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Hope you all having a Good Day and a Good Weekend too....!!!

Today i gonna clear you some of the questions that you think when you remember the topic " TONNE OF REFRIGERATION ", shortly TR.

This is most common term used in Pharma industry when ever projects or chiller plants were concerned. Now, the basic questions that arise when you start on this topic were.,

What is the value of 1 TR ??What does Actually TR means ??How the value for 1 TR derived and how it got to be 3024 KCal, ??

So, before clearing all these doubts and demonstrating the way to calculate required TR, i wish to present you the basics required to understand this topic fluently.

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This term TR or Tonne Of Refrigeration has been started from US, and apart from US the cooling terminology is basically in the format of MW or KW.

And whenever we start to discuss over the units of any figure then the most disgusting thing that leads us to confusion is having two types, one is British units and the other one is English units.

All the British countries will keep on using the British units and the English countries will be using English or SI units, in between there will be some countries conducting debate over the units, what to choose without keeping to something consistently, because we want to know everything in this world, and in turn this will lead to confusion.

What is a Short Ton and Long Ton ??

The British ton is the long ton, which is 2240 pounds, and the U.S. ton is the short ton which is 2000 pounds.

Both tons are actually defined in the same way. 1 ton is equal to 20 hundredweight. It is just the definition of the hundredweight that differs between countries. In the U.S. there are 100 pounds in the hundredweight, and in Britain there are 112 pounds in the hundredweight. This causes the actual weight of the ton to differ between countries.

To distinguish between the two tons, the smaller U.S. ton is called short, while the larger British ton is called long.

There is also an third type of ton called the metric ton, equal to 1000 kilograms, or approximately 2204 pounds. The metric ton is officially called tonne. The SI standard calls it *tonne*, but the U.S. Government recommends calling it

*metric ton*.

So its Clear about Short Ton and Long Ton.

Now, i'll explain you the above mentioned questions,

Actually 1 TR holds the value 3024 KCal. It means that Energy required or to be removed to freeze one ton of water to Ice in one day i.e., 24 hrs, So mathematically 1 TR means 288,000 Btu are required to make one ton of ice, divide this by 24 hours to get 12,000 Btu/h required to make one ton of ice in one day, In Simple words one 1 TR means energy required for change of phase from water at 0°C to Ice at 0°C.

In British terminology, 1 TR is 3.517 KW,

In English terminology, 1 TR is 3024 KCal.

Learn How the value of 3024 KCal or 3.517 KW derived.....!!

The latent heat of ice (also the heat of fusion) = 333.55 kJ/kg = 144 Btu/lb

One short ton = 2000 lb, Heat extracted = 2000 x 144/24 hr = 288000 Btu/24 hr = 12000 Btu/hr = 200 Btu/min,

1 ton refrigeration = 200 Btu/min = 3.517 kJ/s = 3.517 kW = 4.713 HP .

Another unit of measure is the calorie which is the amount of heat removal required to raise or lower the temperature of one gram of water by one °C. A kilo-calorie is the amount of heat required to raise or lower 1 kg of water by 1°C. One ton of refrigeration is equal to 3024 kilo-calories per hour. This is 12,000 BTU/ h divided by 2.204 (pounds per kilogram) divided by 1.8 (°C to °F).

So, i think you got answers for the hidden questions, So now i'll show you how to calculate the TR required for an operation. Let it, i need to cool 350 L toluene vapour which is produced at 35°C under 710 mm Hg vacuum, so how much TR required ??

**Soln.**Volume to be cooled 350 L toluene. Mass to be cooled 350 L x 0.868 Kg/L = 303.8 Kg.

Specific heat for Toluene vapour at 35°C is 1.72 KJ / Kg. °K,

and the Sensible heat change is not in my hand, so i'll consider the temperature change of condensate as 5°C, so the Total Sensible heat load of the vapor can be calculated as

Qs = 303.8 x 1.72 x 5 = 2612.68 KJ = 625.04 KCal.

And now i need to calculate the major thing the Latent heat, QL

QL = 303.8 x 351 = 106633.8 KJ = 25510.48 KCal.

So the Total energy is Q = Qs + QL = 625.04 +25510.48 = 26135.52 KCal.

So Now the required TR is 26135.52 / 3024 = 8.64 TR.

So for the above operation of cooling a 350 L toluene vapour at 35°C which is produced at 710 mm Hg, i need 8.64 TR.

So, simply TR required = Total Heat Load / 3024.

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

Hope you all understand the above topic clearly without any doubts . Still any doubts We are Happy to Help, Contact Us through the Contact us Button available on page top,

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

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**A**bout The Author

Good one.. But my mind was tickling at one point,

ReplyDeleteWhat is the correct definition of TR ??

IS it energy req to freeze one ton of water to ice at 0°c

Or

Energy req to melt one ton of ice to water..

So my doubt is both freezing and melting emphasis same meaning ???

