an infinite number of itty-bitty disks of thickness h up the stream with volume {\pi}(R(y))^{2}h

    Because of gravity and all that, the rate this disk is passing y is:

    F={\pi}(R(y))^{2}h\sqrt{19.6y}

    Where F is the rate the water is leaving the tap.

    Now, solve for R(y).


    Now, maybe we could integrate. Maybe construct a DE. That's all I have for now.

    This does look like an interesting little problem, though.
    The problem of the instability of the stream is more interesting, It is a
    problem related to the (in)stability of broad planetary ring systems to break
    up into ringlets.

    RonL
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    Grand Panjandrum
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    Question 12 Solution

    Quote Originally Posted by CaptainBlack View Post
    Water leaves a tap (faucet) in a steady stream at a speed of 20 cm/s
    which is 1 cm wide.

    How wide is the stream 30 cm below the tap (assuming that the stream
    does not break up due to surface tension induced instability).

    RonL
    Question 12: Solution

    The mass flow rate is a constant so at any point Av is that same
    as at any other point, where A is the crossectional area of the stream
    and v is the speed of the stream.

    A mass element undergoes an accelaration of g m/s^2 (sign convention
    is positive downwards), and so at time t its speed is:

    v(t)=g.t+v(0)

    and it has fallen a distance:

    s= g.t^2/2+v(0)t.

    So the time to fall 0.3 m, is a root of (taking g=9.81 m/s^2):

    0.3=9.81 t^2/2+0.2t,

    the roots of this are t~=-0.2685, and t~=0.22776. The first
    of these roots is unphysical so we are left with t~=0.22776 seconds.

    Its speed when it has fallen 0.3 m is:

    v(0.22776)=9.81 0.22776 + 0.2 ~= 2.434 m/s

    So now we have from the constancy of the mass flow rate past any point:

    pi*(0.01/2)^2*0.2=pi*(D/2)^2*2.434

    which gives:

    D~=0.002872 m, or D~=0.29 cm.

    RonL
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Question 12: Solution

    The mass flow rate is a constant so at any point Av is that same
    as at any other point, where A is the crossectional area of the stream
    and v is the speed of the stream.

    A mass element undergoes an accelaration of g m/s^2 (sign convention
    is positive downwards), and so at time t its speed is:

    v(t)=g.t+v(0)

    and it has fallen a distance:

    s= g.t^2/2+v(0)t.

    So the time to fall 0.3 m, is a root of (taking g=9.81 m/s^2):

    0.3=9.81 t^2/2+0.2t,

    the roots of this are t~=-0.2685, and t~=0.22776. The first
    of these roots is unphysical so we are left with t~=0.22776 seconds.

    Its speed when it has fallen 0.3 m is:

    v(0.22776)=9.81 0.22776 + 0.2 ~= 2.434 m/s

    So now we have from the constancy of the mass flow rate past any point:

    pi*(0.01/2)^2*0.2=pi*(D/2)^2*2.434

    which gives:

    D~=0.002872 m, or D~=0.29 cm.

    RonL
    Might I point out that this is an application of Bernoulli's equation (an extremely simplified version of fluid dynamics: the flow for a non-viscous incompressible irrotational fluid) along with the constant mass flow-rate condition. I typically assign something like it in my Freshman Physics course.

    -Dan
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