## 14 Mechanical Collectors MCC

MCC.l MINIMUM PARTICLE SIZE

A hydrochloric acid mist in air at 25°C is to be collected in a gravity settler. You are requested to calculate the smallest mist droplet (spherical in shape) that will always be collected by the settler. Assume the acid concentration to be uniform through the inlet cross section of the unit and Stokes' law applies. Operating data and information on the gravity settler are given below.

Dimensions of gravity settler = 30 ft wide, 20 ft high, 50 ft long Actual volumetric flowrate of acidic gas = 50ft3/s Specific gravity of acid = 1.6

Viscosity of air = 0.0185 cP= 1.243 x 10~5lb/(ft- s) Density of air = 0.076 lb/ft3

Gravity settlers, or gravity settling chambers, have long been utilized industrially for the removal of solid and liquid waste materials from gaseous streams. Advantages accounting for their use are simple construction, low initial cost and maintenance, low pressure losses, and simple disposal of waste materials. Gravity settlers are usually constructed in the form of a long, horizontal parallelepipeds with suitable inlet and outlet ports. In its simplest form the settler is an enlargement (large box) in the duct carrying the particle-laden gases; the contaminated gas stream enters at one end, the cleaned gas exits from the other end. The particles settle toward the collection surface at the bottom of the unit with a velocity at or near their terminal settling velocity.

Gravity settlers can be designed to ensure capture of some minimum particle size. If Stokes' law applies, one can show that this minimum particle size is given by

Solution where p = gas (fluid) viscosity q — volumetric flowrate g = gravity force pp = particle or mist density B = width of settler L = length of settler

For the intermediate range,

dp = 0A93(gpvf62\BLfM Finally, for Newton's law range,

For the problem at hand, first determine the density of the acid mist:

Calculate the minimum particle diameter both in feet and microns assuming Stokes's law applies:

/(18)(1.243 x lQ-5)(50)\'/2 p V (32.2)(99.84)(30)(50) J

= 4.82 x 10"5 ft There are 3.048 x 105 pm in 1 ft. Therefore, dp = 14.7 pm

The particle diameter calculated above represents a limiting value since particles with diameters equal to or greater than this value will reach the settler collection surface and particles with diameters less than this value may escape from the unit. This limiting particle diameter can ideally be thought of as the minimum diameter of a particle that will automatically be captured for the above conditions. This diameter is denoted by d* or c/p(min).

### MCC.2 GRAVITY SETTLER DESIGN

As a recently hired engineer for an equipment vending company, you have been requested to design a gravity settler to remove all the iron particulates from a dustladen gas stream. The following information is given:

dp = 35 |xm; uniform, i.e., no distribution gas = air at ambient conditions q = 130 ft3/s; throughput velocity, v = lOft/s /9p = 7.62g/cm3

Solution

First convert dp and pp to engineering units:

Pp = (7.62 g/ cm3)(l lb/453.6 g)(28,316 cm3/ft3) = 475.7 lb/ft3

The K value (see Problem, FPD.4) is calculated by

Stokes' law applies and the collection area required can be calculated from the equation

Solving for BL, gPpdj

The cross-sectional area for v= lOft/s is

Based on the minimum required for cleaning purposes, H is usually 3 ft. Then,

L = (142.5)/fi = 142.5/4.33 = 32.9 ft and the total volume of the settler is

The key process design variable for gravity settlers is the capture area, A, which is given by

Once the capture area is calculated, the cost of the gravity settler internals may be assumed fixed. Because of material costs, however, the larger the outer casing of the physical system, the higher will be the total cost, all other factors being equal. These material costs generally constitute a significant fraction of the total cost of the settler. If the thickness of the outer casing is the same for alternate physical designs and if labor costs are linearly related to the surface area, then the equipment cost will roughly be a linear function of the outer surface area of the structural shell of the unit. This essentially means that the cost is approximately linearly related to the perimeter, P, where

To help minimize the cost (by minimizing the perimeter) one can equate the derivative of the perimeter to zero. Thus setting

gives

and dP 2A dL~ "insetting the above derivative equal to zero leads to

so that

since

A=BL

Interestingly, most gravity settlers (as well as electrostatic precipitators) are often designed physically in a form approaching a square box. Thus

BL=L2 = 142.5 ft2 B = L= 11.94ft H = 3ft and, in this case, the velocity of the gas would be

MCC.3 OVERALL COLLECTION EFFICIENCY

A settling chamber is installed in a small heating plant that uses a traveling grate stoker. You are requested to determine the overall collection efficiency of the settling chamber given the following operating conditions, chamber dimensions, and particle size distribution data.

