[BHUVANESH]

Coal combustion reaction taking place inside the reactor, which is exothermic and heat of formation is 393.7 kJ/mol (32808.33 kJ/kg carbon). If 10 kg/hr of coal is fed into the reactor, what will be the insulation required for the reactor? (Assume complete combustion taking place)
Data:
ID of reactor, Di               = 2.067 ’’ = 0.0525018 m
OD of reactor, Do               = 2.375 ‘’ = 0.060325 m
Length of reactor, L                               = 3 m
Thermal conductivity of steel, ks                    = 16.3 W/(m.K)
Thermal conductivity of insulation (Kaowool),kins    = 0.25 W/(m.K)
Temperature inside the cylinder,T1                 = 700 OC
Temperature outside the insulation (desired), Tout   = 70 OC

I tried with normal conduction formula (Q= ΔT/(ln(R0/Ri)/(2πLk))and i'm getting 0.8 mm insulation thickness...

[Enthalpy]

If no condition is given on the accepted heat leak, then the minimum insulation thickness is: zero. With no insulation at all, the mere insulation by air permits a flame hotter than 700°C. Nor do I believe that you accept to lose all the heat through the walls.

As a sidenote, coal is not carbon nor graphite. It's a hydrocarbon mainly, with additional compounds, so the heating value is to be checked.

[BHUVANESH]

If no condition is given on the accepted heat leak, then the minimum insulation thickness is: zero. With no insulation at all, the mere insulation by air permits a flame hotter than 700°C. Nor do I believe that you accept to lose all the heat through the walls.

As a sidenote, coal is not carbon nor graphite. It's a hydrocarbon mainly, with additional compounds, so the heating value is to be checked.

Thank you sir...

Actually i wanted to conduct an experiment for coal combustion, in downer reactor with respect to above said conditions. Without insulation is not possible as heat lose ll be more. What ll be the calculation i have to check for?? (I'm assuming complete combustion), Can i calculate heat loss through heat of formation of reaction C+O₂  CO₂

Help me with this..

[Enthalpy]

You have to decide how much heat you accept to lose through the walls. For instance, how big a temperature difference is allowed to be. The combustion products being mainly nitrogen, the heat capacity varies little with such temperatures, so if you accept 20K difference from end to end, it means 1/35 the heat can be lost.

Different approach: Put some insulation, decide that it suffices.