Page 346 - Air Pollution Control Engineering

06_chap_wang.qxd 05/05/2004 4:10 pm Page 324
324 Lawrence K. Wang et al.

M = negligible
e
P = 760 mm Hg
e
RE = 90%
P = 1 mm Hg
partial
T = 20ºF
con
Solution
Use Eqs. (2)–(7):
1a. Q e = 2000 scfm
)
HAP = Q ( 392 HAP × 10 − 6 (2)
em e e
,
)(
HAP = ( 2000 392 13 000) ×, 10 − 6
,
em
HAP = 0.06633 lb-mol min
,
em
.
1b. P = 1 0 mm Hg
vapor
P = 760 mm Hg
e
−
6
HAP = Q ( 392 1 ) [ −( HAP × 10 )][ P vapor ( P − P vapor)]
, om e e e (3)
) [
. (
−
−
−
×
6
,
.
HAP = ( 2000 392 1 13 000 10 ] [ 1 0 760 1 0)]
, om
HAP o,m = 0.00663 lb-mol min
1c HAP = HAP − HAP (4)
.
con , em , om
HAP con = 0 0663 − 0 00663
.
.
.
= 0 0597 lb-mol min
2a. ∆H = 17 445 Btu lb-mol (see ref. 3)
,
.
2b MW =104 .2 lb lb-mol
HAP
(
C = 24 Btu lb-mol º F extrapolated from data in ref. 7)
p HAP
H = HAP con[ ∆ H + C (T − T con )]
con p HAP e (5)
[
H = . 0 0597 17 445 24 − )]
,
+ (90 20
con
H con =1140 Btu min
(
2c. C p air = . 705 Btu lb-mol ºF see ref. 3 or Appendix for details)
=
HAP noncon [ Q ( e 392 ) − HAP , e m] C ( T − T con ) (6)
p
e
air
. ]
−
= (2000 392
HAP noncon [ ) − . 0 0663 7 05 ×(90 20 )
HAP noncon = , 2 480 Btu min
×
3. H = 1.1 60 H ( + H noncon ) (7)
load con
(
×
H = 1.1 60 1140 + 2480)
load
H = 239,000Btu h
load
Example 3
An air emission stream with its characteristic data shown in Table 7 is to be treated by a
condenser. Determine the following: