What is boiling?

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The conundrum is we are applying roughly 3055 Btu's a min but only seeing 1088 Btu's a min in temperature rise. The effective rate and applied rates are quite a bit apart, but if the applied rate is used the 970 Btu's/Lb value is within about 10% of applied heat. during the testing steam was generated with 3.1Lbs/minute of water flow into boiler, all that was leaving coils in kettle was water not steam. Lack of insulation and a cover over the kettle has skewed the results, but without enough instrumentation we are left to guess the end result.
The observed effective heat applied was equal to 19 Kw/Hr, Boiler output was equal to 53.7 Kw/Hr, we need further research to close applied and observed gap. The plan is to repeat this exercise on an instrumented platform where all the inputs and outputs can be measured and recorded.
 
Are you turning on the element while it is filling? Once its covered.

After the kettle is full (13 gallons) I then turn on the element. 11 minutes later she is a rolling. This is not a matter of turning on the element while the kettle is filling, that is cheating.
 
The conundrum is we are applying roughly 3055 Btu's a min but only seeing 1088 Btu's a min in temperature rise. The effective rate and applied rates are quite a bit apart, but if the applied rate is used the 970 Btu's/Lb value is within about 10% of applied heat. during the testing steam was generated with 3.1Lbs/minute of water flow into boiler, all that was leaving coils in kettle was water not steam. Lack of insulation and a cover over the kettle has skewed the results, but without enough instrumentation we are left to guess the end result.
The observed effective heat applied was equal to 19 Kw/Hr, Boiler output was equal to 53.7 Kw/Hr, we need further research to close applied and observed gap. The plan is to repeat this exercise on an instrumented platform where all the inputs and outputs can be measured and recorded.

Right, I am reading something wrong. 19kWh would get you to a boil, from 50F in 15 minutes flat.

You are dealing with 100 pounds of water right?
You said you raised it 110F in 15 minutes. It takes 11,000 BTUs to get that temp rise, so that being done in 15 minutes (11000/15) is 733 BTUs per minute, right? Where was the 1088 calculated from?

I am perplexed as to why it took 30 minutes to get from 160F to a boil when you were clearly applying 733 BTUs per minute to the boil between 50F and 160F... theoretically you should have reached a boil in 7 more minutes. (100 pounds, 50F temp delta and te 733 BTUs per minute you were applying) Meaning you SHOULD have seen 22 minutes from 50F to a boil...

Unless, heating time (transfer) (and cooling for that matter) will be affected by the DIFFERENTIAL of the two elements... whether it is water and a heating element, or water and water in the case of a chiller.

Could it be that as your 12 gallons approached the temp. of your steam that the effective heat transfer was lessened because the differential was not as great? Think about it, this happens in chillers all the time. Chilling with 70F tap water will get you to say 100F pretty fast from a boil, but as your wort nears the chilling water temp, your transfer takes a hit. I have documented this extensively with closed cooling systems. Once the two mediums get close in temp, the heat transfer is drastically reduced. I think this may be playing in a role in that 30 minutes to reach a boil from 160F. Just a thought.

If you went from 160 to 210F in 30 minutes, you were effectively transferring 166 BTUs per minute. I think this is a case of the heating medium (steam) and the water in your kettle losing the large delta that gave you the faster heating in the beginning. This is simple thermal dynamics, you are losing your delta, and thusly greatly reducing your rate of transfer.

This happens to heating elements too... as the water gets hotter, the heating slows, though to a lesser degree since the elements are so much hotter to begin with.
 
Not quite so it takes 45 minutes to hit a boil at constant input rate, that is the problem, 3.1 Lbs steam a minute in, all liquid condensate out, and 45 minutes later it boils. Not sure how much superheat there was but the assumption for calculations was that it was saturated steam. With 3.1Lbs/Minute dumping 970 Btu's/lb into the water, it is taking more heat than just warmup and 970 Btu's combined should be. Since the condensate exiting the kettle remained a liquid during entire run we believe that the heat is going into the kettle.
 
How hot is your steam?

I think you are losing your delta between the steam and the 12 gallons of water.

Think about it, place a coil in a kettle of cold water, pump HOT water through that coil and you will see a temp. rise, but as your kettle of water heats up, it heats slower and slower until it ALMOST reaches the temp. of the hot water circulating in the coil.

This is the only thing that I can think of that explains the reduction in heat transfer.

All assumptions aside. If I look at your temp. rise, time and weight of water, you were averaging 733 BTUs per minute from 50-160F and averaging 166 BTUs per minute from 160F to a boil... just looking at temp. rise, weight of water, and time.

Are you measuring the steam temps?
 
The problem with the math is in order for the steam to return to water it has to give up 970 Btu's/lb, just as it takes 970 Btu's lb to create. We put the steam in and get water condensate out, the heat has to have gone somewhere as there is no steam coming out with condensate.
 
Bottom line is, you are not creating the heat that you think you are, or there is some efficiency loss somewhere. The 12 gallons temp. and time wont lie though, and I have no idea where the loss is.