Chamber width = 10.8 ft Chamber height = 2.46 ft

Chamber lengths 15.0ft

Volumetric flowrate of contaminated air stream = 70.6 scfs

Flue gas temperature = 446°F

Flue gas pressure = 1 atm

Particle concentration = 0.23 gr/scf

Particle specific gravity = 2.65

Standard conditions = 32°F, 1 atm

Particle size distribution data of the inlet dust for the traveling grate stoker are given in the following table:

Particle Size Range |
Average Particle diameter |
Q |
Vfj |

(pm) |
(pm) |
(gr/scf) |
(wt %) |

0-20 |
10 |
0.0062 |
2.7 |

20-30 |
25 |
0.0159 |
6.9 |

30-40 |
35 |
0.0216 |
9.4 |

40-50 |
45 |
0.0242 |
10.5 |

50-60 |
55 |
0.0242 |
10.5 |

60-70 |
65 |
0.0218 |
9.5 |

70-80 |
75 |
0.0161 |
7.0 |

80-94 |
87 |
0.0218 |
9.5 |

+ 94 |
+ 94 |
0.0782 |
34.0 |

0.2300 |
100.0 |

Assume that the actual terminal settling velocity is one-half of the Stokes law velocity.

Solution

The collection efficiency, E, for a monodispersed aerosol (particulates of one size) in laminar flow can be shown to be

q where vt is the terminal settling velocity of the particle.

The validity of this equation is observed by noting that vtBL represents the hypothetical volume rate of flow of gas passing the collection area, while q is the volumetric flowrate of gas entering the unit to be treated. An alternate but equivalent form of the above equation is where H is the height of the settling chamber and v is the gas velocity.

If the gas stream entering the unit consists of a distribution of particles of various sizes, then frequently a fractional or grade efficiency curve is specified for the settler. This is simply a curve describing the collection efficiency for particles of various sizes. The dependency of E on dp arises because of the vt term in the above equations.

Since the actual terminal settling velocity is assumed to be one-half of the Stokes' law velocity (according to the problem statement),

2 18 n and therefore

E _ gdjPpBL 36 pq

H (446°F) = 1.75 x 10~5 lb/(ft • s) The particle density in lb/ft3 is pp = (2.65)(62.4) = 165.4 lb/ft3

To calculate the collection efficiency of the system at the operating conditions, the standard volumetric flowrate of contaminated air of 70.6 scfs is converted to the actual volumetric flow using Charles' law:

The collection efficiency in terms of dp, with dp in microns is given below. Note: To convert dp from ft to pm, dp is divided by (304,800) pm/ft:

E = VjBL =gppBL^ q 3 6fiq (32.2)(165.4)(10.8)(15) d2p ~ (36)(1.75 x 10-5)(130)(304,800)2 = 1.14 x 10"4 d\

where dp is in microns.

MCC.3 OVERALL COLLECTION EFFICIENCY 291 Thus, for a particle diameter of 10 [xm,

E= 1.14 x 10 ~A d2p = (1.14 x 10"4)(10)2 = 1.1 x 10"2 = 1.1%

The following table provides the collection efficiency for each particle size:

dp (urn) |
d2p (urn2) |
E (%) |

93.8 |
8800 |
100.0 |

90 |
8100 |
92.0 |

80 |
6400 |
73.0 |

60 |
3600 |
41.0 |

40 |
1600 |
18.2 |

20 |
400 |
4.6 |

10 |
100 |
1.1 |

The size efficiency curve for the settling chamber is shown in Figure 65.

