Diff Eq for a conical tank

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NHgirl

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Hello -

I'm trying to come up with a differential equation to model a conical tank with a flow in (Fo) and a flow out (F=K(sqrt(h)))which is a function of the valve coefficient and the height of the liquid in the tank. The differential equation has to be in terms of the height of the liquid in the tank which is what I am unsure about. Should the differential equation be in terms of dh/dt or just h itself?

I am given placeholder parameters like Rmax=max radius of cone, Hmax = max height of cone etc. I am not looking for this diff eq to be solved...just looking for a little assistance coming up with the correct equation. Would the volume forumula for a cone come into play here?

Thanks for any help!
 
NHgirl said:
Hello -

I'm trying to come up with a differential equation to model a conical tank with a flow in (Fo) and a flow out (F=K(sqrt(h)))which is a function of the valve coefficient and the height of the liquid in the tank. The differential equation has to be in terms of the height of the liquid in the tank which is what I am unsure about. Should the differential equation be in terms of dh/dt or just h itself?

I am given placeholder parameters like Rmax=max radius of cone, Hmax = max height of cone etc. I am not looking for this diff eq to be solved...just looking for a little assistance coming up with the correct equation. Would the volume forumula for a cone come into play here?Thanks for any help!
Click to expand...

Yes

First decide - what are you modelling - are you modelling storage (increase in height) and/or flow out or what? (surely you are not modelling the cone - as you stated in your post)

Then you need to do is draw a sketch.

What is the exit point of the fluid? - the vertex of the cone or base of the cone?

Then draw an arbitrary height (h) at any arbitrary time (t) - and analyze what happens in the next time increment dt.

Your post is incomplete - so I could not suggest anything more concrete than this.
 
Attached is the diagram. No, I am not modelling the cone itself. The model is of the height of the water in the cone as the parameters change. The exit point of the fluid is at the very base of the cone and the entry point is the vertex.

[attachment=0:2gybm4gj]tank.JPG[/attachment:2gybm4gj]
 
\(\displaystyle A_{w}=\text{surface area of top of water at height h}\)

By similar triangles \(\displaystyle \frac{r}{h}=\frac{R}{H}\Rightarrow r=\frac{Rh}{H}\)

Therefore, the top of the water at height h has surface area \(\displaystyle A_{w}={\pi}\left(\frac{Rh}{H}\right)^{2}\)

So, \(\displaystyle \frac{dV}{dt}=A_{w}\frac{dh}{dt}\)

Can you finish knowing what dV/dt equals?.
 
Well I have what is perhaps a stupid question...aren't I trying to solve \(\displaystyle \frac{dh}{dt}\)? I'm sure you are correct I'm only curious why we start off by finding \(\displaystyle \frac{dV}{dt}\).
Also, how does the flow come into play? Is it correct to say:
\(\displaystyle \frac{dV}{dt}=A_{w}\frac{dh}{dt}=F_{0}-k_{v}\sqrt{h}\)
 
NHgirl said:
Well I have what is perhaps a stupid question...aren't I trying to solve \(\displaystyle \frac{dh}{dt}\)? I'm sure you are correct I'm only curious why we start off by finding \(\displaystyle \frac{dV}{dt}\).
Also, how does the flow come into play? Is it correct to say:
\(\displaystyle \frac{dV}{dt}=A_{w}\frac{dh}{dt}=F_{0}-k_{v}\sqrt{h}\)
Click to expand...

Correct.

So we start off by saying

\(\displaystyle \frac{dV}{dt}=F_{0}-k_{v}\sqrt{h}\)

and

\(\displaystyle \frac{dV}{dt}=A_{w}(h)\frac{dh}{dt}\)

then

\(\displaystyle A_{w}(h)\frac{dh}{dt}=F_{0}-k_{v}\sqrt{h}\)

This is the DE you are looking for.
 
Ok so now I have a question...

Subhotosh, what happens to h if I change Rmax, Kv, Hmax or Fo? Is the steady state of h affected?

if I change Rmax, why should it effect the rate of change of h?

I know I am missing something silly...

Thank you
 
Anthony28 said:
Ok so now I have a question...

Subhotosh, what happens to h if I change Rmax, Kv, Hmax or Fo? Is the steady state of h affected?

if I change Rmax, why should it effect the rate of change of h?

I know I am missing something silly...

Thank you
Click to expand...

Those are actually good questions...

