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100                                                      Chapter3


                For a final oxygen concentration of  1.0 %  at a nitrogen flow rate of  180 1/s
                    3
           or0.180m /s,
                 40000   0.01
             =_  ———hi —— = 6.766xl0  5                              (3.1.17)
           t
                 0.180   0.21
                                    s
           The purging time is 6.766 x 10  s (188 h).




           Example 3.2  Cooling-Tower Analysis_______________________

            Water,  from  lakes,  rivers,  and  the  sea, is  a  common  coolant.  Because  of  water
            shortages or the environmental effects  of discharging heated water, air may also be
           use  as a coolant, either directly or indirectly.  In the direct method, called the dry
            system, a fan blows air directly over a heat exchanger surface. Because of the low
           heat capacity of air,  a large quantity is required.  In the indirect method, called the
            wet  system,  water  is  the  primary  coolant.  Air  cools  the  water  by  evaporating  a
            small  fraction  of  the  water  in  a tower.  The  cooled  water  is  then  returned  to  the
           process.  A  process  engineer  will  have  to  choose  either  the  dry  or  wet  method.
            Cooling water is not a main part of the process  but an "offsite"  operation, i.e., it is
            generally located off to one side of the process area. We may consider cooling and
            treating the water  to remove dissolved salts as a sub-process.
                Figure 3.2.1 shows the mechanical-draft  crossflow  tower, which is the most
            commonly used  cooling  tower  [11].  Water  enters the  top  of  the tower  and  flows
            downward  over  packing,  called  fill.  The  fill  increases  the  surface  area  for  mass
            transfer  by breaking  up the  water  into  droplets  or  spreading  it into a thin  film.  A
            cooling tower,  like  a packed bed  absorber  or  stripper, must provide  good  contact
           between air and water to promote rapid evaporation. Good contact reduces the size
            of  the  tower  and  also  the  pressure  drop,  called  "draft"  by  cooling-tower  design
            engineers. A fan,  located at the top of the tower and shown in Figure 3.2.1, draws
            air  into  the  tower.  Louvers  distribute  incoming air,  which  then  flows  across  the
            tower, removing evaporated water.
                During the operation of the tower, water is lost by evaporation, water drop-
            lets  entrained  in  the  outgoing air, and in a water purge,  called blowdown. To re-
            duce carry-over of water droplets the air flows  across drift  eliminators.  The  water
            droplets impinge on the drift  eliminators and then flows  down to the bottom of the
           tower.  The  droplet  water  loss  is  about  0.2% of  the  incoming  water  [11]. After
            leaving  the  drift  eliminators,  air  flows  up  and  out  of  the  tower.  Evaporation  of
           water  into air  transfers  heat  from  the  water to the air.  Cooling the  water  requires
            about  1.0  %  evaporation  for every  5.56  °C (10.0 °F)  drop  in  the  water  tempera-
           turefl 1].  To reduce scale formation in the tower because of dissolved calcium or




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