and P = H = v^2 => H(steadystate) = Hmax/4

]]>There are some errors; only for a cylinder tank the Torricelli law can be written exactly:

2 (1-p^2) u u’ = – 2 g p u

2 (1-p^2) u’ = – 2 g p

u = u_0 – g p t/(1-p^2)

this equation is true for each area of cap surface A, for A -> 0 u=u_0=(2 g h)^{1/2}, so that

u = (2 g H)^{1/2}-g p t/(1-p^2)

there is a linear reduction of the velocity, that is null when the tank is empty

t = (2 g H)^{1/2} (1-p^2)/ (g p)

there is a difference with the Torricelli solution (differential equation with a constant velocity), a 2 factor, and it seem right.

]]>I made another mistake. My calculations gives me a steady state v of 1/4 the max, so the height should be 1/16???

]]>It will, but the rate of filling and the rate of emptying also depend on those things, and you’re told something about them.

]]>According to my calculation, you are right. To be sure, I verified with numerical calculations.

]]>[Integration of 1/sqrt(h) is necessary to obtain the expression for the time to empty the tank, or am I missing a simpler approach?] ]]>

When I empty the tank, without tap running; the water has a surface S, and the velocity of the water surface is w, then

w S = u A (like a syringe)

where A is the cap surface, then because of the old Bernoulli

w^2 + 2 g h = u^2

but

w = u A / S = u p

then

p^2 u^2 + 2 g h = u^2

(1-p^2) u^2 = 2 g h

using the reduction of the cylinder tank volume

h’ = -p u

so that

2 (1-p^2) u u’ = – 2 g u’

(1-p^2) u = – g

u = -g /(1-p^2)

the velocity is constant! It is an interesting result.

h = H – p u t = H – g t /(1-p^2)

and the emptying time is

t = H (1-p^2)/g

the filling time is

v B t = H S

t = H S / (B v)

(1-p^2)/g = S / (B v)

p^2 = 1- S g /(B v)

A^2 = S^2 [1-S g/ (B v)]

A(B) = S [1-S g/ (B v)]^{1/2}

and this is the constraint on the opening surfaces.

The rest are calculations, if all is right