[How to] find Reactor Heat Transfer Area Theoretically

Gotta request from a reader asking how to Determine Reactor Heat Transfer area theoretically,

Getting into point, A reactor usually used for carrying out reactions, extractions, distillations, pH treatments, etc, that means a common equipment for multiple operations, and to avoid some problems, reactors comes in different Materials of Construction, out of those many MOC's most commonly used were SSR- Stainless Steel Reactors, GLR- Glass Lined Reactors, MSGLR- Mild Steel Glass Lined Reactors, PPR- Poly Propylene Reactors etc.
Usually SSR can be made with two high proportional Steel grades, those were SS316, SS304,

they both vary in Molybdenum content

Both SS304 and SS316 are austenitic stainless steels.

         CHEMICAL COMPOSITION OF SS304 AND SS316               

              C       Mn       P         S        Si           Cr              Ni             Mo

        

SS304  0.08    2.0    0.045   0.03    1.0    18.0-20.0     8.0-10.0        -

SS316  0.08    2.0    0.045   0.03    1.0    16.0-18.0    10.0-14.0    2.0-3.0 

* From the above chemical composition chart we can see the main difference between SS304 and SS316 is that SS316 contains 2%-3% molybdenum and SS304 has no molybdenum.Now the question arises that we are paying double money for SS316 only for that 2.5% 'Moly'.The answer is yes, because this  "molybdenum" is added to improve the corrosion resistance to chlorides (like sea water).That is why in coastal area plants SS316 is used.

Lets start the Calculation of Heat transfer area of Reactor,

So usually one thumb rule is used for calculation,
* Length to Dia ratio is considered in between 1.2 to 1.6,
L/D=1.3 (i make it 1.3)

and the data i needed is the Batch Reactor Volume, and i take it 200L (0.2KL),

so my reactor volumes compromises of one cylindrical middle part and a torispherical bottom(ASME head),



so Volume of Reactor(0.2)=(Cylindrical Volume+Torispherical Dish volume)
                                          =((Pi*r*r*l)+((Pi/24)*D*D*D))    *here D is Internal dia
                                          =((Pi*(D/2)*(D/2)*(1.3D))+((Pi/24)*D*D*D)))
                                          =(1.022+0.1311)*D*D*D
                                    0.2 =1.153*D*D*D
so now i got the value of D=0.5576 m=0.56 m, L=0.728 m=0.73 m,

So now coming to the Calculation of Heat transfer Area, if you cut off the reactor cylindrical surface and spread it parallel to the ground it will take the shape of a Rectangle, so the area of a rectangle is Length * Breadth, and in this case the length becomes perimeter of cylinder and Breadth becomes length of cylinder,





so Heat Transfer area of cylinder= Perimeter of the Rectangle * Length of Cylinder
                                                    = (2*Pi*r) * (L)
                                                    = (2*3.147*(0.56/2)) * (0.73)
                                                    = 1.2864 Sq.m= 1.29 Sq.m
And now we got the Heat transfer Area of Cylinder, and we still need to calculate the Heat transfer surface of the Torispherical Dish, 

Area of a torispherical dish = (𝝅 / 4) x ((1.147 x D)^2) = (3.141/4) x ((1.147 x 0.5576)^2)
                                                = 0.785 x 0.409 = 0.32 Sq.m

Total heat transfer area of the reactor = Area of cylinder + Area of torispherical dish 
                                                                   = 1.29 + 0.32 = 1.61 Sq.m




That's it, Cheers.

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Hi! I am Ajay Kumar Kalva, Currently serving as the CEO of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.
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