p.1 

OAR Box 1870 

Prepped by Ollie Stewart 

Document Number: 

22) ll-A-34 

Docket Number: 

A-90-19 

Printed 11/28/2008 7:05:04 AM Sheet 22 of 67 
P.2 

A-90-19 

ll-A-34 

4»a 

1 

8? 

EJUULIl 

EPA/625/6-86/014 

September 1986 

n 

J l l RFC - 9 fedZ j»JV IS1 

gPA AIR DOCKET 

1 

Handbook 

Control Technologies for 

Hazardous Air Pollutants 

•S3-

iS 

i_a. 

mihX 

Air and Energy Engineering Research Laboratory 

Office of Research and Development 

U.S. Environmental Protection Agency 

Research Triangle Park, NC 27711 

Center for Environmental Research Information 

Office of Research and Development 

U.S. Environmental Protection Agency 

Cincinnati, OH 45268 

wmm^srms ."<?«• 
P.3 

Example Case 

Using Equations 4.7-2, -3, and -5: 

m = 1.3 

(for the operating conditions in the system, consult reference 21) 

Qe 

Gmol 

Gmol 

L-mol 

Lgal 

Lgal 

= 3,000 scfm 

= 0.155x3,000x60 

= 465 Ib-moles/hr 

= 1.6 x 1.3x465 

= 970 Ib-moles/hr 

= 0.036 x 970 

= 35 gal/min 

Figure 4-16. Correlation for flooding rate in randomly packed 

towers.! 12) 

4.7.3.2 Column Diameter 

Once the gas and liquid streams entering and leaving the absorber column and their concentrations 

are identified, flow rates calculated, and operating 

conditions (type of packing) determined, the physical dimensions of the column can be calculated. 

The column must be of sufficient diameter to accommodate the gas and liquid streams. 

The calculation of the column diameter is based on 

flooding considerations, the usual operating range 

being taken as 60 to 75 percent of the flooding rate. 

One of the commonly used correlations in determining the column diameter is shown in Figure 4-

16.(12) The procedure to calculate the column diameter is as follows: First, calculate the abscissa 

(ABS): 

ABS = (L/G) (PQ/PL)05 W.7-6) 

where: 

L = solvent flow rate, lb/hr 

G = gas stream flow rate, lb/hr 

PG = density of emission stream, lb/ft3 

The values for the variables L and G can be calculated by multiplying Lmoi

 and Gmoi with their respective molecular weights. Then proceed to the 

flooding line in Figure 4-16 and read the ordinate 

(ORD), and solve the ordinate expression for GarM,f 

at flooding: 

ORD = [(G (4.7-7) are8.f>

2

 (a/€3) (nL

0

-

2

)] /PGpLgc 

Thus, 

Garea.f = {[ORD pG P_ge]/[(a/€3) (|__a2)l}°-5 

where: 

Garea.f 

(4.7-8) 

= gas stream flow rate based on column 

cross sectional area (at flooding conditions), Ib/ft^-sec 

a,e = packing factors (see reference 21) 

M-L = viscosity of solvent, centipoises 

gc = gravitational constant, ft/sec2 

Assuming f as the fraction of flooding velocity appropriate for the proposed operation, the gas 

72 

^ . 0.01 

o 

0.001 

0.01 0.045 0.1 ':•<; 

(L/G)WPL)0-

9 

10.0 

Example Case 

Using Equations 4.7-6, -7, -8, -9, -10, and -11: 

L =MW80lvent x U,0. = 18x970 

= 17,460 lb/hr 

G = MW, x Gmoi = 28.4 x 465 = 13,200 lb/hr 

(see Appendix B.1, reference 8, for 

calculating MW.) 

