25 replies to this topic

[Big Nake]

I get the concept... you have a fermenting primary in a room where a thermometer says it's 65°. But the beer could be fermenting at 70° or warmer because of the heat produced by the primary fermentation. But what do you science-heads say about a plastic bucket (or a glass carboy...) that is submerged in 10 gallons of water in a swamp bucket? If a thermometer in that water says that it's 55°, do you think that the beer inside the primary could still be 5° higher or more? I guess my non-scientific mind is saying that the cold water is going to insulate that primary and keep the temp more consistent than air would. Thoughts?

[Humperdink]

Yes.The thermal mass of the water insulates the fermentor. But if you are reading the thermometer then you aren't measuring the ambient, you are measuring the temp of the ferment.

[3rd party JKor]

Do you want to have a detailed heat transfer and fluid mechanics discussion?

[ThroatwobblerMangrove]

Do you want to have a detailed heat transfer and fluid mechanics discussion?

YAY!

[Humperdink]

Do you want to have a detailed heat transfer and fluid mechanics discussion?

Yes please. And I think "insulate" was a bad choice of words on my part without qualifying it. That's what I get when PUI. What I meant was the water in the tub insulates the fermenter from easily fluctuating temperature wise because of the greater thermal mass. Maybe J can correct me here but if you have a 5 gallon fermenter chuggin away submerged in five gallons of water, the temperature change should be halved due to twice the volume of liquid. J care to correct me? That's just what my brain tells me when I used to use a water filled tub.

[3rd party JKor]

The difference between having cool air around the fermenter and cool water is the difference in the heat transfer coefficient. The heat transfer coefficient, denoted by h, tells you how well a fluid convects heat. The higher h is, the more heat is convected. Heat transfer coefficients for liquids are naturally higher than for gases.The heat transfer coefficient varies widely for any particular fluid depending on the temperature, temp differential, flow properties, type of convection, geometry of the system, etc. The range of h for air is about 10 to 100, for water it's 500 to 10,000. We're talking about free convection here (not forced, unless you're agitating the air/water), so we're on the low end of the scale. You can expect the h of air to be around 10 and the h of water to be around 500, i.e. the water transfers heat away from the carboy 50 times as fast as the air.We often see estimates that the fermenting beer can be 5-8°F higher than the surrounding air (or something like that). If we assume that the beer is fermenting at 67°F, then those estimates would tell you that you would need to keep the ambient air temp around 60°F. Knowing this we can use Newton's Law of Cooling to predict the heat loss from the carboy via convection:Q = hairA(Twort-Tair) [Newton's Law of Cooling]Q is the heat flow rate being convected away fromt he fermenter, A is the carboy surface area, the Ts are temp of the air and wort.If we assume the same fermentation is going on in a water bath, we can again write out the convection heat transfer equation:Q = hwaterA(Twort-Twater)Since both fermentations are the same we can assume both are producing the same amount of heat. Also, the surface area of the carboy is the same. Using that information we can combine the two equations to:hwaterΔTwort-water = hairΔTwort-airRearranging gives:hair/hwater = ΔTwort-water/ ΔTwort-airThis tells us that the temperature differential required for the air and water systems, respectively, is proportional to the ratio of convective heat transfer coefficients of the water and air. Since we know the approximate values of hwater, hair and ΔTwort-air we can calculate the ΔTwort-water required to maintain fermentation temperature.ΔTwort-water = hairΔTwort-air/hwater = 10*7/500 = 0.14So if it requires a 7°F air temp differential to maintain temp a specified fermentation temperature, it only requires 0.14°F temperature differential to maintain the same fermentation temperature in a water bath.The glitch in the system is that the water in the tub eventually has to transfer the heat it's absorbing somewhere else. It will just keep heating up if it doesn't. The only place it has to transfer it's heat to is the surrounding air. So, again, the convective heat transfer coefficient of air becomes the limiting factor. The difference in the case of the tub is that you can remove heat from the water in several ways, such as a swamp cooler or adding ice.Any questions?

