Dimensions of a Standard Cyclone

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DIMENSIONS OF A STANDARD CYCLONE

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Dimensions of a Standard Cyclone. Problem. Design of Cyclone Body diameter = 0.75 m Flow rate = 2.75 m 3 /s ρ p = 1600 kg/m 3 ρ g = 1.1 kg/m 3 µ = 2.5*10-5 kg/m-s High throughput H = 0.8 * body diameter W = 0.35 * body diameter Lb = 1.7 * body diameter Lc = 2.0 * body diameter - PowerPoint PPT Presentation

Transcript of Dimensions of a Standard Cyclone

Page 1: Dimensions of a Standard Cyclone

DIMENSIONS OF A STANDARD CYCLONE

Page 2: Dimensions of a Standard Cyclone

PROBLEM Design of Cyclone

Body diameter = 0.75 mFlow rate = 2.75 m3/sρp = 1600 kg/m3

ρg = 1.1 kg/m3

µ = 2.5*10-5 kg/m-sHigh throughputH = 0.8 * body diameterW = 0.35 * body diameterLb = 1.7 * body diameterLc = 2.0 * body diameterCalculate the diameter of particle with 50% efficiency

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SOLUTION Step 1:Inlet Velocity = Q/A

= 17.46 m/s Step 2:Number of effective turns

Ne = (Lb + Lc/2)/H

Ne = 3.375 Step 3: Diameter of particle

dpc = 10 µm

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SCRUBBER Efficiency

where,k = Scrubber coefficient (m3 of gas/ m3 of liquid)R = Liquid-to-gas flow rate (QL/QG)ψ (Psi)= internal impaction parameter

Internal impaction parameter

where,c = Cunningham correction factorρp = particle density (kg/m3)Vg = speed of gas at throat (m/sec)dp = diameter of particle (m)dd = diameter of droplet (m)μ = dynamic viscosity of gas, (Pa-S)

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VENTURI SCRUBBERAbsolute Pressure Drop

Δp = pressure drop ( cm of water)

ug = gas velocity (cm/s)

Qt = liquid volume flow rate

Qg = gas volume flow rate

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PROBLEM Water is introduced into the throat of a venturi scrubber.

The air velocity through the scrubber is 550 fps and the liquid to gas ratio is 8.5 gal/1000 actual ft3. Determine the pressure drop?

SOLUTION Step 1:Absolute pressure drop is given by

Δp = 4.8

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Table Operating characteristics of venturi scrubbers

Pollutant

Pressure drop (Δp)

Liquid-to-gas ratio (L/G)

Liquid-inlet pressure (pL)

Removal efficiency

Gases13-250 cm of

water2.7-5.3 l/m3

< 7-100 kPa

30-60% per venturi,

depending on pollutant solubility

Particles

50-250 cm of water (50-150 cm of water is

common)0.67-1.34 l/m3 90-99% is typical

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DESIGN OF ELECTROSTATIC PRECIPITATORS The efficiency of removal of particles by an Electrostatic

Precipitator is given by

η = fractional collection efficiency

w = drift velocity, m/min.A = available collection area, m2

Q = volumetric flow rate m3/min

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MIGRATION VELOCITY

Where, q = charge (Columbus)

Ep = collection field intensity (volts/m)r = particle radius (m)μ = dynamic viscosity of gas (Pa-S)c = Cunningham correction factor

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Cunningham correction factor

where,T = absolute temperature (°k)dp = diameter of particle (μm)

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PROBLEM An ESP is designed to treat 50,000 m3/min with 97 %

efficiency. Assuming an effective drift velocity of 2.5 m/min, calculate the required plate area and the number of plates. The plate size is 10 m by 5 m (height by length).

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SOLUTION Step 1:Efficiency of an Electrostatic Precipitator is given by

A =-[ (Q/w)*ln(1- η)]A = 70,000 m2

Step 2:Number of plates = total area/plate area

= 1400

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PROBLEM Estimate the net cloth area for a shaker bag house that must

filter 40,000 cfm of air with 10 grams of flour dust per cubic

foot of air. Also specify the number of components to be used

and calculate the total number of bags required if each bag is 8

feet long and 0.5 feet in diameter. The maximum filtering

velocity for flour dust is 2.5 ft/min.

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SOLUTION Step 1: Calculate total area and number of components required.

A = Q/V Step 2:Calculate the area of each bag.

A = Π (d) l Step 3:Calculate the total number of bags required. Number of bags required = Total area / Area per bag

= 1270 bagsNumber of compartments 4000 sq. ft. / compartment