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WASTE HEAT BOILER by Y57f2Z

VIEWS: 33 PAGES: 29

									   WASTE HEAT BOILER


Presented by Madeha
SIR SHAH MUHAMMAD




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GROUP MEMBERS




                             06-CHEM-33
                             06-CHEM-89
                             06-CHEM-101
                             06-CHEM-105
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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
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   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.


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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 within the design
    limitation of the equipment

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    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


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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.

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 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


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    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/ft2 oF
     Heat losses = 2%.
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    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.


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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



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Density at mean temperature,
        ρm = ρ (T/T2)
           = (0.0806) (492)/(1460)
           = 0.027 lb/ft3



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    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


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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


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 Water Flow Rate

∆H = H2 – H1
   = 1015 Btu/lb

m’ = Q \ (∆H + λ)
   = (28.13 X 106)/(1015)
   = 27,710 lb/hr


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LMTD weighted

Log-mean temperature difference

∆T =    (1500 – 366)-(500 -366)
       ln(1500 -366)/(500 – 366)

  = 468 oF

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  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


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Inside Film Coefficient

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
    = 0.208
hi = (2.44 X 0.208 X (167)0.8)/(1.77)1.8
  =10.9 Btu/ft2 hr oF

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  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


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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)


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We get,
          Ui Ai = Uo Ao

Ui = 9.6 X 2/1.77
   = 10.85 Btu/ft2 hr oF



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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.
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Area Calculation

So, the required total area is
Ao = 3.14 X 2 X 600 X (20/12)
   = 6280 ft2
Ai = 5558 ft2




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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
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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


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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
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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)


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∆Pg = 93 X 10-6 X w2f Le /ρgdi5

Where
Le = equivalent length = L+5di
(tube inlet and exit losses)




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∆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




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