Thin Layer Chromatography (TLC) - Analysis and Evaluation

Hiii ....!!! Good day all.

Hope everyone is fine, i'm back with a general post this time. This post is developed based on a request from a process engineer from Ireland. Don't want to publish his name over here.

The query is "How to understand a TLC analysis ? What is the principle behind TLC ? How accurate is the TLC method ?"

Would like to complete the post shortly.

TLC - Thin Layer Chromatography.

TLC is the most basic chromatographic technique and is usually practiced by process engineers and some time production team to know the impurity profile. Coming to my case, i've practiced many time and based on the experience, i'm posting this.





TLC is a qualitative analytical method and can't be used for quantification.

Based on the experience of the test sample and TLC method, some persons might quantify the impurity profile with slight variations from the exact value.

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Principle:  

1. Basic principle involved in this method is separation based on the capillary action of the compounds over the TLC plate.
2. It depends on the relative affinity of the related compounds(sometimes impurities) with the mobile phase and the stationary phase(TLC plate).



Identifications of the compounds:

Based on the affinity of the compounds over the stationary phase, the subject were eluted as spots on the TLC plate.




Minimum requirements for performing the TLC:

1. TLC Plate(Usually a uniform silica plate),
2. TLC chamber (Chamber used for compounds separation),
A TLC chamber will looks like:

3. Mobile phase (Usually a combination of two or more solvents),
4. UV chamber.

Procedure:

Here i can't mention any procedure, as it will be different from analysis to analysis depending on the sample preparation and elution pattern.

1. TLC Plate / Stationary phase preparation: First of all a TLC plate shall be cut off from the available mother sheet with uniform edges, there shouldn't be any rough edges and the edges shall be parallel. Then a straight line shall be marked on the plate for our reference in parallel to the edges of the sheet.

2. Mobile Phase preparation: Based on the method ratio's the mobile phase shall be prepared.

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3. Test Solution, Standard solution Preparation: There will be specific procedure to prepare test(sample), standard solution preparations mentioned in the Method of analysis. Based on the pre-defined procedures, the working standard/impurity standard and the test sample shall be weighed and diluted with a relevant solvent in mentioned composition.




4. Sample spotting on the Stationary phase: Depending upon the limit of the impurity, injection volume shall be calculated and injected on the reference line drawn on the TLC plate at spots with approx. uniform distances as shown below.

5. Sample injection on the Stationary phase: With an adequate syringe standard solution shall be injected on the standard spot, followed by sample spot and the co-spot is a combination of standard and sample qty.'s. Here a precaution that we should take care is, while injecting there will be spots created on the injection location those spot area's depends on the time of injection, if the sample is injected in a single shot the spot would be bigger and the elution will become rough. Hence the samples shall be injected slowly. Below is the screen for a single shot injection and slow injection:

If the sample is injected quickly, then the eluted spots may over lay on each other leading to improper results. Hence care shall be taken during injecting the solutions.




6. Positioning the TLC plate(Stationary phase) inside TLC chamber: This is quite important step and proper handling is preferred. mobile phase shall be transferred into the TLC chamber. The stationary phase after injection shall be placed inside the TLC chamber and one more precaution is the reference line that we draw before injection shall be slightly higher than the mobile phase level. if its immersed then there is probability for abnormal results, and sometimes zero results.

7. Development / Elution of spots : After positioning of the plate inside the mobile phase in TLC chamber, open end of the chamber shall be closed and shall be maintained for some time till complete elution happens and the sheet to be dried with air dryer to knock out the solvent presence.

8. Impurity parameter evaluation: Upon elution, the TLC plate shall be dried with air dryer and placed in a UV cabinet and based on the defined wavelength the spots shall be observed, Mostly the plate shall be kept in 254 nanometer wavelength(short) and then the spots shall be identified.
                                                       

UV Cabinet

From the above plate you can observe a spot against to that of the standard spot and more spots against the test and co-spots.



