# conical tank

Printable View

• April 21st 2009, 02:24 PM
Automaton
conical tank
Use Torricelli's principle to find the time it takes to empty a conical tank of circular cross section standing on its apex whose angle is 45° and has an outlet of cross sectional area 1.0cm². The tank is initially full of water and at time t = 0 the outlet is opened and the water flows out. The initial depth of the water in the tank is 2m.

Torricelli's principle: √2gh
Volume of a Cone: V= (1/3) π r² h

I think you set up a differential equation here
dV/dt = -a√2gh I'm not sure if this is right place to start.
• April 21st 2009, 04:14 PM
skeeter
Quote:

Originally Posted by Automaton
Use Torricelli's principle to find the time it takes to empty a conical tank of circular cross section standing on its apex whose angle is 45° and has an outlet of cross sectional area 1.0cm². The tank is initially full of water and at time t = 0 the outlet is opened and the water flows out. The initial depth of the water in the tank is 2m.

Torricelli's principle: √2gh
Volume of a Cone: V= (1/3) π r² h

I think you set up a differential equation here
dV/dt = -a√2gh I'm not sure if this is right place to start.

for the cone ...

$r = h\tan(22.5)$

$V = \frac{\pi}{3} [h\tan(22.5)]^2 h$

$V = \frac{\pi}{3}\tan^2(22.5) h^3$

$\frac{dV}{dt} = \pi \tan^2(22.5) h^2 \cdot \frac{dh}{dt}$

let $b = \pi \tan^2(22.5)$

$\frac{dV}{dt} = bh^2 \cdot \frac{dh}{dt}$

$\frac{dV}{dt} = -a\sqrt{2gh} = -a\sqrt{2g} \cdot \sqrt{h}$

let $d = a\sqrt{2g}$

$\frac{dV}{dt} = -d\sqrt{h}$

$bh^2 \cdot \frac{dh}{dt} = -d\sqrt{h}$

$h^{\frac{3}{2}} \, dh = -\frac{d}{b} \, dt$

integrate and find h as a function of t.

I get the time to empty to be about 46 minutes.
• April 22nd 2009, 03:21 AM
Automaton
is the height not equal to 2m, if the tank is full and has a depth of 2m?
• April 22nd 2009, 04:43 AM
skeeter
Quote:

Originally Posted by Automaton
The tank is initially full of water and at time t = 0 the outlet is opened and the water flows out. The initial depth of the water in the tank is 2m.

does that answer your question?
• April 22nd 2009, 07:28 PM
Automaton
thats where i got my question from in the first place. I wasn't sure if h was known from that information or if h still needed to be found. If i integrate the left side I get http://img10.imageshack.us/img10/715/95476278.jpg
but i'm still not on how to integrate:
-d/b dt

i understand b is just being used to represent the pi*tan(22.5),
• April 23rd 2009, 04:25 AM
skeeter
Quote:

Originally Posted by Automaton
thats where i got my question from in the first place. I wasn't sure if h was known from that information or if h still needed to be found. If i integrate the left side I get http://img10.imageshack.us/img10/715/95476278.jpg
but i'm still not on how to integrate:
-d/b dt

i understand b is just being used to represent the pi*tan(22.5),

I can't see your image for your integration of the left side. As far as the right side, $-\frac{d}{b}$ is just a constant ...

$
\int -\frac{d}{b} \, dt = -\frac{d}{b}t + C
$
• April 23rd 2009, 09:52 AM
Automaton
left side is [2h^(5/2)]/5
ok thats what i was getting but i wasn't sure because i didn't know what 'd' was.
• April 23rd 2009, 01:53 PM
skeeter
Quote:

Originally Posted by Automaton
left side is [2h^(5/2)]/5
ok thats what i was getting but i wasn't sure because i didn't know what 'd' was.

look at my solution again ... d is defined there.
• April 23rd 2009, 04:24 PM
Automaton
ok now I see it. Sorry about that. I must have over look it. Thanks so much for your help. I was completely lost on this question