Edited by JKoravos, 25 September 2009 - 10:39 AM.

[ColdAssHonky]

The difference between having cool air around the fermenter and cool water is the difference in the heat transfer coefficient. The heat transfer coefficient, denoted by h, tells you how well a fluid convects heat. The higher h is, the more heat is convected. Heat transfer coefficients for liquids are naturally higher than for gases.The heat transfer coefficient varies widely for any particular fluid depending on the temperature, temp differential, flow properties, type of convection, geometry of the system, etc. The range of h for air is about 10 to 100, for water it's 500 to 10,000. We're talking about free convection here (not forced, unless you're agitating the air/water), so we're on the low end of the scale. You can expect the h of air to be around 10 and the h of water to be around 500, i.e. the water transfers heat away from the carboy 50 times as fast as the air.We often see estimates that the fermenting beer can be 5-8°F higher than the surrounding air (or something like that). If we assume that the beer is fermenting at 67°F, then those estimates would tell you that you would need to keep the ambient air temp around 60°F. Knowing this we can use Newton's Law of Cooling to predict the heat loss from the carboy via convection:Q = hairA(Twort-Tair) [Newton's Law of Cooling]Q is the heat flow rate being convected away fromt he fermenter, A is the carboy surface area, the Ts are temp of the air and wort.If we assume the same fermentation is going on in a water bath, we can again write out the convection heat transfer equation:Q = hwaterA(Twort-Twater)Since both fermentations are the same we can assume both are producing the same amount of heat. Also, the surface area of the carboy is the same. Using that information we can combine the two equations to:hwaterΔTwort-water = hairΔTwort-airRearranging gives:hair/hwater = ΔTwort-water/ ΔTwort-airThis tells us that the temperature differential required for the air and water systems, respectively, is proportional to the ratio of convective heat transfer coefficients of the water and air. Since we know the approximate values of hwater, hair and ΔTwort-air we can calculate the ΔTwort-water required to maintain fermentation temperature.ΔTwort-water = hairΔTwort-air/hwater = 10*7/500 = 0.14So if it requires a 7°F air temp differential to maintain temp a specified fermentation temperature, it only requires 0.14°F temperature differential to maintain the same fermentation temperature in a water bath.The glitch in the system is that the water in the tub eventually has to transfer the heat it's absorbing somewhere else. It will just keep heating up if it doesn't. The only place it has to transfer it's heat to is the surrounding air. So, again, the convective heat transfer coefficient of air becomes the limiting factor. The difference in the case of the tub is that you can remove heat from the water in several ways, such as a swamp cooler or adding ice.Any questions?

= Use a water bath?

[ThroatwobblerMangrove]

One thing to keep in mind that you may or may not be accounting for:In the case of air you can almost assume an infinite amount of air. In the case of water though there is usually a pretty limited amount so the water will approach the temperature of the fermenter.

[3rd party JKor]

= Use a water bath?

Depends...

[denny]

One thing to keep in mind that you may or may not be accounting for:In the case of air you can almost assume an infinite amount of air. In the case of water though there is usually a pretty limited amount so the water will approach the temperature of the fermenter.

Unless you do something like add ice packs to the water. That's my MO.

[ThroatwobblerMangrove]

Unless you do something like add ice packs to the water. That's my MO.

Now the math gets calculated

[stellarbrew]

