Cold weather + boil off = ???

When you brew with colder temperatures does your boil off per hour go up or down?

I was gonna ask the same question. When I started boiling yesterday I had 7 gallons. When I finished my boil(90mins) I had a touch over 5 gallons. Do I have the flame up to high or does it really matter?? Sorry for the highjack...

boil rate goes up. IIRC, it has something to do with the ability of cold air vs. hot air to hold moisture. Cold air has less moisture in it, which increases the rate at which moisture is transfered from the boil pot to the air.

hope this helps...I'm sure someone else will chime in with a better worded explanation.

I did a brew one time in Arizona in the summer. It was about 105 with no humidity. My boil rate was much higher than normal. The humidity in the air will affect the rate not just the temp.

It has to do with the fact that in the winter the air is usually much drier and therefor will hold and absorb more moisture. Also I believe that cooler air has the potential to hold more moisture just because the air is denser.

Warmer air holds more moisture. When it gets cold, the vapor condenses and forms a liquid again. It's why there's dew on everything in the morning.

i'm in the same boat! lost way more than I was planning on with my boil yesterday. for some reason, the cooler temps hadn't dawned on me as a potential reason. DUH. :cross:

this would explain my weekend session as well

Ok so I will have to take into account that my boil off per hour will be higher then during the summer months. How much more do you think the cold temps affect it? up to 2 gals per hour?

hmm mods can you move this thread meant to put it in General Beer discussion,since this isn't just AG, must have clicked the wrong forum yesterday...thanks

I'm not so sure that ambient air temperature in and of itself gives a higher evaporation rate. The main factors that effect it are relative humidity and air movement (breeze, wind). There is a "boundary layer" at the surface of the liquid being boiled. At this layer, the liquid is evaporating. The colder the air temperature, the more the evaporating liquid at the boundary layer has a chance to condense and fall back into the liquid. So, by that logic, all other things being equal, colder air temperature would cause less evaporation.

But still, I think humidity and air movement have a bigger impact here. The highest evaporation rate would be seen in a hot, dry, windy desert-like condition. If you took these same conditions (keeping air movement and relative humidity the same) and only lowered the air temperature, you'd actually get less evaporation because of more condensation occuring at the boundary layer. It just so happens that for those of us in a temperate climate, it's common to see the breeze pick up and the relative humidity go down with the seasonally cooler temperatures... giving us a higher evaporation rate than on a calm, relatively humid summer day.

If I'm way off here, call me out... I pulled that out of my a$$.

That sounds about right. It's basic science that hotter air holds more moisture, and the lower the relative humidity, the more potential there is for evaporation. This would explain why I hate Florida in the summer...

What about putting a lid on it before the end of your boil but when you reach the level you should be at i.e. I lose about 1 gal/hr in the summer but 2 weeks ago I lost 2.5/gal in 90 minutes. What would be the down falls of putting a cover on if any?

When you lid the pot during the boil, you have to monitor more closely for boil overs depending on the size of the pot. The lid will also allow you to maintain a good boil with a lower flame on the burner and force more condensate to fall back into the mix, so less evaporation.

weizen said:

What about putting a lid on it before the end of your boil but when you reach the level you should be at i.e. I lose about 1 gal/hr in the summer but 2 weeks ago I lost 2.5/gal in 90 minutes. What would be the down falls of putting a cover on if any?

Click to expand...

DMS not being boiled off... = cooked cabbage or corn flavor. Not sure how long it takes to get rid of the sulfur compounds that create it and to boil it off, but I think it's a better idea just to start with more if possible. I know for some of us it's hard, my pot is always at max capacity when I brew.

https://www.homebrewtalk.com/f36/boil-off-48883/index2.html

uwmgdman had a great explaination of this. Cold air = more boil off.

Thanks that post really cleared things up....I did a search before posting, but i must have been using the wrong key words.

What uwmgdman seems to be getting at there is "Humidity Ratio". HR is a measurement of humidity in the air that excludes temperature as a factor. Here is the calculation for humidity ratio (for degrees Fahrenheit):

HR =15432.3584*0.62*10^-5*0.01*Relative Humidity*(610.78*EXP(17.2694*((Degrees Fahrenheit-32)/1.8)/(((Degrees Fahrenheit-32)/1.8)+238.3))/2.2046).

You can plug that into an excel spreadsheet replacing Degrees Fahrenheit and Relative Humidity with cell addresses in which you've entered values for dF and RH.

This explanation still does not indicate temperature alone as a factor. In most cases, the HR of cold air is lower than the HR of warm air, hence the higher evaporation in cold air. But, if you compare the following 2 scenarios:

90dF, with 20% relative humidity = 41.47 humidity ratio
60dF, with 80% relative humidity = 61.10 humidity ratio

In this case, the warmer air has a lower humidity ratio and will allow for less condensation at the boundary layer (less water falling back into the pot) due to the lesser difference in boil temp vs. air temp. So, the 90dF air temp will result in a higher evaporation rate.

Like I said before, in most cases, colder air has a lower relative humidity and humidity ratio and will result in a higher evaporation rate, but colder air temperature alone does not result in a higher evaporation rate.

menschmaschine said:

HR =15432.3584*0.62*10^-5*0.01*Relative Humidity*(610.78*EXP(17.2694*((Degrees Fahrenheit-32)/1.8)/(((Degrees Fahrenheit-32)/1.8)+238.3))/2.2046).

Click to expand...

+1 on HR =15432.3584*0.62*10^-5*0.01*Relative Humidity*(610.78*EXP(17.2694*((Degrees Fahrenheit-32)/1.8)/(((Degrees Fahrenheit-32)/1.8)+238.3))/2.2046) except I would use f(x)=22/7 - radical 2 for this particular calculation...