does anyone think this is important and why?
btw-why do brewer's call it evaporation when we are really boiling?
Water is changing from liquid phase to vapor phase.btw-why do brewer's call it evaporation when we are really boiling?
Boiling creates evaporation.
Water is changing from liquid phase to vapor phase.
Well, yes it does. This is going to be a much greater influence than, say, humidity which would have no influence at all. Consider my kettle filled to height h and to which I can deliver y BTU/hr. At boiling I am losing 2*π*r*h*(100 - Tamb)*G (r is the radius of the kettle, Tamb the ambient temperature and G the thermal conductivity to ambient per unit area) units of heat per unit time and, therefore, the potential to evaporate ∆H*2*π*r*h*(100 - Tamb)*G mass per unit time where ∆H is the enthalpy of vaporization units of mass per unit time. If I fill the kettle to 2*h obviously more heat will be lost through the sides and my evaporation mass/time will go down by ∆H*2*π*r*h*(100 - Tamb)*G. Thus both the percentage an the absolute rate depend on fill level. Practically speaking, where the level changes by a inches and the volume may go from 50 gal to 48, it is fine to speak of % rates as the calculations don't warrant better accuracy and, in my case anyway, I always top up to the desired volume anyway.It is properly specified in gal/hr, rather than %/hr, as the boil off rate does not depend on how many gallons you have in the Bk, but %/hr does depend on volume in BK.
There is only one process - the state change.I thought boiling and evaporation were 2 different kind of mechanisms for the vaporization of a liquid, but I guess both are probably happening during the boil process.
At temperatures below the boiling point the hydrostatic pressure at the bottom of the kettle is less than the saturated vapor pressure of water so that any pocket of vapor than forms into a bubble collapses (you can hear these collapses as a sort of knocking sound as the BP is approached). The BP is defined as the temperature at which the vapor pressure of water is equal to the hydrostatic pressure so that the hydrostatic pressure cannot collapse a bubble. The bubble rises, as it is lighter than the liquid, and as long as the liquid temperature above the point of formation is high enough that the vapor pressure remains at or above the hydrostatic pressure, the bubble continues to rise and ebullition occurs.yeah, but during the boil isn't most of it by boiling, Ithink of evaporation as a slower process based on concentration gradient.
evaporation
The physical process by which a liquid substance is converted to a gas or vapour. This may occur at or below the normal boiling point of the liquid (the temperature at which a liquid boils at 1 atmosphere pressure) and the process is endothermic.
Well, yes it does. This is going to be a much greater influence than, say, humidity which would have no influence at all. Consider my kettle filled to height h and to which I can deliver y BTU/hr. At boiling I am losing 2*π*r*h*(100 - Tamb)*G (r is the radius of the kettle, Tamb the ambient temperature and G the thermal conductivity to ambient per unit area) units of heat per unit time and, therefore, the potential to evaporate ∆H*2*π*r*h*(100 - Tamb)*G mass per unit time where ∆H is the enthalpy of vaporization units of mass per unit time. If I fill the kettle to 2*h obviously more heat will be lost through the sides and my evaporation mass/time will go down by ∆H*2*π*r*h*(100 - Tamb)*G. Thus both the percentage an the absolute rate depend on fill level. Practically speaking, where the level changes by a inches and the volume may go from 50 gal to 48, it is fine to speak of % rates as the calculations don't warrant better accuracy and, in my case anyway, I always top up to the desired volume anyway.
What I don't know how to estimate is how much heating of the un-wetted sidewall will occur due to condensation of steam. Some of this has to occur, but how much? If we knew how much steam condensed, it would be possible to calculate the un-wetted sidewall temp and estimate how much heat is lost thru that portion of the wall, and compare that to how much is lost if the BK is full. If the difference is small, then fullness of the BK would not affect boil off rate much. But if the difference is large, then fullness would affect boil off significantly.
My thoughts? You are expending a lot of effort trying to calculate something that you probably won't be able to because of unknowns some of which you have already considered. For example, condensation on the 'dry' parts of the vessel side wall. The rate of condensation will depend on how much heat flows through the wall to the ambient and that will depend on things like how free the convection is around the kettle and it's color (need to take radiation into account too).Your thoughts?
My thoughts? You are expending a lot of effort trying to calculate something that you probably won't be able to because of unknowns some of which you have already considered. For example, condensation on the 'dry' parts of the vessel side wall. The rate of condensation will depend on how much heat flows through the wall to the ambient and that will depend on things like how free the convection is around the kettle and it's color (need to take radiation into account too).
Now if you are doing this in order to enhance your understanding of how things work I applaud your efforts. From the practical POV I think you will be better off pressing the pedal to the metal and seeing how fast the temperature of a known mass of water rises in your boil kettle. Do this with an insulated lid on the pot. That will give you data on how much power you are delivering to the liquid. When the water boils measure the rate of decrease in level (don't forget to compensate for volume change with temperature). This will give you an idea of how much heat is going into enthalpy of vaporization and an estimate of how much isn't at each fill level at which you take a measurement.
The steam coming out of the BK makes the atmosphere inside the kettle oversaturated with water vapor (>100% relative humidity). If the temperature of the sidewall goes below the temperature of this vapor which is ~100-105°C, condensation will occur instantly. This saturation point at 100°C is quite high ~0.6 kg/m3. The heat of condensation of water is 2.26 kJ/g.
Now, let's take as an example 15 gallon BK D=40cm, Hfull=47cm, Wort boiling temp=102°C, Ambient temp=24°C, Air velocity=0
For these conditions, I estimate the heat loss to be 14 watts per cm of height of BK.
Let's imagine you have 6 gallons in BK, so you get 29cm un-wetted.
The heat loss through the un-wetted area (assuming its temp=wort temp=102°C) is 29*14= 406W
Let's assume your boil-off rate is 1 gallon/hour=0.001 L/sec=1 g/sec.
If all that vapor condenses, it'll release 2.26 kW, and you only need 0.406 kW to sustain 102°C wall temperature. Thus, you need 0.406/2.26*100%=18% of that vapor to be around to condense.
Ok, now I'm thinking of the local conditions, only vapor near the wall will condense, while the rest will escape. What can I do to estimate this local humidity gradient? The vapor density 1.6 kg/m3, the velocity of vapor coming out of BK = 0.005 m/s, kinematic viscosity=22/10^6 m2/sec, Re=90 laminar flow. That tells me not a lot of mixing will occur. But at what distance from the wall of BK will the vapor condense? I don't know.
And, the condensing vapor at the wall would create a slight pressure drop, which would cause more steam to flow towards the wall.
Brew on
Brewing science will certainly supply you with plenty of those. I too use brewing related problems for what most turn to Sudoku as I am also a retired engineer (but not a recently retired engineer).Yeah, I'm doing it more for the intellectual exercise. I'm a recently retired engineer, so I have to make up my own problems to solve now (and try to keep my mind sharp.)
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