Chapter 5, Problem 5.75P

To determine

The solidification time for tripled height and diameter.

Answer to Problem 5.75P

Time required to solidify the metal if the height is tripled is 6.612 min.

Time required to solidify the metal if the diameter is tripled is 12.96 min.

Explanation of Solution

Given Information:

  $\begin{array}{l}\text{Diameter to heght ratio:}\frac{d_1}{h_1}=1\\ \text{Then, Solidification time, }t=4\text{min}\end{array}$

The volume of the cylinder is,

  $V=\frac{\pi }{4}{d}^{2}h$

The area of the cylinder is,

  $\begin{array}{l}A=\pi dh+2\times \frac{\pi }{4}{d}^2\\ A=\pi dh+\frac{\pi }{2}{d}^2\end{array}$

The solidification is given as,

  $\begin{array}{l}t=C{\left(\frac{V}{A}\right)}^2\\ t=C{\left(\frac{\frac{\pi }{4}{d}^{2}h}{\pi dh+\frac{\pi }{2}{d}^2}\right)}^2\\ t=C{\left(\frac{dh}{4h+2d}\right)}^2\mathrm{.......}(1)\end{array}$

From the given data, $h_1=d_1\&t_1=4\text{min}$

Then, solidification becomes,

  $\begin{array}{l}{t}_1=C{\left(\frac{d_1{h}_1}{4h_1+2d_1}\right)}^2\\ 4=C{\left(\frac{d_1^2}{4d_1^2+2d_1^2}\right)}^2\\ 4=C{\left(\frac{d_1}{6}\right)}^2=C{\left(\frac{h_1}{6}\right)}^2\mathrm{.....}(2)\end{array}$

Now if the height is tripled

  $\begin{array}{l}{h}_2=3h_1\\ h_2=3d_1\\ \&\\ d_2=d_1\end{array}$

  $\begin{array}{l}{t}_2=C{\left(\frac{d_2{h}_2}{4h_2+2d_2}\right)}^2\\ t_2=C{\left(\frac{d_1\times 3h_1}{4\times 3h_1+2\times d_1}\right)}^2\\ t_2=C{\left(\frac{3d_1{h}_1}{12h_1+2d_1}\right)}^2\\ t_2=C{\left(\frac{3d_1^2}{12d_1+2d_1}\right)}^2\\ t_2=C{\left(\frac{3}{14}{d}_1\right)}^2\mathrm{......}(3)\end{array}$

From

Equation (2) & (3)

  $\begin{array}{l}4=C{\left(\frac{d_1}{6}\right)}^2\\ t_2=C{\left(\frac{3}{14}{d}_1\right)}^2\\ \text{Dividing both the equations,}\\ \frac{4}{t_2}=\frac{C{\left(\frac{d_1}{6}\right)}^2}{C{\left(\frac{3}{14}{d}_1\right)}^2}\\ \frac{4}{t_2}=\frac{{\left(\frac{1}{6}\right)}^2}{{\left(\frac{3}{14}\right)}^2}\\ \frac{4}{t_2}={\left(\frac{7}{9}\right)}^2\\ \frac{4}{t_2}=\frac{49}{81}\\ t_2=\frac{81\times 4}{49}\\ t_2=6.612\text{min}\end{array}$

Hence, the time required to solidify the metal if the height is tripled is 6.612 min.

Now, if the diameter is tripled then,

  $\begin{array}{l}{d}_3=3d_1\\ \&\\ h_3=h_1\\ h_3=d_1\end{array}$

  $\begin{array}{l}{t}_3=C{\left(\frac{d_3{h}_3}{4h_3+2d_3}\right)}^2\\ t_3=C{\left(\frac{3d_1\times d_1}{4d_1+2\times 3d_1}\right)}^2\\ t_3=C{\left(\frac{3d_1^2}{4d_1+6d_1}\right)}^2\\ t_2=C{\left(\frac{3d_1^2}{10d_1}\right)}^2\\ t_3=C{\left(\frac{3}{10}{d}_1\right)}^2\mathrm{......}(4)\end{array}$

From

Equation (2) & (4)

  $\begin{array}{l}4=C{\left(\frac{d_1}{6}\right)}^2\\ t_3=C{\left(\frac{3}{10}{d}_1\right)}^2\\ \text{Dividing both the equations,}\\ \frac{4}{t_3}=\frac{C{\left(\frac{d_1}{6}\right)}^2}{C{\left(\frac{3}{10}{d}_1\right)}^2}\\ \frac{4}{t_3}=\frac{{\left(\frac{1}{6}\right)}^2}{{\left(\frac{3}{10}\right)}^2}\\ \frac{4}{t_3}={\left(\frac{5}{9}\right)}^2\\ \frac{4}{t_3}=\frac{25}{81}\\ t_3=\frac{4\times 81}{25}\\ t_3=12.96\text{min}\end{array}$

Hence, the time required to solidify the metal if the diameter is tripled is 12.96 min.