The ULTIMATE chiller calculator thread

Hey guys, sup.

I have noticed many people on the forums talking about how their chillers don’t preform. Here is a super simple formula you can use to calculate the cooling capacity needed for a project to make sure you aren’t undersizing.

Take the specific heat of your solvent in BTU/lb F – you can find it in this link - Butane - Thermophysical Properties (we are using butane in this example which is equal it .39 BTU/lb F)

Multiple by pounds of solvent desired to be cooled – here, we use 100 lbs

Multiple by your temperature differential - here, we are going from 80F to -40 F

Divide your desired target temp time into hours for your time multiple (i.e. 15 min would be a BTU/hr multiple of 4). In this case we want to take 100lbs of butane from 80F to -40F in 15 min.

Then convert the BTU to Kw@ your desired temp. Here is a simple converter link - BTU/hr to kilowatts (kW) conversion

Once you have the answer in Kw reduce by your estimated drop do to loss. (normally 20-30% or below for a jacketed vessel and non-vacuum insulated lines). To do this we add 20% to our cooling capacity by multiplying by 1.20. Hot environment about 70 degrees? Add more. Non vacuum insulated vessels? Add way more. Conditions are paramount.

So, in our example 100lbs of butane cooled from 80f to -40f in 15 min would be -

.39 (specific heat) * 100 (lbs of butane) * 120 (degrees of change) * 4 (multiple of 1 hr (15min)) = 18,720 BTU/hr @-40F

Same thing but in one hour instead of 15 min = 4,680 BTU/hr @-40F

Convert 18,720 BTU/hr to Kw - 5.48Kw @ -40F

Add losses from thermal fluid, hoses, jacket, and shell. (20-30%)

Add the difference 1+.20 (added 20% over power for losses)

5.48KW*1.20 = 6.57KW @ -40F

Remember that cooling capacities of chillers generally drop as they get cooler. The deeper you go, the more expensive it gets. Be sure to check your chillers cooling curve before spec’ing a chiller.

Alternatively, you can just call me. Questions and comments welcome. I hope this helps.


Thanks @precisionnick!

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I made my clients a neat little “plug and play” excel sheet for butane and ethanol. They love it. I tried teaching the “why.” It rarely works that way…lol


Could you use this to figure out in-line flow cooling?

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Hi there I used your formula and am a bit confused

4lbs x .39btu = 1.56btu x 120 = 187.2° change x 60 (1 min) = 11,220btu/h x 0.000293 (btu/h to kw) = 2.93kw

That’s to change 4 pounds of butane from -40° to 80° Per minute but when BizzyBee999 did the math he came out to 2.92 kw need to condense or heat 1LB/min

Maybe I missed something but I’m kinda wondering why this is ( both mine and his equations account for no loss)


You’re talking about specific heat capacity (energy required to change temperature).

Bizzybee is talking about heat of vaporization (the energy required for a liquid/vapor phase change - aka vaporizing or condensing).

You just made a minor math error at the end (aside from units being all over the place you still got them right in the end). Bizzybee’s calculations should have also included the energy required to cool the butane, not solely condense it. Though it’s a small value relative to the energy for vaporization/condensing, it still makes a difference as you scale up.

Was half way through sending you a private message but figured I’d share these tidbits here too, to clear up the picture.

187.2 is the energy required in BTU’s, not a temperature change in degrees. And your units are all over the place. You also multiplied degrees by seconds and ended up with BTU/hr.

I’m not saying this to shame you, but to actually get you to think about why the math is working the way it is (or why you messed up)…hate that I even have to make that disclaimer…

Start at the top: Write everything down in fractions so you can more easily keep track of units and visualize how they “cancel out.”

4 lbs × .39 btu/(lb • °F) × 120 °F = 187.2 BTU = total energy required

187.2 BTU/min = 11,232 BTU/hr …now I follow you again.

1 kilowatt = 3412.14 BTU/hr

(x kW) / (1 kW)= (11,232 BTU/hr) / (3412.14 BTU/hr)
x = 3.29

3.29 kW is the power required.

Look at that last conversion from BTU/hr to kW that you did. The digits are the same in both values (2,9,3, respectively). You could only get that by multiplying by a factor of 10 (aka you just shifted the decimal placement). I didn’t follow much of your math until this point haha. You almost had it, just messed up that final multiplication.

Now, as far as Bizzybee’s calculations, he has a typo in his units. He multiplied (mol/lb) by (kJ/mol) and ended up with (kJ/min). It should be 175 kJ/lb. Guess he just shorthanded the conversion…Otherwise, he’s right. Here is a more explicit unit conversion:

3600 kJ = 1 kWh
175 kJ = 0.0486 kWh

[ 0.0486 kWh ] / [ (1/60) hr ] = 2.916 kW

This is what happens when I eat 300+ mg THC and it somehow wakes me up. I’ll be wired all night😂. That being said, I’m pretty stoned and typing this all from my phone. Someone double check me for what it’s worth.


You’re a real life saver, thank you for looking over my math this is all a little above my knowledge but my college classes will catch up with this eventually. In bizzys how much more energy is required to bring down the temperature after the phase change? If chilling the vapor to liquid which equation would be more valuable bizzys or mine? Hope you don’t say up too long man! Amazing how a sleep aid can be so stimulating lmao :laughing:

heat (energy) = mass × specific heat × |change in temperature|

Add that to the energy needed to vaporize/condense. “Recovering” solvent requires both cooling and condensing the solvent.


