1 / 31

MAJED AL-SHAHRANI ID#992967

King Fahd University of Petroleum & Minerals Mechanical Engineering Department COOP PROGRAM ME351. MAJED AL-SHAHRANI ID#992967. ADVISOR Dr. MUMMER KALYON. CONTENTS. INTRODUCTION WORK ENVIRONMENT HEAT EXCHANGERS 2 CASE STUDIES CONCLUSION. Introduction.

havyn
Download Presentation

MAJED AL-SHAHRANI ID#992967

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. MAJED AL-SHAHRANI ID#992967

  2. MAJED AL-SHAHRANI ID#992967

  3. King Fahd University of Petroleum & Minerals Mechanical Engineering Department COOP PROGRAM ME351 MAJED AL-SHAHRANIID#992967 ADVISOR Dr. MUMMER KALYON MAJED AL-SHAHRANI ID#992967

  4. CONTENTS • INTRODUCTION • WORK ENVIRONMENT • HEAT EXCHANGERS • 2 CASE STUDIES • CONCLUSION MAJED AL-SHAHRANI ID#992967

  5. Introduction MAJED AL-SHAHRANI ID#992967

  6. Cooperative Training PrograminSaudi AramcoConsulting Services Department(CSD) Heat Exchangers Heat Transfer Group MAJED AL-SHAHRANI ID#992967

  7. Work Environment Consulting Services Department(CSD) ESD M&CED ME&CCD RED The mission of consulting services department (CSD) is to provide technical consultations on field problems and maintaining quality standards. PEU P&VU CEU HUG The responsibility of Process Equipment Unit (PEU) is to offer technical consultations on the parts related to pressure vessels, storage tanks and heat transfer equipment. Pressure Vessels & Storage Tanks boilers HeatTransferEquipments Fired heaters Heat exchangers Insulation & refractory materials MAJED AL-SHAHRANI ID#992967

  8. HEAT EXCHANGERS MAJED AL-SHAHRANI ID#992967

  9. HEAT EXCHANGERS HEAT EXCHANGERS transfer heat from a hot fluid to a colder fluid through the combined mechanisms of conduction and forced convection. ALL HEAT EXCHANGERS are similar in their principle of operation; however, heat exchangers may differ in the specific fluids that are used in the heat transfer process, the layout of the metal tubes, and the configuration of the enclosure. • THE PURPOSES of the heat exchanger are: • change the temperature of fluid. • change the phase of a fluid. MAJED AL-SHAHRANI ID#992967

  10. HEAT EXCHANGERS HEAT EXCHANGERS 2.Air cooled heat exchanger. • Shell and tube heat exchangers. 4.Plate and frame heat exchanger. 3.Double pipe heat exchanger. • Shell and tube heat exchangers. • Air cooled heat exchanger. • Double pipe heat exchanger. • Plate and frame heat exchanger. MAJED AL-SHAHRANI ID#992967

  11. shell-and-tube heat exchanger: The shell-and-tube heat exchanger is the type that is most commonly used in process plants. SHELL CHANAL THE BUNDEL MAJED AL-SHAHRANI ID#992967

  12. Air-Cooled Heat Exchangers: FANS Types of Air-Cooled Heat Exchangers • Forced draft. • Induced draft. • Humidified forced draft. BUNDEL MAJED AL-SHAHRANI ID#992967

  13. CASE STUDY MAJED AL-SHAHRANI ID#992967

  14. CASE STUDY#1 Tube insertion: Tube insertion is a new technology that used to repairing the tube failures by insert a tube with thin wall inside the damaged tube. This method can be done without replacing the existing tube. This case study is about a problem in one of Saudi Aramco air cooled heat exchangers. Problem:- The tube of a fin fan cooler at Shaybah Producing Facilities has a problem which is tube failure. • This new method will lead to: • Protecting damaged tube inlets. • Restoring plugged leaking tubes to active service. • Restoring original compressive strength to weakened tube to tube sheet joints. MAJED AL-SHAHRANI ID#992967

