DESIGNING OF CRACKING FURNACE

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 (Q_n)        = 92698183.06 /0.89                                           

                                                          = 104015016.9 Btu/hr

Radiant section heat duty (Q_R)            = 36801986.83 Btu/hr

Convective Section heat duty (Q_C)      = 55896196.23 Btu/hr

Radiant Flux

(Q_R/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 (G_f) = (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 

A_CP= (exposed tube length)*(center to center spacing)*(number of tubes except shield tubes)

                                                                   = 708.58 ft2

Inside Surface of Shell

Dia of Furnace (D_F) = 12 ft

A_s= 1677.6144ft2

Refractory Surface

A_w=A_s - A_CP

                                 = 969.0335 ft2

Effective Absorbitiviy

X = (center to center) / (outer Dia)

             = 1.5

α= 0.963

α A_R =A _shield+ α A_cp        = 913.346 ft²   

A_w/ α A_{R =} 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 (T_t) = 1355 ^oF

Conclusion

*By hit and trail we found T_G=2181^oF is the temperature at which radiant zone heat absorption and radiant zone heat balance become equal

At F=0.566

Radiant zone Heat transfer

(Q_R/ α A_RF)=1730((T_G+460/1000)⁴-(T_t+460/1000)⁴) +7(Tg-Tt)

                 =71187

Radiant Zone heat Balance

(Q_R/ α A_RF)=Qn/ α A_RF (1-.02-Qg/Qn)

                =71187

QR =36801986.83 Btu/hr