BM3 asked in Science & MathematicsEngineering · 8 years ago

# 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

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• 8 years ago

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.