But, read below... this blows my mind sorta.
 
I have to ask...

How are you creating 3-4 POUNDS of steam PER MINUTE? That is really incredible. That is 22-30 GALLONS of vapor per hour. That is like having a boil off rate of 22 gallons per hour...

970BTUs per pound of vapor
2,910 BTUs per minute
174,600 BTU/hr heat source

What in the heck do you have that is effectively transferring 175,000 BTUs to your boiler?

If you go to 4 pounds per minute
that is over 232,000 BTU/hr heat source

That is 232,000 BTUs effectively transferred, not rated, but actually transferred....

I have to work this out in my head... I am astonished...

175,000 BTU burner, if it could transfer 100% of its BTUs (which it cant) running full tilt for 45 minutes. A pound of propane would yield about 22k BTUs, so this would take 6 pounds of propane to run that burner full tilt for 45 minutes, and even then it would not create 3 pounds of vapor per minute. 6 pounds is 1/3 of a 20lb tank, so that is about $5 or more, just to reach a boil, theoretically.

Nothing is adding up here...

My 9000W kettle is only 30k BTUs...
So at 100% eff. and at FULL TILT it can only create .5 pounds not 3 or 4 pounds, but .5 pounds of vapor per MINUTE. But it will heat 12 gallons from 50F to a boil in 35 minutes... with 30K BTUs, which can only create .5 POUNDS of vapor per MINUTE.

It is also running 11 minutes from sparge temp. (160F) to reach a boil. Your test too 3x as long and you are presumably creating 6-8x more BTUs than I am.
 
Take a look at GreenMontis flash boiler and MK-I threads to see the hardware and testing done.https://www.homebrewtalk.com/f51/flash-boiler-153116/,https://www.homebrewtalk.com/f51/mk-i-161799/ With a flowmeter on the water into the boiler we calculated the steam yeild from the boiler, without any leakage between boiler and kettle the heat had only one place to end up. Accuracy of the flow meter at that range could account for some variation in calculated results but bottom line is this moves quite a bit of heat from shrouded burner to kettle. Burner firing may be beyond rating on burner but is not observable as it is inside tube
 
Okay, after sifting through a few threads that he has running...

He is using a 170,000 BTU burner.
What size is your burner?

I also see that he increased his flow rate in another thread from a max of .25 gallons/min. to about .5 gallon per minute.

So, this begs the question, because I see a lot of numbers being thrown out, but every time I read another page, the numbers change...

How fast is GreenMonti reaching a boil with his 170,000 BTU fired boiler?
 
EDIT:

There are too many conflicting numbers in the several threads I read on GMs boiler. So, can we post the time to boil and in how many gallons, etc... to get a starting point?

Then I can at least compare that to what you are seeing Kalude. He is using a 170,000 BTU burner, what is yours rated at.

I keep reading that GMs boiler is faster at reaching a boil than electric by a long shot, one referecen is 4x faster than a 4kWh heating element, but I cannot find the actual times, just theoretical BTU transfer.
 
I admit, this is some really cool stuff...

So 42 minutes in 14 gallons from 54F to a rolling boil.

So that is an effective transfer of 26,026 BTU/hr or 7.6kWh

The question then becomes, how much gas is the burner expending during that time. Obviously the burner is running at more than 27,000 BTUs

I cant heat much faster than that, I am just under 40 minutes to boil the same qty at the same temps. at a cost of $.57 to reach the boil. It then costs me $.40 to boil off 1.5 gallons over an hour. Total cost to boil then comes to $.97 from a cold start (which none of us really do)

Like I said it is really cool what you all have built, I am curious however how many pounds of LP it will take to perform a 60 minute boil, the resulting BTU usage and the overall cost (eff.) of the setup.

Yall have done a crap ton of work here, and great workmanship... I hope you don't mind me trying to wrap my head around the #s.

EDIT:
After piecing some numbers together from a coupe threads that talked about turns on the regulator, the resulting gas pressure and the associated BTU output of the burner...

It appears that the afore mentioned test was run with the burner at between 75K and 80K BTUs, 1.5 - 2 turns, the resulting heating times and temps indicate about 26K to 27K BTUs to the kettle for heating.

That is an eff. of about 36% which is pretty darned good for a gas fired anything in HBing.

Taking those numbers:

75K BTU/hr for 40 minutes to reach a boil (36% eff, effective BTUs approx. 27,000)
56,000 BTUs expended
2.55 pounds of LP to produce those BTUs
$.92 per pound of LP
$2.34 to reach a boil

Now boiling off the 1.5 gallons per hour will take
1.5 gallons x 8.34lbs per gallon
12.51 pounds of vapor
at 970 BTUs per pound
12,134 BTU/hr heat transfer to boil off 1.5 gallons

Assuming that the eff. of the burner is linear, and that Monti's previous posts are accurate with pressure and BTU useage, 36% assumed eff, the burner output would have to be about 33,000 BTUs over an hour. This is an additional 1.5 pounds of propane.

