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WASTE HEAT BOILER

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  1. WASTE HEAT BOILER Presented by Madeha SIR SHAH MUHAMMAD engineering-resource.com

  2. GROUP MEMBERS 06-CHEM-33 06-CHEM-89 06-CHEM-101 06-CHEM-105 engineering-resource.com

  3. Heat Recovery In Process Plants • Competitive market conditions on the most products make it essential to reduce processing cost • The cost of fuels keeps rising • Limited fuel availability is already causing plant interruptions • There is restriction in using some of the lower-cost fuels because of environmental pollution engineering-resource.com

  4. Increasing emphasis is being placed on the minimizing thermal pollution • Increasing amounts of elevated-temperature flue gas streams are becoming available from gas turbines, incinerators, etc. engineering-resource.com

  5. Advantages Of Steam Generation • Results in relatively compact heat recovery system • Usually the lowest initial installation cost • Fewer operating problems • Rapid response rate • Will permit some adjustability in heat removal capacity, by raising or lowering the steam side operating pressure withinthe design limitation of the equipment engineering-resource.com

  6. Disadvantages Of Steam Generation • Must operate at high pressure to ensure economic justification (650-150 psig) • Cannot cool elevated temperature streams through as wide a range as other heat recovery techniques • The cost of water treatment may significantly reduce the economics advantage • Has limited flexibility in utilizing the recovered energy engineering-resource.com

  7. Applications • For process heating. (Steam usually generated at 125-650 psig) • For power generation. (usually generated at 650-1500 psig and will require superheating) • For use as a diluents or stripping medium in a process. This is a low-volume use. engineering-resource.com

  8. Problem Determine the size of a fire tube waste heat boiler required to cool 100,000 lb/h of flue gases from 1500oF to 500oF. Gas analysis is (vol%) CO2=12, H2O=12, N2 =70, and O2 =6; gas pressure is 5 in.WC. Steam pressure is 150 psig, and feed water enters at 220oF. Tubes used are in 2 in. OD*1.77 in. ID engineering-resource.com

  9. Data Given • Fouling factors are Gas side (ft) = 0.002 ft2 h oF/Btu Steam side (ff) = 0.001 ft2 h oF/Btu • Tube metal thermal conductivity, km =25 Btu/ft2 h oF • Steam side boiling heat transfer coefficient, ho = 2000 Btu/ft2oF • Heat losses = 2%. engineering-resource.com

  10. At the average gas temperature of 1000oF, the gas properties can be shown to be • Cp =0.287 Btu/lb oF • µ=0.084 lb/ft h • k =0.0322 Btu/ft h oF. engineering-resource.com

  11. Density Calculations MWmix = ∑ (MWi Xi) =(0.12)(44)+(0.12)(18)+(0.70)(28)+(0.06)(32) = 28.96 lb/lbmole Density at standard temperature, ρ = 28.96/359 = 0.0806 lb/ft3 engineering-resource.com

  12. Density at mean temperature, ρm =ρ (T/T2) = (0.0806) (492)/(1460) = 0.027 lb/ft3 engineering-resource.com

  13. Heat Duty Boiler duty Q = Wg CP(T1 –T2)(1-L\100) = 100,000 X 0.98 X 0.287X (1500 -500) = 28.13 X 106 Btu/hr engineering-resource.com

  14. From steam tables Enthalpies of saturated steam H1= 1195.5 Btu/lb Enthalpies of feed water H2 = 180 Btu/lb Latent heat of steam, λ = 857.8 Btu/lb engineering-resource.com

  15. Water Flow Rate ∆H = H2 – H1 = 1015 Btu/lb m’ = Q \ (∆H + λ) = (28.13 X 106)/(1015) = 27,710 lb/hr engineering-resource.com

  16. LMTD weighted Log-mean temperature difference ∆T = (1500 – 366)-(500 -366) ln(1500 -366)/(500 – 366) = 468 oF engineering-resource.com

  17. Flow per tube Typically w ranges from 100 to 200 lb/hr for a 2 in tube. Let us start with 600 tubes, hence w = 100,000/600 = 167 lb/hr engineering-resource.com

  18. Inside Film Coefficient hi = 2.44 X w0.8 XC/di1.8 C = (CP/µ)0.4X k0.6 = (0.287/0.084)0.4 X (0.0322)0.6 = 0.208 hi = (2.44 X 0.208 X (167)0.8)/(1.77)1.8 =10.9 Btu/ft2 hr oF engineering-resource.com

  19. Overall Heat Transfer Coefficient 1/U = (do/di)/ hi + ffo + ffi (do/di) + do ln(do/di)/24Km +1/ho = 0.10+0.001+0.00226+0.00041+0.0005 = 0.10417 Hence, Uo= 9.6 Btu/ft2 hr oF engineering-resource.com

  20. If U is computed on the basis of tube inner surface area, then Ui is given by the Q= Ui Ai (LMTD) (1) If U is computed on the basis of tube outer surface area, then Ui is given by the Q= Uo Ao (LMTD) (2) engineering-resource.com

  21. We get, Ui Ai = Uo Ao Ui = 9.6 X 2/1.77 = 10.85 Btu/ft2 hr oF engineering-resource.com

  22. Putting back in eq.2 Ao = (28.13X106)/(468 X 9.6) = 6261 ft Ao= л nt d L 6261 = 3.14X2X600(L/12) L = 19.93 ft so required length L of the tubes=19.93 ft. Use 20 ft. engineering-resource.com

  23. Area Calculation So, the required total area is Ao = 3.14 X 2 X 600 X (20/12) = 6280 ft2 Ai = 5558 ft2 engineering-resource.com

  24. Thickness Of Shell Ts = P(D+2C)/ [(2fJ-P)+C] Where, P = design pressure D = inner diameter of shell C = corrosion allowance f = permissible stress factor J = welded joint factor engineering-resource.com

  25. From literature we know that, for Carbon steel C= 1/8 of an inch f= 13400psi J=0.75 - 0.95 We get, Ts = 0.6584 in engineering-resource.com

  26. Outer diameter of tube bundle = 1.32 X do X (nt)½ = 64.66 in Providing allowances for corrosion, = 64.66 + 6 = 70.66 in Shell diameter, DS = 70.66 X 1.20 = 84.8 in engineering-resource.com

  27. PRESSURE DROP CALCULATIONS Tube side pressure drop: V = 0.05 W/diρg V = 19520 ft/ hr Re = ρgdiV/µ = 890.12 f = 0.02 (from graph) engineering-resource.com

  28. ∆Pg = 93 X 10-6 X w2f Le /ρgdi5 Where Le = equivalent length = L+5di (tube inlet and exit losses) engineering-resource.com

  29. ∆Pg = 93 X 10-6X 1672X 0.02 X (20+5 X 1.77) 0.0267 X (1.77)5 = 3.23 in. WC engineering-resource.com