Steel can of H height and A area, full of water.
Hole in the bottom of the can of \( A_2 \) area.
Find time it takes to drain the can.

Yadda yadda Bernouli's Eq applies.

V = velocity, P = pressure, z = height, Row = density.
\(
P_1 + 1/2RowV^2 + Row*g*z = P_2 + 1/2RowV_2^2 + Row*g*z_2
\)
Pressures are equal outside the can, Row cancels, V_1 is negligible and \( z_2 \) = 0.
\(
gz=1/2V_2^2
V_2 = sqrt{2gz}
\)

Relationship between volume out of the can and change in height, feel like I'm wrong here:
\(
A_1 H/t=A_2 V_2
H = A_2 /A_1 *V_2 t
dh/dt = A_2 /A_2 V_2
\)

Regardless of whether that is right, I think this is my differential equation:
\(
dh/dt = -A_2 /A_1 V_2
dh/dt = -A_2 /A_1 sqrt{2gh}
\)
Separate Variables
\(
dh/sqrt{h}=-A_2 /A_1 sqrt{2g}dt
\int 1/sqrt{h} dh = \int -A_2 /A_1 sqrt{2g}dt
2sqrt{h} = -A_2/A_1 sqrt{2g} t +C
h = ((-A_2/A_1 sqrt{2g} t)/2 +C)^2
\)

Solve for t when h = 0.

\(
0 = ((-A_2/A_1 sqrt{2g} t)/2 +C)^2
2C*A_1/A_2 * 1/sqrt{2g} = t
\)

Doesn't feel right, and my calculator is graphing it with an absurdly long time until empty.

 

Did you mean Bernouli or Torricelli?

 

Hi,

Maybe he meant Tortellini or Vermicelli, as the stream coming out might look like a strand of spaghetti

 

Steel can of H height and A area, full of water.
Hole in the bottom of the can of \( A_2 \) area.
Find time it takes to drain the can.

Yadda yadda Bernouli's Eq applies.

V = velocity, P = pressure, z = height, Row = density.
\(
P_1 + 1/2RowV^2 + Row*g*z = P_2 + 1/2RowV_2^2 + Row*g*z_2
\)
Pressures are equal outside the can, Row cancels, V_1 is negligible and \( z_2 \) = 0.
\(
gz=1/2V_2^2
V_2 = sqrt{2gz}
\)

Relationship between volume out of the can and change in height, feel like I'm wrong here:
\(
A_1 H/t=A_2 V_2
H = A_2 /A_1 *V_2 t
dh/dt = A_2 /A_2 V_2
\)

Regardless of whether that is right, I think this is my differential equation:
\(
dh/dt = -A_2 /A_1 V_2
dh/dt = -A_2 /A_1 sqrt{2gh}
\)
Separate Variables
\(
dh/sqrt{h}=-A_2 /A_1 sqrt{2g}dt
\int 1/sqrt{h} dh = \int -A_2 /A_1 sqrt{2g}dt
2sqrt{h} = -A_2/A_1 sqrt{2g} t +C
h = ((-A_2/A_1 sqrt{2g} t)/2 +C)^2
\)

Solve for t when h = 0.

\(
0 = ((-A_2/A_1 sqrt{2g} t)/2 +C)^2
2C*A_1/A_2 * 1/sqrt{2g} = t
\)

Doesn't feel right, and my calculator is graphing it with an absurdly long time until empty.

Way too long since I did Differential Calculus,---but as the water empties the can,the "head of water" drops,& with it,the flow rate,until it is flowing so slowly that theoretically it will take forever to empty.
(This is quite apart from such esoterica as surface tension which will rise their ugly heads when the layer of water becomes very thin.)

That is why the graphing calculator is showing such a long time.

The same applies to the discharge of a capacitor through a resistor,which is why we use approximations like t=CR for the time to discharge to approx. 63% of the initial charge,& use 5CR as a "rule of thumb" for "complete" discharge.

 

One limiting case that needs to be explored is the case of the hole at the bottom being the same size as the can. In that situation, you simply have a slug of water that is dropping in free fall (sans friction and surface and adhesion effects) through a height H. If your equation doesn't limit to that, then you need to understand why not. Note that I haven't looked at your equation to see if it does or not.

 

I think I figured it out. The confusion was due to an example I found online. The example EQ was giving me a time of 52 seconds, my equation was saying 26. Redid the examples algebra, found where they added a 2, and decided to trust my instinct. We'll see if I'm correct next week.

 

Hi,

From what i remember there is a simple approximation to this kind of problem where the tank is much larger diameter than the leakage hole diameter. Then because the hole is much smaller than the tank diameter the top of the fluid is assumed to be not actually dropping with time (so v1=0). This means there is a simple equation because the whole system is in steady state.

The solution is something like (height is initial height of the water column):
v=sqrt(2*g*height)

and so the volumetric flow rate for hole radius 'r' is:
q=v*pi*r^2

Thus you can check your results by assuming the hole is small and see if your calculated results come close to this.
Of course this is for thin liquids too not maple syrup

 

The crux of the problem is that velocity and volumetric flow is a function of height, thus the differential equation.