PG = 0.071 lb/ft3 (from ideal gas law at 85°F) 

(see reference 21 for calculating pG) 

pL = 62.18 lb/ft3 (reference 18; at 85°F) 

ABS = (17,460/13,200) (0.071/62.18)°

 5

 = 0.045 

From Figure 4-16, at ABS = 0.045, the value of 

ORD at flooding conditions is about 0.15. For 2-

inch ceramic Raschig rings, from reference 21: 

a = 28 

€ =0.74 

Also, 

ge = 32.2 ft/sec2 

OL = 0.85 cp (reference 18, at 85°F) 

Thus, 

G.„..f = |(0.15 x 0.071 x 62.18 x 32 

(28/(0.74)3] (0.85)° 

G,re_.f = 0.56 Ib/sec-ft2 (at flooding) 

I2.2)10S 

..2 J 

Gar,. = 0.60 x 0.56 = 0.34 Ib/sec-ft2 

Thus, 

Aeoiumo =13,200/(3,600x0.34) 

Acolumo = 10.8 ft2 

Dcofumn = 1.13 (10.8)°

 6

 = 3.7 - 4 ft 

V. 
P.4 

Example Case 

1. Calculation of N0G 

Using Equation 4.7-13: 

HAPe = 20,000 ppmv 

HAP0 = 20,000 (1 - 0.98) = 400 ppmv 

NOG = ln[(20,000/400) 0.375 + 0.625]/0.375 

N0 G =7.9 

Using Figure 4 -17: 

HAPe/HAP0 = 20,000/400 = 50 

At AF = 1.6, 1/AF = 0.63, and NOG = 8 

2. Calculation of H 0G: 

Using Equations 4.7-14, -15, -16, and -17: 

L" = 17,460/10.8 = 1,617 Ib/hr-ft2 

3,600 Garea = 1.224 lb/hr-ft 2 

From reference 12, the packing factors are: 

b = 3.82 

c = 0.41 

d = 0.45 

Y = 0.0125 

s = 0.22 

Although 1,224 lb/hr-ft2 is outside the range 

shown in the table, assume that the packing 

factors are applicable and the error introduced into the calculations will be negligible. 

From reference 12: 

ScG = 0.66 

ScL = 570 

Also, 

p.L" = 0.85 x 2.42 = 2.06 Ib/ft-hr 

(The factor 2.42 is used to convert from centipoise to Ib/ft-hr.) 

Hence, 

HG = [3.82(1,224)°-41/(1,617)°-45](0.66)os 

= 2.06 

HL = 0.0125(1,617/2.06)a22(570)a5 

= 1.29 3 

Using AF = 1.6, 

HOG = 2.06 + (1/1.6) 1.29 - 2.87 -2.9 "7 

3. Calculation of Ntcoitmn: _ 

Using Equation 4.7-12: ~ "* 

Htc0lumn = 7.9 x 2.9 = 22.9 ^23 ft ts 

4. Calculation of Httotal: 

Using Equation 4.7-18: 

Httotal = 23 + 2 + (0.25 x 4) = 26 ft 7~7 

For costing purposes, it is necessary to calculate 

the column weight. Use the-fOHowing equation:(24) 

Wtcoiumn = (48 Deoi u m n X Httotat) + 395(0^^5,) ? 

^ —*<«7-19 ) 

where: 

Wteoiumn = column weight lb 

Also, to determine packing costs, volume occupied' 

-bythe packing material (Vpack.ng)hasto be calculated. 

Use the following expression: 

'packing

 = ^'^l^coiumni * Ht c o | u m n 

v column

 = "•'85(DC O | umn ) X n^Qiumn 

(4.7-20) 

Example Case 

Using Equation 4.7-19: 

<-»column = 4 rt 

Httota. = 26 ft 

Wtc0tumn = (48 x 4 x 26) + 39(4)2 

Wtc0lumn =5,600 lb 

Using Equation 4.7-20: 

Htc0|umn = 23 ft 

r

 packing 

'packing 

= 0.785 x (4)2 x 23 

= 290 ft3 

4.7.3.4 Pressure Drop Through the Column 

The pressure drop through a packed column for 

any combination of liquid and gas flows in the 

operable range is an important c <nomic consideration in the design of such colu <s. For a particular packing, the most accurate data will be those 

available from the manufacturer. For purposes of 

estimation, use the following correlation:(12) 

AP, = (g x 10~8) [10°j H (3,600 Ga .)2

/pG (4.7-21) 

where: i^^Xk» ' L" 

APa = pressur&drop, Ib/fr'-ft 

g, r = packing constants (see reference 12) 

The total pressure drop through the column is then 

expressed as: . 

