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  • Wednesday, 14 September 2016

    [How To] Calculate Rate of Distillation in a Batch Reactor

    Hello guyzzz..........!!!

    Back after many months and sorry for not being regular here, i've received a query from one of the visitor, and i think many of you too have the same question i.e., How to calculate the rate of distillation in a batch reactor....??

    And for this query today i gonna put a end mark, i'll explain you with an example here,
    but, before that you need to know some basics like, when are we going to do distillation in a batch reactor during a product manufacturing......!!
    Most commonly there are 2 cases,

    1) after completion of reaction, for impurity removals we will do some workups with solvents like water, which is followed by product layer concentration, simply called as product concentration,

    2) The 2nd one will be removal of Moisture from the raw materials, for this the raw materials should be made soluble in a suitable solvent and then the solvent medium is distilled off under vacuum, which will remove the water molecules from the raw materials and make them suitable for carrying out reaction, this is usually done for moisture sensitive products.

    So, here now i'll demonstrate an example case study,

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    Our case is, we need to calculate the distillation rate in a reactor having a product layer of Ethyl Acetate, and the distillation should be carried out with vacuum NLT 650 mmHg, and temperature NMT 40 deg C,

    Let suppose our reactor capacity is 10 KL with MOC SS316, and the product layer volume is around 5 KL say,

    Now we need to collect some raw data like HT area of SSR, Specific heat & Density of Solvent, etc.

    10 KL SSR will have an HT area of 19.82 Sq.m, but the effective HT area for this operation will be equals to Total HT area x Occupancy of reactor,

    Occupancy of reactor = Reaction mass volume / Reactor volume = 5 KL / 10 KL = 0.5,

    Effective HT Area = 19.82 x 0.5 = 9.91 Sq.m,

    For More info on calculating HT area of reactors, Read this:

    [How To] Calculate Reactor HT Area

    Density of Ethyl Acetate is 897 Kg/cu.m, Specific heat : 0.454 KCal/Kg. K , Latent Heat of vaporisation is 87.6 KCal/Kg,

    As the MOC of the reactor is SS316, the Overall Heat Transfer Coefficient can be considered as 250 KCal/Sq.m. Hr. K.

    If you want to calculate the Overall Heat Transfer of a reactor with more accuracy, then read this post,

    [How To] Calculate Overall Heat Transfer Coefficient 

    So now, getting into our calculation we need to calculate the amount of energy that our HT area is going to transfer per hour, and that we can calculate as follows,

    Q = U x A x LMTD,

    Sorry guys, i forgot to calculate the LMTD previously, anyway now i'll do it for you,

    for calculating the LMTD we need two streams, one is utility stream and the other one is batch volume,

    So , coming to Hot water utility, now you may have one doubt that why i've used Hot water as utility, the reason is we have a Specification that reaction mass temperature shouldn't exceed 40 deg C, thats why i've considered Hot water instead of Steam, and now another question may arise in your mind, what can be the hot water set point, usually we will maintain a Delta T of 10 deg C, so the Hot Water Set point will 40 deg C + 10 = 50 deg C.

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    so now Th1 = 50 deg C, Th2 = 45 deg C (say), Trxn = 25 deg C [ i've considered the reaction mass temperature as 25 deg C because at 650 mmHg of vacuum, the boiling point of ethyl acetate will be hardly 25deg C, this can be calculated from antonie equation],

    LMTD = ( ( Th1 - Trxn ) - ( Th2 - Trxn ) ) / Ln [ ( Th1 - Trxn ) / ( Th2 - Trxn ) ]
                =  ( ( 50 - 25 ) - ( 45 - 25 ) ) / Ln [ ( 50 -25 ) / ( 45 - 25 ) ]
                =  5 / Ln [ 1.25 ] = 22.727 Deg C,

    Now, lets resume our calculation,

    Q = U x A x LMTD

        =  250 x 9.91 x 22.727 = 56306.143 KCal.

