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In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENT

In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENT. TOPIC OF PRESENTATION. DESIGN OF SHELL AND TUBE HEAT EXCHANGER…. GROUP MEMBERS MUHAMMAD IMRAN 2006-CHEM-35 OSAMA AKRAM 2006-CHEM-39 WAQAS AHMAD 2006-CHEM-15

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In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENT

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  1. In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENT Shine

  2. TOPIC OF PRESENTATION DESIGN OF SHELL AND TUBE HEAT EXCHANGER….. Shine

  3. GROUP MEMBERS MUHAMMAD IMRAN 2006-CHEM-35 OSAMA AKRAM 2006-CHEM-39 WAQAS AHMAD 2006-CHEM-15 TAIMOOR RAI 2006-CHEM-81 SUMITTED TO: Prof. Shah muhammad Shine

  4. Shell And Tube Heat Exchanger Main Parts 1.Tubes 2.Shell 3.Baffles 4.Tube Sheets 5.Head 6.Tube Bundle Shine

  5. Problem Statement: Dry ammonia gas at 83 psia and at a rate of 9872 lb/hr is to be cooled from 245oF to 95oF using cooling water from 85oF to 95oF. Design a shell and tube heat exchanger to perform the above duty. Shine

  6. Design of a Shell And Tube Exchanger: • In this presentation following • topics would be presented: • General Design Steps. • Thermal Design. • Hydraulic Design. Shine

  7. Part (A) General Design Steps of Shell and Tube Heat exchanger: Shine

  8. General Design Steps: Step 1. Perform the energy balance and calculate heat exchanger duty. Step 2. Obtain the necessary thermo physical properties of hot and cold fluid streams at their mean temperature. Step 3. Select the tentative number of shell and tube passes. Step 4. Calculate the LMTD and the correction factor FT . Step 5. Assume a reasonable value of the overall coefficient on outside tube area designated as Udo . Shine

  9. General Design Steps: Step 6. Select the tube diameter, it's wall thickness (In Terms Of BWG) and the tube length. Calculate the number of tubes required to provide the area calculated above. Step 7. Select the tube pitch. Select the shell diameter that can accommodate the required number of tubes. Step 8. Select the type, size (e.g. percentage cut) number and spacing of baffles. Shine

  10. General Design Steps: Step 9. Estimate the shell side and tube side heat transfer coefficients. Step 10. Select the dirt factors Rd applicable to the system. Step11. Calculate the overall coefficient Udo Step 12. Calculate the area based on this Udo. Shine

  11. General Design Steps: Step13. Compare the Udo and A values with those assumed Step 14. If area is in excess of 10% of that calculated then the design is acceptable. This excess area is sometimes required and sometimes not. If area calculations do not agree assume a new value of Udo and proceed in a similar way Step 15. Calculate the shell side and tube side pressure drops Shine

  12. Part (B) Thermal Design of Shell and Tube Heat exchanger: Shine

  13. Physical Properties: • Prior entering to thermal • design calculations; • Tabulating the physical properties of • Hot fluid (Ammonia) • And cold fluid (Water) Shine

  14. Physical Properties for Ammonia: Properties: Values: Mass Flow Rate mh 9872 lb/hr Entering Temperature T1 245 oF Leaving Temperature T2 95 oF Average temperature Tavg 170 oF Specific Heat Capacity Cph 0.53 Btu/lb-oF Viscosity Of Ammonia µh 0.027 lb/ft-hr Thermal Conductivity kh 0.0176Btu/hr-ft-oF Density ρh 0.02079lb/ft3 Shine

  15. Physical Properties for water: Properties: Values: Mass Flow Rate mc 78482.4 lb/hr Entering Temperature t1 85 oF Leaving Temperature t2 95oF Average temperature tavg 90oF Specific Heat Capacity Cpc 1 Btu/lb-oF Viscosity Of water µc 1.846 lb/ft-hr Thermal Conductivity kc 0.358 Btu/hr-ft-oF Density ρc 62.11 lb/ft3 Shine