:slight_smile: I thought I was the only one who bounced on 300mg.

100mg makes me wobbly, 200mg makes me fall down. Damned if 300mg doesn’t knock me over, then keep me up all night. varies a little by strain I think.



When I look at my Julabo -50C chiller specs, it has a chart that says:

Cooling capacity (Medium Ethanol)
°C 20 0 -20 -50
kW 0.9 0.8 0.5 0.16

What does this mean exactly and why does it matter?

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I believe those specs let you know how much power the killer will need to use to achieve the -50 you are looking for. if the medium you are trying to chill has an initial temperature of -50 and you want to maintain that -50 it will use .16KW if your initial temperature is 20 degree C, it will need to use 0.9KW to achieve your desired temperature (-50)

Please correct me if i am wrong @precisionnick I haven’t seen that layout of numbers really but it would use less energy to bring an already chilled medium to a close desired temperature whereas, it would require more energy the bigger that delta T is.

As to why it matters, i believe you should be able to relate these specs with time. this would give you an idea of how fast you would be able to chill to a desired temp?

I’m pumped on this thread because i’m trying to hone my chiller spec skills so i am sorry if this is incorrect :grimacing:

@Verchedabby @FRESHcoastextracts

The cooling capacity of any chiller generally diminishes the colder you get.

What you are seeing is the 900W on your chiller is rated at 20C. What you are looking as is the drop in chilling capacity as the temperature drops. So your 900W becomes 160W at -50C.

When formulating your needed chilling capacity, you should do so at your desired temp.

For example, if you needed 500W (based upon the above needed capacity calculation), your chiller would be fine to reach -20C. If you wanted to reach -50C, you would be under-capacity by 360W.


@precisionnick @Verchedabby

Thanks for both of your reply’s, they were helpful in understanding more about the chiller specs.

Let’s ask another question. If my chiller readout says the alcohol running through the lines and through my jacketed solvent tank are at -35°C, what do you think the solvent in the solvent tank is actually reading at? If you do know, how did you determine it?

Whats the best way to transfer that heat between the solvent in the solvent tank and the alcohol running through the chiller? Does it make more sense to use a plate style heat exchanger?

It may require more engineering to achieve, and the design of the solvent tank may need to change, but would it be more efficient then cooling a jacket around the outside?

Street cred:



There are many answers to your questions. First, a just around the outside of the vessel is always going to be the most inefficient because 50% of that jacket (the outside) isn’t touching the solvent. This can be combated by a dual jacket with the outer layer being a vacuum jacket, but it still leaves room for improvement and you won’t reach 100% efficiency with just a jacket on your vessel.

Plate, tube ans shell, tube in tube etc. heat exchanges are also very common for an in line application. A common example of an inline application is a post pumping condenser coil (either with a chiller or dry ice bath).

The problem with inline cooling is the flow rates are often fast with our applications and therefore the retention time is very low (contact of solvent to heat exchanger). Using the very first calculation that I posted, you would have to adjust your retention time accordingly. You can calculate your retention time in your heat exchanger by taking the volume of solvent and dividing it by the flow rate of your heat exchanger (make sure to account for any pressure as it will change the flow rate). My preferred method if using an inline heat exchanger is a tube in tube or tube in bath. The reason being we can engineer the tube to be 100+ft, coil it up, and achieve a longer retention time of solvent in the heat exchanger.


@precisionnick answered the rest but I thought i’d pipe in one more time.

If you don’t have a probe telling you the solvent temp…

There is a conductive heat transfer equation you can use to determine the temperature of the solvent inside the tank without a probe but I know there’s a few “assume negligible” variables which would give you a different answer than the actual temp. Here’s a link if you’d want to learn more and give it a go.

That equation would tell you the temperature on the inside of the solvent tank of which is in contact with the solvent. you could then determine the temperature of the solvent from there.

Hope it helps.

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Class 1 Division 1 UL listed thermocouple. They normally mount with a 1/4 compression or FNPT.

Ill see if I can find you a link to purchase one.

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@precisionnick What are you using as a display for those?

Normally with the EXProof thermocouples they have the display built into the head and intrinsically sealed for use in the rated environment.

This isn’t our vendor but this will give you an idea of what I am talking about.



So, I am hoping someone here will check these calculations. It would appear I need a massive chilling set-up.
Not clear to me why Cp or Cv is used as BTU/lb F. They are 20% different.
Here’s my equation;
10usg per batch EtOH, from 70F to -40F, 4x per hr, Cv=.52 BTU/lb F, 20% loss to hoses, no data for chiller efficiency.
As I understand it
10usg sg .789 = 7.89 lbs x .52 x 120 * 4 = 17,330 BTU/hr
COnvert to kW 17330 / 3412 = 5.08kW
5.08 / losses (.2) = 6.09kw
Then add whatever loss from the chiller’s efficiency at set point?

Thanks for the link to the thermocoupler. I’m going to check out the price on that. Would be nice to know the actual temperature of the solvent entering the material column. It sounds like jacketing a column isn’t the way to go, but rather your suggesting this: Use a jacket’d and cooled solvent tank but then when you go to flood the material column, pump that liquid from the solvent tank through a couple hundred feet of SS 3/8" tubing stuffed in dry ice/alcohol slurry before it enters the material column?

@Verchedabby thanks for the equation information.

When it comes to extractions, what’s the difference between different SS types anyways: 304, 316?