Chapter 5, Problem 5.13P

To determine

(a)

The recovery of plastic B in the overflow.

Answer to Problem 5.13P

The recovery of plastic B in the overflow is $92.1\%$ .

Explanation of Solution

Given:

The feed rate is $38\text{kg}/\text{h}$ .

The overflow rate is $35\text{kg}/\text{h}$ .

Concept Used:

Write the equation to calculate percentage of recovery

  $P_{\text{RB}}=\left(\frac{x_1}{x_2}\right)\times 100$   ...... (I).

Here, the percentage of recovery of plastic B is $P_{\text{RB}}$ , overflow rate is $x_1$ and the feed rate is $x_2$ .

Calculation:

Calculate the percentage of recovery of plastic B.

Substitute for $35\text{kg}/\text{h}$

  $x_1$ and $38\text{kg}/\text{h}$ for $x_2$ in Equation (I).

  $\begin{array}{c}{P}_{\text{RB}}=\left(\frac{35\text{kg}/\text{h}}{38\text{kg}/\text{h}}\right)\times 100\\ =0.921\times 100\\ =92.1\%\end{array}$

Conclusion:

Thus, the recovery of plastic B in the overflow is $92.1\%$ .

To determine

(b)

The purity of plastic B in the overflow.

Answer to Problem 5.13P

The purity of plastic B in the overflow is $60.3\%$ .

Explanation of Solution

Given:

The overflow rate of plastic A is $5\text{kg}/\text{h}$ .

The overflow rate of plastic B is $35\text{kg}/\text{h}$ .

The overflow rate of plastic C is $18\text{kg}/\text{h}$ .

Concept Used:

Write the equation to calculate percentage of purity.

  $P_{\text{PB}}=\left(\frac{x_1}{x_1+y_1+z_1}\right)\times 100$   ...... (II).

Here, the percentage of recovery of plastic B is $P_{\text{PB}}$ , overflow rate of plastic A, B and C are $x_1$ , $y_1$ and $z_1$ respectively.

Calculation:

Calculate the percentage of purity of plastic B.

Substitute $35\text{kg}/\text{h}$ for $x_1$ , $5\text{kg}/\text{h}$ for $y_1$ and $18\text{kg}/\text{h}$ for $z_1$ in Equation (II).

  $\begin{array}{c}{P}_{\text{PB}}=\left(\frac{35\text{kg}/\text{h}}{\left(35+5+18\right)\text{kg}/\text{h}}\right)\times 100\\ =0.603\times 100\\ =60.3\%\end{array}$ .

Conclusion:

Thus, the purity of plastic B in the overflow is $60.3\%$ .

To determine

(c)

The time taken by the plastic B to reach the top.

Answer to Problem 5.13P

The time taken by plastic B to reach the top is $220.19\text{s}$ .

Explanation of Solution

Given:

The density of fluid is $1.2\text{gm}/{\text{cm}}^{\text{3}}$ .

The viscosity of fluid is $0.015\text{poise}$ .

The diameter of plastic is $0.5\text{mm}$ .

The distance is $2\text{m}$ .

The density of material is $1.1\text{gm}/{\text{cm}}^{\text{3}}$ .

Concept Used:

Write the equation to calculate the time taken.

  $t=\frac{L}{v}$   ...... (III).

Here, the time is $t$ , distance is $L$ , and terminal velocity is $v$ .

Calculation:

Calculate the terminal velocity by using Stock’s Law.

  $v=\frac{d^{2}g\left(\rho -{\rho }_s\right)}{18\mu }$   ...... (IV).

Here, the diameter of particle is $d$ , the acceleration due to gravity is $g$ , the density of fluid is $\rho$ , the density of material is ${\rho }_s$ , and the viscosity of the fluid is $\mu$ .

Calculate the terminal velocity.

Convert the unit of viscosity from $\text{poise}$ to $\text{kg}/\text{m}/\text{s}$ .

  $\begin{array}{c}\mu =\left(0.015\text{poise}\right)\left(\frac{0.1\text{kg}/\text{m}/\text{s}}{1\text{poise}}\right)\\ =1.\text{5}\times {\text{10}}^{-3}\text{kg}/\text{m}/\text{s}\end{array}$

Substitute $0.5\times {10}^{-3}\text{m}$ for $d$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $1.2\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ , $1.1\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_s$ and $\text{1}\text{.5}\times {\text{10}}^{-3}\text{kg}/\text{m}/\text{s}$ for $\mu$ in Equation (IV).

  $\begin{array}{c}v=\frac{{\left(0.5\times {10}^{-3}\text{m}\right)}^2\times 9.81\text{m}/{\text{s}}^{\text{2}}\times \left(1.2\times {10}^3\text{kg}/{\text{m}}^{\text{3}}-1.1\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)}{\left(18\times \text{1}\text{.5}\times {\text{10}}^{-3}\text{kg}/\text{m}/\text{s}\right)}\\ =\frac{2.4525\times {10}^{-4}\text{kg}/{\text{s}}^{\text{2}}}{0.027\text{kg}/\text{m}/\text{s}}\\ =9.083\times {10}^{-3}\text{m}/\text{s}\end{array}$

Calculate the time taken by the plastic B to reach the top.

Substitute $9.083\times {10}^{-3}\text{m}/\text{s}$ for $v$ and $2\text{m}$ for $L$ in Equation (III).

  $\begin{array}{c}t=\frac{2\text{m}}{9.083\times {10}^{-3}\text{m}/\text{s}}\\ =220.19\text{s}\end{array}$

Conclusion:

Thus, the time taken by plastic B to reach the top is $220.19\text{s}$ .