The standard spot indicates the impurity that need to be checked, and against to that in the test spot there is a spot at relevant height, that is the actual content of the subject impurity at that position.

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and when the two spots were compared, the standard spot is more darker/thicker than that of the spot in the test sample. From that we can say the particular impurity is somewhat low in the test sample. If the standard is prepared with 5% solution, then we can say that the test sample is having less than 5%. But based on the experience, we can quantify the parameter with slight variations.

Now you may get a doubt that what are the other spots in the test sample, and the answer is the standard is a pure compound solution but the test sample may contain some excess impurities along with the one desired we are checking.

Lets say the cement spot in the test sample is product spot, then the above spots to the product spot can be referred as non-polar impurities and below the product spots are polar-impurities sometimes.

And in the half of the discussion, you might got a doubt that what is use of co-spot, that additional spot is attributed by standard and test solution in 50-50% composition, the co-spots gives us an idea about the variation in impurities and sometimes it is helpful to confirm whether a spot is actually from standard solution or not during abnormal mappings.

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

Hope the above description is understandable to all......!!!

If any queries, feel free to comment or message me......!!!

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.
Follow Me on Twitter AjaySpectator & Computer Innovations

Vent Sizing's for Reactor's & Storage vessel

Hello all....!!!

Have taken a gap.....  because of my profession!!

After many days gone through the mail and comments, and most of the requests were related to vent sizing, ejectors. And even in my circle many were expressing the same query related to vent line sizing.

Also recently have seen a disaster in a chineese chemical plant (Xiangshui), nearly 78 persons were dead and many went missing, which is majorly due to lack of proper safety measures and may be avoidable if adequate planning for relief is available. And especially it can be considered as a black day in every process engineer's career. See Below video:

As per every management, Process safety is everyone's responsibility, but as per me a Process Engineer holds major accountability than others. Hence being an process engineer, its quite necessary to evaluate every hidden risk before going further with a process. As in a manufacturing company process engineer would be like a head light defining right direction to the operators. 

Today here i gonna add a small notes related to that, 

Actually its a very simple task if we have all the data which is required for calculation, as based on assumption its quite waste to calculate because the calculation would be something like satisfying ourselves.

I've performed the calculation for one of my products, so with that experience i'll elaborate the calculation by altering the values little bit.

Before that lets have some standard definitions.

Set pressure: Activation pressure of a relief device 

Maximum allowable working pressure: Maximum gauge pressure at a given/designated temperature.

Operating pressure: Gauge pressure, usually 10-12% below the MAWP.

Over pressure: Pressure above the set pressure during relief process.

Back pressure: Outlet pressure of the relief device during pressure knocking. Usually it will be equal to atmospheric pressure.

Accumulated pressure: Pressure raise above MAWP during relief process.

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Now lets go into the calculation part.

Vent Size calculation for a reactor:


Usually the vent size shall be calculated from the data obtained from ARC(Accelerated rate calorimetry), RSST(Reactive System Screening Tool), ARSST(Advanced Reactive System Screening Tool).

Below is the data required for calculation:

1. Rate of temperature raise (T, C/min),

2. Rate of pressure raise (P, psi/min).

Here is the formula,

A =                            ( 3.5 x 10^-3 )                        x  V ( T + P )

                 p [1 + (1.98x10^3)/(p^1.75)]^0.285                Cd

where,

p - Set pressure of relief device (psig),

V - Volume of gas / reactant (m3),

Cd - discharge coefficient = Actual flowrate / theoretical flowrate,

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Now, i'll demonstrate with a case study,

Details of the reaction from the ARC study in run away case:

Rate of temperature raise = 0.2 C/min,

Rate of pressure raise      =  0.05 bar/min = 0.73 psi/min,

Other details:

Set pressure of the relief device = 4.0 Kg/cm2 = 56.89 psig,

Volume of the gas = 30 m3, (This may seems higher, but it is quite regular as the gas densities were very low),

Tentative discharge coefficient (Lets say) = 0.7,

Then using above formula,

A =  ( 3.5 x 10^-3 ) / [49.78 x (1 + (1.98*10^3 / 49.78^1.75)^0.285]    x [30*(0.2+0.73)/0.7]

   = 0.00203 m2.

