Chapter 10, Problem 51P

To determine

The factor by which his time be cut if he uses another diameter hose, is should be calculated.

Answer to Problem 51P

The time factor is $87.04\%$ .

Explanation of Solution

Given info:

Diameter

  $\begin{array}{l}{d}_1=\frac{3}{8}inch\\ d_2=\frac{5}{8}inch\end{array}$

Formula used:

According to the Poiseuille's formula

  $V=\frac{\pi p{r}^4}{8\eta l}$

Where, $V$ is the volume discharge flow

  $p$ , Pressure difference between any two ends of object

  $\eta$ Viscosity of fluid

  $r$ Is the inner radius

  $l$ Is the length of the pipe

Calculation:

According to the question, all other conditions remain unchanged which means that pressure is constant, viscosity coefficient, and length same but radius is different.

According to this formula time is inversely varies with the square of the radius. So,

  $\begin{array}{l}V=K{r}^4\left(putK=\frac{\pi p}{8\eta l}\right)\\ \frac{dv}{dt}=K{r}^4 \end{array}$

  $\begin{array}{l}\frac{V}{t_1}=K{r}_1{}^{4}and\frac{V}{t_2}=K{r}_2{}^4\\ \Rightarrow \frac{\frac{V}{t_1}}{\frac{V}{t_2}}=\frac{K{r}_1{}^4}{K{r}_2{}^4}=\frac{r_1{}^4}{r_2{}^4}\Rightarrow \frac{t_2}{t_1}={\left(\frac{3/8}{5/8}\right)}^4={\left(\frac{3}{5}\right)}^4=\frac{81}{625}=0.1296\\ \Rightarrow t_2=t_{1}0.1296\end{array}$

Now the factor is

  $factor=\frac{t_{1}0.1296}{t_1}\times 100=12.96\%$

And the factor which will be cut is

  $factor=100\%-12.96\%=87.04\%$

Conclusion:

The time factor is $87.04\%$ .