Chapter 5, Problem 5.11P

To determine

(a)

The critical speed of the trommel screen.

Answer to Problem 5.11P

  $27.3\text{rpm}$ .

Explanation of Solution

Given:

The diameter of trommel screen is $2.4\text{m}$ .

The length of trommel screen is $7.6\text{m}$ .

The angle of inclination is $15°$ .

Concept Used:

Write theexpression to calculate the critical speed of trommel screen.

  $n_c=\sqrt{\frac{g}{4{\text{π}}^{2}r}}$  ..... (I)

Here, the acceleration due to gravity is $g$ , the radius of trommel is $r$ and the critical speed of trommel screen is $n_c$ .

Calculation:

Calculate the critical speed of trommel screen.

Substitute $980\text{cm}/{\text{s}}^{\text{2}}$ for $g$ and $120\text{cm}$ for $r$ in Equation (I).

  $\begin{array}{c}{n}_c=\sqrt{\frac{980\text{cm}/{\text{s}}^{\text{2}}}{4{\left(\text{π}\right)}^2\left(120\text{cm}\right)}}\\ =\left(0.45\text{rotation}/\text{sec}\right)\left(\frac{60\text{rpm}}{1\text{rotation}/\text{sec}}\right)\\ =27.3\text{rpm}\end{array}$

Conclusion:

Thus, the critical speed of trommel screen is $27.3\text{rpm}$ .

To determine

(b)

The type of movement of MSW in the trommel screen.

Answer to Problem 5.11P

The movement of MSW in trommel screen is cataracting.

Explanation of Solution

Given:

The speed of trommelis $4\text{rpm}$ .

The feed rate is $10\text{tons}/\text{h}$ .

Calculation:

The movement of MSW is determined by using the relation $n<n_c$ .

Here, $n=4\text{rpm}$ , $n_{}=27.3\text{rpm}$ .

  $n<n_c$

  $4<27.3$

Hence the movement of MSW is either cascading or cataracting.

The critical speed and the fraction of the screen occupied by the refuse are related which gives the process of movement of MSW.

Write the equation to calculate the bulk volume.

  $F=\frac{S}{V}$   ..... (II).

Here, the bulk volume of trommel screen is $F$ , volume occupied by the solids and the air spaces between the solids is $S$ and total volume inside the trommel screen is $V$ .

Consider a single particle in the screen.

The following figure shows the force analysis diagram.

  

  Figure-(1)

Here, gravity is $G$ , supporting force is $M$ , friction force is $F_f$ , acceleration is $a$ and the inclined angle is $\beta$ .

Write the equilibrium equation for the x and y axis as shown in Figure-(1).

  $\begin{array}{l}M-mg\cos \beta =0\\ mg\sin \beta -F_f=ma\\ F_f=\mu M\end{array}$

Here, the coefficient of friction is $\mu$ .

Write the equation to calculate acceleration.

  $a=g\sin \beta -\mu \cos \beta$  ..... (III)

Here, the acceleration is $a$ .

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $15°$ for $\beta$ , 0.2 for $\mu$ in Equation (III).

  $\begin{array}{c}a=\left(9.81\text{m}/{\text{s}}^{\text{2}}\times \sin 15°\right)-\left(0.2\times 9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \cos 15°\\ =2.539\text{m}/{\text{s}}^{\text{2}}-1.895\text{m}/{\text{s}}^{\text{2}}\\ =0.643\text{m}/{\text{s}}^{\text{2}}\end{array}$

Calculate time for particle to move from end to end.

  $t=\sqrt{\frac{2L}{a}}$  ..... (IV).

Here, the length is $L$ and the time is $t$ .

Substitute $7.6\text{m}$ for $L$ , $0.643\text{m}/{\text{s}}^{\text{2}}$ for $a$ in Equation (IV).

  $\begin{array}{c}t=\sqrt{\frac{2\times 7.6\text{m}}{0.643\text{m}/{\text{s}}^{\text{2}}}}\\ =\sqrt{23.63\text{1}/{\text{s}}^{\text{2}}}\\ =4.86\text{s}\end{array}$

Calculate the volume occupied by the solids and the air spaces between the solids.

  $S=\frac{Q_m}{\rho }\times t$  ..... (V).

Here, the mass flow rate is $Q_m$ , the density is $\rho$ .

Convert the unit of $Q_m$ to $\text{kg}/\text{h}$ .

  $\begin{array}{c}10\text{tons}/\text{h}=\left(10\text{tons}/\text{h}\right)\left(\frac{1000\text{kg}}{1\text{ton}}\right)\\ =1\times {10}^4\text{kg}/\text{h}\end{array}$

Convert the unit of time to $\text{h}$ .

  $\begin{array}{c}4.86\text{s}=\left(4.86\text{s}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\\ =1.35\times {10}^{-3}\text{h}\end{array}$

Substitute $1\times {10}^4\text{kg}/\text{h}$ for $Q_m$ , $1.35\times {10}^{-3}\text{h}$ for $t$ and $120\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in Equation (V).

  $\begin{array}{c}S=\left(\frac{1\times {10}^4\text{kg}/\text{h}}{120\text{kg}/{\text{m}}^{\text{3}}}\right)\times \left(1.35\times {10}^{-3}\text{h}\right)\\ =\left(83.33{\text{m}}^3/\text{h}\right)\times \left(1.35\times {10}^{-3}\text{h}\right)\\ =0.1125{\text{m}}^{\text{3}}\end{array}$

Calculate the total volume inside the trommel screen.

  $V=\text{π}{r}^{2}h$  ..... (VI).

Substitute $1.2\text{m}$ for $r$ and $7.6\text{m}$ for $h$ in Equation (VI).

  $\begin{array}{c}V=\text{π}{\left(1.2\text{m}\right)}^{2}7.6\text{m}\\ =34.38{\text{m}}^3\end{array}$

Calculate the bulk volume of trommel screen.

Substitute $34.38{\text{m}}^3$ for $V$ and $0.1125{\text{m}}^{\text{3}}$ for $S$ in Equation (II).

  $\begin{array}{l}F=\frac{0.1125{\text{m}}^{\text{3}}}{34.38{\text{m}}^3}\\ F=3.27\times {10}^{-3}\end{array}$

The following graph shows critical speed and the fraction of the screen occupied by the refuse are related which gives the process of movement of MSW.

  

  Figure-(2)

Since, $F$ is small the movement of particle is cataracting.

Conclusion:

The type of movement is calculated by equating time, bulk volume, mass flow, density, speed and feed rate.