Chapter 5, Problem 5.6P

To determine

The diameter of cyclone at various air flow rates.

Answer to Problem 5.6P

The diameter of cyclone at different air flow rate is classified into a table below.

Terminal velocity$\left(\text{m}/\text{sec}\right)$	Diameter of the Cyclone
$5$	$2153\text{m}$
$10$	$748\text{m}$
$15$	$411\text{m}$
$25$	$199\text{m}$
$40$	$105\text{m}$
$150$	$21\text{m}$
$300$	$10.54\text{m}$

Explanation of Solution

Calculation:

Write the expression for pressure changes in the cyclone.

  $\frac{dP}{dr}=\rho \times \frac{V_t{}^2}{r}$   ...... (I)

Here, the change of pressure is $dP$ , the change of rotation radius is $dr$ , the tangential velocity is $V_t$ , the air density is $\rho$ and the rotational radius is $r$ .

Calculate the tangential velocity at radius $r$ .

  $V_t=\omega r$

Here, the rotational velocity is $\omega$ .

Write the expression for pressure drop in the cyclone.

  $\Delta P=\rho \times V_t\times \ln \left(R\right)$   ...... (II)

Substitute $\omega r$ for $V_t$ in Equation (II).

  $\begin{array}{l}\Delta P=\rho \times \omega R\times \ln \left(R\right)\\ \omega =\frac{\Delta P}{\rho R\times \ln \left(R\right)}\end{array}$   ...... (III)

Write the expression for the terminal velocity.

  $v={\left[\frac{4\left({\rho }_g-\rho \right){\omega }^{2}Rd}{3C_D\rho }\right]}^{\frac{1}{2}}$ ..... (IV)

Substitute $\frac{\Delta P}{\rho \times \ln \left(R\right)}$ for $\omega$ in Equation (IV).

  $\begin{array}{c}v={\left[\frac{4\left({\rho }_g-\rho \right){\left(\frac{\Delta P}{\rho \frac{D}{2}\times \ln \left(\frac{D}{2}\right)}\right)}^{2}Rd}{3C_D\rho }\right]}^{\frac{1}{2}}\\ ={\left[\frac{8\left({\rho }_g-\rho \right){\left(\Delta P\right)}^{2}Rd}{3C_D{\rho }^{3}D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right]}^{\frac{1}{2}}\end{array}$ ..... (V)

Here, the diameter of the cyclone is $D$ .

Convert $1\text{in}$ into metric system.

  $1\text{in}=0.0254\text{m}$

Substitute $2700\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_g$ , $1.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ , $249\text{Pa}$ for $\Delta P$ , $0.0254\text{m}$ for $d$ and $2.5$ for $C_D$ in Equation (V).

  $\begin{array}{l}v={\left[\frac{8\left(2700\text{kg}/{\text{m}}^{\text{3}}-1.2\text{kg}/{\text{m}}^{\text{3}}\right){\left(249\text{Pa}\right)}^{2}0.0254\text{m}}{3\times 2.5\times {\left(1.2\text{kg}/{\text{m}}^{\text{3}}\right)}^3\times D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right]}^{\frac{1}{2}}\\ v={\left(\frac{2623542.5}{D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right)}^{\frac{1}{2}}\end{array}$   ...... (VI)

Calculate the diameter at an air flow rate of $5\text{m}/\text{sec}$ .

Substitute $5\text{m}/\text{sec}$ for $v$ in Equation (VI).

  $5\text{m}/\text{sec}={\left(\frac{2623542.5}{D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right)}^{\frac{1}{2}}$

Solve for the value of $D$ .

  $D=2153\text{m}$

Calculate the diameter at an air flow rate of $10\text{m}/\text{sec}$ .

Substitute $10\text{m}/\text{sec}$ for $v$ in Equation (VI).

  $10\text{m}/\text{sec}={\left(\frac{2623542.5}{D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right)}^{\frac{1}{2}}$

Solve for the value of $D$ .

  $D=748\text{m}$ .

Calculate the diameter at an air flow rate of $15\text{m}/\text{sec}$ .

Substitute $15\text{m}/\text{sec}$ for $v$ in Equation (VI).

  $15\text{m}/\text{sec}={\left(\frac{2623542.5}{D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right)}^{\frac{1}{2}}$

Solve for the value of $D$ .

  $D=411\text{m}$

Calculate the diameter at an air flow rate of $25\text{m}/\text{sec}$ .

Substitute $25\text{m}/\text{sec}$ for $v$ in Equation (VI).

  $25\text{m}/\text{sec}={\left(\frac{2623542.5}{D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right)}^{\frac{1}{2}}$

Solve for the value of $D$ .

  $D=199\text{m}$

Calculate the diameter at an air flow rate of $40\text{m}/\text{sec}$ .

Substitute $40\text{m}/\text{sec}$ for $v$ in Equation (VI).

  $40\text{m}/\text{sec}={\left(\frac{2623542.5}{D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right)}^{\frac{1}{2}}$

Solve for the value of $D$ .

  $D=105\text{m}$

Calculate the diameter at an air flow rate of $150\text{m}/\text{sec}$ .

Substitute $150\text{m}/\text{sec}$ for $v$ in Equation (VI).

  $150\text{m}/\text{sec}={\left(\frac{2623542.5}{D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right)}^{\frac{1}{2}}$

Solve for the value of $D$ .

  $D=21\text{m}$

Calculate the diameter at an air flow rate of $300\text{m}/\text{sec}$ .

Substitute $300\text{m}/\text{sec}$ for $v$ in Equation (VI).

  $300\text{m}/\text{sec}={\left(\frac{2623542.5}{D{\left(\ln \left(\frac{D}{2}\right)\right)}^2}\right)}^{\frac{1}{2}}$

Solve for the value of $D$ .

  $D=10.54\text{m}$

Classify the calculated data.

Terminal velocity$\left(\text{m}/\text{sec}\right)$	Diameter of the Cyclone
$5$	$2153\text{m}$
$10$	$748\text{m}$
$15$	$411\text{m}$
$25$	$199\text{m}$
$40$	$105\text{m}$
$150$	$21\text{m}$
$300$	$10.54\text{m}$

Conclusion:

Thus, the diameter at different air flow rate is classified into a table below.

Terminal velocity$\left(\text{m}/\text{sec}\right)$	Diameter of the Cyclone
$5$	$2153\text{m}$
$10$	$748\text{m}$
$15$	$411\text{m}$
$25$	$199\text{m}$
$40$	$105\text{m}$
$150$	$21\text{m}$
$300$	$10.54\text{m}$