  15. CASE STUDY#1 staggered type INDUCED DRAFT MAJED AL-SHAHRANI ID#992967

  16. THREE BAYS ONE BUNDLE ONE BAYS MAJED AL-SHAHRANI ID#992967

  17. This table represents the Properties of water at 130 ºF This table represents the Properties of air at 90 ºF MAJED AL-SHAHRANI ID#992967

  18. THE TUBE AND BUNDLE PARAMETER MAJED AL-SHAHRANI ID#992967

  19. Area calculations Surface area of fins: (AF) AF= ((N*L*Pi)/(s+w))*(0.5*(Df²-Dr²) +Df*w) 36944.02 ft² Surface area b/w fins: (Aw) Aw= ((N*L*Pi)/(s+w))*(Dr*s) 1785.4 ft² Surface area of tubes with fins: (A) A= Aw+AF 38729.4 ft² Surface area of tubes without fins: (AT) 2072.623 ft² AT= N*L*Pi*Dr Frontal area: (Ao) Ao= L * cooler width 684.9 ft² MAJED AL-SHAHRANI ID#992967

  20. Minimum flow area: (Amin) A1= NT*L*(St-Dr-((2*w*h)/ (w+s))) = 201.13 ft² Or SD= (SL²+ (ST/2)²)½ = (0.25²+ (0.25/2)²)½ = 0.2792 ft A2=2*Nt*L*(SD-Dr-((2*w*h)/ (w+s))) = 479.254 ft² 201.13 ft² Then (Amin) = MAJED AL-SHAHRANI ID#992967

  21. To find the air side Coefficient of heat transfer: ha =681.12ft/ min V max = V approach/б The maximum velocity =595463.99Ib/hr The Mass flow rate Ma = V max* Amin * ρa = 5513.3743 Rea = (V max* Dr * ρa)/ µa The Reynolds number = 0.708 Pr = (µ Cp/ kt) The Prandtl number = 34.428 Nu = 0.242*(Re ^0.658)*((s/h) ^0.297)*((St/Sl) ^-0.091)*(Pr^ (1/3))*F1*F2 The Nusselt number F1= factor of fluid property variation. Assume it F1=1. F2= factor of number of tube rows. When the number of tube rows=6 and staggered type. The Coefficient of heat transfer: ha F2 =0.95. = 6.285 (Btu/ft².hr. ºF) ha = Nu * kt / Dr MAJED AL-SHAHRANI ID#992967

  22. To Find Fin efficiency:ηf ηf = (tanh (((2*ha/ (w*λf)) ½)* ψ)/ (((2*h/ (w*λf)) ½)* ψ) ψ= (Dr/2)*((Df/Dr)-1)*(1+0.35*ln (Df/Dr))= 0.0669 ηf = 0.8891 Effective Air Side Heat Transfer Coefficient Based on Total Surface Area h'a = ((ηf*AF+Aw)/A)*(ha) = 5.628 (Btu/hr.ft². ºF) Effective Air Side Heat Transfer Coefficient Based on ExternalSurface Area without Fins ha'r = h'a *(A/AT) = 105.14 (Btu/hr.ft². ºF) MAJED AL-SHAHRANI ID#992967

  23. To find the tube side Heat Transfer Coefficient: ht = 26.1566 in² The Crosse sectional area: At At= (pi*ID²/4)*(N/Ntp) = 95.736 Ib/s. ft² Gt= mt/At Mass flux: Gt = 92.475 ft/min Ub, t = Gt/ρt Tube Side Velocity: Ub,t = 13431.9 Ret = (Gt* ID)/µt Tube Side Reynolds Number: Ret = 5.184 Pr = µt*CP/K The Prandtl number = 0.006873 ft = 0.046*((Ret) ^ (-0.2)) Friction factor: ft = 89.15134 Nut = 0.023*((Ret) ^ (0.8))*((Pr) ^ (0.4)) Tube Side Nusselt Number: Nut = 440.8534(Btu/hr.ft².F) ht = Nu*(k/ID) Tube Side Heat Transfer Coefficient: ht MAJED AL-SHAHRANI ID#992967

