Chapter 4, Problem 4.11P

To determine

(a)

The characteristic size of the product.

Answer to Problem 4.11P

The characteristic size of the product is $2.08\text{in}$ .

Explanation of Solution

Calculation:

Draw the table to calculate the corresponding $Y$ value for use in the Rosin Rammler equation from the sieve analysis result and the linearized $Y$ values.

Sieve Size (inches)$\left(a\right)$	Percent by weightFiner than Feed$\left(Y\right)$$\left(b\right)$	$X=\ln \left(a\right)$	Linearised $Y$$f=\log \left(\ln \left(\frac{1}{1-Y}\right)\right)$
$4$	$80\%$	$1.39$	$0.21$
$2$	$15\%$	$0.69$	$-0.79$
$1$	$5\%$	$0$	$-1.29$
$0.5$	$0\%$	$-0.69$	$0$

The slope calculated by the slope function in the excel sheet is,

$n=0.165$

The intercept calculated by the intercept function in the excel sheet is,

$i=-0.525$

Write the expression to calculate the characteristic size.

$x_0=e^{\left(-\frac{i}{n}\right)}$ ...... (I)

Here, the characteristic size is $x_0$ , the intercept is $i$ and the slope is $n$ .

Substitute $-0.525$ for $i$ and $0.165$ for $n$ in Equation (I).

$\begin{array}{c}{x}_0=e^{\left(-\frac{-0.525}{0.165}\right)}\\ =e^{3.182}\\ =24\text{in}\end{array}$

Calculate the characteristic size for product.

Sieve Size (inches)$\left(a\right)$	Percent by weightFiner than Product$\left(Y\right)$$\left(b\right)$	$X=\ln \left(a\right)$	Linearised $Y$$f=\log \left(\ln \left(\frac{1}{1-Y}\right)\right)$
$4$	$95\%$	$1.39$	$0.476$
$2$	$65\%$	$0.69$	$0.02$
$1$	$25\%$	$0$	$-0.54$
$0.5$	$10\%$	$-0.69$	$-0.98$

The slope calculated by the slope function in the excel sheet is,

$n=0.70$

The intercept calculated by intercept function in the excel sheet is,

$i=-0.511$

Calculate the characteristic size for the product.

Substitute $-0.511$ for $i$ and $0.70$ for $n$ in Equation (I).

$\begin{array}{c}{x}_0=e^{\left(-\frac{-0.511}{0.70}\right)}\\ =e^{0.73}\\ =2.08\text{in}\end{array}$

Conclusion:

Thus, the characteristic size of the feed is $24\text{in}$ .

The characteristic size of the product is $2.08\text{in}$ .

To determine

(b)

If both the distributions fit the Rosin-Rammler particle size distribution function.

Answer to Problem 4.11P

Only the product distribution fits the Rosin-Rammler distribution function.

Explanation of Solution

Only the distribution for the product fits the Rosin-Rammler distribution function as in the feed distribution the characteristic size is of very high order of $2\text{ft}$ which is not possible.

Thus, only the product distribution fits the Rosin-Rammler distribution function.

To determine

(c)

The diagram for the particle size distribution curve for the given cases.

Answer to Problem 4.11P

1. The feed becomes wetter (higher moisture content).

The particle size distribution curve is shown below.

    

The shredder is run at a higher speed.

The particle size distribution curve is shown below.

    

Explanation of Solution

The feed becomes wetter (higher moisture content).

    

    Figure (1)

When the moisture content is increase the product size also increases considerably.

The shredder is run at a higher speed.

    

    Figure (2)

When the shredder speed is increased the product size decreases considerably.

To determine

(d)

The effective power requirement.

Answer to Problem 4.11P

The effective power requirement is $4.445\text{kWh}/\text{ton}$ .

Explanation of Solution

Calculation:

Write the expression to calculate the effective power requirement.

$E=\frac{10\times E_1}{{\left(x_0{\left(1.61\right)}^{\left(\frac{1}{\text{n}}\right)}\right)}^{\frac{1}{2}}}-\frac{10E_1}{{\left(L_{\text{F}}\right)}^{\frac{1}{2}}}$ ...... (II)

Here, the bond work index is $E_1$ , the effective power requirement is $E$ and the diameter of particle in micrometer is $L_{\text{F}}$ .

Substitute $10\text{in}$ for $L_{\text{F}}$ , $400$ for $E_1$ , $0.70$ for $n$ and $2.08\text{in}$ for $x_0$ in Equation (II).

$\begin{array}{c}E=\left[\frac{10\times 400}{{\left(\left(2.08\text{in}\right)\left(\frac{25400\text{μm}}{1\text{in}}\right)\times {\left(1.61\right)}^{\left(\frac{1}{0.70}\right)}\right)}^{\frac{1}{2}}}-\frac{10\times 400}{{\left(\left(10\text{in}\right)\left(\frac{25400\text{μm}}{1\text{in}}\right)\right)}^{\frac{1}{2}}}\right]\text{kWh}/\text{ton}\\ =\left(12.385-7.94\right)\text{kWh}/\text{ton}\\ =4.445\text{kWh}/\text{ton}\end{array}$

Conclusion:

Thus, the effective power requirement is $4.445\text{kWh}/\text{ton}$ .