Tank Draining Problem
David Moran
The problem reads: At time t=0 the bottom plug (at the vertex) of a full
conical tank of water 16 ft high is removed. After 1 hour, the water in the
tank is 9 ft deep. When will the tank be empty?
I know that my initial conditions are f(0)=16 and f(1)=9 and I think that
the volume of a cone will probably be part of it too, but I just can't see
how to set the Differential Equation up. Any help would be greatly
appreciated.
Dave
Scott J. McCaughrin
The 'f' you mention sounds like a function of time giving the water's
height left in the conical tank. If so, then you are asked to find t:
f(t) = 0 ft. One way to solve this is to determine the rate at which
the water's depth decreases from 16 ft to 0 ft: this will relate the
water's height (h) to time (t) via: dh/dt, and you will find the cone's
volume formula useful for this.
Is this a right-circular cone? If so, its volume is: V = (1/3)PI*r^2*h.
In sci.math David Moran <ktulwx...@hotmail.com> wrote:
: Hi all, I am working on a problem that I just can't seem to work out.
David Moran
Dave
"Scott J. McCaughrin" <sjmc...@bluestem.prairienet.org> wrote in message
news:ca0epe$oui$1...@wildfire.prairienet.org...
Lynn Kurtz
<ktulwx...@hotmail.com> wrote:
>I am assuming it's a right circular cone.
>
I doubt the fact that it is a "right" cone matters, and perhaps not
even that it's circular. I think the question is what you are using to
model this type of problem. Are you assuming water drains at a
constant rate? Or that the rate of change of the depth of the water is
proportional to the depth? Or maybe that the rate of change of volume
is proportional to the depth? Different assumptions will give
different answers.
--Lynn
Peter Webb
"David Moran" <ktulwx...@hotmail.com> wrote in message
news:2ihrauF...@uni-berlin.de...
The problem needs a statement of the rate at which the water is removed
(flow rate), or it cannot be solved.
If the flow rate is a constant, then its pretty simple.
Usually in these problems the rate is given as an explicit function of the
depth, and you have to use the chain rule for differentiation. You state you
want a Differential Equation, which may mean that you are supposed to invent
some arbitraray function for the flow rate, presumably as a function of
water depth - say dV/dt = g(x), where x is the water depth and produce a
general differential equation to solve (which won't in general be solvable).
However, the fact that you have been given 16 and 9 as boundary conditions
and asked to work out the time that the tank is empty suggest its not that
complicated.
So what's the statement of the flow rate in the problem?
And BTW, the answer to the question is t=4.
If f(0) = 4^2 and f(1) = 3^2 then I can safely predict that f(2)=2^2,
f(3)=1^2 and f(4)=0^2, so the answer is t=4.
You could use that to work out what the flow rate formula is, and hence what
is missing from the question!
BTW, you may not need the angle of the cone, as this could well drop out of
the equation.
David Moran
"Lynn Kurtz" <kurtzDEL...@asu.edu> wrote in message
news:lMnDQKfCwbIvLC...@4ax.com...
I'm not sure. The problem that I posted is word for word out of the book.
Dave
David Moran
"Peter Webb" <wabbfamily...@yahoo.com> wrote in message
news:40c3cbf8$0$1583$afc3...@news.optusnet.com.au...
I just stated what was in the book. I really think there's some crucial
information missing.
Dave
Rick Decker
David Moran wrote:
The pressure at the opening will be proportional to the depth
of the water, so if we let V represent the volume of water we have
dV/dt = -kh
where h is the depth and k > 0 is some constant.
At any time, the water will be effectively a cone with the
radius proportional to the height. If we write r = sh for
some constant s, we'll have
V = (\pi r^2 h)/3
= ((\pi s^2)/3) * h^3
so
dV/dt = (\pi s^2)h^2 dh/dt
giving us
(\pi s^2)h^2 dh/dt = -kh
from which it's easy enough to get the time t at which h = 0.
[As a reality check, it'll take roughly an hour and a half to
empty the tank.]
Regards,
Rick
Robert Israel
Rick Decker <rde...@hamilton.edu> wrote:
>The pressure at the opening will be proportional to the depth
>
>of the water, so if we let V represent the volume of water we have
>
> dV/dt = -kh
>
>where h is the depth and k > 0 is some constant.
Nope. Neglecting friction in the opening, according to conservation
of energy the kinetic energy of the water coming out should be equal
to the drop in potential energy from the surface to the opening, and
the kinetic energy is proportional to the square of the speed, so
dV/dt = k sqrt(h).
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Virgil
"David Moran" <ktulwx...@hotmail.com> wrote:
I suspect that you are meant to assume that the rate of flow through the
drain is proportional to the pressure at the bottom of the tank , which,
of course, will be proportional to the depth of the water.
Have you tried to do the problem using this assumption?
David Moran
"Virgil" <ITSnetNOTcom/vir...@COMCAST.com> wrote in message
news:ITSnetNOTcom/virgil-C82D8D.20425706062004@[63.218.45.211]...
The only equation that the book gives is Torricelli's Law.
