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  • Friday, 18 March 2016

    [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)))









     (10% of 1.29Sq.m)





    Specially for my Visitors or Readers, we have simulated the Calculation of heat transfer area calculation, Download this
    About The Author


    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.
    Follow Me on Twitter AjaySpectator & Computer Innovations

    18 comments:

    1. Very helpful one..keep going@ jaii

      ReplyDelete
      Replies
      1. sure mate, if anything needed then just write it in comments or send it through WHAT SHOULD I WRITE NEXT column.
        :)

        Delete
    2. Very helpful one..keep going@ jaii

      ReplyDelete
    3. Mole balences n stoichometry basics;Material balence of some 'x' process and calculating theoretical yield,Actual yield.

      ReplyDelete
    4. Sme query regarding dat "write abt next column"
      It is displaying error whenever im posting my queries..so help me out

      ReplyDelete
      Replies
      1. ok fine, please message me from contact us page

        Delete
    5. how to calculate the plate heat exchanger no of plates
      pls do the calculation.
      thapliyalashish48@gmail.com

      ReplyDelete
      Replies
      1. ok Ashish, but first of all please mention me what is the mass flow rate/ volumetric flowrate of the two fluids and also what material medium you are using as baffles for structural support and also whats the volume you want to install the plate heat exchanger....!!!

        Delete
      2. ok first of thnks for ur prompt reply.
        here is the detail
        utility fluid
        service fluid 420m3/hr
        utility fluid approx 470m3/hr

        titanium grade 2 plate double pass plate heat exchanger.
        kindly mail me ur mail id if possible will inform u further .
        i have calculated although but need to rectify it.
        pls mail ur personal id .
        thapliyalashish48@gmail.com

        Delete
      3. ajaykalva823@gmail.com, u possibly mention your fluids temperatures also, and also what is the utility fluid, whether its brine[what temp.°C], service fluid properties....!!!

        Delete
      4. brine/bittern specifications
        temp inlet 30 degree
        outlet temp 98 degree
        effluent inlet temp 108
        effluent outlet 40 we need to have
        cp of brine 3.28kj/kg c
        cp of effluent 3.5 kj/kg c
        density of bittern 1258 kg/m3
        density of effluent 1238 kg /m3
        calculate the no of plates required .pls do step wise calculation

        Delete
    6. @Ashish, sorry for the late reply, what ever data you have mentioned is something un natural to be found, as this won't apply to any case upto my knowledge, as the temperature of the brine should be less than that of RT water but you have mentioned it to be 30 deg, and also whatever the utility outlet temperature you used shouldn't cross a standard difference of 5 degree, but sometime practically it will cross 5 degree and reach upto 10-12degree maximum, but here you mentioned 68 degree difference, so this will cause some severe effect to the utility plants as they can't recover back from this delta T value, so finally i want to tell you that some error lies in the data you provided, for successful calculation the inlet and outlet temperatures of cold fluid should be less than that of the final temperature of the hot fluid,

      Hope you understand this....... Still if you want an spreadsheet for calculating the plates number and other parameters, we are happy to help you, thanks for commenting

      ReplyDelete
    7. Dear ajay you did'nt get my actual point .
      let me explain you .we are having a stripping column in which the brine is feeded from the top and steam is used for stripping purpose .the vapour bromine gets condensed and get recovered from the condensers .the brine is feed normally at 98 degree centigrade and the effluent at the bottom of the column reached upto 108 degree and is passed through phe and feed brine which is at 30 degree is pass through the phe and utilize the heat of effluent and get 98 in degree centigrade.and outlet effluent gets 45in degree after heat transfer through phe to bittern / brine
      hope u get the exact idea about the flow across phe .

      ReplyDelete
    8. @Ashish, then here is the calculation that will extend upto calculation of heat transfer area, coz u didnt mentioned me the volume dimensions that you have for installing this PHE,

      First of all we need to check the feasibility of the case whether the data meets our requirement or not, so as per your data,
      Heat to be taken off from effluent to cool from 108 to 45= 420 x 1238 x 3.5 x ( 108-45 ) = 114651180 Kj/hr = 31847550 Watts,
      Heat that our brine need to take from the effluent by sensible heat change from 30 to 98 = 470 x 1258 x 3.28 x ( 98-30 ) = 131874630.4 Kj/hr = 36631841.8 Watts,

      As our brine can take upto 36,631,841.8 Watts of energy, there wont be any problem in execution of out plan using PHE, and now lets coming to Area calculation, the most probable appropriate U value will be 2500 watt/Sq.m K (this is the minimum U value considered for a PHE),

      So now Area A= 36631841.8 / ( 2500 x ( ((108-30)-(98-45))/Ln ((108-30)/(98-45)) ) ) = 226 Sq.m, if you can consider the maximum U Value as 4000-4500 W/Sq.m. K, then the required area would be 141.5 Sq.m,

      Now coming to the thickness calculation, Thickness = ( 23 x 226 x (98-30) ) / 36631841.8 = 9.64mm, here figure 23 is the thermal conductivity of the titanium grade 2 in W/m. K,

      As usually the chevron angle should be in between 25 degree to 65 degree, and the enlargement factor will be taken in between 1.1 and 1.25 as this is indirectly means some extended surface,

      And if you can tell me the volume, where the PHE is supposed to be installed, then i can make the number of plates calculation.

      Thats it, if anything you got wrong comment here, thanks

      ReplyDelete
    9. Q1. The topic is well explained for jacket type ,but heat transfer area if practically we will see in jacket reactor is of cylinder + torrispherical bottom dish only ,nothing is present in top torrispherical dish , so we can take area 5% instead of 10%
      Q2. what will we take if it is limpet type?
      Q2. in attached excel sheet if you enter 2 m3 ,l/d=1.3 result comes to be 20m2 which is very high.

      ReplyDelete
      Replies
      1. Hey anshu, thanks for reminding me of the wrong simulation, and i've updated it, now try to download and work it out, and what ever the case may be with Jacket having Annular surface or limpet coil because the only thing that changes is developing of air pockets in case of annular jackets, which is a advantage in limpet coil, for a HT area of torispherical dish we will consider 11% of the cylinder HT area.

        Delete
    10. Hey Ajay,
      I have a question please: Why did you say the L/D is between 1.2 to 1.6 ? Do you have reference for that? and do you know what is the L/D for plug flow reactors I mean can I use 1.2-1.6 for PFR ?

      Thanks

      ReplyDelete
      Replies
      1. Hey Mr.Anonymous,

        there wont be any reference for that, i've considered it based on many design, that i've seen, and also mostly, it will be considered in between 1.05 to 1.1,

        PFR dont have an agitator, and composition will vary along the length, then L/D doesn't make any sense.

        Regards,
        PHARMA ENGINEERING

        Delete

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