00:01
So in this question, we've got a cylindrical tank with diameter d and height h0.
00:14
And at time t equals zero, a small stopper of diameter small d is removed from the bottom.
00:23
Now, benoulli's equation tells us that the energy density of fluid is the same everywhere in the fluid.
00:30
So what we have is that let's pick a point here.
00:36
The pressure is the atmospheric pressure.
00:40
We've got the velocity of the fluid here is equal to zero.
00:47
We've got the height is h0.
00:53
So y equals h0.
00:55
And now let's think of a point down here.
00:58
The pressure is also the atmospheric pressure, because this is open to the elements.
01:01
The speed is some speed, when it's call it v, and the height, y equals zero.
01:14
Now bernoulli's equation tells us the energy density is constant, and the energy density is the pressure, plus the kinetic energy density, a half row v squared, plus the gravitational potential energy, row g times y.
01:34
And this is constant if there are no losses.
01:40
So what we have is that p -0, and then there's no kinetic energy density, and then we've got, so p -0 plus row g -h -0, well, at the top, so as it drains, this h -0 is actually going to be h -t, and the speed is going to be dh by d -t.
02:06
So actually what we've got is p -naut plus a half row dh by d t squared plus row g h is going to be equal to p -nought down here plus a half row v squared as the water is shooting out the bottom of the tank.
02:38
And then there's no gravitational potential energy.
02:41
So what we have is a half dh by d t squared plus gh equals a half row v squared.
02:50
But we can relate v here to dh by d t because the speed that it comes out times the cross -sectional area.
03:00
So v times pi d over two squared...