AP«ot_. = AP. x Htcolum„ / 5 • 2. (4.7-22) 

Example Case 

Using Equation 4.7-21: 

From reference 12: 

g = 11.13 

r = 0.00295 

Also, 

L" 

3,600 G a r u 

PG 

Pt 

Thus, 

- 1,617 lb/hr-ft2 

= 3,600x0.34 = 1,224 lb/hr-ft2 

= 0.071 lb/ft3 

= 62.18 lb/ft3 

CL 

.•0 

AP. = 

[11.13 X IO * X 1Q">00a9»»<'>617/82.18)(1 224) 2) 

(0.071) 

AP, = 2.8 Ib/f^-ft 

Using Equation 4.7-22: 

Htcolumn = 23 ft 

AP«o_ii = 2.8 x 23 = 64.4 lb/ft2 

AP«ot_i - 64.4/5.2 = 12 in H20 

(The factor 5.2 is used to convert from lb/ft2 to in 

H20.) 

74 
r 

mimmmmmmm i __HB___ •Baa 

P.5 

Figure 5-6. Prices for custom carbon adsorber*, laaa 

carbon.(5)-

o 

s 

o 

a. 

E 

o 

to 

3 

u 

£ 

.o 

3 20 40 60 80 100 120 140 160 180 200 

Carbon Weight. C^ , (1,000 lb) 

Price includes beds, instrumentation and controls, 

and a steam regenerator. 

All auxiliary equipment (ductwork, fan, and stack) 

purchased costs and freight and taxes must be added to obtain the total purchased cost. 

The cost curve for absorbers (Figure 5-6) does not 

include the cost of packing, platforms, and ladders. 

The cost of platform and ladders (Figure 5-7) and 

packing (Table 5-4) must be added to obtain the 

major equipment purchased cost. All auxiliary 

equipment (ductwork, fan, and stack) purchased 

costs, the cost of instrumentation and controls, and 

freight and taxes must be added to obtain the total 

purchased cost. 

10 so 100 500 1.000 

Column Weight WU**TM (1,000 lb) 

Price includes manholes, skirts, and painting. 

Figure 5-7. Pricsa for absorber platform and ladder, ,6) 

(A lla 

o •o 

m-6/8 

40 

30 

20 

*. 15 

o 

•o 

•o 

<a 

_f 

•o 

c 

IS 

E 

w 1 

a: 

10 

9 

8 

7 

6 

-

-

1 

'y^ • H0 = 40 ft 

' • Hc = 33 ft 

4 Hc = 27 ft 

1 « i J 

4 5 6 7 8 9 10 

Column Diameter, t>ootum„ 

15 

(ft) 

Table S-4. Pries of Packing tar Abaorber Systems (61 

Packing Type and 

Material 

Packing Diameter, ii 

Pail Ringa 

Carbon steel 

Stainless steel 

Polypropylene 

Bert Saddlea 

Stoneware 

Porcelain 

Intalox Saddles. 

Polypropylene 

Porcelain 

Stoneware 

Peaking Ringa 

Carbon steel 

Porcelain 

SUttftiSMStsel 

iches 

Cost/Ft* 

(#81 Dollars) 

1 1.5 

24.3 16.5 

92.1 70.3 

21.9 14.8 

28.1 21.7 

34.5 26.6 

21.9 — 

19.4 14.8 

18.2 13.3 

30.3 19.8 

13.2 10,6 

109.0 82.6 

2 

15.1 

0.8 

13.8 

— 

— 

13.6 

13.3 

12.2 

- "N 

<O7.0 

"9.7 

22.9 

20 

3 

— 

— 

— 

— 

— 

7.0 

12.2 

11.0 

13.9 

8.1 

— 

Oy. 

The cost curve for condensers (Figure 5-8)/yields 

the total capital cost for cold water condenser systems. For systems needing refrigerant (ethylene 

glycol), the applicable cost from Figure 5-9 must be 

added to the cost obtained from Figure 5-8. Since a 

total capital cost is determined, no additional cost 

estimates are necessary; therefore, proceed to Section 5.3 to calculate annualized operating costs. 

The cost curve for a negative pressure fabric filter 

(Figure 5-10) does not include the the cost of bags 

(Table 5-5), which depend upon type of fabric used. 

This cost must be added to obtain the major equipment purchased cost. All auxiliary equipment 

(ductwork, fan, and stack) purchased costs, the cost 

100