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Process heat transfer

Process heat transfer. Spiral plate heat exchanger. Group members. 2006-chem-01 2006-chem-37 2006-chem-77. construction. Two long strips of plate, which are spaced apart and wound around an open, split center to form a pair of concentric spiral passages.

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Process heat transfer

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  1. Process heat transfer Spiral plate heat exchanger engineering-resource.com

  2. Group members • 2006-chem-01 • 2006-chem-37 • 2006-chem-77 engineering-resource.com

  3. construction • Two long strips of plate, which are spaced apart and wound around an open, split center to form a pair of concentric spiral passages. • Spacing is maintained uniformly along the length of the spiral by spacer studs welded to the plates. engineering-resource.com

  4. Material of construction • Any material that can be cold worked and welded, such as carbon steel, stainless steels, nickel and nickel alloys, aluminum alloys, titanium and copper alloys. engineering-resource.com

  5. Operating conditions • Limitations of material of construction govern design temperature. • Because the turns of the spiral are of relatively large diameter, each turn must contain its design pressure and thickness of plate id somewhat restricted –for these reasons the maximum design pressure is 150 psi. engineering-resource.com

  6. Flow arrangements • Both fluids in spiral flow • One fluid in spiral flow and the other in axial flow • One fluid in spiral flow and the other in a combination of axial and spiral flow engineering-resource.com

  7. Spiral flow in both channels • The spiral assembly includes flat covers at both sides. • The fluids flow counter-currently. • Cold fluid entering at the periphery and flowing toward the core. • Hot fluid entering at the core and flowing toward the periphery. • Can be mounted horizontally or vertically engineering-resource.com

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  9. Application • Liquid-liquid service • For gases or condensing vapors if the volumes are not too large engineering-resource.com

  10. Spiral flow in one channel and axial in the other • The spiral assembly contains conical covers, dished heads or extensions with flat covers. • The passage for axial flow is open on both sides and the spiral flow channel is welded on both sides. engineering-resource.com

  11. Application • When there is large difference in volumes of the two liquids • For heating or cooling gases, condensing vapors, or as reboilers. engineering-resource.com

  12. Combination flow • A conical cover distributes the fluid into its passage. • Part of the spiral is closed at the top, and the entering fluid flows only through the center part of the assembly. • A flat cover at the bottom forces the fluid to flow spirally before leaving the exchanger. engineering-resource.com

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  14. Application • Often used for condensing vapors. • Vapors first flow axially until their volume is reduced sufficiently for final condensing and sub-cooling in spiral flow. engineering-resource.com

  15. Advantages • Single-flow makes it ideal for cooling and heating sludge and slurries. • Good distribution • Low rate of fouling because of single-flow channel and curved-flow path. • If it fouls it can be easily cleaned. It can be easily cleaned with high pressure water or steam. engineering-resource.com

  16. Continued • Well suited for viscous fluids. • Flow can be assumed completely counter-current , when the fluids are flowing in spirals. No need for correction factor. • In axial flow , a large flow area affords a low pressure drop , which becomes especially important when condensing under vacuum. engineering-resource.com

  17. Limitations • Repairing is difficult. A leak cannot be plugged. However the possibility of leakage is small because of the thickness of the plate. • For spiral-axial flow , the temperature difference must be corrected. engineering-resource.com

  18. “DESIGN STEPS” engineering-resource.com

  19. Steps followed: • Calculation of heat duty • LMTD • Assume any value of design co-efficient • Calculation of heat transfer surface area • Reynold’s number • Selection of type of flow • Selection of equations • Calculation of S.O.P • Pressure drop calculations engineering-resource.com

  20. DESIGNING OF A LIQUID-LIQUID SPIRAL PLATE HEAT EXCHANGER engineering-resource.com

  21. PROBLEM: engineering-resource.com

  22. CONTINUED Material of construction is stainless steel having k=10 For it plate thickness Þ may be taken as 1/8’’, 3/16’’, ¼’’, 5/16’’ • HEAT DUTY: Q=m*Cp*∆T =6225×.71×(200-120) ×1.8 =636400 BTU/hr • LMTD(∆TM): {60-49.4}/ln{60/49.4}=54.5°C engineering-resource.com

  23. FOR FIRST TRIAL: • The approximate surface can be calculated using an assumed overall heat transfer coefficient. U=50 btu/(hr sq. ft. °F) so A=Q/(U*A*∆Tm*1.8) =130 sq. ft. engineering-resource.com

