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FURNACES




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Problem:
• A furnace is to designed for a total duty of
  50,000,000Btu/hr.The overall efficiency is to be
  75%(lower heating value). Oil fuel with lower
  heating value of 17,130Btu/lb is to be fired with
  25% excess air (corresponding to 17.44lb of
  air/lb fuel), and air preheated to 400oF. Steam
  for atomizing the fuel is 0.3lb/lb of oil. The
  furnace tubes are to be 5in.OD on 8.5in.centers,
  in a single row arrangement. The exposed tube
  length is to be 38.5 ft. the average tube
  temperature in the radiant section is estimated to
  be 800oF.
• Design the radiant section of a furnace having a
  radiant section average flux of 12,000Btu/(hr)ft2

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Solution:
As in all trial and error solutions, a starting point
  must be assumed and checked. For orientation
  purposes, one can make an estimate of the
  number of tubes required in the radiant section
  by assuming that, overall exchange factor is
                  F=0.57
         Q/αΑcpF =2*average flux/0.57
                    =42,000Btu/(hr)ft2


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• From fig we find the exit temperature of
  gas with the help of tube temperature and
  ΣQ/αAcpF, that is 1730oF.the duty in
  cooling the furnace gases to 1730oF can
  be calculated, and from it required number
  of tubes determined for the first
  approximation of the design.



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Heat liberated by the fuel
                 QF=50,000,000/0.75
                    =66,670,000 Btu/hr
       Fuel quantity=QF/17130
                    =3890 lb/hr
        Air required=fuel quantity*17.44 lb
                    =67,900lb/hr
Steam for atomizing=fuel quantity*0.3lb
                    =1170lb/hr


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• Q=QF+QA+QS-QW-QG
 Q=Total radiant section duty, Btu/hr
QF=Heat liberated by fuel, Btu/hr(lower heat value)
QA=Sensible heat above 60oF in combustion air.
QG=Heat leaving the furnace radiant section in the
   flue gases.
QS=Sensible heat above 60oF in steam used for
   oil atomization.
QW=Heat loss through furnace walls.

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QF=66,670,000Btu/hr
QA=air required*82Btu/lb at 400oF(above
 60OF)
QA=67,900*82=5,560,000Btu/hr
QS=negligible
QW=2% of QF
   =0.02*66,670,000=1,330,000Btu/hr
QG
Heat out in gases at 1730oF, 25% excess
 air,476Btu/lb of flue gas
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• QG=mass flow rate*heat of flue gas
     =476Btu/lb(3890+67,900+1170)lb/hr
     =34,500,000Btu/hr

  Q=QF+QA+QS-QW-QG
   =66,670,000+5,500,000+0-1,330,000-
    34,500,000
   =36,500,000Btu/hr

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The surface area per tube:
        A=π*D*L
         =3.14*5/12*38.5
         =50.4 ft2
The estimated number of tubes:
      Nt=Q/flux*area
        =36,400,000/12000*50.4
        =60.1
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• Try for 60 tubes

The layout of the cross section of the
 furnace may be as shown in fig,




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Equivalent cold plane surface, Acp:
Center to center distance=8.5 in
α=factor by which Acp must be reduced to
obtain effective cold surface, dimensionless

Acp per tube =8.5/12*38.5
             =25.7 ft2
Ratio=center to center/OD
     =8.5/5=1.7
Calculate “α” from figure for single row
    α=0.973

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• αAcp/tube=25.7*0.973
      αAcp=60*25=1500ft 2
 Refractory surface:
  end walls=2*L*W=2*20.46*14.92
           =611 ft2
 Side wall=14.92*38.5=575ft2
Bridge wall=9.79*38.5=377ft2
Floor and arch=2*20.46*38.5=1575ft2
     AT=3138ft2
     AR=AT-αAcp=1638
AR/αAcp=1638/1500=1.09


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• Dimension ratio=38.5*20.46*14.92
                 =3:2:1
          L=2/3(cube root of volume)
          L=15 ft


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• Gas emissivity: from the analysis of the
  fuel quantity ,and the assumption that the
  humidity of the air is 50% of the saturation
  at 60oF, the partial pressure of Co2 and
  H2o in the combustion gases with 25%
  excess air are
Pco2=0.1084
pH2o=0.1248
Pco2*L=0.1084*15=1.63
pH2o*L=1.87

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• εG=100-%/100*[(qco2+qH2o)TG- (qco2+qH2o)TS]/(qb)TG
     -(qb)TS
 % correction=8%
 From graph we take the values of q for
 water and Co2,
qco2=6500 at TG
qH2o=14500 at TG
qco2=650 at TS
qH2o=1950 at TS
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• εG=0.489
 overall exchange factor F :
 F at εG=0.49 and AR/αAcp=1.09
 from figure
 F=0.635
Check of gas temperature required to effect
 assumed duty on assumed surface:
  ΣQ=36,400,000Btu/hr assumed
αAcp=1500ft2 assumed

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 ΣQ/αAcpF=36,400,000/1500*0.635
          =38200
TG required from graph (at TS=800oF)
          =1670 oF
Compared with 1730oF




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• According to that exit temperature the duty
  would be 37,050,000Btu/hr, assuming that
  F does not change (it will go up slightly)
  ΣQ/αAcpF=39,000 requiring a “driving”
  temperature of 1695oF which is close
  approximation. the circumferential flux will
  be 37,050,000/60*50.4=12280Btu/(hr)ft2
  as compared with the 12,000 flux specified
  such a difference is negligible.

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