    Now we need to know what is the Heat load of the Reaction mass i.e., summation of Sensible heat & Latent heat,

    Qhl = Sensible heat + Latent heat

          =  M x Cp x dT   +   M1 x Lambda

          = ( V x Rho )  x Cp x dT   + M1 x Lambda

          =  5 x 897 x 0.454 x ( 30-25)  +  M1 x  87.6

          =  10180.95 + M1 x 87.6

    Now, we need to equate the overall heat load to the reactor supply load i.e., 56306.143 Kcal,

    10180.95 + M1 x 87.6  =  56306.143

    M1 x 87.6  =  46125.192

    M1 = 526.54 Kgs of Ethyl Acetate = ( 526.54 / 0.897 ) Lts of ethyl acetate

    M1 =  587 Lts of Ethyl Acetate.
    If you guys understood the above done calculation clearly, Then say Cheers :)
    And if any doubts were there, please ask us, we are happy to help you,

    Comments are most appreciated :)

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    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.
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    1. You cannot equate Q with Qhl as unit of Q is Kcal/Hr and not Kcal

      1. ok, but can you tell me what are the units for U.....??

        And for your king information Qhl will also be having KCal/hr only as the variable M in eqn of Qhl is having the units of Kg/Hr, and this whole posts describes about rate of distillate per hour, so the value of time is taken as unity and anything multiplied with unity returns the same, thats why i've neglected that Hr in whole post.

        Still any queries, feel free to comment

    2. Please explain LMTd a bit clear example.
      Here im giving the e.g,
      EA distillation T<25°C,Hotwter as medium
      Here Temp of hot body will be hot water and cold body will be Rxn mass(10°C)
      So why cant we make difference of (Temp of reactor- Temp of jacket) instead of lmtd.
      Why should we use LMTD
      Xplain bit clearly.
      2streams are existing,but pls clearly justify me wid a xample

      1. If you consider the difference between reaction mass and utility, then the answer will be a crap one, because whenever you need to calculate the energy of a stream then you need to consider both temperatures of that stream only, never mess with temperatures of two different streams, unless it is a LMTD, LMTD is used while calculating overall heat transfer rate

    3. You are taken 5 kl but 500 kgs not understood

      1. yup, i've taken 5KL and i got 526 Kgs, thats mass flowrate, if u need in Lts then divide the 526 with solvent density.

    4. I getting confusion about using LMTD and DT. When we should use LMTD and DT??

      1. whenever you calculate heat load of a single fluid then then you may go with dT, but whenever the notation overall heat transfer coefficient arrives, then automatically LMTD should be used as there will be two fluids involving at that time


    5. But latent heat of ethyl acetate under vacuum will change.. Can u calculate that latent heat and then rate of distillation.. Eg. If u take ethyl acetate and heat under atmospheric condition and in another heat under vacuum... Boil up will increase under vacuum due to change in latent heat..

      1. that major difference between boil up is due to relative lowering of vapour pressure, and the latent heat will also have some impact for sure, but not that much....... simply below limit of consideration.

        Ajay Kumar

    6. can you cal the distillation time required to distill out 587 lits of EA for the same conditions you have explained.

    7. I want the distillation rate per hr. not the quantity to be distilled....\

    8. it means 587 lits of EA/hr???

    9. What ever the 587Lts i've derived indicates the distillation rate but not the qty to be distilled

    10. I am getting confusion in sensible heat part of your heat load calculation....why you have taken (30-25) as delta T....since rxn mass is already at boiling point did'nt we have to just consider latent heat part??

      1. Hey buddy, i've considered that the distillation is happening in RT, so i've considered 30 degC, and the Rxn mass is at 25 degC, so in order to raise the temp of rxn mass to 30degC from 25 degC, the sensible heat is required, and also the sensible heat load will be far low than the latent heat when compared, if required you can neglect it.

      2. Liquid Ammonia boiling point is -33.34 degC.

      3. I cant understand sensible heat part of your heat load calculation....why you have taken (30-25) as delta T...please clarify

        i read your previous comment but still i am confuse .

      4. Hey, that 30-25 means, your reaction mass is at RT, and that reaction mass temperature should be raised upto 30 degC from 25 degC, then under vacuum it will start boiling, thats why i've considered it as sensible ehat load, if required you can ignore that part.

        AJAY K

    11. is it applicable for multi component mixture

      1. Applicable but if the required distillate is a low boil.

        AJAY K



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