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gas ("dry air" + H2O) density

gas ("dry air" + H2O) density

gas ("dry air" + H2O) density

(OP)
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
 

RE: gas ("dry air" + H2O) density

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

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