Question

Inverted Conical Tank Suppose that the conical tank in Problem 13(a) is inverted, as shown in Figure 3.2.5, and that water leaks out a circular hole of radius 2 inches in the center of its circular base. Is the time it takes to empty a full tank the same as for the tank with vertex down in Problem 13? Take the friction/contraction coefficient to be $\mathit{c}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{6}$and $\mathit{g}\mathbf{=}\mathbf{32}\mathbf{\text{ft}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{2}}$

Short answer

The approximate time at which the tank is empty$t\approx 38.16\text{min}$ which is the difference from 13(a)

Step-by-step solution

Define the equation of height of the water

The height of water in this tank is described by the $\frac{\mathbf{d}\mathbf{h}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\frac{{\mathbf{A}}_{\mathbf{h}}}{{\mathbf{A}}_{\mathbf{w}}}\sqrt{\mathbf{2}\mathbf{g}\mathbf{h}}$where ${\mathit{A}}_{\mathbf{w}}$is the cross-sectional area of water, with the condition

Obtain the amount of water in the cylinder at time t as the following:

Here $\frac{\text{d}h}{\text{d}t}=-\frac{5\sqrt{h}}{6{(20-h)}^2}$

$\text{h(0) = 20}$

Now separate the variable

$$\begin{gathered} \frac{\text{d}h}{\text{d}t}=-\frac{5\sqrt{h}}{6{(20-h)}^2} \\ \int \frac{6{(20-h)}^2}{\sqrt{h}}\text{d}h=\int -5\text{d}t \end{gathered}$$

Simplify

$$\begin{gathered} \int \frac{6{(20-h)}^2}{\sqrt{h}}\text{d}h=-5t+C \\ \int \left(2400h^{-\frac{1}{2}}-240h^{\frac{1}{2}}+6h^{\frac{3}{2}}\right)\text{d}h=-5t+C \\ 4800h^{\frac{1}{2}}-160h^{\frac{3}{2}}+\frac{12}{5}{h}^{\frac{5}{2}}=-5t+C \end{gathered}$$

Calculate the value of C when t=0 and n(0)=20 

evaluate

$$\begin{gathered} 4800(20{)}^{\frac{1}{2}}-160(20{)}^{\frac{3}{2}}+\frac{12}{5}(20{)}^{\frac{5}{2}}=-5·0+C \\ 9600\sqrt{5}-6400\sqrt{5}+1920\sqrt{5}=C \\ 5120\sqrt{5}=C \end{gathered}$$

Now put $C=5120\sqrt{5}$

$$\begin{gathered} 4800h^{\frac{1}{2}}-160h^{\frac{3}{2}}+\frac{12}{5}{h}^{\frac{5}{2}}=-5t+5120\sqrt{5} \\ 4800(0{)}^{\frac{1}{2}}-160(0{)}^{\frac{3}{2}}+\frac{12}{5}(0{)}^{\frac{5}{2}}=-5t+5120\sqrt{5} \\ 0=-5t+5120\sqrt{5} \end{gathered}$$

The approximate time at which the tank is empty$t\approx 38.16\text{min}$