  16. How should I start ? • An obvious answer would be from making • an overall • Energy And • Material Balance. Shine

  17. HOW TO CALCULATE HEAT DUTY? By using Formula Q = m Cp ∆T Using Values Of m, Cp and ∆T from ammonia Q = 784824 Btu/hr. From Which Mass Flow Rate of Water would be calculated using Q =m Cp ∆T Yields m = 78482.4 lb/hr. Shine

  18. Calculation of Volumetric Flow Rate of Water: Volumetric Flow Rate of Water q = mc /ρc = 1263.6 ft3/hr. Shine

  19. Overall energy balance: Calculating LMTD … t2 = 95oF t1 = 85oF T1= 245oF T2 = 95 oF LMTD = [(T1- t2) - (T2-t1)] [ln(T1- t2)/(T2-t1)] LMTD = 51.7oF Shine

  20. Calculation Of Heat Transfer Area : Assume Udo=27 Btu/hr-ft2-oF Using the formula Q = Udo × A× LMTD Calculate Heat Transfer Area A = 562.235 ft2 Shine

  21. Select A Suitable Size Tube: A Iterative Selection is 3/4" and 16 BWG Select A suitable length L = 16 ft Outside Diameter of Tube = O.D = 0.0625 ft Inside Diameter of Tube = I.D = 0.0516 ft Wall Thickness = xw = 0.005416 ft Shine

  22. 1 – 1 Pass Arrangement calculation: Outer Surface Area of Tube = 3.142×O.D×LOuter Surface Area of Tube = 3.142 ft2 No. of Tubes Required = Area/ Outside Surface Area of One Tube Number of Tubes Required = 178.94 Flow Area = π× (I.D) 2×Number of Tubes 4 Flow Area = 0.3742 ft2 Linear Velocity within the Tubes = Volumetric flow Rate/Flow Area Linear Velocity within the Tubes = 0.937 ft/sec Deduction: According to rule of thumbs and conventions it is well known that the velocity in the tubes should be between 3-10 ft/sec. So 1-1 pass is rejected and we go for a 1-4 pass exchanger that might bring the velocity in the range. Shine

  23. 1 – 4 Pass Arrangement calculation: Take 180 tubes Number of tubes per pass = 180/4Number of tubes per pass = 45 Flow Area Per Pass = π× (I.D)2×Number Of Tubes Per Pass 4 Flow Area per Pass = 0.0941ft2 Linear Velocity within the Tube = Volumetric flow Rate/Flow Area Linear Velocity within the Tubes = u = 3.73 ft/sec Deduction: This velocity is in the allowable range so we proceed further with 1-4 Pass Exchanger. Shine

  24. 1 – 4 Pass Arrangement calculation: Calculation of reynolds no for water: NRe = (density×velocity*Di)/µ NRe = 23302.51 Calculation of Prandtl No for water: Npr = (Cp×µ)/k Npr = 5.16 Calculation of L/Di = (16 ft / 0.0516 ft) =310 Calculation Of jh factor Which from chart comes to be = 80 Using value of jh to calculate hi by using correlation jh = [hi (Di× Npr1/3)] / k hi = 959.115 Shine

  25. Summary of tube side Calculation... Nre = 23302.51 Npr = 5.16 µ/µw = 1 Jh = 80 hi = 959.11 (Btu/hr- ft2-oF) Shine

  26. SHELL SIDE CALCULATIONS: 1. Selecting Tube Arrangement:Triangular (0.75 inch O.D tubes on 1 inch triangular pitch) 2. Shell Diameter: From the tables we see that the shell which can accommodate 180 tubes have I.D =17.25 inches. Inside Diameter of Shell=Ds=1.4375ft Pitch=PT=0.0833ft 3. Selecting Baffle: Select 25% cut segmental baffles. Shine

  27. SHELL SIDE CALCULATIONS: 4. Selecting Baffle Spacing: According to the TEMA standards the allowed baffle spacing is 0.2Ds--------Ds we consider Baffle Spacing B = 0.66×Ds Baffle Spacing = 0.95ft 5. Calculating tube clearance: Tube Clearance=Pitch - O.D Tube Clearance= PD = 0.0208ft Shine

  28. SHELL SIDE CALCULATIONS: 6. Calculating flow area:Flow Area = PD× B ×DsPTFlow Area = 0.3409 ft2 7. Calculating mass velocity for ammonia: Mass Velocity of Ammonia Gs = Mass Flow Rate of Ammonia Flow Area Mass Velocity of Ammonia = 28950.45 lb/hr-ft2 Shine

