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#1 Mike Green

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Posted 11 October 2017 - 12:17 PM

I want to start fermenting in pin lock kegs inside my mini fridge.My question is How long will it take to get 4.5 gallons of wort down to pitching temp. I'm looking for a formula so I can plug different values in such as Starting temp, fridge temp, different volumes of wort.Hopefully, someone has a link to an online calculator.


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#2 JKor

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Posted 11 October 2017 - 12:29 PM

Just sticking it a 5 gal keg in the fridge?  That's actually a fairly complex heat transfer calculation.  You can do a very basic calc if you make some assumption but I don't know how accurate it will be.  you'd probably be better off just doing it and seeing how it takes.


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#3 Mike Green

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Posted 11 October 2017 - 12:47 PM

Just sticking it a 5 gal keg in the fridge?  That's actually a fairly complex heat transfer calculation.  You can do a very basic calc if you make some assumption but I don't know how accurate it will be.  you'd probably be better off just doing it and seeing how it takes.

My plan is to bring down the temp with the plate chiller and then pop the keg into the fridge @ 35F until the wort reaches 65f.I can program my fermenting fridge with BrewPi to alert me when the wort temp reaches 65F but I need to know within 12 hours when it hits its target temp so I can start my SNS starter.Trial and error seem like the best approach but I was hoping someone had already worked out a formula or knows of an online calculator.  

EDIT: My groundwater will only bring down the temp to low 80's high 70's and I'm not interested in messing around with pre chillers and ice.


Edited by Mike Green, 11 October 2017 - 12:50 PM.

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#4 SchwanzBrewer

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Posted 11 October 2017 - 09:08 PM

OK, here's the deal. The equation for finding how much heat is required to be removed from the system. The difficulty is estimating the convection coefficient. q = hA(T1 - T2). q is the heat that needs to be removed (in Watts, W), A is the surface area of the keg (square meters), T1 is the initial temp (in kelvin), T2 is the final temp (in kelvin). h is the convection coefficient (in W/m^2K ... watts per (square meter * Kelvin)). Trust me it's a million times easier to do these calculations in metric than in english units. h is the big question mark. If the air is swirling around pretty good in the fridge then h is likely around 20 W/m^2K. Your delta T is 8.3K...

 

so...

 

q = 20W/m^2K x 0.997m^2 x 8.3K = 165.5W

 

That's what it takes to cool the wort from 80F to 65F. Now, you need to know what the cooling rate is for the fridge. Google tells me a mini fridge can handle about 150 BTU/hr = 1055.06 J

 

1 W = 1 J/s

 

So...

 

1055.06J/165.5J/s = 6.4 seconds... Well that's wrong so most likely we estimated h wrong or I screwed up a conversion. (Edit: I realized what I did, q is the rate of heat transfer, 665107.2 J is the energy required to cool the wort, so according to this math it would take 4018.7s to cool the wort, which is 11.2 hours. h is probably still wrong).

 

 

Simpler...

 

A BTU is the amount of energy to raise (or lower) 1 lb of water the temperature of 1 deg F. We have 4.5 gal of 1.050 wort that weighs 39.4 lb. That means it takes 39.4 BTU to lower the temperature 1 deg F. That means we need 15 x 39.4 = 630.4 BTU to lower the temp from 80 to 65. If the fridge puts out 150 BTU/hr then it takes 4.2 hours to cool down the wort assuming it's 100% cooling at that rate and it doesn't have a duty cycle that shuts down the compressor. That sounds much more reasonable though.

 

I hope this was helpful. It was bugging me. I'm an engineer.


Edited by SchwanzBrewer, 11 October 2017 - 09:16 PM.

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#5 JKor

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Posted 12 October 2017 - 03:26 AM

OK, here's the deal. The equation for finding how much heat is required to be removed from the system. The difficulty is estimating the convection coefficient. q = hA(T1 - T2). q is the heat that needs to be removed (in Watts, W), A is the surface area of the keg (square meters), T1 is the initial temp (in kelvin), T2 is the final temp (in kelvin). h is the convection coefficient (in W/m^2K ... watts per (square meter * Kelvin)). Trust me it's a million times easier to do these calculations in metric than in english units. h is the big question mark. If the air is swirling around pretty good in the fridge then h is likely around 20 W/m^2K. Your delta T is 8.3K...

