Air Cooler

Designing of Air Coolerengineering-resource.com

engineering-resource.com

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)engineering-resource.com


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 FCp = 0.55 btu/lb FUvis = 0.51 cpInlet temperature of air = t1 = 100 F



Heat dutyQ = m Cp dT = 15,015,000 BTU/hr

Assuming Ux to be 4.5 according to the table 5.4Temperature difference for air dT = ((Ux + 1)/10) x ((T1 + T2)/2 t1) = 54 FdT = t2 t1t2 = 154 F



LMTD=(dt1 dt2)/ log(dt1/dt2)=70.5 F

Assuming number of passes is three, therefore Ft is 1Corrected LMTD = 70.5 F

Tube outside extended areaAx = Q/(Ux x LMTD) = 48356 ft2



Face Area CalculationFA = Ax/ APSFAPSF value is obtained from the table 5.3FA = 407 ft2/ft2WidthW = FA/LSuppose length is equal to 30 ft thenW = 13.6 ftFans required = 2



Number of tubesNT = Ax / (APF x L)APF value is noted from the values mentioned in table 5.3 = 300 tubes Modified Reynold Number CalculationAi = 3.1416 (Di/2)2 = 0.5945 in2GT = (144 x Wt x Np)/ (3600 x NT x Ai) = 184 lb/sec ft2NRe = (Di x GT)/ Uvis = 314



J = exp[-3.913+3.968 x (log NR)-0.5444 x (log NR)2+0.04323 x (logNR)3 0.001379 x (log NR)4] = 2027

Tube Inside Film CoefficientHT= J * K * (Cp * Uvis/K)1.3/Di= 275 BTU/hr ft2 FAIR FLOWWA = Q/(Cp x dt)Cp= 0.24 WA = 1157400 lb/hr



Mass Flow RateGA = WA/FA = 2844 lb/hr ft2Extended Tube Film Coefficient HA = exp[ -7.1519+1.7837 x (log GA) 0.076407 x (log GA)2] = 9 BTU/ hr ft2 F

Overall Heat Transfer CoefficientUx = 1/[(1/HT)*(AR*Do/Di)+ RDt * (AR*Do/Di) + (1/HA)]AR = 21.4 From table 5.3RDt = Fouling Factor = 0.001 Ux = 4.44 btu/ h ft2 F



Fan Area per FanFAPF = 0.4 x FA/ number of fans= 84 ft2Fan diameterDfan = (4 FAPF/3.1417)0.5 = 11 ftAir side static pressure loss( DPAT) CalculationTav = dt/2 + t1 = 127 F



Static pressure loss of air flow across each single tube row DPADPA = exp[1.82 * log GA - 16.58]= 0.122 inch of water per tube rowDPAT = 4 x DPA/DRWhere DR = 0.9 at 127 F from table 5.5DPAT = 0.54 inch of water



Fan inlet actual volumetric flowACFM = WA/[ DR * 60 * 0.075]

Where DR = 0.95 at 100 F from table 5.5ACFM = 271000



Pforce = DPAT + [(4* ACFM)/(4009 * 3.1417 * Dfan 2)]2= 0.67 inch of water

Fan horse power

Bhp = [ACFM per fan * Pforce]/ (6387*Efficeincy) Assume Eff= 70%Bhp = 20.3 Hp



Investor Chemical Process Design by Dougles & Ervinengineering-resource.com


Date post:	26-Nov-2015
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Air Cooler

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