Solving a DE for a Leaking Tank's Height over Time

[KillerZ]

Homework Statement

A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. When friction and contraction of water at the hole are ignored, the height h of the water in the tank is described by:

\frac{dh}{dt} = -\frac{A_{h}}{A_{w}}\sqrt{2gh}

where Aw and Ah are the cross-sectional areas of the water and the hole, respectively.

a) Solve the DE if the inital height of the water is H. By hand, sketch the graph of h(t) and give its interval I of definition in terms of the symbols Aw. Ah, and H. Use g = 32 ft/s^2.

b) Suppose the tank is 10 feet high and has radius 2 feet and circular hole has radius 1/2 inch. If the tank is initially full, how long will it take to empty?

Homework Equations

\frac{dh}{dt} = -\frac{A_{h}}{A_{w}}\sqrt{2gh}

The Attempt at a Solution

I want to make sure I did this part right before attempting b) as it needs the answer for a).

a)

\frac{dh}{dt} = -\frac{A_{h}}{A_{w}}\sqrt{2gh}

\frac{dh}{dt} = -\frac{A_{h}}{A_{w}}\sqrt{64H}

\frac{dh}{\sqrt{64H}} = -\frac{A_{h}}{A_{w}}dt

\int\frac{dh}{\sqrt{64H}} = -\int\frac{A_{h}}{A_{w}}dt

\frac{h}{\sqrt{64H}} = -\frac{A_{h}}{A_{w}}t

 

[Mark44]

Homework Statement

A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. When friction and contraction of water at the hole are ignored, the height h of the water in the tank is described by:

\frac{dh}{dt} = -\frac{A_{h}}{A_{w}}\sqrt{2gh}

where Aw and Ah are the cross-sectional areas of the water and the hole, respectively.

a) Solve the DE if the inital height of the water is H. By hand, sketch the graph of h(t) and give its interval I of definition in terms of the symbols Aw. Ah, and H. Use g = 32 ft/s^2.

b) Suppose the tank is 10 feet high and has radius 2 feet and circular hole has radius 1/2 inch. If the tank is initially full, how long will it take to empty?

Homework Equations

\frac{dh}{dt} = -\frac{A_{h}}{A_{w}}\sqrt{2gh}

The Attempt at a Solution

I want to make sure I did this part right before attempting b) as it needs the answer for a).

a)

\frac{dh}{dt} = -\frac{A_{h}}{A_{w}}\sqrt{2gh}

In the next line, don't put H in yet; leave it as h, the variable.

\frac{dh}{dt} = -\frac{A_{h}}{A_{w}}\sqrt{64H}

I have changed what you had to the following. Can you carry it through?

\frac{dh}{8\sqrt{h}} = -\frac{A_{h}}{A_{w}}dt

(1/8) \int h^{-1/2}dh = -\int\frac{A_{h}}{A_{w}}dt

 

[KillerZ]

how is this:

(1/8) \int h^{-1/2}dh = -\int\frac{A_{h}}{A_{w}}dt

(1/8) 2h^{1/2} = -\frac{A_{h}}{A_{w}}t

\frac{\sqrt{h}}{4} = -\frac{A_{h}}{A_{w}}t

 

[Mark44]

how is this:

(1/8) \int h^{-1/2}dh = -\int\frac{A_{h}}{A_{w}}dt

Don't forget the constant of integration in the next and following lines.

(1/8) 2h^{1/2} = -\frac{A_{h}}{A_{w}}t

\frac{\sqrt{h}}{4} = -\frac{A_{h}}{A_{w}}t

This is fine, except for the missing constant, but why don't you go ahead and solve for h, instead of its square root? I.e., h(t) = ...

