Designing of Air Cooler

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PROCESS HEAT TRANSFER

• Designing of Air Cooler

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PRESENTERS

• Anas Javaid (06-CHEM 51)
• Mohammed Saqib (06-CHEM-57)
• Malik Qadeer (06-CHEM-91)
• Ali Raza (06-CHEM-93)
• Mohammad Iqbal (06-CHEM-97)

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STATEMENT

• Assume that a typical hydrocarbon naphtha liquid
  from a fractionation tower-side cut stream is to
  be cooled to 150 F the naphtha stream enters the
  air cooler at 250 F at a flow rate of 273,000
  lb/hr. the physical tube side properties at the
  average temperature of 200 F are
• K 0.0766 btu ft/hr ft2 F
• Cp 0.55 btu/lb F
• Uvis 0.51 cp
• Inlet temperature of air t1 100 F

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Solution Begins

• Heat duty
• Q m Cp dT 15,015,000 BTU/hr
• Assuming Ux to be 4.5 according to the table 5.4
•  
• Temperature difference for air
• dT ((Ux 1)/10) x ((T1 T2)/2 t1)
• 54 F
• dT t2 t1
• t2 154 F

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• LMTD (dt1 dt2)/ log(dt1/dt2)
• 70.5 F
• Assuming number of passes is three, therefore Ft
  is 1
• Corrected LMTD 70.5 F
• Tube outside extended area
• Ax Q/(Ux x LMTD)
• 48356 ft2

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• Face Area Calculation
• FA Ax/ APSF
• APSF value is obtained from the table 5.3
• FA 407 ft2/ft2
• Width
• W FA/L
• Suppose length is equal to 30 ft then
• W 13.6 ft
• Fans required 2

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• Number of tubes
• NT Ax / (APF x L)
• APF value is noted from the values mentioned in
  table 5.3
• 300 tubes
• Modified Reynold Number Calculation
• Ai 3.1416 (Di/2)2
• 0.5945 in2
• GT (144 x Wt x Np)/ (3600 x NT x Ai)
• 184 lb/sec ft2
• NRe (Di x GT)/ Uvis
• 314

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• J exp-3.9133.968 x (log NR)-0.5444 x (log
  NR)20.04323 x (logNR)3 0.001379 x (log NR)4
• 2027
• Tube Inside Film Coefficient
• HT J K (Cp Uvis/K)1.3/Di
• 275 BTU/hr ft2 F
• AIR FLOW
• WA Q/(Cp x dt)
• Cp 0.24
• WA 1157400 lb/hr

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• Mass Flow Rate
• GA WA/FA
• 2844 lb/hr ft2
• Extended Tube Film Coefficient
• HA exp -7.15191.7837 x (log GA) 0.076407 x
  (log GA)2
• 9 BTU/ hr ft2 F
• Overall Heat Transfer Coefficient
• Ux 1/(1/HT)(ARDo/Di) RDt (ARDo/Di)
  (1/HA)
• AR 21.4 From table 5.3
• RDt Fouling Factor 0.001
• Ux 4.44 btu/ h ft2 F

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HYDROLIC DESIGN

• Fan Area per Fan
• FAPF 0.4 x FA/ number of fans
• 84 ft2
• Fan diameter
• Dfan (4 FAPF/3.1417)0.5
• 11 ft
• Air side static pressure loss( DPAT) Calculation
• Tav dt/2 t1
• 127 F

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• Static pressure loss of air flow across each
  single tube row DPA
• DPA exp1.82 log GA - 16.58
• 0.122 inch of water per tube row
• DPAT 4 x DPA/DR
• Where DR 0.9 at 127 F from table 5.5
• DPAT 0.54 inch of water

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• Fan inlet actual volumetric flow
• ACFM WA/ DR 60 0.075
• Where DR 0.95 at 100 F from table 5.5
• ACFM 271000

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POUND FORCE CALCULATION

• Pforce DPAT (4 ACFM)/(4009 3.1417 Dfan
  2)2
• 0.67 inch of water
• Fan horse power
• Bhp ACFM per fan Pforce/ (6387Efficeincy)
• Assume Eff 70
• Bhp 20.3 Hp

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REFERENCE

• Investor Chemical Process Design by Dougles
  Ervin

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THANK YOU VERY MUCH FOR BEING SOO INDIFFERENT