  24. CONTINUED • Because this a small exchanger assume, plate width W=24 in.=2ft therefore L=A/W*2=130/2*2 =32.5ft assume channel spacing=3/8in for both fluids engineering-resource.com

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  26. CONTINUED • Reynolds number : NRe=10,000{w/W*µ} Where w=mass flow rate W=channel plate width=24in µ=viscosity engineering-resource.com

  27. Hot side NRe={10,000*6.225}/{24*3.35} =774 Cold side NRe={10,000*5.925}/{24*8} =309 as laminar flow so spiral flow is selected for heat exchanger design engineering-resource.com

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  30. HEAT TRANSFER CALCULATIONS • Using eq. 3 hot side: ∆Th/∆Tm=32.6*[M.222/S.889]*[Zf/Zb].14*[w2/3* T2-T1]/∆Tm*ds/LW2/3] =32.6*[200.4.222/.843.889]*[6.2252/3*80/ 54.5]*[.375/242/3*32.5]=0.848 engineering-resource.com

  31. Cold side: ∆Tc / ∆Tm=0.927 FOR FOULING, eq{12}: It is assumed that there is no fouling on hot fluid side. So for cold fluid, ∆Ts/ ∆Tm=6000*[Cp/h]*[w{T2-T1}/ ∆Tm]*1/LW =6000*[.66/1000]*[5.925*90.4/54.5]*[1/32.5*24] =0.050 engineering-resource.com

  32. For plate: ∆Tw/ ∆Tm=500*[Cp/k]*[w*{T2- T1}/ ∆Tm]*[Þ/L*W] =500*[.66/10]*[5.925*90.4]/54.5*[.125/32.5*24]= 0.052 engineering-resource.com

  33. SUM OF PRODUCTS(SOP): • SOP=0.848+0.927+0.050+0.052 =1.877 Because SOP>1, assumed heat exchanger is inadequate . Surface area must be enlarged by increasing L or W. As in all equations L applies directly so new length adopted is 1.877*32.5=61ft. engineering-resource.com

  34. PRESSURE DROP CALCULATIONS: • HOT SIDE (Eq. 15): ∆P = 0.001*(L/s)*[W/(ds*H)]*[1.035*Z0.5/(ds+0.125)]* *[(Zf/Zb)0.17*(H/W)0.5+1.5+16/L] ∆P = [(0.001*61)/0.843]*[6.225/(0.375*24)]* *[{(1.035x3.350.5x1x240.5)/ (0.375+0.125)*6.2250.5} +1.5+16/61] = 0.461 psi COLD SIDE: ∆P = 0.65 psi Because the ∆P is less than allowable, ds can be decreased. engineering-resource.com

  35. SECOND TRIAL: • For 2nd trial, ¼’’ spacing for channels is adopted. • Because the heat transfer Eqn. for every factor except the plate varies directly with ds, a new SOP can be calculated. ∆Th/∆Tm = 0.848*(0.25/0.375) = 0.565 ∆Tc/∆Tm = 0.927*(0.25/0.375) = 0.618 ∆Ts/∆Tm = 0.050 ∆Tw/∆Tm = 0.052 SOP = 0.565+ 0.618+ 0.050+ 0.052 = 1.285 L = 1.285*32.5 = 41.8 ft. A = 41.8 x 2 x 2= 167 sq. ft. engineering-resource.com

  36. NEW PRESSURE DROP IS: HOT SIDE: ∆P = 0.605 psi COLD SIDE: ∆P= 0.861 psi The ∆Ps are less than the maximum allowable. The plate spacing can not be less than ¼’’ for 24’’ plate width; decreasing the plate width would result in a higher than allowable ∆P. Therefore, the design is acceptable. DIAMETER OF THE OUTSIDE SPIRAL: Using the eqn. Ds = [15.36 X L*( dsc + dsh + 2p) + C2]0.5 Ds = [15.36 X 41.8)[0.25+0.25+2(0.125)] + 82]0.5 Ds = 23.4’’ engineering-resource.com

  37. For a spiral plate exchanger, the best design is often that in which The outside diameter approx. equals the plate width. DESIGN SUMMARY: Plate width 24’’ plate length 41.8 ft. channel spacing ¼’’ (both sides) spiral dia 23.4’’ heat trasfer area 167 sq. ft. hot side ∆P 0.607 psi cold side ∆P 0.861 psi U 38.8 Btu/(hr.)(sq.ft.)(F) engineering-resource.com

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