The collection efficiency of each particle size may be read from the size-efficiency curve. The product, w^, is then calculated for each size. The overall efficiency is equal to 'LwiE[. These calculations are provided in the following table.

Average dp (|rm) |
Weight Fraction vv. |
Ei (%) |
WjSj (%) |

10 |
0.027 |
1.1 |
0.030 |

25 |
0.069 |
7.1 |
0.490 |

35 |
0.094 |
14.0 |
1.316 |

45 |
0.105 |
23.0 |
2.415 |

55 |
0.105 |
34.0 |
3.570 |

65 |
0.095 |
48.0 |
4.560 |

75 |
0.070 |
64.0 |
4.480 |

87 |
0.095 |
83.0 |
7.885 |

+94 |
0.340 |
100.0 |
34.000 |

Total |
1.000 |
58.7 |

The overall collection efficiency for the settling chamber, E, is 58.7%.

MCC.4 COMPLIANCE CALCULATION FOR A GRAVITY SETTLER

A gravity settler is 15 ft wide by 15 ft high by 40 ft long. In order to meet required ambient air quality standards, this unit must remove 90% of the fly ash particles entering the unit. Planned expansion will increase the flowrate to 4000 acfm with a dust loading of 30 gr/ft3. The specific gravity of fly ash is 2.31 and the process gas stream is air at 20°C and 1.0 atm. The inlet size distribution of the fly ash is given below:

Size range (|rm) |
Mass Percent |

0.0-10.0 |
1.0 |

10-20 |
1.0 |

20-30 |
3.0 |

30-40 |
15.0 |

40-60 |
20.0 |

60-80 |
25.0 |

80-100 |
20.0 |

100-150 |
15.0 |

Will the unit meet the specification?

MCC.4 COMPLIANCE CALCULATION FOR A GRAVITY SETTLER 293

The following data are provided:

5= 15ft

Loading = 30 gr/ft3

Specific gravity = 2.31; pp = 144 lb/ft3

Solution

The throughput velocity, u, is u = q/BH = 4000/[(15)(15)] = 17.778 ft/min = 0.296 ft/ s The fractional efficiency equation may be written as

= / (32.2X144) \ / (40)(15) \ 2 \(18)(1.23 x 10"5)/ V(4000/60)/ P

The following table may now be generated:

Size Range (|_im) |
Average Particle Size (pm) |
Mass Fraction w. |
E, (%) |
(%) |

0.0-10.0 |
5 |
0.01 |
5.1 |
0.051 |

10-20 |
15 |
0.01 |
45.45 |
0.455 |

20-30 |
25 |
0.03 |
100 |
3.0 |

30-40 |
35 |
0.15 |
100 |
15 |

40-60 |
50 |
0.20 |
100 |
20 |

60-80 |
70 |
0.25 |
100 |
25 |

80-100 |
90 |
0.20 |
100 |
20 |

100-150 |
125 |
0.15 |
100 |
15 |

£ = |
98.51 |

The overall efficiency is 98.51%. As expected, the efficiency is high because of the coarse dust and the size of the unit. The reader should also check to ensure that

Stokes' law does in fact apply to those size ranges where the efficiency is less than 100%.

### MCC.5 CHECK ON EFFICIENCY OF A GRAVITY SETTLER

A salesman from Bogus, Inc. suggests a gravity settler for a charcoal dust-contaminated air stream that you must preclean. Your supervisor has provided the particle size distribution shown below. The inlet loading is 20.00 gr/ft3 and the required outlet loading is 5.00 gr/ft3. Will the settler that the salesman has suggested do the job?

