*how to determine the theoretical distillation time cycle in a batch reactor,*and today i'm gonna demonstrate you a simple way to determine outta many critical ways to do that,

Today this determination of Time required to carry out distillation is Based upon the Reactor Heat transfer area.

So Please read

How to find heat transfer area before proceeding to determination of distillation time

So, initially we need some of the data regarding

* The reactor mechanical design which includes some parameters like shell thickness, inner dia, length of reactor,

* The batch input details like how much amount of volume we need to handle, how much amount we need to distill out,

* The properties of the material we need to handle like density, specific heat, composition etc, and on the other end we need some properties like thermal conductivity of materials which out heat transfer area is composed off.

**Properties of Materials:**Usually for make of Heat Transfer surface we will be usually going for SS316 Grade and for other purposes like for preparing the limpet jacketed coil and the insulation coating, we will be using SS304, and occasionally we will use GLR for Ph sensitive distillation, so below are the thermal conductivities of these materials

Material Thermal Conductivity,K [W/m.K]

SS304 10-13

SS316 13-17

MSGLR 1.2-1.5

* Latent heat calculation,

Lh = M * Lambda

Latent heat load = M x Lambda = 1 x 790 x 264 = 208560 KCal.

Total Heat load = 265440 KCal.

Amount of heat transfer that a MSGL surface can transfer = K x A x dT = 0.029 x 9.93 x (64-25)

i.e., 11.52 KCal

**An automated excel sheet specially designed for our readers/visitors.**

We need theoretical as well as practicality..that mainly emphasis a chem engineer... jaisai

ReplyDeletepractical means everything will remains same in both except the utilities, that means we considered ideal one means the heat required to heat that heat transfer surface is given a constant but practically if it comes true then it will be OK, and also there is chance of scalling and some oxidation over the heat transfer area that might delay the process of heat transfer, so possibly we cant determine it practically, just based on thumb we have to calculate as redcuing 10-20% of the heat transfer rate per sec and calculating the time will give us practical time cylce determination, use the simulated excel sheet for calculations.

DeleteHai ajay..I am satish..for sensible heat and latent heat you are taking M as different values...wheather it is correct or not...please clarify

DeleteHii Sir,

DeleteM x Cp x dT, M is the total reaction mass, because if we want vapour 100 L of reaction mass then we have to heat the total mass to boiling point, then the remaining heat supply will be helpful for vaporising(latent heat). But only certain amount of qty will be vaporised and not whole reaction mass, so both M's will have different figures.

Best Regards,

AJAY K

Cannot download the excel file please help

ReplyDeleteLink is working fine, share me ur mail id, i'll forward u

DeleteCannot download the excel file please help

ReplyDeleteDear Vin,

DeleteKindly share your mail id, so that i can forward you,

Regards,

AJAY K

In your above calculation temp is in 0c convert into k

ReplyDeleteAbove calculation what about temp. Unit conversion

ReplyDeleteDear Pradeep,

DeleteEven though if i've converted the temperature to K from C, the difference remains same, so it wont be a big deal.

Also dont comment same content more than once, the filters will divert your comments as spam.

Regards,

AJAY K

Dear ajay,

Deletehow can u take equation amount of heat transfer in msgl, here convection heat transfer takes place than use UA deltaT

Dear,

DeleteHere i've taken the amount of heat required for distillation and i've equated it to the amount of heat that we can transfer through an SS surface using hot water. thats it,

Regards,

AJAY K

if u use steam inplace of hot water

ReplyDeleteRate of distillation will be high when compared to hot water and during LMTD calculation both hot fluid temperatures shall be considered as same.

DeleteRegards,

AJAY K

If I consider the use of steam as utility, would it be correct to say:

ReplyDeleteMCdT + m*(lambda) = UAdT = KAdT/x = m x lambda (for condensation of saturated steam to saturated liquid)

Dear Natasha,

DeleteEven though you are using steam for heating, it doesn't mean us to consider dT in place of LMTD, it shall be U x A x LMTD. apart from that everything remains same.

LMTD for steam = ( ( Ts - Ta ) - ( Ts - Tb ) ) / Ln ( ( Ts - Ta ) / ( Ts - Tb ) ).

Ts is input steam temperature,

Ta - initial reaction mass temperature,

Tb - Final reaction mass temperature.

Regards,

AJAY K

Thanks for the prompt reply!

ReplyDeleteIs there some way to know the temperature of saturated steam when I know only it’s pressure?

Dear Natasha,

DeletePl refer to Steam Tables.

Anyway in my next post, i'll update them.

Regards,

AJAY K

(Pressure)^(1/4)*100 will give you temperature

DeleteAlmost near to steam tablet data

Amount of heat transfer that MSGL can transfer ,why we are not using the formula Q= UAdt plz reply

ReplyDeleteDear Kavi,

Deleteif you can look into the formula deeply, it includes U, it is the combination of many factors, like scaling, fouling etc., but what i've demonstrated is KAdT, this indicates the pure potential heat transfer rate of a surface, but where as while using UAdTln, you need to consider the utility also, and somewhere somehow it depends on the input utility also, if doesn't give pure potential of the surface.

Regards,

AJAY K

I am unable to download this formula excel sheet can u provide me??

ReplyDeletereach me through the contact page, so that i can share you through mail.

DeleteDear Ajay Kumarji,

ReplyDeleteWould you pls comment on energy conservation opportunities for Pharma industry such as Bulk drugs, Formulation, etc.

Thanks & regards,

Hemant Kumar..

If I have a condenser of capacity 4M Sq. Then how I get valu of vapor condensation rate of condenser means how much vapor he condens in a hr

ReplyDeleteDear ,

DeleteSome more data is required for calculating the condensate. I'll consider those missing data,

Area is 4 m2, OHTC be 450 KCal/hr.m2.C, Let the LMTD be 30 C,

Heat load that the condenser can have in one hour is 4 x 450 x 30 = 54000 KCal.

So now you can equate this load to the solvent latent heat, if its water, latent heat would be 540 KCal, the condensate will be 54000 / 540 = 100 Kgs/hr.

If its methanol, 54000 / 264 = 204.5 Kgs/hr.

Hope you understand.

Best Regards,

AJAY K