Solid Waste Engineering
Solid Waste Engineering
3rd Edition
ISBN: 9781305635203
Author: Worrell, William A.
Publisher: Cengage Learning,
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Chapter 5, Problem 5.6P
To determine

The diameter of cyclone at various air flow rates.

Expert Solution & Answer
Check Mark

Answer to Problem 5.6P

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

    Terminal velocity(m/sec) Diameter of the Cyclone
    5 2153m
    10 748m
    15 411m
    25 199m
    40 105m
    150 21m
    300 10.54m

Explanation of Solution

Calculation:

Write the expression for pressure changes in the cyclone.

  dPdr=ρ×Vt2r   ...... (I)

Here, the change of pressure is dP , the change of rotation radius is dr , the tangential velocity is Vt , the air density is ρ and the rotational radius is r .

Calculate the tangential velocity at radius r .

  Vt=ωr

Here, the rotational velocity is ω .

Write the expression for pressure drop in the cyclone.

  ΔP=ρ×Vt×ln(R)   ...... (II)

Substitute ωr for Vt in Equation (II).

  ΔP=ρ×ωR×ln(R)ω=ΔPρR×ln(R)   ...... (III)

Write the expression for the terminal velocity.

  v=[4( ρ g ρ) ω 2Rd3 C Dρ]12 ..... (IV)

Substitute ΔPρ×ln(R) for ω in Equation (IV).

  v=[ 4( ρ g ρ ) ( ΔP ρ D 2 ×ln( D 2 ) ) 2 Rd 3 C D ρ]12=[ 8( ρ g ρ ) ( ΔP ) 2 Rd 3 C D ρ 3 D ( ln( D 2 ) ) 2 ]12 ..... (V)

Here, the diameter of the cyclone is D .

Convert 1in into metric system.

  1in=0.0254m

Substitute 2700kg/m3 for ρg , 1.2kg/m3 for ρ , 249Pa for ΔP , 0.0254m for d and 2.5 for CD in Equation (V).

  v=[ 8( 2700 kg/ m 3 1.2 kg/ m 3 ) ( 249Pa ) 2 0.0254m 3×2.5× ( 1.2 kg/ m 3 ) 3 ×D ( ln( D 2 ) ) 2 ]12v=( 2623542.5 D ( ln( D 2 ) ) 2 )12   ...... (VI)

Calculate the diameter at an air flow rate of 5m/sec .

Substitute 5m/sec for v in Equation (VI).

  5m/sec=( 2623542.5 D ( ln( D 2 ) ) 2 )12

Solve for the value of D .

  D=2153m

Calculate the diameter at an air flow rate of 10m/sec .

Substitute 10m/sec for v in Equation (VI).

  10m/sec=( 2623542.5 D ( ln( D 2 ) ) 2 )12

Solve for the value of D .

  D=748m .

Calculate the diameter at an air flow rate of 15m/sec .

Substitute 15m/sec for v in Equation (VI).

  15m/sec=( 2623542.5 D ( ln( D 2 ) ) 2 )12

Solve for the value of D .

  D=411m

Calculate the diameter at an air flow rate of 25m/sec .

Substitute 25m/sec for v in Equation (VI).

  25m/sec=( 2623542.5 D ( ln( D 2 ) ) 2 )12

Solve for the value of D .

  D=199m

Calculate the diameter at an air flow rate of 40m/sec .

Substitute 40m/sec for v in Equation (VI).

  40m/sec=( 2623542.5 D ( ln( D 2 ) ) 2 )12

Solve for the value of D .

  D=105m

Calculate the diameter at an air flow rate of 150m/sec .

Substitute 150m/sec for v in Equation (VI).

  150m/sec=( 2623542.5 D ( ln( D 2 ) ) 2 )12

Solve for the value of D .

  D=21m

Calculate the diameter at an air flow rate of 300m/sec .

Substitute 300m/sec for v in Equation (VI).

  300m/sec=( 2623542.5 D ( ln( D 2 ) ) 2 )12

Solve for the value of D .

  D=10.54m

Classify the calculated data.

    Terminal velocity(m/sec) Diameter of the Cyclone
    5 2153m
    10 748m
    15 411m
    25 199m
    40 105m
    150 21m
    300 10.54m

Conclusion:

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

    Terminal velocity(m/sec) Diameter of the Cyclone
    5 2153m
    10 748m
    15 411m
    25 199m
    40 105m
    150 21m
    300 10.54m

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