Dia, d = ((4 / 3.141) x A)^0.5 = 0.0507 m = 5.07 cms = 1.999 in ~2 in.

Vent size calculation for a storage vessel:

This is quite lengthy and requires most of the data from the vendor side,

the details include the constants like discharge coefficient, back pressure correction factor.

And other details include the vapour/gas compressibility factor, heat capacity ratio, Mass flowrate.

Here is the formula for the vent size of a storage vessel,


A = [ w / C x Kd x P1 x Kb ] x ( T x Z / M )^0.5.

w - mass flowrate of the solvent vapour,

C = [ j x g / R ] x [ (2 / ( j + 1 )) ^ (( j+1)/(j-1))],


j - Heat capacity ratio of vapour or gas,


R - universal gas constant,


P1 (psia) - Pmax (psig) + 14.7,

Kd - discharge coefficient (~ 0.7),

Kb - back pressure correction factor (~1.0),

Now, let me demonstrate with a case study by calculating the required vent size for a storage vessel of 60 KL capacity containing toluene,

Heat capacity ration of the vapour be 1.2, Solvent be methanol - mol wt.: 92 g/mole,

Vessel dia be 2.5 m, height of the vessel be 3 m (torispherical vessel),

Wetted surface area of the vessel (Awet) = ( 𝜋 d l ) + ( 1.084 x D^2 ) 

= 30.33 m2 = 326.5 ft2.

Total heat absorption, Q = 34500 x (Awet)^0.82 = 3973775.31 BTU / hr.

Let the rate of vaporization be 20 Kgs/sec [the value is quite lower than the actual, usually for hydrocarbons it will vary between 30-50 Kg/sec as they possess lower latent heats] ,

P1 = Pmax  + 14.7 = (Pmawp x 1.21) + 14.7 = (4 x 1.21) + 1.013 

= 5.853 bar abs = 5.853 x 10^5 N/m2.

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C = [ (1.3 x 1 / 8314) x (2/(1.3+1))^(1.3/0.3)]^0.5 = 7.32*10^-3,

A = ( 20 / (7.32*10^-3 x 0.7 x 5.853 x 10^5 x 1)) x ((273.15+110.65) x 1 / 92)^0.5

   

    =  0.0067 x 2.042 = 0.0136 m2.

Dia, d = ( 4 x 0.0136 / 3.141 )^0.5 = 0.132 m = 13.2 cms = 5.185 in ~5.5 / 6 in.

The above demonstrated calculation would work great if relevant authenticated safety reports are available. If not, then as an engineer we should be in a position to provide a temporary way forward without giving a major break in production.

Vent Size Calculation without Safety Data:

In such cases we should depend on the theoretical material balance,

I'll give you a case study with a random chemical reaction,

Yeah, i got a Balz Schiemann reaction, where a Aniline undergoes diazotization with Nitrous acid followed by reaction with fluoroboric acid giving aryl diazonium salts and thermal decomposition of dry aryl diazonium salts gives aryl fluoride and bi-product of the reaction is Nitrogen gas and boron trifluoride.

Below is the scheme of the reaction:

C6H5NH2 + HNO2 + HBF4  ----- > C6H5F + N2 + BF3

Note: Property of Pharma Engineering

Now lets suppose, we are manufacturing fluorobenzene with a batch size of 100 Kgs.

Below are the raw materials quantities used in the batch manufacturing:

Aniline                       -   100 Kgs,

HNO2                        -   50 Kgs,

Fluoroboric acid      -    150   Kgs.