Freezing employs liq to solid phase and melting employs solid to liq..phase.. so what's the relation of both terms in TR

Pls clear my query.

thats what we called energy balance, latent heat of freezing=latent heat of melting, but those operations should be at 0 degrees only

ReplyDeleteNow it makes me sense ☺ keep goin

ReplyDeleteNice job man

ReplyDeleteThanks buddy

DeleteI think the way TR is described is not correct . The Term TR is applicable for cooling or Heating irrespective of ICE formation . In room airconditioner capacity is defined in TR where the formation of ICE is not happening . TR means removal of 12000 BTU / Hr .

ReplyDeleteA. K. BAKSI

Here, 1 TR = 3024 kcal, which can be calculated from the energy to be removed to freeze 1 ton of water in a day (alternatively energy required to melt 1 ton of ice in a day)

DeleteThis is only giving us a value of the heat energy that our system can extract. Ice formation is not necessary. The heat transfer is also limited by the design and operating parameters.

Dear Natasha,

DeleteExactly, but if you gonna extract 3024 KCal of energy from the subject system, the phase will change.

Regards,

AJAY K

Usually TR is used to define cooling rates, and being frank until now i haven't heard that TR is applicable for Heating also.

ReplyDeleteCan I used TR to calculate for the cooling tube capacity with information of dT of product temperature hot (in) to cold (out), weight of product pass through and time needed for all the product to pass through?

ReplyDeleteThanku ajay

ReplyDeleteSir plz provide PUMP CALCULATION ALSO ,IN A VERY SIMPME AND SYSTEMATIC MANNER

ReplyDeleteGo through this, if you need any help mail me or comment here.

Deletehttps://pharmacalc.blogspot.in/2016/04/how-to-select-pump-and-motor-line.html

Ajay Sir, please let help to calculate the TR require to cool down petrolium jelly from 65 deg.C to 12 Deg.C. in 5 minutes. Total volme is 60 kg.

DeleteCould you calculate tr in SI unit

ReplyDeleteBrine solution METHANOL 33% and minus chiller (-20C)

Temperature delta T-5°C

Flow - 100 M3/hr

Specific heat - 3.744 kj/kgk

Just tell about the unit conversion also

Dear Prabakaran,

DeleteTR = ((100 x 1000 x (3.744/4.187) x 5) + (100 x 1000 x 540)) / 3024.12 = 18004.72 TR,

Since you haven't given any solvent, i've considered water latent heat which is 540 KCal,

& TR itself is a unit,

Unit conversions for TR :

1 TR = 12000 BTU / Hr = 3024.12 KCal / Hr = 3517.2 Watts.

Still if any queries please reply back.

Regards,

AJAY K

Dear sir,

ReplyDeleteThank for your quick response,

But I am not able to get your calculation

Actual formula is Q=M*Cp*delta T*specific gravity

But how you are calculate I don't know could you expalin little deeper.

I am a beginner so kindly guide me,

TR = ((100 x 1000 x (3.744/4.187) x 5) + (100 x 1000 x 540)) / 3024.12 = 18004.72 TR,

Note: using solvent is METHANOL by 33%

Specific heat of methanol=3.744 kj/kgk

Regarding units you are used for each terms

Q= kW or kcal

Cp= kj/kg k

Delta T= °C

M= M3/ hr

Just correct me if my units used is worng

Thanks & Regards

PRABAKARAN

Dear Prabhakaran,

DeleteYou have given input as for utility side, so there latent heat can be eliminated, so

Q = M x Cp x dT = 100 x 920 x ( 3.744 / 4.187 ) x 5 = 411330 KCal.

M = Mass flowrate = Volumetric flow rate x density = 100 x 920.

Specific heat, Cp = 3.744 KJ/Kg.K = ( 3.744 / 4.187 ) KCal/Kg.K.

dT = 5 degK.

So, Q will be in KCal, since i've converted specific heat from Joules to Calories,

So, now Q = 411330 KCal.

TR = 411330 / 3024.12 = 136.016 TR.

Note: Earlier your query is not that much clear, so i've considered calculating Total energy including Latent heat. Now its clear for me.

Still any queries feel free to comment / message.

Regards,

AJAY K

Density i've considered as 920 Kg/Cu.m, since -20 degC brine will have a specific gravity of around 0.92.

DeleteRegards,

AJAY K

Super sir

ReplyDeleteNow I understand

One more doubt

I need COP calculation for compressor(refrigeration compressor)

Same parameters units and all others in details.

Give the formula along with any excercise, it will be good.

Dear Prabhakaran,

DeleteMay be you already know refrigeration means driving away the heat from low temperature medium to high temperature medium with some external work done on the system.

COP = Cold reservoir heat energy / Work done = Tc / ( Th - Tc )

Lets suppose, our system is at 50 degC(323 degK), and i wanna move it to 5 degC(278 degK), then the COP shall be,

COP = 278 / ( 323 - 278 ) = 6.17.

COP is unit less.

Regards,

AJAY K

In your above writings in cooling of 350L toluene vapour problem you have written "QL = 303.8 x 351 = 106633.8 KJ = 25510.48 KCal."..How it is so sir?