Dave
Gordo
The potential energy of a small mass of fluid at the top (height h) must
match the kinetic energy of the same mass of fluid exiting at the bottom:
mgh=mv^2/2
Hence, ideal exit velocity = v = sqrt(2gh) [Torricelli's law]
Gordo
The potential energy of a small mass m of fluid at the top (height h) must
match the kinetic energy of the same mass of fluid exiting at the bottom:
mgh=mv^2/2, where v is velocity, g is gravity.
Hence, ideal v = sqrt(2gh) [Torricelli's theorem]
Phil Smith
> In article <40C3D14D...@hamilton.edu>,
> Rick Decker <rde...@hamilton.edu> wrote:
>
> >The pressure at the opening will be proportional to the depth
> >
> >of the water, so if we let V represent the volume of water we have
> >
> > dV/dt = -kh
> >
> >where h is the depth and k > 0 is some constant.
>
> Nope. Neglecting friction in the opening, according to conservation
> of energy the kinetic energy of the water coming out should be equal
> to the drop in potential energy from the surface to the opening, and
> the kinetic energy is proportional to the square of the speed, so
> dV/dt = k sqrt(h).
which on dimensional grounds is better written
dV/dt = - C (2pi/15) tan(alpha) H^(5/2) T^(-1) sqrt(h) [0]
where C is a positive nondimensional constant[1], H = 16 ft,
T = 1 hr, and alpha is the angle of the cone.
[0]: The (2pi/15) tan(alpha) factor is included so that all known
non-dimensional coefficients drop out at the earliest possible stage.
[1]: This constant can be determined a posteriori from the given initial
conditions (there are two, and the ODE is first order).
--
P.A.C. Smith
replying by email: s/NOSPAM//
'My duke of the blood royal [Gloucester] can beat up your duke of the blood
royal [York], and you know it.'
- Me, ars.userfriendly.org, 23/4/04
Richard L Walker
wrote:
>Today Robert Israel wrote in alt.math.undergrad and 2 other newsgroups:
>
>> In article <40C3D14D...@hamilton.edu>,
>> Rick Decker <rde...@hamilton.edu> wrote:
>>
>> >The pressure at the opening will be proportional to the depth
>> >
>> >of the water, so if we let V represent the volume of water we have
>> >
>> > dV/dt = -kh
>> >
>> >where h is the depth and k > 0 is some constant.
>>
>> Nope. Neglecting friction in the opening, according to conservation
>> of energy the kinetic energy of the water coming out should be equal
>> to the drop in potential energy from the surface to the opening, and
>> the kinetic energy is proportional to the square of the speed, so
>> dV/dt = k sqrt(h).
>
>which on dimensional grounds is better written
>
>dV/dt = - C (2pi/15) tan(alpha) H^(5/2) T^(-1) sqrt(h) [0]
>
>where C is a positive nondimensional constant[1], H = 16 ft,
>T = 1 hr, and alpha is the angle of the cone.
>
>[0]: The (2pi/15) tan(alpha) factor is included so that all known
>non-dimensional coefficients drop out at the earliest possible stage.
>
>[1]: This constant can be determined a posteriori from the given initial
>conditions (there are two, and the ODE is first order).
Is the fact it is filled with water (rather than molasses) a factor?
Water has a certain weight per volume (like 1 gm per cubic centimeter
-- or something similar)
Greg
In the absence of anything more specific, I would use 'flow-rate = k*h'.
Where k is a constant and h the height (or head) of liquid. The radius will
be an arbitrary constant which you can express in terms of h when the tank
is part full.
--
Greg
(remove BALL from address to reply)
S.Sriharan
The rate of flow, measured in volume per unit time, is K'*sqrt(h)
This is derived from Torricelli's Law for the ideal velocity of
discharge. Friction is not neglected but it is assumed that the effect
of friction and other imperfections can be accounted for by a reduction
of the velocity by a constant factor, irrespective of the height (head)
of water.
K' = coefficient of discharge * area of orifice * sqrt(2*g)
Since the volume V, of water in the cone, decreases with the flow,
dV/dt = - K'*sqrt(h) ......................... (1)
But V = K"*h^3
because the volume of similar solids (cones) is proportional to the cube
of the linear dimension. It is not necessary for the cone to have a
circular base.
dV/dh = 3*K"*h^2
dh/dt = (dh/dV)*(dV/dt)
= (1/(3*K"*h^2))*(-K'*sqrt(h))
i.e. dt = - (3*K"/K')*h^(3/2)*dh
Integrating, t = - K*h^(5/2) + C ....................... (2)
where K = (2/5)*3*K"/K' and C is the integration constant
At t=0, h=16
i.e. 0 = - K*16^(5/2) + C
C = K*1024
t = K*(1024 - h^(5/2)) ......................(3)
At t=1, h=9, using the hour as the unit of time
1 = K*(1024 - 9^(5/2))
= K*(1024 - 243)
= K*781
t = (1024 - h^(5/2))/781 ....................(4)
For h=0 t = 1024/781 = 1.3111396 hours
= 1 hour 18 minutes 40 seconds
--
S.Sriharan