The performance of an evaporator(fire tube or water tube,bare or finned) may be obtained by solving the following equations:(The performance evaluation of superheater,economizers are more involved and readers may see my books,"Waste Heat Boiler Deskbook" or "Steam Plant Calculations Manual").
tg1,tg2,ts=gas inlet and exit temperatures and saturation steam temperature,F
Cpg=gas specific heat,btu/lbF
hl=heat loss,factor.1 % heat loss means hl=0.99
U=overall heat transfer coefficient,Btu/ft2hF
Simplifying the above,we have:
of a Fire Tube Boiler
Let us assume that a fire tube boiler is operating as shown below in clean conditions:
gas inlet temperature=1500 F
gas flow=100,000 lb/h
gas exit temperature in clean conditions=500 F
steam pressure=150 psig (ts=366 F)
feed water temperature=230 F
blow down=5 %
tubes: 2x1.77in,600 tubes,20 ft long,surface area=6280 ft2(OD basis)
Duty=100,000x.98x.287x1000=28.1 Mm Btu/h and steam=27,710 lb/h
Let us compute U.
ln[(1500-366)/(500-366)]=Ux6280/(100,000x0.287x0.98) or U=9.56 Btu/ft2hF
Let us assume that a scale has formed and the equivalent additional fouling=0.05 ft2hF/Btu. What happens to the performance?
This will decrease U to: 1/(1/9.56)+.05)=6.5 Btu/ft2hF
This in turn will affect the exit gas temperature as follows:
ln[(1500-366)/(tg2-366)]=6.5x6280/(100,000x.98x0.287) or tg2=630 F
The duty=100,000x0.98x0.287x(1500-630)=24.4 Mm Btu/h and steam generation=24,100 lb/h
Table below summarises the results.(For more detailed calculations of tube wall temperatures,heat flux,see my books) Note that even though the heat flux has reduced due to lower U in fouled condition,the tube wall temperature is significantly higher!
|drop across scale,F||6||207|
|tube wall temp,F||377||577|
|exit gas temperature,F||500||630|
Effect of scale conductivity on tube wall temperatures
|scale||ther cond,Btu in/ft2h||fouling resistance||temperature rise,F|
|magnetic iron oxide||20||0.001||10|