Volume of the Reactor
Volumetric Flow (Vo) = 61.164 ft3 /sec
Space Time (T) = 0.65 sec T=V/Vo
Reactor Volume (V) = T*Vo
= 0.65*61.164
= 39.7566 ft3
Heat duty of Convection & Radiation sections
Heat duty of Furnace (Q) = 92698183.06 Btu/hr
Efficiency = 0.89
Heat release by Combustion (Qn) = 92698183.06 /0.89
= 104015016.9 Btu/hr
Radiant section heat duty (QR) = 36801986.83 Btu/hr
Convective Section heat duty (QC) = 55896196.23 Btu/hr
Radiant Flux
(QR/A) = 25000 Btu/hr.ft2
Radiant Surface
A = (Radiant section heat duty) / (Radiant Flux)
= 1472.08 ft2
Number of Tubes
Outer Dia of Tubes (D) = 0.1979 ft
A=πDL
Length of Tubes (L) = A/πD
= 1472.08/0.62172
= 2367.743 ft
*Length of a single tube is 40 ft with 38.5 ft exposed length
So Number of tubes = 2367.743/38.5
= 61.5 tubes
Let N = 68 *of which six are shield tubes
Flue Gases
Let Excess Air = 19 %
Then
Flue gas rate (Gf) = (Heat release byCombustion/106)*(822*(780*Fraction of excess air)
= 100915.3694 Lb/hr
Mass velocity (G) = 0.3617 Lb/sec*ft2
Cold Plane Area
ACP= (exposed tube length)*(center to center spacing)*(number of tubes except shield tubes)
= 708.58 ft2
Inside Surface of Shell
Dia of Furnace (DF) = 12 ft
As= 1677.6144ft2
Refractory Surface
Aw=As - ACP
= 969.0335 ft2
Effective Absorbitiviy
X = (center to center) / (outer Dia)
= 1.5
α= 0.963
α AR =A shield+ α Acp = 913.346 ft2
Aw/ α AR = 1.06
Other required data
Mean beam length = 12 ft
Partial pressure of Radiating Gases (CO2+H2O)= 0.24774 atm
PL=0.24774*12
=2.973 atm*ft
Emissivity of Radiating gases = 0.42
Tube skin temperature (Tt) = 1355 oF
Conclusion
*By hit and trail we found TG=2181oF is the temperature at which radiant zone heat absorption and radiant zone heat balance become equal
At F=0.566
Radiant zone Heat transfer
(QR/ α ARF)=1730((TG+460/1000)4-(Tt+460/1000)4) +7(Tg-Tt)
=71187
Radiant Zone heat Balance
(QR/ α ARF)=Qn/ α ARF (1-.02-Qg/Qn)
=71187
QR =36801986.83 Btu/hr
good explained
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