**A**bout 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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Very helpful one..keep going@ jaii

ReplyDeletesure mate, if anything needed then just write it in comments or send it through WHAT SHOULD I WRITE NEXT column.

Delete:)

I want to know....the area that u have calculated is on the inner side of reactor?

DeleteI've considered only ID.

DeleteVery helpful one..keep going@ jaii

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

ReplyDeleteSme query regarding dat "write abt next column"

ReplyDeleteIt is displaying error whenever im posting my queries..so help me out

ok fine, please message me from contact us page

Deletehow to calculate the plate heat exchanger no of plates

ReplyDeletepls do the calculation.

thapliyalashish48@gmail.com

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....!!!

Deleteok first of thnks for ur prompt reply.

Deletehere 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

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....!!!

Deletebrine/bittern specifications

Deletetemp 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

@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,

ReplyDeleteHope 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

Dear ajay you did'nt get my actual point .

ReplyDeletelet 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 .

@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,

ReplyDeleteFirst 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

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%

ReplyDeleteQ2. 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.

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.

DeleteHey Ajay,

ReplyDeleteI 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

Hey Mr.Anonymous,

Deletethere 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

how can we directly consider 10% of torisherical HT?

ReplyDeletethat's just a thumb rule buddy.

DeleteRegards,

AJAY K

Hi seems there is an error in the calculation : 1.29 + 10/100 x 1.29 = 1.29 + 0.129 = 1.419 sq.m and not 1.3.

ReplyDeleteHey buddy,

DeleteThanks for correcting it......!!!!

Thanks & Regards,

AJAY K

How to calculate centrifuge feed capacity used in api induStry

ReplyDeleteDear,

DeleteUsually there will be some standard basket capacities for the centrifuges, lets suppose a centrifuge of size 36" will have a basket capacity of around 150 L, and the material you want to filter in that centrifuge is having a bulk density of 0.4, then you can filter a reaction mass having material weight 150 x 0.4 = 60 Kgs.

That's it.....!!!

Regards,

AJAY K

Dear sir,

ReplyDeleteCan u plz provide the path for determining the desired heat transfer area of condensers (any utility)for the distillation of any specific solvent

Dear Ashish,

Deletepl follow the link and go through the post,

https://pharmacalc.blogspot.in/2016/04/how-to-design-condenser.html

Still any queries pl feel free to comment,

For your information, i'm leaking out a thumb rule, if you want to select a condenser for a reactor just double the reactor capacity, that's it......!!!

If its 1 10 KL reactor, simply go with 20 Sq.m condenser.

Regards,

AJAY K

Dear Ajay,

ReplyDeleteNice information....

Dear Ajay,

ReplyDeleteHow to do energy balance for FBD & RCVD or RVPD??

Dear Mahesh,

DeleteUsually wherever there is a solvent rich content, there energy balance can be effectively applied, but during drying most of the times there will be falling rate period cases, hence we cannot apply energy balance approximately.

And coming to your query, you have mentioned two types of dryer,

1) Indirect contact dryer - RCVD,

2) Direct contact dryer - FBD.

In RCVD most of the times we cannot predict the energy required for drying, and the same is the case of FBD too.

Because we wont have any provisions for monitoring the material temperatures.

If its a ATFD / ATFE we can perform the energy balance, as the input itself will be solvent rich.

Regards,

AJAY K

hi ajay,

ReplyDeleteI need 0.5 KL SS REACTOR boil up rate and condenser area calculation

Dear Sir,

DeletePl find the below calculation steps, considered water as reaction mass,

Remaining parameters considered as worst case,

Occupancy 60%

HTA 2.68 Sq.m

Effective HTA 1.61 Sq.m

U 200 KCal/Sq.m.hr.K

Steam 110 C

Initial Final

Rxn. Mass 30 90

Mix. Density 1000 Kg/Cu.m

LMTD 43.28 C

Boil up 27.84 Kg

U for condenser 300 KCal/Sq.m.hr.K

in out

Utility 10 18

LMTD 75.93 C

Condenser Capacity 0.61 Sq.m

Safety consideration 30% excess 0.79 Sq.m

Any queries, pl reply back,

Regards,

AJAY K

Can I caluculation a reactor TR value with only heat transfer area.

ReplyDeleteLike that reactor heat transfer area is 11.89 sq.mtr, what is TR for that.

Dear ,

DeleteIf only Heat transfer area is known, then lets consider the remaining things as default.

I'll consider water as solvent with 80% occupancy, and i think the 11.89 Sq.m of reactor will have 5 KL volume. so the effective volume will be 4000 L,

Heat energy = 4000 x 1 x 1 x (100-0) = 800000 Kcal.

TR = 400000 / 3024 = 132.27 TR.

100 is the boiling point of water,

0 is the freezing point of water.

1, 1 are the densities and specific heat.

Hope you understand,

Next time kindly comment with your good name plz.

Best Regards,

AJAY K

Dear sir,

ReplyDeleteHow have you taken the L/D value Whether it is constant for all reactor volumes and also i have not understood what is boilup. And i know only the volume of reactor how can i find its ID and Thickness

Dear ,

DeleteI've taken the L/D in general, surely it will vary from reactor to reactor,

Boilup means the amount of vapour that is generated by heat transfer through the reactor surface.

You can get the inner dia and thickness from equipment GA(General Arrangement) drawings.

Next time kindly comment with your good name plz.

Best Regards,

AJAY K

Dear sir,

ReplyDeleteWill this calculation be applicable for agitated vessel. And also how can we say this heat transfer area for all, because we may have internal coil or limpet coil or simple jacket at that time also do we have same area. Kindly clarify it, am very confused.

Thanks and regards

Rakeshkumar

Dear Rajesh,

DeleteMost probably for annular jackets, you can use it, for limpet coil also you can use it because limpet coil is designed for effective heat transfer and the gap between the coil rings acts as extended surfaces which attributes to effectiveness.

So, as per my knowledge, in case of limpet coil gaps between rings there will be heat transfer but it will be slightly less than that of the ring area's, so you may take it as ~90% of the total heat transfer area.

If its a double limpet then there wont be any issue.

If nay queries, feel free to comment / message me through the contact me page.

Best Rgards,

AJAY K

Hai, here D is a dia of either torispherical or cylindrcal

ReplyDeleteCylindrical

DeleteAll topics are very good and helpful .

ReplyDeleteThank you Sir

Perimeter of cylinder= 2(pid+ l)

ReplyDeleteDear,

DeleteIn the above calculation i've used equilateral dimensions, i.e., if we dont know the perimeter of a cylinder directly then we may use that of a rectangle by visualizing that the cylinder is cut open at one point.