Time to reduce volume of water by boiling?

[Insanity]

Mayhaps anyone provide assistance in determining how long it would take to reduce a specific volume of water (well, a solution, mostly water) to a smaller volume by boiling it?

i.e. boil 15 gallons down to 5 gallons?

Rule of thumb I've used is 5% of volume per hour, but I know it varies based on surface area, yet have little idea to figure it out. By this I roughly figure close to 48hrs.

Actual task at hand would be reducing ~30ish gallons to 4 gallons via boiling in a container with a surface area of ~254 sq. in.

 

[Borek]

It depends more on the speed at which you can deliver heat, than on the surface area.

 

[Topher925]

It depends more on the speed at which you can deliver heat, than on the surface area.

I would say it almost entirely depends on the heat flux. If you're vaporizing water at a relative fast rate then the surface area of the volume of water is irrelevant.

You can easily calculate the amount of heat required to boil off a given amount of water as long as you know its heat capacity and heat of vaporization.

 

[Insanity]

70k BTU/hr propane burner

 

[Borek]

If I read it correctly, 70k BTU is 73.9 MJ, assume water latent heat of evaporation of 2,270 kJ/kg. That's all you need to calculate mass of water evaporated per hour. Surface doesn't matter (unless you have some pathological situation with almost closed tank and pressure build up or something like that; I assume you don't).

--

 

[Insanity]

so then 73,853.9 kJ/hr divided by 2,270 kJ/kg is roughly 32 kg/hr or ~ 8.4 gallons/hr.
if at full flame.

 

[Borek]

Something like that. Some of the heat will get lost to sides, and not everything will get absorbed by the container, so don't expect 100% efficiency.

 

[thanushka]

Only depend on the rate of heat provided and the pressure on the mixture. If you can boil it using 1500 watts heater is more efficient and less time than 1000 watts heater. Another thing is you have to insulate the container, so that it will not loss heat to the surrounding. Third is, the pressure. If you boil the mixture at less pressure, you need less heat. So, you can apply vacuum; then the boiling point will be reduced. Water evaporate at 100 celcius at 1 atm.

 

[Insanity]

Something like that. Some of the heat will get lost to sides, and not everything will get absorbed by the container, so don't expect 100% efficiency.

Sure, a good portion of the heat is lost around the sides of the kettle anyways, and I usually don't have the burner at full flame.

Been pondering a really high gravity ale, and the boil time needed to reduce the volume. Wanted some ballpark figure that was better then guesstimating so I could plan for time.

If starting at 35gallons, and wishing to reduce to 4 gallons. Assuming the flame is at 50%, and mayhaps 75% efficiency.

(8.4/50%)*75%=~3.2 gallons per hour. Need to reduce by 31 gallons, so 9.6 hours.

 

[russ_watters]

I'd think 50% would be the absolute max. If the 70 Mbh is input you lose 20% right off the bat in combustion efficiency.

 

[SKinDt]

I think there is also a practical side to this which has to do with surface area. That is that this is not pure water it's a wort and if you get too much heat in you will get boil over. And the longer he's boiling the thicker the mixture is going to be. Going from 35 Gal to 4 Gal he may be close to the consistency of honey by the end.

So while BTU's in is the only factor for evaporation the diameter of the vessel determines how many BTU's you will be able to put in before boil over.

I was pondering this because in home-brew circles it's generally assumed that diameter is the biggest factor and in a way in this specific situation it probably is.

One way to mitigate it that other brewers have done is to use a fan on the surface. That will let you pump more heat in without boiling over.

However boiling is probably not a good choice to power up your wort as this extended a boil will have other nasty affects on it.