Below is the material balance for the above scheme with the raw material Qty.'s:

So based on the material balance, it can be said that

29.80 Kgs of Nitrogen and 72.13 Kgs of Boron trifluoride is liberating.

As mostly the Diazonium salts are highly unstable at lower temperatures, the preferable temperature of addition would be below 0° C. Hence the first reaction would be formation of Diazo salts, and nitrogen would be liberating in this point and it depends on the addition rate of HNO2, as the Nitrogen formation is attributed by Nitrous acid addition.

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Reaction - 1: Formation of Diazonium salts

Hence in the first step, Nitrogen would be liberating and the Qty.'s is 29.80 Kgs.

Converting into volumes, it would be 29.80 / 1.25 = 23.84 m3, which is quite higher.

and let the addition of nitrous acid took 2 hours, 

In 1st case, lets consider Nitrogen is liberating uniformly throughout the addition,

i.e., 23.84 / (2 x 60) = 0.199 m3/min = 199 L /min.

But while performing the calculations through material balance, we need to consider the worst scenario's, as liberation would go in 10 minutes with a peak liberation of 90%.

The worst scenario would be 23.84 x 0.9 / 10 = 2.15 m3/min.

Now to position ourselves in a safe state during addition, we shall make the system ready to knock out 2.15 m3 of nitrogen per minute,

So, now one more assumption we need to consider is velocity of nitrogen, i.e., based on my experience i'll consider as 10 m/sec.

So, the relief device(vent) area shall be (A) = (2.15/60) / 10 = 0.0036 m2.

Dia, d = ( 0.0036 x 4 / 3.141 )^0.5 = 0.0675 m = 6.75 cms = 2.66 in ~3 in.

Hence for the first reaction we should provide a vent size of 3 inch.

Reaction - 2: Formation of aryl fluoride

Now enter the dragon, lets move to next reaction,

i.e., formation of fluorobenzene, where boron trifluoride is a bi-product and the gas is quite toxic and the same can't be left into atmosphere, it need to be scrubbed.

If you need a scrubber design, pl refer below link, in this subject post i'll describe only about vent size.

[How To] Design a Scrubber ?

72.13 Kgs of Boron trifluoride is being liberated,

Lets follow the same procedure like above,

Converting it to volume,  it will be 72.13 / 2.76 = 26.13 m3.

Same 90% of peak load in 10 minutes,

it will be   26.13 x 0.9 / 10 = 2.35 m3 /min.

And as this time, i'll be supported by scrubber, hence i'll consider the velocity as 10 m/sec,

Here you might get a doubt that, "why am i considering velocity as 10 m/sec even having a scrubber advantage?".

Answering your query, "dear engineers, if you can see the density above, it is 2.76 Kg/m3 which is more than double of nitrogen(1.25 Kg/m3)". Hope you got it.

Let's get back into calculation, 2.35 m3 of BF3 to be knocked out in one minute.

Relief device(vent) area shall be (A) = (2.35/60) / 10 = 0.0039 m2.

Dia, d = ( 4 x 0.0039 / 3.141 ) ^ 0.5 = 0.071 m = 7.065 cms = 2.78 in ~ 3 inches.

Hence, from the reaction 2 it is observed that a minimum size of 3 inches is required for both reactions.

And here the lucky fact is both are liberating in different reactions, or else it would be a complicate case to solve.

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Reaction - 3: Knocking out the Unreacted compounds

And also we need to check the vent sizing during quenching operations, as in the above case, there is an excess reagent used in process i.e., HBF4,

If water is used for quenching the unreacted HBF4, then the reaction would be as below:

So here, in this case, lets suppose 100 lts of water used for quenching the unreacted Fluoroboric acid. products will be boric acid (Solid) and hydrogen fluoride (gas).

And the sane calculation shall be repeated. [I'll leave it and let it be an exercise to the readers].

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

Hope everything is understandable to all.......!!!

Any queries feel free to comment / message....!!!

Comments are most appreciated.....!!!

This post is dedicated to my Dearest brother Govindarajan V.

Finally, I'm expressing my view here, "If you are unable to Knockout a process risk, you'll be knocked out by it".

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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.
Follow Me on Twitter AjaySpectator & Computer Innovations