a differential equation problem?

Question

hey guys...
i am aminah...

can you help me with my problem?

here it goes...

FIND THE TIME TO EMPTY A CYLINDRICAL TANK, AXIS VERTICAL, RADIUS 1/2 FT., HEIGHT 1.25 M, INITIALLY FULL OF WATER,THROUGH AN ORIFICE 1/50 FT. IN DIAMETER IN ITS BOTTOM....

hope you can find a marvelous solution to my problem...

thanks in advance....

Answer 1

The goal here is to equate the volumetric flow rate, with some function of the water's height:

dV/dt = f(y).

First, to make things simpler, let's convert all the units to metric. The radius of the tank is 0.152 m , and the orifice has a radius of 0.0061 m.

In this case, it's helpful to use "Torricelli's law" which states that the flow velocity out of an orifice in a container is:

q = sqrt(2gy)

where g is the acceleration due to gravity, and y is the height of the fluid above the orifice.

In practice, because of boundary layer effects, the actual flow tends to be less than predicted by the above equation. Usually, an "orifice flow coefficient" (C) is introduced;( for a sharp, circular hole, a good approximation is probably about 0.6:)

q = C sqrt(2gy).

Now, the volumetric flow rate can be related to the area of the orifice (Ao) by

dV/dt = Ao q;

therefore,

dV/dt = Ao C sqrt(2gy).

Lastly, if a differential volume of water dV leaves the tank, then the the change in height of the water is

dV = - At dy,

where At is the cross-sectional area of the tank(provided the tak has a constant cross section...) Then, finally:

At dy/dt = -Ao C sqrt(2gy), or

dy/dt = -(Ao C / At) sqrt(2gy).

This is a simple, separable, non-linear equation of first order. It can be solved by rearranging slightly:

dt = -(At / (Ao C)) dy / sqrt(2gy);

which can then be directly integrated,

∫dt = -(At / (Ao C)) ∫dy / sqrt(2gy).

Then, using the substitution:

u = 2gy
dy = du/2g

t = -(At / [2g C Ao]) ∫du / sqrt(u)

t = -(At / [2g C Ao]) 2 sqrt(u)

t = -(At / [g C Ao]) sqrt(2gy) + D

where D is some constant of integration.

This is a "classic" initial value problem. In order to find the value of D, you can use the fact that when t = 0, then y = y0, or the original height of the water. Then:

0 = -(At / [g C Ao]) sqrt(2g y0) + D;

so, obviously,

D = (At / [g C Ao]) sqrt(2g y0),

which changes the expression for t to:

t = (At / [g C Ao]) ( sqrt[2g y0] - sqrt[2gy] ).

Another, perhaps, "obvious" fact, is that when the water is done draining, y = 0. So the total amount of time (after all our mathematical acrobatics) is simply:

t = (At / [g C Ao]) sqrt(2g y0).........!

From here, I think I'll just let you actually "plug in" the numbers for At, Ao, C, G, and y0 yourself. Remember that,

At = tank x-section area
Ao = orifice area
g = acceleration due to gravity,
C = orifice flow coefficient = ~ 0.6
y0 = original height of water.

Good luck, hope that helps,
~W.O.M.B.A.T.

Answer 2

Try this method.

dN/dt = RN - K
The total change is related to the initial amount by a constant. K is subtracted because water is being lost.

dN/dt = (RN - K)
dN/dt -RN = -K
solve this using the magic integration factor for first order method ( the mu)

Solve for constants using initial conditions.

This may be useful.

volume = 269 cm^3
Water has density = 1g/ cm^3
so grams of water is 269 g
hole size = 6*10^-3 m

Tell me if it works. I'm really sleepy. Good luck.

Answer 3

This is a linear D.E. Arrange in this form. y' + 8y = 32 p(x) = 8; q(x) = 32 I.F. = e^integral of p(x)dx I.F. = e^8x Therefore: ye^8x = integral(32e^8xdx) + C ye^8x = 32/8 e^8x + C ye^8x = 4e^8x + C y = 4 + C e^(-8x); then y = 7 as x = 0 7 = 4 + Ce^0 C = 3 So, y = 4 +3e^(-8x)...end

Answer 4

depends on how pure the water is...if there is any impurity it will take longer