Question: Draining Tank problem

[matthewslaby]

A cylidrical tank (R= tank radius) filled with a liquid with an opening for the insertion of pressurized gas on top and an exit hole on bottom (r=exit hole radius).

Plot the Height of the liquid as a function of time and exit radius.

Gas pressure=const.
No swirl (1D analysis)

Any advice is welcome: matthewslaby1645@comcast.net

 

[matthewslaby]

I want to make this problem more realistic by adding friction at the fluid exit plan and possibly swirl of the fluid (3D flow), and maybe even non-adiabatic conditions.

Please let me know if you could guide me on this. At least the solution for the simple form of this problem should be known and documented.

 

[Rayquesto]

sounds like a calculus problem I had. Except, I'm clueless in terms of adding that gas factor into the equation.

so, volume of a cylinder is pie r^2 h and deriving that you get

the change in volume over the change in time=2 pie r dr/dt + pie r^2 dh/dt but dr/dt=0 since the radius is constant in a cylinder.

I mean to me, the only way that I can picture an exit hole is by adding a cone into the equation, or something cut out, but if it were cut out, then the rate going through that small opening will depend on that small opening's radius squared times the change in height referring to the speed of the volume of the liquid.

 

[Rayquesto]

well radius is constant in the cylindrical tank.

 

[pmacias]

Thank you for your question, Matthew. This is a classic problem in fluid mechanics known as the draining tank problem. To solve this problem, we can use the equations of fluid mechanics, specifically the continuity equation and Bernoulli's equation.

First, let's define some variables. Let h(t) be the height of the liquid in the tank at time t, R be the tank radius, r be the exit hole radius, and P be the constant gas pressure. We will also assume that the liquid is incompressible and the flow is steady.

Using the continuity equation, we can relate the velocity of the liquid at the exit hole, v, to the height of the liquid in the tank, h, and the exit hole radius, r:

v = (r/R)^2 * sqrt(2*g*h)

where g is the acceleration due to gravity. This equation shows that as the height of the liquid decreases, the velocity at the exit hole will increase.

Next, we can use Bernoulli's equation to relate the pressure at the surface of the liquid, P, to the pressure at the exit hole, P0, and the height of the liquid, h:

P + 1/2 * rho * v^2 + rho * g * h = P0

where rho is the density of the liquid. This equation shows that as the height of the liquid decreases, the pressure at the exit hole will decrease.

Now, we can combine these two equations to get an expression for the height of the liquid as a function of time and exit hole radius:

h(t) = (P0 - P) / (rho * g) + (R/r)^2 * sqrt(2*g*(h0 - h(t))) * t

where h0 is the initial height of the liquid in the tank.

To plot the height of the liquid as a function of time and exit radius, we can fix the time t and vary the exit hole radius r. This will give us a curve showing how the height of the liquid changes as the exit hole radius changes. We can also fix the exit hole radius r and vary the time t to see how the height of the liquid changes over time.

In terms of advice, it is important to note that this is a simplified 1D analysis and there may be other factors at play in a real-world scenario. These could include the effects of viscosity, turbulence, and the shape of the tank. It is always