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 1500 o F to 500 o F. Gas analysis is (vol%) CO 2 =12, H 2 O=12, N 2 =70, and O 2 =6; gas pressure is 5 in.WC. Steam pressure is 150 psig, and feed water enters at 220 o F. Tubes used are in 2 in. OD*1.77 in. ID engineering-resource.com

9. Fouling factors are Gas side (ft) = 0.002 ft 2 h o F/Btu Steam side (ff) = 0.001 ft 2 h o F/Btu Tube metal thermal conductivity, k m =25 Btu/ft 2 h o F Steam side boiling heat transfer coefficient, h o = 2000 Btu/ft 2 o F Heat losses = 2%. Data Given engineering-resource.com

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

11. MW mix = (MW i X i ) =(0.12)(44)+(0.12)(18)+(0.70)(28)+(0.06)(32) = 28.96 lb/lb mole Density at standard temperature, ρ = 28.96/359 = 0.0806 lb/ft 3 Density Calculations engineering-resource.com

12. Density at mean temperature, ρ m = ρ (T/T 2 ) = (0.0806) (492)/(1492) = 0.027 lb/ft 3 engineering-resource.com

13. Boiler duty Q = W g C P (T 1 –T 2 )(1-L\100) = 100,000 X 0.98 X 0.287 X (1500 -500) = 28.13 X 10 6 Btu/hr Heat Duty engineering-resource.com

14. Enthalpies of saturated steam H 1 = 1195.5 Btu/lb Enthalpies of saturated water H 2 = 338 Btu/lb Latent heat of steam, λ = 857.8 Btu/lb From steam tables engineering-resource.com

15. H = H2 – H1 = 1015 Btu/lb m = Q \ (H ) = (28.13 X 10 6 )/(1015) = 27,710 lb/hr Water Flow Rate engineering-resource.com

16. LMTD weighted Log-mean temperature difference T = (1500 – 366)-(500 -366) ln(1500 -366)/(500 – 366) = 468 o F 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. hi = 2.44 X w 0.8 X C/di 1.8 C = (C P /µ) 0.4 X k 0.6 = (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/ft 2 hr o F Inside Film Coefficient engineering-resource.com

19. Overall Heat Transfer Coefficient 1/U = (d o /d i )/ hi + ff o + ff i (d o /d i ) + do ln(d o /d i )/24Km +1/ho = 0.10+0.001+0.00226+0.00041+0.00 05 = 0.10417 Hence, U o = 9.6 Btu/ft 2 hr o F engineering-resource.com

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

21. We get, U i A i = U o A o U i = 9.6 X 2/1.77 = 10.85 Btu/ft 2 hr o F engineering-resource.com

22. Putting back in eq.2 A o = (28.13 X 10 6 )/(468 X 9.6) = 6261 ft A o = л n t d L 6261 = 3.14 X 2 X 600(L/12) L = 19.93 ft so required length L of the tubes=19.93 ft. Use 20 ft. engineering-resource.com

23. So, the required total area is A o = 3.14 X 2 X 600 X (20/12) = 6280 ft 2 Ai = 5558 ft 2 Area Calculation 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, T s = 0.6584 in engineering-resource.com

26. Outer diameter of tube bundle = 1.32 X d o X (n t )½ = 64.66 in Providing allowances for welding, = 64.66 + 6 = 70.66 in Shell diameter, D S = 70.66 X 1.20 = 84.8 in engineering-resource.com

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

28. P g = 93 X 10 -6 X w 2 f L e /ρ g di 5 Where L e = equivalent length = L+5di (tube inlet and exit losses) engineering-resource.com

29. P g = 93 X 10 -6 X 167 2 X 0.02 X (20+5 X 1.77) 0.0267 X (1.77) 5 = 3.23 in. WC engineering-resource.com