Evaporation Rate
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flars
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When you are boiling water or wort evaporation takes place. If you need to end up with a certain volume after a 60 or 90 minute boil, this volume needs to be added pre-boil.
Knowing how much water to start with to get a desired post-boil volume is some times trial and error. There are calculators that will help you get close, but factors like humidity, boil intensity, kettle dimensions, etc., can create variations.
Boiling creates evaporation.
Water is changing from liquid phase to vapor phase.
I usually see it referred to as "boil off rate." It's important to know if you want to be able to predict how much wort you will end up with at the end of a brew day (unless you are doing extract, and just topping up with additional water to make your volume.) Boil off rate is dependent upon your KB diameter (surface area of wort really), heat input during boil (how vigorous a boil), and to some extent weather conditions (temp, humidity, wind velocity.) 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.
Other important volumes or volume rates are grain absorption rate (gal or qt per lb), hop absorption rate (usually less important than grain abs.), MLT dead volume (unless doing BIAB), BK trub loss. You need to know all these so you know how much water to start with to obtain a desired amount in the fermenter.
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Other important volumes or volume rates are grain absorption rate (gal or qt per lb), hop absorption rate (usually less important than grain abs.), MLT dead volume (unless doing BIAB), BK trub loss. You need to know all these so you know how much water to start with to obtain a desired amount in the fermenter.
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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.
yeah, but during the boil isn't most of it by boiling, Ithink of evaporation as a slower process based on concentration gradient.
It's funny how the terms we use are so anecdotal
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.
There is only one process - the state change.
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.
At first glance, it would seem that evaporation and boiling might be the same process, because both start with a liquid and end up with a gas. However, there are some important macroscopic and microscopic differences that help make the distinction.
Evaporation happens only at the surface of a liquid and occurs at any temperature (so long as the substance is a liquid at that temperature). However, as most people are aware, liquids evaporates faster at a higher temperature.
Boiling, on the other hand, happens throughout the bulk of a liquid, usually starting from some site on the inside of the container and rising in a bubble to the surface. It only happens when the temperature is above the boiling point of that substance.
Boiling occurs when the average motion of particles is fast enough to overcome the forces holding them close together. This happens evenly throughout a boiling liquid because the temperature is uniform throughout.
Evaporation happens for the following two reasons:
Not all particles in the liquid are moving at the same speed and, as a result, the faster particles are more likely to overcome the forces they feel from their neighbors.
The particles at the surface of the liquid are only held in place by forces from the neighboring particles beneath them, whereas particles in the middle of the liquid have forces holding them on all sides. Thus, particles at the surface find it easier to break away from the liquid.
In the cases of both boiling and evaporation, the force between two particles is always present. The greater the space between the particles becomes, however, the weaker the force is between them. To break the bond between two particles, one particle has to be moving fast enough to overcome the pull of the other, until it gets so far away that pull is diminished. An analogy would be if you tried to jump off the Earth: we cannot jump fast enough to get to where Earth's gravity has less pull, but a rocket can.
copied from the interweb
Evaporation happens only at the surface of a liquid and occurs at any temperature (so long as the substance is a liquid at that temperature). However, as most people are aware, liquids evaporates faster at a higher temperature.
Boiling, on the other hand, happens throughout the bulk of a liquid, usually starting from some site on the inside of the container and rising in a bubble to the surface. It only happens when the temperature is above the boiling point of that substance.
Boiling occurs when the average motion of particles is fast enough to overcome the forces holding them close together. This happens evenly throughout a boiling liquid because the temperature is uniform throughout.
Evaporation happens for the following two reasons:
Not all particles in the liquid are moving at the same speed and, as a result, the faster particles are more likely to overcome the forces they feel from their neighbors.
The particles at the surface of the liquid are only held in place by forces from the neighboring particles beneath them, whereas particles in the middle of the liquid have forces holding them on all sides. Thus, particles at the surface find it easier to break away from the liquid.
In the cases of both boiling and evaporation, the force between two particles is always present. The greater the space between the particles becomes, however, the weaker the force is between them. To break the bond between two particles, one particle has to be moving fast enough to overcome the pull of the other, until it gets so far away that pull is diminished. An analogy would be if you tried to jump off the Earth: we cannot jump fast enough to get to where Earth's gravity has less pull, but a rocket can.
copied from the interweb
I've bolded the key phrase. This says, as I indicated in the earlier posts, that the only difference between boiling and slower evaporation is that in boiling the saturated vapor pressure has reached the ambient pressure.
The quote is from the IUPAC Gold Book.
Edited to remove snarky comments. If any one has seen them I apologize.