The difference between having cool air around the fermenter and cool water is the difference in the heat transfer coefficient. The heat transfer coefficient, denoted by h, tells you how well a fluid convects heat. The higher h is, the more heat is convected. Heat transfer coefficients for liquids are naturally higher than for gases.The heat transfer coefficient varies widely for any particular fluid depending on the temperature, temp differential, flow properties, type of convection, geometry of the system, etc. The range of h for air is about 10 to 100, for water it's 500 to 10,000. We're talking about free convection here (not forced, unless you're agitating the air/water), so we're on the low end of the scale. You can expect the h of air to be around 10 and the h of water to be around 500, i.e. the water transfers heat away from the carboy 50 times as fast as the air.We often see estimates that the fermenting beer can be 5-8°F higher than the surrounding air (or something like that). If we assume that the beer is fermenting at 67°F, then those estimates would tell you that you would need to keep the ambient air temp around 60°F. Knowing this we can use Newton's Law of Cooling to predict the heat loss from the carboy via convection:Q = hairA(Twort-Tair) [Newton's Law of Cooling]Q is the heat flow rate being convected away fromt he fermenter, A is the carboy surface area, the Ts are temp of the air and wort.If we assume the same fermentation is going on in a water bath, we can again write out the convection heat transfer equation:Q = hwaterA(Twort-Twater)Since both fermentations are the same we can assume both are producing the same amount of heat. Also, the surface area of the carboy is the same. Using that information we can combine the two equations to:hwaterΔTwort-water = hairΔTwort-airRearranging gives:hair/hwater = ΔTwort-water/ ΔTwort-airThis tells us that the temperature differential required for the air and water systems, respectively, is proportional to the ratio of convective heat transfer coefficients of the water and air. Since we know the approximate values of hwater, hair and ΔTwort-air we can calculate the ΔTwort-water required to maintain fermentation temperature.ΔTwort-water = hairΔTwort-air/hwater = 10*7/500 = 0.14So if it requires a 7°F air temp differential to maintain temp a specified fermentation temperature, it only requires 0.14°F temperature differential to maintain the same fermentation temperature in a water bath.The glitch in the system is that the water in the tub eventually has to transfer the heat it's absorbing somewhere else. It will just keep heating up if it doesn't. The only place it has to transfer it's heat to is the surrounding air. So, again, the convective heat transfer coefficient of air becomes the limiting factor. The difference in the case of the tub is that you can remove heat from the water in several ways, such as a swamp cooler or adding ice.Any questions?

I agree. Good explanation.I will add that JK makes the reasonable assumption that the outside surface temperature of the fermenter is equal to the temperature of the wort, which isn't exactly true. However, it is a valid assumption in order to show the much greater delta-T which must exist when the surrounding media is air versus when it's water. To be strictly rigorous, there is a temperature gradient across the wall of the fermentor to account for, based on the fermentor wall's thickness, and the conductive heat transfer coefficent for the particular material the fermenter is made of. Whithout resorting to heat transfer calculations for the glass wall of a carboy, the plastic wall of a bucket, or the steel wall of a keg or conical, I'd say intuitively that the temperature gradient wouldn't be more than one degree F, and would be practically zero for a thin-walled steel fermenter.

[3rd party JKor]

It's far from an accurate calculation. It's a fairly complex system and I left a lot of stuff out for simplicity, but it gives you in idea of the effect of the main driver in the system, which is the convective heat transfer coefficient.

[Big Nake]

