Chapter 1.10, Problem 69E

To determine

Find the required velocity to carry particles 0.12 inch in diameter

Answer to Problem 69E

The required velocity to carry particles 0.12 inch in diameter is 0.61 mile per hour.

Explanation of Solution

Given:

Given that diameter $d$ of the largest particle that can be moved by a stream varies approximately directly as the square of the velocity $v$ of the stream. Therefore

  $d=k{v}^2$

Here the value $k$ is called constant of variation.

Substitute the known values for $d$ and $v$ and solve for the unknown value of $k$ .

Given equation: $d=k{v}^2$

Substitute $d=0.02,v=\frac{1}{4}:0.02=k\cdot {\left(\frac{1}{4}\right)}^2$

Using ${\left(\frac{1}{4}\right)}^2=\frac{1}{16}:0.02=\frac{k}{16}$

Multiply both sides by $16:0.02(16)=\frac{k}{16}\cdot 16$

Divide out common factor : $0.32=\frac{k}{16}\cdot 16$

Simplify: $0.32=k$

If $a=b$ then $b=a:k=0.32$ : $k$

Hence the constant of variation $k$ is equals 0.32 . Now with value of $k$ known let's write variation model as $d=0.32v^2$

Now to find the required velocity to carry particles 0.12 inch in diameter, substitute $d=0.12$ in model equation $d=0.32v^2$ and solve it for $v$ .

  $0.12=0.32v^2$

  $v^2=\frac{0.12}{0.32}$

  $\begin{array}{c}{v}^2=0.375\\ v\approx 0.61\end{array}$

Therefore the required velocity to carry particles 0.12 inch in diameter is 0.61 mile per hour.