I did some calculations (with the help of http://www.engineeringtoolbox.com/overall-heat-transfer-coefficient-d_434.html). I came up with 10,103 BTU/hr required to boil off 1.25 gal/hr. Then I looked at losses thru the wall of the BK. Depending on air velocity, the sidewall losses range between 1690 and 16823 BTU/hr for a full pot with 6.4 ft^2 of sidewall area (my 15 gal pot) @ an ambient of 62°F. So those numbers are definitely large enough to affect boil off if the losses drop significantly if the pot is less than full. The thermal conductivity of SS is fairly low (compared to Cu & Al), so heating of the un-wetted sidewall due to vertical conduction in the wall can probably be ignored.
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.
Your thoughts?
As far as using %/hr as a boil off rate, I don't like if for the following reason: I can brew 5 gal (5.5 to fermenter) or 10 gal (11 fermenter) batches in my kettle. If I'm targeting 1.25 gal/hr boil off for a 1 hr boil, then for a net 5.5 gal I need to start with 6.75 gal, and my boil off is 1.25/6.75 = 18.5%. For a net 11 gal I need to start with 12.25 gal, and my boil off is 1.25/12.25 = 10.2%. And if the gal/hr go up due to a less full pot, my %/hr boil off rate diverge even more.
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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.
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.
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.)
Detailed experiments with my current propane burner set up would be all over the place, as I don't have a windscreen on my burner yet, and my regulator is very inconsistent. Too many variables I can't control to be able to make any sense out of the data. Once I go electric, and come in out of the wind, such experiments would make more sense. Right now I just go with a rough estimate of my boil off.
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Thanks for your contributions to this question @qmax.
Based on your calculations that we need to condense ~20% of the vapor to maintain a wall temp of ~100°C, and assuming the 20cm kettle radius, we would only need to condense the vapor within 2.1 cm of the wall. And, the condensing vapor at the wall would create a slight pressure drop, which would cause more steam to flow towards the wall.
It appears that the condensing steam will make a significant contribution towards equalizing the boil off rates for different fill levels in the same kettle, when heat input is constant. As I note in my response to AJ, I can't control my heat inputs well enough to do any kind of a verification at this time.
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And, the condensing vapor at the wall would create a slight pressure drop, which would cause more steam to flow towards the wall.
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Why would it create a pressure drop though?
I've just attempted to put the wall thermal conductivity into numbers.
Heat flux in, W = π*(D^2-d^2)/4*λ*(Twort - Twall)
Heat flux out, W = π*D*∂h*(Twall - Tambient)*α
λ=17 W/m/°C
For the heat transfer coefficient 'α' I use an empirical equation for generic steel pipes used in heating systems with water as the heat-transfer fluid.
α(t°C, V) = 9.3+0.047*t+7*V^0.5
V = Air velocity, m/s
I do not have a formula that accounts for specific radiative properties of different colors.
I also do not know how to turn this into a differential equation in the form of ∂t/∂h or something like that and solve it, so I just used incremental wall height and Excel Solver with new flux and deltaT values calculated for each step.
I'm not entirely confident in my results, but they seem to indicate that the wall temperature should drop from 102°C to <40°C within 1 cm from the surface level. If this is correct, it means the high temperature of the un-wetted wall is primarily driven by condensation or convective heat transfer by the rising steam.
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).
Being retired one has the luxury of saying "This is interesting. Let's go do an experiment." So I put about 17 gal of water, enough to fill a 55 gal drum about
1/3 full and boiled it. I was interested in the condensation contribution to loss because I had never noticed condensation on the interior of a kettle during boil. That's probably because I usually don't have 2/3 freeboard but the other reason is that not much condensation occurs and what does occur happens mostly, in the water experiment, on the top third of the drum (where the metal is cooler). There is very little reflux (condensate that actually makes it back into the volume being boiled and that is because it is re-evaporated as it gets close to the surface of the boiling water. The explanation for this is that the steam leaving the surface is super heated and that is because the pressure at the bottom of the water column (where the steam coil is) is higher than at the surface (each foot of water contributes corresponds to 1 in Hg hydrostatic pressure) and because the steam in the coil is at 5 - 10 psig.
It takes quite a while for enough condensation to appear before any runs down the sides and in this earlier time one can plainly see the condensation occur (the stainless looks dark when it is wet) and they disappear again (it turns lighter color again). Is the energy required to re-evaporate this condensate wasted? Yes, it is. Assuming that the only reason for expenditure of energy here is to remove water if we have to do that twice we have to supply the enthalpy of vaporization to as much as condenses a second time. But that enthalpy is wasted anyway as we have supplied the enthalpy of evaporation and that required to raise the temperature of the steam above boiling. Boiling in itself is a very inefficient way of removing water as we have to heat the wort to the point where the saturated vapor pressure of water equals one atm. Our bretheren in Denver are much better off as they don't have to apply as much of that penalty heat.
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