  24. To find the Overall heat transfer coefficient: Ur 1/Ur = I/har + (OD/ (2*λf))*(ln (OD/ID)) + (1/ht)*(OD/ID) 1/Ur = 0.012826 ft².F.hr/Btu Uclean Ur = 77.965 Btu/hr. ft².F rf, d =Ff, t+ (ID/OD)*Ff, s= 0.002741 ft².F.hr/Btu 1/Ud = (I/Ur) + rf, d= 0.015567 ft².F.hr/Btu Uservice Ud = 64.238 Btu/hr. ft².F MAJED AL-SHAHRANI ID#992967

  25. OD=1 in ID= 0.87in Thick= 0.065in OD=1in NEW ID=0.8149in Insert tube thick=0.02756in NEW thick=0.09255in MAJED AL-SHAHRANI ID#992967

  26. BEFORE AFTER Crosse sectional area: At At=26.16in² Mass flux: Gt Gt= 5744.16 Ib/min. ft² Tube Side Velocity: Ub,t Ub, t =92.475 ft/min Tube Side Reynolds Number: Ret Ret = 13431.9 Pr =5.184 Friction factor: ft ft = 0.006873 Tube Side Nusselt Number: Nut Nut = 89.15 Tube Side Heat Transfer Coefficient: ht ht = 440.85(Btu/hr.ft².F) Crosse sectional area: At At=22.95in² Mass flux: Gt Gt= 6547.18 Ib/min. ft² Tube Side Velocity: Ub,t Ub, t =105.403 ft/min Tube Side Reynolds Number: Ret Ret = 14340.034 Pr =5.184 Friction factor: ft ft = 0.0067834 Tube Side Nusselt Number: Nut Nut = 93.94 Tube Side Heat Transfer Coefficient: ht ht = 495.9 (Btu/hr.ft².F) Uservice Ud = 63.87 Btu/hr. ft².F Uservice Ud = 64.238 Btu/hr. ft².F The tube insertion would cost only one third of the retubing MAJED AL-SHAHRANI ID#992967

  27. CASE STUDY#2 This case study is about a new channel cylinder of shell and tube heat exchanger. The out side diameter of the channel cylinder is 44.75” then the outside radius is 22.375”. The inside diameter of the channel cylinder is 38.500” then the inside radius is 19.25”. The thickness of the channel cylinder is 3.125”. The internal design pressure is 2296 Psi. The design temperature is 875ºF. This channel cylinder has44 3/4"O.D,32" lengthand 3 1/8" wall. After machining process the inspection found different thicknesses when he measures it by using ultra sonic 3.084” and 3.078” tmin = (P*R)/(S*E-(0.6*P)) The stress of this material: SA-336F22CL3 forging at875ºF can be found by using The American Society of Mechanical Engineers (ASME) section II part D. P= internal design pressure, psi R= the inside radius, in S= maximum allowable stress value, psi E= joint efficiency. S= stress S=16100psi @ T=875ºF 3.084” and 3.078” tmin = (2296*19.25)/(16100*1-(0.6*2296)) = 3.002 in MWAP = S*E*t/(R+0.6t) = 16100*1*3.125/ (19.25+0.6*3.125) = 2381.66 psi MAJED AL-SHAHRANI ID#992967

  28. Conclusion: • I Got hands-on experience on heat exchangers. • In my first case staduy, the method of tube inserts is acceptable from thermal point of view. • I did some Field Trips: 1. AlQatif Project 2. Uthmaniyah Gas Plant 3. Juaymah Gas Plant 4. The Saudi Aramco Shell Refinery Company 5. Juaymah Heat Exchanger Shop • The co-operative training program (co-op) is very important issue through university studying. It is helpful for the applied engineering students because it gives them an idea and background about the field work and that will be helping students after the graduation. MAJED AL-SHAHRANI ID#992967

  29. MAJED AL-SHAHRANI ID#992967

  30. Thank you for your attention MAJED AL-SHAHRANI ID#992967

  31. DONE BY: MAJED SAAD AL-SHAHRANI MAJED AL-SHAHRANI ID#992967

More Related