1 / 15

PROCESS HEAT TRANSFER

PROCESS HEAT TRANSFER. Designing of Air Cooler. 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). STATEMENT.

jens
Télécharger la présentation

PROCESS HEAT TRANSFER

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. PROCESS HEAT TRANSFER Designing of Air Cooler engineering-resource.com

  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) engineering-resource.com

  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 engineering-resource.com

  4. Solution Begins Heat duty Q = m CpdT = 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 engineering-resource.com

  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 engineering-resource.com

  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 engineering-resource.com

  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 engineering-resource.com

  8. 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 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 engineering-resource.com

  9. Mass Flow Rate GA = WA/FA = 2844 lb/hr ft2 Extended 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 Coefficient Ux= 1/[(1/HT)*(AR*Do/Di)+ RDt * (AR*Do/Di) + (1/HA)] AR = 21.4 From table 5.3 RDt = Fouling Factor = 0.001 Ux = 4.44 btu/ h ft2 F engineering-resource.com

  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 engineering-resource.com

  11. Static pressure loss of air flow across each single tube row DPA DPA = exp[1.82 * 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 engineering-resource.com

  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 engineering-resource.com

  13. POUND FORCE CALCULATION Pforce = DPAT + [(4* ACFM)/(4009 * 3.1417 * Dfan2)]2 = 0.67 inch of water Fan horse power Bhp = [ACFM per fan * Pforce]/ (6387*Efficeincy) Assume Eff= 70% Bhp = 20.3 Hp engineering-resource.com

  14. REFERENCE Investor Chemical Process Design by Dougles & Ervin engineering-resource.com

  15. THANK YOU VERY MUCH FOR BEING SOO INDIFFERENT engineering-resource.com

More Related