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WASTE HEAT BOILER. GROUP MEMBERS. 06-CHEM-06 06-CHEM-46 06-CHEM-48. INTRODUCTION. Waste heat boiler:

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WASTE HEAT BOILER


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waste heat boiler
WASTE HEAT BOILER

engineering-resource.com

group members
GROUP MEMBERS

06-CHEM-06

06-CHEM-46

06-CHEM-48

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introduction
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.

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slide4
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. 

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heat recovery in process plants
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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slide6
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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applications
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
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
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%.

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slide10
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
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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slide12
Density at mean temperature,

ρm =ρ (T/T2)

= (0.0806) (492)/(1492)

= 0.027 lb/ft3

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heat duty
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
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

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water flow rate
Water Flow Rate

∆H = H2 – H1

= 1015 Btu/lb

m’ = Q \ (∆H )

= (28.13 X 106)/(1015)

= 27,710 lb/hr

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lmtd weighted
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
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
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

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overall heat transfer coefficient
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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slide20
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)

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slide21
We get,

Ui Ai = Uo Ao

Ui = 9.6 X 2/1.77

= 10.85 Btu/ft2 hr oF

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slide22
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
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
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
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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slide26
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

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pressure drop calculations
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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slide28
∆Pg = 93 X 10-6 X w2f Le /ρgdi5

Where

Le = equivalent length = L+5di

(tube inlet and exit losses)

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slide29
∆Pg = 93 X 10-6X 1672X 0.02 X (20+5 X 1.77)

0.0267 X (1.77)5

= 3.23 in. WC

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