conservation of mass finding time?

a cylindrical tank being drained through a duct whose cross-sectional area is 3 X10^-4 m^2. The velocity of the water at the exit varies according to (2gz)^(1/2), where z is the water level, in m, and g is the acceleration of gravity, 9.81 m/s^2. The tank initially contains 2500 kg of liquid water. Taking the density of the water as 10^3 kg/m^3, determine the time, in minutes, when the tank contains 900 kg of water.

m(initial)=2500 kg

density=10^3 kg/m^^3

a=1 m^3

2 Answers

  • 8 years ago
    Favorite Answer

    The reason why people aren't answering this is that

    we don't understand "a=1 m^3".

    I'm going to hazard a guess that what you meant is

    that the cross-sectional area of the tank is 1 m^2.

    Without knowing the x-sectional area of the tank,

    it turns out that this problem cannot be solved.

    dV/dt = -v (0.0003 m^2)

    If the cross-sectional area of the tank is 1 m^2,

    then dz/dt = dV/dt divided by 1 m^2 = -0.0003 v

    = -0.0003 sqrt(2gz)

    and z^(-1/2) dz/sqrt(2g) = -0.0003 dt

    2z^(1/2) / [sqrt(2g)] = -0.0003 t + C [ EQUATION S ]

    The initial height is 2.5 m if the x-sec area is 1 m^2;

    plugging in z = 2.5 and t = 0 gives

    C = (5/7) second...makes some sense when you realize that

    t has to get up to about 2000 s to make RHS of EQN S a zero.

    When the tank contains 900 kg of water,

    the z = 0.9 m, so the LHS of EQUATION S is 3/7 of 1 second,

    3/7 = 5/7 - 0.0003 t

    2/7 (seconds) = 0.0003 t

    t = 2381 seconds = 39.7 minutes

    Do check my arithmetic.

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  • 4 years ago

    No, when the iron rusted, it went through a chemical reaction and in doing so added oxygen to the iron. So the rust is really the mass of the iron it was before it combined with oxygen from the air. The increase in mass is the added oxygen to the original iron.

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