Please I need help with this Fluids Mechanics problem?
A 10-cm jet of water issues from a 1-m diameter tank, as shown in the figure. Assume that the velocity in the jet is 2gh m/s. How long will it take for the water surface in the tank to drop from h0 = 2 m to hf = 0.50 m?
( The figure is in this link: http://img412.imageshack.us/img412/6426/... )
Thanks!
Answer:
http://en.wikipedia.org/wiki/navier-stok...
http://en.wikipedia.org/wiki/reynolds-av...
I suppose we are to understand that the jet of water is 10 cm in diameter; let's call it Dj.
The instantaneous debit of water is given by : v(t) * pi*(Dj/2)^2
with v(t) = sqrt(2 g h(t)) (you need the square root there because else it's not a velocity).
The volume of water inside the tank varies as such :
dV/dt = - debit = -v(t) * pi * (Dj/2)^2
The volume can be expressed as : V(t) = h(t) * pi * (Dt/2)^2 where Dt = 1m is the diameter of the tank.
If you use all this data, then you obtain :
pi * (Dt/2)^2 * dh/dt = -sqrt(2 g h(t)) * pi * (Dj/2)^2
which simplifies into the differential equation :
dh/dt + (Dj/Dt)^2 sqrt(2 g h) = 0
Now you need to integrate this between h0 and hf...
After some research, I found that the solution to that kind of diff equ is something like h(t) = h0 - a * (t-tf)^2. A little more calculus should do the trick.
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( The figure is in this link: http://img412.imageshack.us/img412/6426/... )
Thanks!
Answer:
http://en.wikipedia.org/wiki/navier-stok...
http://en.wikipedia.org/wiki/reynolds-av...
I suppose we are to understand that the jet of water is 10 cm in diameter; let's call it Dj.
The instantaneous debit of water is given by : v(t) * pi*(Dj/2)^2
with v(t) = sqrt(2 g h(t)) (you need the square root there because else it's not a velocity).
The volume of water inside the tank varies as such :
dV/dt = - debit = -v(t) * pi * (Dj/2)^2
The volume can be expressed as : V(t) = h(t) * pi * (Dt/2)^2 where Dt = 1m is the diameter of the tank.
If you use all this data, then you obtain :
pi * (Dt/2)^2 * dh/dt = -sqrt(2 g h(t)) * pi * (Dj/2)^2
which simplifies into the differential equation :
dh/dt + (Dj/Dt)^2 sqrt(2 g h) = 0
Now you need to integrate this between h0 and hf...
After some research, I found that the solution to that kind of diff equ is something like h(t) = h0 - a * (t-tf)^2. A little more calculus should do the trick.
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