waste heat boiler engineering-resource.com. group members 06-chem-06 06-chem-46 06-chem-48...
TRANSCRIPT
WASTE HEAT BOILER
engineering-resource.com
GROUP MEMBERS
06-CHEM-0606-CHEM-4606-CHEM-48
engineering-resource.com
INTRODUCTIONWaste 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
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
Heat Recovery In Process Plants
Competitive market conditions on the most products make it essential to reduce processing cost
The cost of fuels keeps risingLimited fuel availability is already
causing plant interruptionsThere is restriction in using some
of the lower-cost fuels because of environmental pollution
engineering-resource.com
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
ApplicationsFor 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
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
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/ft2 oF
Heat losses = 2%.
Data Given
engineering-resource.com
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
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
Density Calculations
engineering-resource.com
Density at mean temperature, ρm = ρ (T/T2)
= (0.0806) (492)/(1492)
= 0.027 lb/ft3
engineering-resource.com
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
Heat Duty
engineering-resource.com
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
From steam tables
engineering-resource.com
∆H = H2 – H1 = 1015 Btu/lb
m’ = Q \ (∆H ) = (28.13 X 106)/(1015) = 27,710 lb/hr
Water Flow Rate
engineering-resource.com
LMTD weighted
Log-mean temperature difference
∆T = (1500 – 366)-(500 -366) ln(1500 -366)/(500 – 366)
= 468 oF
engineering-resource.com
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
hi = 2.44 X w0.8 X C/di1.8
C = (CP/µ)0.4 X k0.6
= (0.287/0.084)0.4 X (0.0322)0.6 = 0.208hi = (2.44 X 0.208 X (167)0.8)/(1.77)1.8
=10.9 Btu/ft2 hr oF
Inside Film Coefficient
engineering-resource.com
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
= 0.10417Hence, Uo= 9.6 Btu/ft2 hr oF
engineering-resource.com
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
We get, Ui Ai = Uo Ao
Ui = 9.6 X 2/1.77
= 10.85 Btu/ft2 hr oF
engineering-resource.com
Putting back in eq.2
Ao = (28.13X106)/(468 X 9.6) = 6261 ftAo= л nt d L6261 = 3.14X2X600(L/12) L = 19.93 ftso required length L of the tubes=19.93
ft. Use 20 ft.
engineering-resource.com
So, the required total area isAo = 3.14 X 2 X 600 X (20/12)
= 6280 ft2
Ai = 5558 ft2
Area Calculation
engineering-resource.com
Thickness Of ShellTs = P(D+2C)/ [(2fJ-P)+C]Where,P = design pressureD = inner diameter of shellC = corrosion allowancef = permissible stress factorJ = welded joint factor
engineering-resource.com
From literaturewe know that, for Carbon steelC= 1/8 of an inchf= 13400psiJ=0.75 - 0.95We get,Ts = 0.6584 in
engineering-resource.com
Outer diameter of tube bundle = 1.32 X do X (nt)½ = 64.66 inProviding allowances for welding, = 64.66 + 6 = 70.66 inShell diameter, DS = 70.66 X 1.20 = 84.8 in
engineering-resource.com
PRESSURE DROP CALCULATIONS
Tube side pressure drop:V = 0.05 W/diρg
V = 19520 ft/ hrRe = ρgdiV/µ
= 890.12 f = 0.02 (from graph)
engineering-resource.com
∆Pg = 93 X 10-6 X w2f Le /ρgdi5
WhereLe = equivalent length = L+5di
(tube inlet and exit losses)
engineering-resource.com
∆Pg = 93 X 10-6 X 1672 X 0.02 X (20+5 X 1.77)
0.0267 X (1.77)5
= 3.23 in. WC
engineering-resource.com