Size Range (pm) |
Weight Percent |

0-10 |
5 |

10-20 |
11 |

20-40 |
10 |

40-60 |
9 |

60-90 |
22 |

90-125 |
23 |

125-150 |
10 |

150+ |
10 |

Use the critical diameter (d* = 80 pm) to calculate the size-efficiency data from the equation

Solution

First calculate k:

Thus, for dp = 5 pm (average diameter for the first size range),

The following table may be generated using the above approach:

Size Range (nm) |
Average Particle Size (nm) |
Weight Percent |
Ei (%) |
W;Ei (%) |

0-10 |
5 |
5 |
0.39 |
— |

10-20 |
15 |
11 |
3.5 |
0.4 |

20-40 |
30 |
10 |
14 |
1.4 |

40-60 |
50 |
9 |
39 |
3.5 |

60-90 |
75 |
22 |
88 |
19.4 |

90-125 |
107.5 |
23 |
100 |
23 |

125-150 |
137.5 |
10 |
100 |
10 |

150+ |
130+ |
10 |
100 |
10 |

E = 67.7 |

The required efficiency, EKg, is

= (20 - 5)/20 = 75% Since 67.7 < 75%, the gravity settler will not do the job.

### MCC.6 CYCLONES IN SERIES

As a graduate student you have been assigned the task of studying certain process factors in an operation that employs three cyclones in series to treat catalyst-laden gas at 25°C and 1 atm. The inlet loading to the cyclone series is 8.24 gr/ft3 and the volumetric flowrate is 1,000,000 acfm. The efficiency of the cyclones are 93, 84, and 73%, respectively. Calculate the following:

1. Daily mass of catalyst collected (lb/day)

2. Daily mass of catalyst discharged to the atmosphere

3. Whether or not it would be economical to add an additional cyclone (efficiency = 52%) costing an additional $300,000 per year (The cost of the catalyst is $0.75 per pound.)

4. Outlet loading from the proposed fourth cyclone

Solution

The mass entering, m{, is

_ /106 ft3\ /60 min\ /24h\ /8.24gr\ / 1 lb \ ~ V min ) V h J [day) V ft3 J 1,7000grJ

The mass collected, mc, is mc = 1,695,086 (0.93) + 0.84 [1,695,086 (1 - 0.93)]

+ 0.73 [1,695,086 (1 - 0.93)(1 - 0.84)] = 1,689,960 lb/day Thus the mass discharge, md, is md = 1,689,960- 1,695,086 = 5,126 lb/day With a fourth cyclone, the additional mass collected, mA, is m4 = 5126 (0.52) = 26661b/day and 5126 — 2666 = 2460 lb/day is discharged. The savings S is

Since the cyclone costs $300,000 annually, purchase it. The outlet loading (OL) is

OL = (2460 lb/day)(l day/24 h)(l h/60min)(l min/106ft3)(7000gr/lb) = 0.012 gr/ft3

MCC.7 PARTICLE SIZE DEPOSITION FROM A MALFUNCTIONING CYCLONE

A cyclone on a cement plant suddenly malfunctions. By the time the plant shuts down, some dust has accumulated on parked cars and other buildings in the plant complex. The nearest affected area is 700 ft from the cyclone location, and the furthest affected area measurable on plant grounds is 2500 ft from the cyclone. What is the particle size range of the dust that has landed on plant grounds?

On this day, the cyclone was discharging into a 6-mph wind. The specific gravity of the cement is 1.96. The cyclone is located 175 ft above the ground. Neglect effects of turbulence.

Solution

A diagram representing the system is provided in Figure 66 (S = smaller, L = larger). For air at ambient conditions, p = 0.0741 lb/ft3

MCC.7 PARTICLE SIZE DEPOSITION FROM A MALFUNCTIONING CYCLONE

MCC.7 PARTICLE SIZE DEPOSITION FROM A MALFUNCTIONING CYCLONE

Particle traveling times are given by s — ut where 5 = horizontal distance traveled u = horizontal velocity t = travel time

For the smaller particle, ts = 2500/31,680 = 0.07891 h and for the larger particle, tL = 700/31,680 = 0.02210 h Settling velocities may now be calculated from v = H/t where v = vertical velocity

H = vertical distance traveled fg = 175/[(0.07891 )(3600)] = 0.6161 ft/s vL = 175/[(0.02210)(3600)] = 2.200 ft/s

To calculate dp, assume Stokes' law to apply. For the smaller particle size,

_ /(18)(1.23 x 10~5)(0.6161)\05 ~ V (32.2X1.96X62.4) )

Checking the value of K,

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