
BTU's
HI All:
I'm trying to improve my evaporation rate during the boil;
Now, I calculate the BTU's at the heating of water as follows:
BBL X 259 (lbs/bbl) X Temp Diff per hour (T1  T2) = BTUs/hr
I know I'm not using the specific density of the H2O heating.
Any advise
Txs,
Fred

Fred,
Yeah......you've got it.
Mass x Temperature Change x Specific Heat Capacity = Heat Added or Lost
M x dT x Sp = Ht
Dividing the "delta T" by a time period will produce a rate of heat added or lost. Using English units, you would use Lbs, degrees F, and BTUs for heat. The Specific Heat Capacity for water in English Units is 1 (pretty simple), which shows why a BTU is the amount of energy it takes to raise 1 Lb of water 1 degree F........... 1 x 1 x 1 = 1.
In the Metric system, Calories are used and 1 Calorie is the amount of heat required to raise 1 gram of water by 1 degree C.
The Heat of Vaporization for water is 1060 BTU/Lb, and is the amount of energy required to change 1 Lb of water from a liquid state to steam. Simply heating water up to 212F doesn't quite get you there. Taking your 1 Bbl of water example, Fred, and assuming you start at 160F after sparge:
259 Lbs x (212F  160F) x 1 = 11,024 BTUs
Seems like a lot........however, it will get to 212F and just shimmer unless some serious energy is poured into the system to change the state to steam......like 1060 BTUs/Lb.
For every Bbl to get converted to steam, you need:
259 Lbs x 1060 BTU/LB = 274,540 BTUs
This is the amount of heat you need to get out of your conversion burner or steam kettle jacket..........plus a little heat transfer fudging for conduction and convection, which REALLY makes this a science project. You can see that timing is everything, and if you're trying to boil off that 259 Lbs in 1 hour, you need quite a bit of fuel. Note that while you're using the 274K BTUs to make steam, you also need to maintain the 212F for the rest of the liquid to keep it hot, and that's based in large part to what the heat loss is through conduction through the unheated sidewall to unjacketed parts of the kettle. Predicting those losses is kinda dicey since it depends on so many factors. Also, if you're running a smaller kettle (our 7 Bbl kettle at the first Brewery will boil off somewhere around 1/4 Bbl per hour) you won't want to boil off nearly as much mass as a barrel of liquid so your energy requirements would be less.
I hope all this helps, Fred.

BTU's
Thanks Brian, one can altimes count on you for replys.
Now, you are using 1060 BTU/lb as Heat of vaporization; we use 970 BTU/lb.
I know the difference is not a lot, but how come the difference?
Also, you said your 7 BBL kettle at your first brewery had ~ 1/4 bbl boil off/hr.
Our is only 0.16 BBL (~ 5 gallon/hr). Was yours gas fired, and what pressure was your incoming pressure.
Also, our burner sais is rated for a minimum of 200,000 BTU's/hr, but I don't see anywhere that the pressure of the gas has an input.?
If you use the water density calculations in the scenerio, this # does not change a lot in water heating:
H2O density (kg/L) = 1.002  0.0001227 * T  0.0000032872 * T2
Now, using my prior formula, see the following example and tell me if its wrong:
10 BBL water heated in 1 hr from 40*F to 80*F
BTU's/hr = 10*259*40 = 103,600 BTU'shr.
At a rating of the boiler (200,000 BTU/hr), this would only be 51.80% efficiency.
Is that correct?
Again, thanks for your help
Fred

Ahhhh, Fred...........ya got me.
Heat of Vaporization (aka Latent Heat of Vaporization) is temperature dependant.
Your value at 970 BTU/Lb is indeed correct for liquids at 212F...........which I should have quoted. Sorry 'bout that.
The 1060 BTU/LB value applies to water at 60F. Believe it or not, there are applications using that value..........ventilation systems being one of the primary examples that comes to mind. The example I reviewed for your answer was of that kind.
Our 7 Bbl is a gas fired kettle with a Natural Gas conversion burner (MidCo, I believe) and was set for 300,000 BTU/Hr. We collect about 7.6 Bbls and get an almost explosive boil! The wort literally leaps out of the top of the kettle.
Looking at your example, it's true that you would get that efficiency if you actually measured that many BTUs in that amount of time. Looking at it another way..........
1 Hr to heat that amount requires 103,600 BTUs/Hr, but you're inputing 200,000 BTUs/Hr. All things considered to be equal, then you could look at it like:
(103,600 / 200,000) x 60 min = 31.1 min
You are dumping in more energy, and Energy In = Energy Out + Energy Stored, so the water heats at a faster rate.

Brian wrote:
(103,600 / 200,000) x 60 min = 31.1 min
You are dumping in more energy, and Energy In = Energy Out + Energy Stored, so the water heats at a faster rate.
HI Brian:
Can you explain this a bit more for me?
Thanks
Fred

Basically, and very fundamentally, the comment regarding Energy In and Energy Out is a somewhat cheeky derivation of the 1st Law of Thermodynamics........the Law of the Conservation of Energy.
Energy can't be created or destroyed, only converted to other various forms. If we naively assume the kettle is a "closed" system, and any heat energy coming in must either heat the water or leak off as waste heat that heats the air, the kettle, the substructure, the walls, burns your arm, etc.
Energy In is the BTUs you dump in from a burner or steam boiler.
Energy Out is waste heat (very simpley in this case).
Energy Stored is the heat energy trapped in the water.
To heat the 10 Bbl of water from your prior example, you derived it would take 103,600 BTUs/Hr. The timing is important because you selected it as a criteria that also coincided with the BTU rating duration of the kettle heating method (200,000 BTUs/Hr). All things being perfect, that is all it will take.......103,600 BTUs..........no more and no less......to simpley heat 10 Bbls of water from 40  80F.
Since the energy is trapped within the water as heat energy, it can't go anywhere (we're keeping this simple, but in reality there are waste heat losses as I mentioned above). So, if we had a burner we tuned to exactly put out 103,600 BTUs (burner BTU ratings are per hour), it would take 60 min to heat the water to 80F.
However, since we are inputing 200,000 BTUs, which is 96,400 BTUs more than what is required to heat the water in 60 min, that energy has to go somewhere..........into the water. Rather than absorb 1,726.7 BTU/min, the water is absorbing 3,333.3 BTU/min. Since this is a higher rate of heat input, and absorbtion, the water heats faster.
Granted, I have ignored kettle geometry and other factors, but it's was the principle I was trying to explain. To increase your evaporation amount, you will need to either lengthen the boil or increase the amount of heat input to the kettle. Reducing the volume being boiled has a lesser similer effect due to maintaining the 212F I mentioned before because you are expending less energy maintain the 212F of a large volume and can expend that heat energy into generating steam.
Basically, you need more combustion BTUs to vaporize a greater volume of water in 1 hour.


No problem, Fred. Anytime.
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