  29. SHELL SIDE CALCULATIONS: 8.Calculating hydraulic diameter: For Triangular Pitch hydraulic diameter can be calculated by using formula Hydraulic Diameter DH = 4(0.43×PT2-.39275×O.D2)/1.571×O.D So; Hydraulic Diameter = 0.059 ft 9. Calculating shell side reynold's number: Shell Side Reynold’s Number= DH×Gs/µh = 63262.09 Shine

  30. SHELL SIDE CALCULATIONS: 10.Calculating shell side Prandtl number:Shell Side Prandtl Number = Cph×µh/kh = 0.81 11. Evaluating Colburn's Factor from chart: Colburn’s Factor (jh)=140 As jh= ho× (Pr) -0.33 ×DH/kh Thus ho= jh/(Pr) -0.33 ×DH/kh Outside Heat Transfer Coefficient = 39.063 Btu/hr-ft-oF Shine

  31. SHELL SIDE CALCULATIONS: 12. Deducing inside and outside heat transfer Dirt co-efficient from data tables: Now from the tables we see that for ammonia water systems Inside Heat Transfer Dirt Coefficient=hi,d=1007.516Btu/hr-ft2-oF Outside Heat Transfer Coefficient=ho,d=3526.306 Btu/hr-ft2-oF Shine

  32. SHELL SIDE CALCULATIONS: 13. Selecting compatible Materials Of Construction Material selected Carbon steel. Thermal Conductivity of Carbon Steel = kw =29.913Btu/hr-ft-oF Shine

  33. Calculation Of Overall Heat TransferCoefficient: The overall heat transfer coefficient can be calculated from the following formula 1 = O.D + O.D . Udo ((hi ×I.D) (hi.d × I.D) + O.D × Xw × ln(O.D/I.D) + 1 + 1 . [kw× (O.D-I.D)] ho ho.d Overall Heat Transfer Dirt Coefficient Udo = 35.03 Btu/hr-ft2-oF Shine

  34. Calculation of LMTD Correction Factor As R = (T1-T2)/(t2-t1) R = 15 P = (t2-t1)/(T1-t1) P = 0.0625 Now there is a well known equation for LMTD correction factor F= (R2+1)0.5ln (1-P/1-RP) . (R-1) × ln [2-P(R+1-(R2+1)0.5] 2- P[R+1+ (R2+1)0.5] F = 0.838 The value of F is acceptable as it is above 0.75. Shine

  35. AREA VERIFICATION: Area Required AR= Q/Udo × F × LMTD Area Required = 517.487ft2 Area Available AA = 3.142×O.D×Length×Number Of tubes Area Available = 565.56ft2 Percentage excess area: % excess area =100× (AA-AR)/AR % excess area = 9.29% Note: Since an excess area of 10% is allowed so our design is acceptable Shine

  36. Part (C) Hydraulic Design involves: • Pressure Drop calculation across: • Tube Side • Shell Side Shine

  37. Pressure Drop Calculations across Tube side… Using The Relation ∆Pt = f x G2t x L x n . 2 x g x p x Di x φt ΔPt = 3194.77 N/m2 = 0.463 lbf/in2 Where: f = friction factor. Gt = Mass velocity of tube side fluid. n = no. of tube passes. φt = Dimensionless Viscosity ratio. Shine

  38. Pressure Drop Calculations across Tube side… Return Losses Can Be Calculated As ∆Pr = 4n( V2) = 1.495 lbf/in2 2 g Total Pressure Losses: ΔPT = ΔPt + ΔPr = 1.958 lbf/in2 Note:Pressure Drop is in allowable limit. Shine

  39. Pressure Drop Calculations across shell side… Using The Relation ∆Ps = f sx G2s x Ds x (N b+1) . 2 x g x ps x DHx φs ∆Ps = 352.46 Kg/m2 = 0.5 lbf / in2 Nb = Tube Length - 1 = 16 Baffle Spacing Note: This pressure drop is in allowable limit. Shine

  40. THANK’S … Shine

  41. Any Question ? Shine

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