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Designing of Air Cooler

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Designing of Air Cooler engineering-resource.com Anas Javaid (06-CHEM 51) Mohammed Saqib (06-CHEM-57) Malik Qadeer (06-CHEM-91) Ali Raza (06-CHEM-93) Mohammad ... – PowerPoint PPT presentation

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Title: Designing of Air Cooler


1
PROCESS HEAT TRANSFER
  • Designing of Air Cooler

2
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)

3
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

4
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

5
  • 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

6
  • 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

7
  • 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

8
  • 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

9
  • 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

10
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

11
  • 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

12
  • Fan inlet actual volumetric flow
  • ACFM WA/ DR 60 0.075
  • Where DR 0.95 at 100 F from table 5.5
  • ACFM 271000

13
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

14
REFERENCE
  • Investor Chemical Process Design by Dougles
    Ervin

15
THANK YOU VERY MUCH FOR BEING SOO INDIFFERENT
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