WASTE HEAT BOILER

1. WASTE HEAT BOILER engineering-resource.com

2. GROUP MEMBERS 06-CHEM-06 06-CHEM-46 06-CHEM-48 engineering-resource.com

3. INTRODUCTION Waste heat boiler: A heat-retrieval unit using hot by-product gas or oil from chemical processes; used to produce steam in a boiler-type system is known as waste heat boiler. It is also known as gas-tube boiler. engineering-resource.com

4. Waste heat boilers may be horizontal or vertical shell boilers or water tube boilers. They would be designed to suit individual applications ranging through gases from furnaces, incinerators, gas turbines and diesel exhausts. The prime requirement is that the waste gases must contain sufficient usable heat to produce steam or hot water at the condition required. Waste-heat boilers may be designed for either radiant or convective heat sources.  engineering-resource.com

5. 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

6. 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

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)/(1492) = 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 saturated water H2 = 338 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 ) = (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 Uo 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 welding, = 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