Thermosiphon Reboiler Design

1 CommentTechnical
During Resent days, I was trying to design thermosiphon reboiler!

After long days of calculation, rewriting calculations on pages! on excel! finally I come to some ‘lesson learn’ which I’ll  be sharing now.

But before that, following are two good references to design a thermosiphone reboiler
1. Applied Process Design Volume-3 by Ludwig
2. Process Heat Transfer – D.Q. Kern

Design of heat exchanger step :2 : which type?

5 CommentsTechnical
The next step is to decide which type of exchanger to be used?
The industrial heat exchanger are varies depending upon
1. Transfer process.
2. Number of fluids
3. Compactness.
4. Constriction.
5. Flow arrangement.
6. Type of heat transfer.
See attached figure for more details.

Heat exchanger: step 1: application

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<p>The first step dorm designing a heat exchanger is to understand where it will be used for.</p>
<p>Application is most important criteria and mostly its input from client.</p>
<p>Application demands understanding of functionaries expected, duty needed, properties of both side fluids, system information e.g presuredrop expected, cleaning method of the equipment, hazard category, local laws pertaining to pressure vessel and hazard handling.

Some time the application demand ‘soft’ heat transfer due to the media e.g yeast cooler due to delicate yeast cells are always co-current type which lead to higher surface area and are more costly than counter current heat exchangers; but due to application those are designed that way.

Cooling – Thermal Calculation

Technical

Lets Start with Diffrent Loading Calculation
Step : 1
1. Heat Loss to Atmosphere : Surface Area m² X Delta (Tavg – Troom)ºC x 0.32 = Q1 kcal/ hr
2. Cooling Load : Volume of liquid (lits) x Delta (Tstart-Tend)ºC/Time (hr) = Q2 kcal/hr
3. Heat evolved during Chemical process/ Fermentation : Q3 kcal/hr

Total Heat to be Removed = Cooling Load = Q = (Q1 + Q2 + Q3) X factor of safety

Please note Unit of Measurement carefully

for Q3, what process do you follow in your vessel, you need to find out the Load.

Step : 2
Calculate LMTD {Log Mean Temperature}
DT1 = TmediaStart – TcoolenEnd
DT2 = TmediaEnd – TcoolentStart
LMTD = (DT1-DT2)/ ln(DT1-DT2)

Step : 3
Calculate Overall heat Transfer co-efficient &amp; Surface Area Required
Q = U A LMTD
Hence, A = Q / (U x LMTD)

U, Over all heat transfer co-efficient : 100-250 Kcal/ hr ºC m², this is mainly depends uppon what type of jacket do you use, limped coil or limped jacket or profile sheets or laser welded jacket etc. you need to have some co-relation to evaluate this if the construction of jacket is non-standard. for standard type of jacket, you can get more info on following link
http://www.cheresources.com/jacketed_vessel_design.shtml

and if nothing works, ask me 🙂

Step : 4
Calculate amount of Cooling media needed for cooling
Now here is a trick, their are two type of coolent, one, they cool by their latent heat (e.g Ammonia) or two, they cool by dropping their temperature (e.g glycol water)

for First option, we need to consider Enthalpy of latent heat, devide Q by this, and you will get mass flow, for Second type follow following equation
Q = m . cp. DT (Tstartcoolent – TendCoolent)

Make sure you get these enthalpies at the working temperature, refer Perry’s handbook, or ask your cooling supplier for detail physical properties, as these properties changes with pressure &amp; temperature, for water mix coolent like glycol, the properties changes with concentation of media

I hope this is sufficient in-sight, in case of help, let me know

Cooling Performance – Overview

Technical

Ok, So you want to design cooling for a vessel.. how to?

Step : 1 : What you should know, and what you should ask
What you should know
a) You should know what type of Vessel Construction your are offering
b) You should know what Jacket for cooling you have
c) You should know your Jacket, Heat transfer co-eff co-releation very well
What you should ask
a) Media to be cooled
b) Media for cooling
c) with agitation or w/o agitation
d) Location (in-house or out-side)
e) Cooling performance needed
Step : 2 : Calculated heat Duty/ Cooling Duty
a) Duty due to Surface Loss
b) Duty due to Fermentation/ Process, i.e to remove heat arising from the process
c) Duty due to Cooling withing given time frame
d) Factor of safety : 10 to 20% depending upon your confidence (prooven history will help to reduce this factor of safety)

Step : 3 : Calculate Amount of cooling media needed

Step : 4 : Calculate Area for Cooling jacket

Step : 5 : Calculate Pressure Drop

Step : 6 : Summary

Next post : How to!

Milk Property fact!

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Just, I got this info & sharing with you
Density of Milk
Whole milk : ρ = 1035.0 − 0.358 T + 0.0049 T^2 − 0.00010 T^3
Skim milk : ρ = 1036.6 − 0.146 T + 0.0023 T^2 − 0.00016 T^3
Buffalo milk : ρ = 923.84 − 0.44 T
Cows’s milk : ρ = 923.51 − 0.43 T

Good, know you know what will be wait of 1 Liter Pouch 🙂

Cooling of Big tanks

Technical

Dear All,

Recently i was working on designing tanks with cooling jacket.
In brewery in particular, there is lots of need of cooling of tank, to take care of heating load generated out of fermentation, surface losses & to get temperature profile.

and as a motive to open this blog, is to share what i learn, in my next post I’ll discuss this in very detail.

See you all soon

PS : if you need any more info ask me, i can add that too in my post!