Question

A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height h of water in the tank in feet after t seconds is described by dh dt = − Ah Aw 2gh , (b) Suppose the tank is 12 feet high and has radius 4 feet and the circular hole has radius 1 2 inch. If the tank is initially full, how long will it take to empty?

Answer 1

The question is incomplete so, here is the complete question.

A tank in 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 water in the tank is described by

$\frac{dh}{dt} = - \frac{A_{h} }{A_{w} } \sqrt{2gh}$, where $A_{h}$ and $A_{w}$ are cross-sectional areas of the hole and the water, respectively. (a) Solve for h(t) if the initial 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 = 32ft/s². (b) Suppose the tank is 12 feet high and has radius 4 feet and the circular hole har radius 1/2 inch. If the tank is initially full, how long will it take to empty?

Answer: a) h(t) = $(\sqrt{H} - 4 \frac{A_{h} }{A_{w} }t )^{2}$, with interval 0≤t≤$\frac{A_{w}\sqrt{H} }{4A_{h} }$

b) It takes 133 minutes.

Step-by-step explanation: a) The height per time is expressed as

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

Using g=32ft/s²: $\sqrt{2gh}$ = $\sqrt{2.32.h}$  $= 8\sqrt{h}$

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

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

$\int\limits^a_b {\frac{dh}{\sqrt{h} } } = - \int\limits^a_b {8.\frac{A_{h} }{A_{w} } } \, dt$

Calculating the indefinite integrals:

2$\sqrt{h}$ = - 8$\frac{A_{h} }{A_{w} }$.t + c

According to the question, when t=0 h₀ for water is H. so

2$\sqrt{H}$ = - 8$\frac{A_{h} }{A_{w} }$.0 + c

c = 2$\sqrt{H}$

2$\sqrt{h}$ = - 8$\frac{A_{h} }{A_{w} }$.t + 2$\sqrt{H}$

Dividing each term by 2, then, the equation h(t) is

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

Now, to find the interval, when the tank is empty, there is not water height so:

0 = ($\sqrt{H} -$$4\frac{A_{h} }{A_{w} } .t$)²

0 = ($\sqrt{H} -$$4\frac{A_{h} }{A_{w} } .t$)

4t$\frac{A_{h} }{A_{w} }$ = $\sqrt{H}$

t = $\frac{A_{w} \sqrt{H} }{4.A_{h} }$

Thus, the interval will be :  0 ≤ t ≤ $\frac{A_{w}\sqrt{H} }{4A_{h} }$

b) $r_{w}$ = 4 ft $h_{w}$ = 12 ft and $r_{h}$ = 1/2 in

Area of the water:

$A_{w} = \pi r^{2}$

$A_{w} = \pi .4^{2}$

$A_{w}$ = 16πft²

Area of the hole:

1 in = $\frac{1}{12}$ft

so, $r_{h}$ = $\frac{1}{2}.\frac{1}{12}$ = $\frac{1}{24}$

$A_{h} = \pi .r^{2}$

$A_{h} = \pi .\frac{1}{24} ^{2}$

$A_{h}$ = $\frac{\pi }{576}$ ft²

As H=10 and having t = $\frac{A_{w} \sqrt{H} }{4.A_{h} }$ :

t = $\frac{16.\pi .\sqrt{12} }{4.\frac{\pi }{576} }$

t = 4.576.$\sqrt{12}$

t = 7981.3 seconds

t = $\frac{7981.3}{60}$ = 133 minutes

It takes 133 minutes to empty.