Chapter 10, Problem 56P

To determine

The decreasing factor of the radius is to be calculated according to the given information.

Answer to Problem 56P

The radius is decreased by the factor of $29\%$ .

Explanation of Solution

Given info:

Flow reduced factor, $75\%$

Formula used:

According to the Poiseuille's equation,

  $Q=\frac{\pi r^4\left(P_2-P_1\right)}{8\eta l}$

Where, $Q$ is the volume discharge flow

  $\left(P_2-P_1\right)$ , 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:

Substitute the values in the given equation, for both cases, except the radius all terms are constant in this formula.

  $Q_1=\frac{\pi r_1{}^4\left(P_2-P_1\right)}{8\eta l}$

For the second case,

  $Q_2=\frac{\pi r_2{}^4\left(P_2-P_1\right)}{8\eta l}$

Now divide these both case with each other, such that

  $\begin{array}{l}\frac{Q_1}{Q_2}=\frac{\frac{\pi r_1{}^4\left(P_2-P_1\right)}{8\eta l}}{\frac{\pi r_2{}^4\left(P_2-P_1\right)}{8\eta l}}=\frac{r_1{}^4}{r_2{}^4}\\ \Rightarrow \frac{r_1}{r_2}={\left(\frac{Q_1}{Q_2}\right)}^{1/4}={\left(\frac{100\%}{100\%-75\%}\right)}^{1/4}\\ ={\left(\frac{100\%}{25\%}\right)}^{1/4}={\left(4\right)}^{1/4}={\left(\sqrt{2}\right)}^{4/4}=\sqrt{2}=1.4142\\ r_1=1.4142r_2\end{array}$

Now the value of radius in the second case,

  $r_2=0.717r_1$

The percentage of the second radius is

  $r_2=71\%$

Now, the decreasing factor is,

  $100\%-71\%=29\%$

Conclusion:

The radius is decreased by the factor of $29\%$ .