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GROUP MEMBERS :. 06-CHEM-90 06-CHEM-62 06-CHEM-63 06-CHEM-100. FURNACES. Problem :.

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  1. GROUP MEMBERS : • 06-CHEM-90 • 06-CHEM-62 • 06-CHEM-63 • 06-CHEM-100 engineering-resource.com

  2. FURNACES engineering-resource.com

  3. 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 engineering-resource.com

  4. 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 engineering-resource.com

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  6. 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. engineering-resource.com

  7. 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 engineering-resource.com

  8. 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. engineering-resource.com

  9. 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 engineering-resource.com

  10. 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 engineering-resource.com

  11. 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 engineering-resource.com

  12. Try for 60 tubes The layout of the cross section of the furnace may be as shown in fig, engineering-resource.com

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  14. 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 engineering-resource.com

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  17. 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 engineering-resource.com

  18. Dimension ratio=38.5*20.46*14.92 =3:2:1 L=2/3(cube root of volume) L=15 ft engineering-resource.com

  19. 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 engineering-resource.com

  20. 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 engineering-resource.com

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  23. 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 engineering-resource.com

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

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  27. 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. engineering-resource.com

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