gas ("dry air" + H2O) density 1

[luolmagu]

Hi -
The following partial calculations are for the selection of a wet scrubber. Please can anyone explain and, if possible, provide a source of reference for the formula shown at step 5 - and in particular explain the terms 0.379 and 353 as shown in the formula?

Given gas volume = 13,400 Ncu.m/hr (0deg C, 760 mm Hg)
Temp. 180 deg C
Moisture 40.5% (by vol.)

Step 1 ACTUAL CONDITIONS AT SCRUBBER INLET
Partial gas volumes (0 deg C, 760 mm Hg)
Dry air 0.595 x 13,400 = 7,975 N cu. m/hr.
H2O 0.405 x 13,400 = 5,425
------
13,400 N cu. m/hr.

Step 2 WEIGHT ANALYSIS
Dry Air 1.293 x 7,975 = 10,310 Kg/hr
H2O 0.804 x 5,425 = 4,360
------
14,670 Kg/hr

Step 3 MOISTURE BY WEIGHT IN %
4360 x 100 = 29.72%
----
14670

Step 4 MOISTURE IN Kg/Kg DRY AIR = w
4360 = 0.423 Kg/Kg
----
10310

Step 5 GAS ("DRY AIR" + H2O) DENSITY @ 180 deg.C
x = 353(1-0.379 w )
----- ---------------
T w + 0.622

Where T = 180 + 273 = 453 deg. Kelvin
w = 0.424 Kg/Kg

Any help would be much appreciated.

Many thanks.

luolmagu

 

[willard3]

Get Keenan and Keyes steam tables....all this stuff is tabularized.