 

so...

 

q = 20W/m^2K x 0.997m^2 x 8.3K = 165.5W

 

That's what it takes to cool the wort from 80F to 65F. Now, you need to know what the cooling rate is for the fridge. Google tells me a mini fridge can handle about 150 BTU/hr = 1055.06 J

 

1 W = 1 J/s

 

So...

 

1055.06J/165.5J/s = 6.4 seconds... Well that's wrong so most likely we estimated h wrong or I screwed up a conversion. (Edit: I realized what I did, q is the rate of heat transfer, 665107.2 J is the energy required to cool the wort, so according to this math it would take 4018.7s to cool the wort, which is 11.2 hours. h is probably still wrong).

 

 

Simpler...

 

A BTU is the amount of energy to raise (or lower) 1 lb of water the temperature of 1 deg F. We have 4.5 gal of 1.050 wort that weighs 39.4 lb. That means it takes 39.4 BTU to lower the temperature 1 deg F. That means we need 15 x 39.4 = 630.4 BTU to lower the temp from 80 to 65. If the fridge puts out 150 BTU/hr then it takes 4.2 hours to cool down the wort assuming it's 100% cooling at that rate and it doesn't have a duty cycle that shuts down the compressor. That sounds much more reasonable though.

 

I hope this was helpful. It was bugging me. I'm an engineer.

 

 

 

I think your h is reasonable for natural convection, probably a bit high.  This link shows an h of 5 W/m2-K for a vertical plate with 30C temp difference, close to what we have here. The delta T will be 80F-35F and go until it's 65-35.  You need to integrate over time to get the temp change.  

 

You also need to take into account internal convection in the keg, if the wort near wall of the keg is cooler than the core the heat transfer will be slower.  That's why I said it's a complex problem, you are dealing with convection on both sides of the keg wall and need to make assumptions on h for both.  But lets just say that we can make a bulk assumption on the inside of the keg (uniform temp), I get a time of 2.5hrs with an h of 5.  The h is probably a bit lower, but not that much.  This also doesn't take into account that the keg warms up the fridge which also slows the heat transfer down.  

 

I'd say the absolute minimum we're talking about here is 2.5 hrs assuming the fridge can keep up with the cooling rate, which it most likely can't.


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#6 pickle_rick

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Posted 12 October 2017 - 03:50 AM

I agree with try it and see.

 

in my chest freezer I took ~11 gal of 75F wort down to 64F in about 3 hours.  my chest freezer is obviously colder.  I also had a small fan circulating air in there.


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#7 Mike Green

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Posted 12 October 2017 - 05:54 AM

Thanks, GuysI was thinking it would take at least a day.I need at least 12 hours for the starter to chooch.I can make my starter on brew day and be able to pitch the next day.


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#8 SchwanzBrewer

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Posted 12 October 2017 - 06:03 AM

I think your h is reasonable for natural convection, probably a bit high.  This link shows an h of 5 W/m2-K for a vertical plate with 30C temp difference, close to what we have here. The delta T will be 80F-35F and go until it's 65-35.  You need to integrate over time to get the temp change.  

 

You also need to take into account internal convection in the keg, if the wort near wall of the keg is cooler than the core the heat transfer will be slower.  That's why I said it's a complex problem, you are dealing with convection on both sides of the keg wall and need to make assumptions on h for both.  But lets just say that we can make a bulk assumption on the inside of the keg (uniform temp), I get a time of 2.5hrs with an h of 5.  The h is probably a bit lower, but not that much.  This also doesn't take into account that the keg warms up the fridge which also slows the heat transfer down.  

 

I'd say the absolute minimum we're talking about here is 2.5 hrs assuming the fridge can keep up with the cooling rate, which it most likely can't.

 

I thought he said 80 to 65? Air temp... ack. Been a while since I've done this. Also  probably shouldn't try this so late at night after working 11 hours. Second analysis is probably closer.

 

Also, I realize there's convection in the keg, but I wasn't about to try and tackle that too.


Edited by SchwanzBrewer, 12 October 2017 - 06:04 AM.

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#9 JKor

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Posted 12 October 2017 - 09:08 AM

It's a way better idea to just try it and see.


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