ReplyDeleteIt is latent heat, toluene having around 86.1 KCal/Kg, convert it into K Joules/Kg

DeleteRegards,

AJAY K

Oh sorry it is the latent heat of vaporization of toluene..

ReplyDeleteDear Sir,

ReplyDeleteQL = 303.8 x 351 = 106633.8 KJ = 25510.48 KCal.

in this step how did you taken 351?

Dear Ganesh,

Delete351 is Latent heat of toluene in KJoules.

Regards,

AJAY K

Dear Sir,

ReplyDelete8.64 TR is it per day or per hr.?

TIA

Dear Sri,

DeleteIt depends upon the vapour generation & condensation,

Regards,

AJAY K

Flow rate is 15cu.mtr,

ReplyDeleteReactor jacket heat transfer is 11.89sq.mtr, required cooling is -40C for process, How much TR required with 40% methanol circulation to rector jacket.

Dear ,

DeletePl make it clear, the TR is irrespective of utility composition,

Based on the utility composition the freezing point varies and it depends on the Raoults law, Theory name is Depression in freezing point, it can be calculated as

dT = Kf x m, m is molality of solute, Kf is the molal freezing point lowering constant.

If any queries pl comment or reach me through contact me page.

Best Regards,

AJAY K

Hello sir ...

ReplyDeleteHow to calculate tr for reactor..

I wann to decide chilling plant for our plant we have 10 as reactor of 10 kl .. to reduce temp of batch from 30def to -5deg.. (brine use) then how we calculate...pls give mi a calculation.. u take any one example

Hey Deepak,

DeleteGlad to hear from you, that's just a basic calculation, but i'll consider one excess parameter i.e., occupancy which you haven't mentioned in your comment,

total of 10 reactors, each 10 KL, Occupancy of 80% each.

For one reactor, lets suppose we need to cool solvent from 30 to -5 C, and i'll consider specific heat as 2 KCal/Kg.C, density is 1 Kg/L,

So, it will be =8000 x 1 x 2 x (30 - (-5) ) = 560000 KCal.

For 10 reactors, it will be 560000 x 10 = 5600000 Kcal.

Now TR = 5600000 / 3024 = 1851.8 TR ~1900 TR.(which is so heavy).

Best Regards,

AJAY K

hello mr ajay could you please clarify me whether should i consider ML in calculations for Calculating TR because im confused totally when to use total energy (Qs + Ql) and only qs i have seen answers on one of your post where you consider only Qs and in your explanation you took both

ReplyDeletehello mr ajay could you please clarify me bit deeper when to consider Qs+QL and only Qs because in your explination you added both latent heat and sensible but in one of your comments you consider only sensible part of it so could you please clarify me with that

ReplyDeletebecause in my case i need to calculate for methanol distillation 70-80 degree c feed is 747.9 kg in this case latent heat is too high 1165 so i am getting TR VALUE OF 510 FOR THE ABOVE CALCULATION IS IT TRUE COULD YOU PLEASE HELP ME

ReplyDeleteDear Surya,

DeleteMail me your query at pharmacalc823@gmail.com, i'll help you,

Regards,

AJAY K

Dear Surya,

ReplyDeleteUsually Qs will be very low when compared to that of Ql, so thats why sometimes i'll be considering only Ql,

Theoretically we need to consider only Ql, but practically while cooling a vapour, it will condense followed by cooling too, thats why physically we need to consider the summation of both.

hope you under stand

Regards,

AJAY K

Hello Sir, please help me to calculate to cool down the temperature of petrolium jelly from 65 degC to 12 degC in just five minutes. Total volume is 150 x 400 jar = 60 kg. Kindly reply me on pramod.malgaonkar@gmail.com

ReplyDeleteHey....,

DeleteWe can calculate the TR required,

Total heat load = 60 x 0.72 x (65-12) = 2289.6 KCal.

Total TR required is 2289.6/3024 = 0.76 TR.

Now tell me what are the utilities available to cool the jelly and what type of vessel are you using for cooling the jelly.

Best Regards,

AJAY K

How to calculate per day production rate of a plant.

ReplyDeleteIs this calculation is for cooling tower capacity calculation? If YES . then what is relation between TR & tower volume capacity? please explain.

ReplyDeleteDear Ajay,

ReplyDeleteFirst of all, I want to thank you for your immense support in clearing off all the queries and doubts. Wish u the best regarding your great work!

Regarding the TR problem you solved,

Latent heat of vaporization of toulene, you took the value as 351 kj/kg. Its value is applicable under atm pressure.

But under vacuum latent heat of vaporization will not be the same. How can you take the same value under vacuum? Bcz under vacuum, boiling point gets lower that means, latent heat gets lower as well bcz vaporization taking place at 35°C.

Also you took mass of liquid for calculation. We should use mass of vapor and specific heat of vapor in MCp∆T

I appreciate your response. Thank you!

Regards,

Sunil Kumar

Dear Sunil,

DeleteThanks for the compliment,

Latent heat will increase if the boiling point decreases but the variation wont be that much,

Mass will remain same for vapour or its liquid, but volume changes.

Best Regards,

AJAY K