The difference between having cool air around the fermenter and cool water is the difference in the heat transfer coefficient. The heat transfer coefficient, denoted by h, tells you how well a fluid convects heat. The higher h is, the more heat is convected. Heat transfer coefficients for liquids are naturally higher than for gases.The heat transfer coefficient varies widely for any particular fluid depending on the temperature, temp differential, flow properties, type of convection, geometry of the system, etc. The range of h for air is about 10 to 100, for water it's 500 to 10,000. We're talking about free convection here (not forced, unless you're agitating the air/water), so we're on the low end of the scale. You can expect the h of air to be around 10 and the h of water to be around 500, i.e. the water transfers heat away from the carboy 50 times as fast as the air.We often see estimates that the fermenting beer can be 5-8°F higher than the surrounding air (or something like that). If we assume that the beer is fermenting at 67°F, then those estimates would tell you that you would need to keep the ambient air temp around 60°F. Knowing this we can use Newton's Law of Cooling to predict the heat loss from the carboy via convection:Q = hairA(Twort-Tair) [Newton's Law of Cooling]Q is the heat flow rate being convected away fromt he fermenter, A is the carboy surface area, the Ts are temp of the air and wort.If we assume the same fermentation is going on in a water bath, we can again write out the convection heat transfer equation:Q = hwaterA(Twort-Twater)Since both fermentations are the same we can assume both are producing the same amount of heat. Also, the surface area of the carboy is the same. Using that information we can combine the two equations to:hwaterΔTwort-water = hairΔTwort-airRearranging gives:hair/hwater = ΔTwort-water/ ΔTwort-airThis tells us that the temperature differential required for the air and water systems, respectively, is proportional to the ratio of convective heat transfer coefficients of the water and air. Since we know the approximate values of hwater, hair and ΔTwort-air we can calculate the ΔTwort-water required to maintain fermentation temperature.ΔTwort-water = hairΔTwort-air/hwater = 10*7/500 = 0.14So if it requires a 7°F air temp differential to maintain temp a specified fermentation temperature, it only requires 0.14°F temperature differential to maintain the same fermentation temperature in a water bath.The glitch in the system is that the water in the tub eventually has to transfer the heat it's absorbing somewhere else. It will just keep heating up if it doesn't. The only place it has to transfer it's heat to is the surrounding air. So, again, the convective heat transfer coefficient of air becomes the limiting factor. The difference in the case of the tub is that you can remove heat from the water in several ways, such as a swamp cooler or adding ice.Any questions?

I like beer.

[3rd party JKor]

I like beer.

That's not a question, Ken.

[BrewerGeorge]

I can't improve on JK's explanation, but it does seem to me that we have two different situations here. When the outside medium is hotter than the ferment and when the medium is cooler. People tend to use water baths when they're trying to keep the ferment cooler than the air in the room. In that case, your original use of the word 'insulate' does have some merit. The increased mass of the whole system does give your protection from swings in temperature because of its higher thermal inertia. You will have to continue to remove energy from the system with ice or a swamp cooler or some such, and you will have to take out more total energy by using the water bath. But doing so will make a slower, more stable temperature change for the fermenting beer.

[dmtaylor]

I avoid all these issues by using a wet t-shirt and fan. The t-shirt is wicking up water from a bin with 2 inches water in it, so it's constantly wet, and as long as your humidity is less than 100%, you will always have evaporative cooling. Just be sure to refill the water every day or two to maintain a few inches water -- but not too much or you will end up with the immersion situation and need to add ice or whatever.

[ThroatwobblerMangrove]

I avoid all these issues by using a wet t-shirt and fan. The t-shirt is wicking up water from a bin with 2 inches water in it, so it's constantly wet, and as long as your humidity is less than 100%, you will always have evaporative cooling. Just be sure to refill the water every day or two to maintain a few inches water -- but not too much or you will end up with the immersion situation and need to add ice or whatever.

That would certainly do a good job of keeping things cool but you miss out on the temperature stabilization aspects of submerging the fermenter in water. I don't really know how big of an impact that has on small batches though honestly. I rarely do it.

[3rd party JKor]

I avoid all these issues by using a wet t-shirt and fan. The t-shirt is wicking up water from a bin with 2 inches water in it, so it's constantly wet, and as long as your humidity is less than 100%, you will always have evaporative cooling. Just be sure to refill the water every day or two to maintain a few inches water -- but not too much or you will end up with the immersion situation and need to add ice or whatever.

+1Evaporative cooling is the shizz. Evaporation of water sucks a boatload of heat from the system. If you're real fancy you can get a temperature controller to turn the fan on and off.

[Big Nake]

Now take all of this into consideration and add the fact that my swamp cooler is sitting on my cool basement floor. It seems to me that the contact with the cool floor keeps the contents of the swamp cooler (and primary) cooler in general. I assume that most people who are trying to keep temps lower without the use of refrigeration are doing so on the floor. I really just need to use my draft fridge as a lager-primary fridge and get the 4-tap chest freezer I've been talking about!