 

[KillerZ]

\frac{\sqrt{h}}{4} = -\frac{A_{h}}{A_{w}}t + c

\sqrt{h} = -4\frac{A_{h}}{A_{w}}t + c

h = (-4t\frac{A_{h}}{A_{w}})^{2} + c

h - (-4t\frac{A_{h}}{A_{w}})^{2} = c

h = (-4t\frac{A_{h}}{A_{w}})^{2} + h - (-4t\frac{A_{h}}{A_{w}})^{2}

 

[Mark44]

\frac{\sqrt{h}}{4} = -\frac{A_{h}}{A_{w}}t + c

Technically, the c in the next line is a different constant from the one in the previous line, but it's still a constant.

\sqrt{h} = -4\frac{A_{h}}{A_{w}}t + c

You can't do what you've done in the next line. You squared the left side to get h; you have to square the entire right side as well.

h = (-4t\frac{A_{h}}{A_{w}})^{2} + c

h - (-4t\frac{A_{h}}{A_{w}})^{2} = c

h = (-4t\frac{A_{h}}{A_{w}})^{2} + h - (-4t\frac{A_{h}}{A_{w}})^{2}

 

[KillerZ]

\frac{\sqrt{h}}{4} = -\frac{A_{h}}{A_{w}}t + c

\sqrt{h} = -4\frac{A_{h}}{A_{w}}t + 4c

h = (-4t\frac{A_{h}}{A_{w}} + 4c)^{2}

c = \sqrt{h} + t\frac{A_{h}}{A_{w}}

h = (-4t\frac{A_{h}}{A_{w}})^{2} + \sqrt{h} + t\frac{A_{h}}{A_{w}}

 

[Mark44]

\frac{\sqrt{h}}{4} = -\frac{A_{h}}{A_{w}}t + c

\sqrt{h} = -4\frac{A_{h}}{A_{w}}t + 4c

h = (-4t\frac{A_{h}}{A_{w}} + 4c)^{2}

Or just call it C instead of 4c.

Now, use your initial condition, that h(0) = H.
So, when t = 0, you have
h(0) = H = C^{2}
From this, you can easily get the constant.

c = \sqrt{h} + t\frac{A_{h}}{A_{w}}

h = (-4t\frac{A_{h}}{A_{w}})^{2} + \sqrt{h} + t\frac{A_{h}}{A_{w}}

This last step didn't do you much good, since you haven't really solved for h; it appears on both sides of your equation.

 

[KillerZ]

h(0) = H = C^{2}

c = \sqrt{H}

h = (-4t\frac{A_{h}}{A_{w}} + \sqrt{H})^{2}

 

[Mark44]

OK, you have done part of the a part of this problem. For the b part, replace the symbols A_h, A_w, and H with the given information in this part, and use your functions to find when the tank is empty; i.e., at the time t for which h(t) = 0.

 

[KillerZ]

b)

h = (-4t\frac{A_{h}}{A_{w}} + \sqrt{H})^{2}

h = (-4t\frac{\pi (0.5/12)^{2}}{\pi 2^{2}} + \sqrt{10})^{2}

t = 1821.47 s

 

[Mark44]

I haven't checked your answer. You do a great job at formatting your work in LaTeX, but you could explain what you're doing a little better.

You're dealing with a function here, so it would be better to indicate that.
h(t) = (-4t\frac{\pi (0.5/12)^{2}}{\pi 2^{2}} + \sqrt{10})^{2}
This gives the height of water in the specified tank at time t.

To find when the tank is empty, you're solving h(t) = 0 for t.
h(t) = 0 \Rightarrow (-4t\frac{\pi (0.5/12)^{2}}{\pi 2^{2}} + \sqrt{10})^{2} = 0
from which you get t = whatever.

 

[KillerZ]

I think this should do it:

h(t) = 0 = (-4t\frac{\pi (0.5/12)^{2}}{\pi 2^{2}} + \sqrt{10})^{2}

4t\frac{\pi (0.5/12)^{2}}{\pi 2^{2}} = \sqrt{10}

t(0.001736111) = \sqrt{10}

t = \sqrt{10}/(0.001736111)

t = 1821.47 s

thanks for the help.

 

[Mark44]

Sure, you're very welcome.