Question

Is it possible to design a tank such that the depth of the liquid, h, will be a linear function of time, t?
question related to Torricelli's law

Answer 1

yes

Answer 2

haha ok , plz elaborate

Answer 3

\(h = h_0 + mt\) should do right ?

Answer 4

Oh you want to design a tank that honors a linear function, interesting !

Answer 5

\(p + \frac{1}{2} \rho v^2 + \rho g y\) = constant

Answer 6

wow

Answer 7

ROFL

Answer 8

Taking \(y=0\) at the bottom and noticing that both the surfaces are exposed to atmosphere, do we get :

\[p_0+\frac{1}{2}\rho v_0^2+\rho g h = p_0+\frac{1}{2}\rho v^2+\rho g (0) \]

where \(v_0\) = speed at which top surface is moving
\(v\) = speed of leaking water

Answer 9

That simplifies to
\[v^2- v_0^2= 2g h \]

Answer 10

you need to include the diameter of the "hole in the bottom" as a parameter

after that, it is Bernoulli and continuity equations

Answer 11

pressure only depends on height, so it seems the geometry of tank is of no importance ?

Answer 12

Alas, the rate of height change is dependent on volume so it appears it does depend on shape (unless i've misinterpreted the question)

Answer 13

Oh, are you saying we can adjust the diameter of the hole and make the leaking rate somehow linear ?

Answer 14

but the size of the hole matters?

Answer 15

@IrishBoy123 - yes hole size matters in a non-ideal system, but I think head loss is overkill for this problem...

Answer 16

The change in the depth of liquid is related to the volumetric flow out of the tank. Assuming the hole doesn't change shape, the volumetric flow is related directly to the velocity of the flow.

If we use Toricelli's law, we know that v=sqrt(2gz) or that v is proportional to the square root of the height. If you vary the cross section such that the cross sectional area changes with the height then i think it would be possible in an ideal system

Answer 17

Though please correct me if I've made any egregious assumptions or mistakes

Answer 18

@ LifeEngineer

yep!!

but look at the OP's question

it is pretty meaningless

unless you are doing other things like keeping the volume constant.

Answer 19

I think we need to have \(v\propto \sqrt{t}\)... does below geometry work ? not so sure..