If you change Rmax - in general you'll change ratio of Rmax/Hmax (cone angle) - and that'll in turn affect dh/dt.

For steady state dh/dt = 0 ? storage is 0 ? flow in = flow out. So the effect will be dependent the Fo and kv and their functional relationship to the geometry of the cone.
 
So if Rmax increases h increases because it takes longer for the tank to fill up. As Rmax decreases h increases.
And I Fo increases so do h, if Fo decrees so h decrees as well.

So obviously those must be the two major players here.

I get to these conclusions by logic when you made me think about the shape, but how do I prove it mathematically?

Sorry, if I insets on this problem. If you get tiered trying to explain it ...there will be no hard feelings.

Thanks
 
Anthony28 said:
So if Rmax increases h increases because it takes longer for the tank to fill up. As Rmax decreases h increases.
And I Fo increases so do h, if Fo decrees so h decrees as well.

So obviously those must be the two major players here.

I get to these conclusions by logic when you made me think about the shape, but how do I prove it mathematically?

Sorry, if I insets on this problem. If you get tiered trying to explain it ...there will be no hard feelings.

Thanks
Click to expand...

"h" is a variable - what do you mean exactly by increase and decrease of this parameter. Are you meaning H - which is really H[sub:eek:284gsqb]max[/sub:eek:284gsqb]?
 
I mean h not Hmax.

"How is the stedy state of h affected by changing Fo, Kv, Rmax and Hmax?"

Well if I change Rmax and Hmax i am essentially changing the shape of the cone, and therefor its angle, because

Rmax/Hmax = r/h (because of similar triangles)

then : h= Hmax*r / Rmax

so if Rmax becomes bigger and nothing else is changed h will become smaller affecting the steady state of h? ( I think....)
if Hmax becomes bigger and nothing else is changed h will rise affecting the steady state of h? (I think....)

Now I know that at a steady state means that Fo=F1 (actually I think....)
and it was given to us that h(0)=2[ft] ( so I am guessing at time 0 when h is steady)
Fo= 2[ft^3/s]
Kv=1 [ft^2 ft^1/2/s]
so :

Fo=Kv?h ; h=(Fo/Kv)^2

but (2/1)^2 is 4 not 2. Or wait...does it maybe have to do with the units? :?

Anyway... if I increase the rate of flow in the tank Fo disturbing the steady state of h, h should go up. (I think...)
if I increase Kv, then h should go down (I think...)

Am I getting warmer colder or am i just far off?

thank you for your patience.
 
Anthony28 said:
I mean h not Hmax.

"How is the stedy state of h affected by changing Fo, Kv, Rmax and Hmax?"

Well if I change Rmax and Hmax i am essentially changing the shape of the cone, and therefor its angle, because

Rmax/Hmax = r/h (because of similar triangles)

then : h= Hmax*r / Rmax

so if Rmax becomes bigger and nothing else is changed h will become smaller affecting the steady state of h? ( I think....)
if Hmax becomes bigger and nothing else is changed h will rise affecting the steady state of h? (I think....)

Now I know that at a steady state means that Fo=F1 (actually I think....)
and it was given to us that h(0)=2[ft] ( so I am guessing at time 0 when h is steady)
Fo= 2[ft^3/s]
Kv=1 [ft^2 ft^1/2/s]
so :

Fo=Kv?h ; h=(Fo/Kv)^2

but (2/1)^2 is 4 not 2. Or wait...does it maybe have to do with the units? :?

Anyway... if I increase the rate of flow in the tank Fo disturbing the steady state of h, h should go up. (I think...)
if I increase Kv, then h should go down (I think...)

Am I getting warmer colder or am i just far off?

thank you for your patience.
Click to expand...

Yes you are correct.

If you increase the in-flowrate - keeping other parameters constant - h[sub:1eqk04jo]steady[/sub:1eqk04jo] will increase

If you increase the out-flowrate - keeping other parameters constant - h[sub:1eqk04jo]steady[/sub:1eqk04jo] will decrease

.
 
Attachment

NHgirl said:
Attached is the diagram. No, I am not modelling the cone itself. The model is of the height of the water in the cone as the parameters change. The exit point of the fluid is at the very base of the cone and the entry point is the vertex.

[attachment=0:2gybm4gj]tank.JPG[/attachment:2gybm4gj]
Click to expand...

Can I get the attachment for reference? I am trying to understand a similar problem and I couldn't see the attachment in the post. It would be of great help if you could re post the attachment. Thank You.
 
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