Total propane useage:
4.05 pounds based on the above information from several threads
$.92 per pound LP
$3.72 for a boil from 54F to 212F and then boiling for an hour with 1.5 gallons/hr boil off.

That is better performance than a straight up gas burner under a kettle from what I have typically seen.

Total BTU usage, using the information at hand, to reach a boil and boil off 1.5 gallons in one hour:
Gas fired steam boil used 89,000 BTUs
DIRECT fired gas boil with typical 25% burner eff utilizing the same burner would use 99,000 BTUs
Electric heating utilizing 9000W to reach a boil and 4000W to maintain a boil used 34,000 BTUs
 
I admit, this is some really cool stuff...

So 42 minutes in 14 gallons from 54F to a rolling boil.

So that is an effective transfer of 26,026 BTU/hr or 7.6kWh

The question then becomes, how much gas is the burner expending during that time. Obviously the burner is running at more than 27,000 BTUs

I cant heat much faster than that, I am just under 40 minutes to boil the same qty at the same temps. at a cost of $.57 to reach the boil. It then costs me $.40 to boil off 1.5 gallons over an hour. Total cost to boil then comes to $.97 from a cold start (which none of us really do)

Like I said it is really cool what you all have built, I am curious however how many pounds of LP it will take to perform a 60 minute boil, the resulting BTU usage and the overall cost (eff.) of the setup.

Yall have done a crap ton of work here, and great workmanship... I hope you don't mind me trying to wrap my head around the #s.

EDIT:
After piecing some numbers together from a coupe threads that talked about turns on the regulator, the resulting gas pressure and the associated BTU output of the burner...

It appears that the afore mentioned test was run with the burner at between 75K and 80K BTUs, 1.5 - 2 turns, the resulting heating times and temps indicate about 26K to 27K BTUs to the kettle for heating.

That is an eff. of about 36% which is pretty darned good for a gas fired anything in HBing.

Taking those numbers:

75K BTU/hr for 40 minutes to reach a boil (36% eff, effective BTUs approx. 27,000)
56,000 BTUs expended
2.55 pounds of LP to produce those BTUs
$.92 per pound of LP
$2.34 to reach a boil

Now boiling off the 1.5 gallons per hour will take
1.5 gallons x 8.34lbs per gallon
12.51 pounds of vapor
at 970 BTUs per pound
12,134 BTU/hr heat transfer to boil off 1.5 gallons

Assuming that the eff. of the burner is linear, and that Monti's previous posts are accurate with pressure and BTU useage, 36% assumed eff, the burner output would have to be about 33,000 BTUs over an hour. This is an additional 1.5 pounds of propane.

Total propane useage:
4.05 pounds based on the above information from several threads
$.92 per pound LP
$3.72 for a boil from 54F to 212F and then boiling for an hour with 1.5 gallons/hr boil off.

That is better performance than a straight up gas burner under a kettle from what I have typically seen.

Total BTU usage, using the information at hand, to reach a boil and boil off 1.5 gallons in one hour:
Gas fired steam boil used 89,000 BTUs
DIRECT fired gas boil with typical 25% burner eff utilizing the same burner would use 99,000 BTUs
Electric heating utilizing 9000W to reach a boil and 4000W to maintain a boil used 34,000 BTUs

Like I have mentioned before (maybe not this thread) I am just learning about all this stuff. BTUs and steam. The info that I posted in my threads is most likely half right. I can't throw this stuff around yet and I am not aware of all the little quirks and what not.You are probably not too far off on those numbers. It certainlly goes along with what is in this thread.

I figure during the actual boil, right now I am using 1 lb for the hour. As you may very well know by reading that info, I ran it for 1 hour at the boiling firing rate and weighed the tank before during and after the run. I have to see what I use for the heating time. From what I understand propane flow rates change when there is back pressure on the system. I can't even fathom that math yet.

Not to worry, I am nipping at your heels. I am working on getting the efficiency up. I would like to get real close to the electric crowd. After all, what's 50 more percent??:rolleyes:
 
Here is a test run of the coil calandria and 14 gallons of water from cold to boil https://www.homebrewtalk.com/f51/boiling-fb-166785/, maybe this will help. GreenMontis boiler is the current design in testing, my previous boilers have been built from SS tubing and perforance is poor compared to the copper coil version.

Thank you for posting this. I couldn't find it last night for some reason. It was something to see 14 gallons heat at 6* min at the end.
 
This thing is really cool Monti, I have actually learned A LOT by reading up on all of this and researching what you are doing over there. I have gained a lot of knowledge not only about your system, but mine as well.

I like to work numbers, numbers are real and solid and mathematics are hard to debunk. Much of this is me trying to get my head around what you have going on over there.

I have found that the eff. of a gas burner increases as you reduce the gas flow rate, so if you used 1 pound of LP for the hour boil, that is 22,000 BTUs and I calculated 34,000 because I do not think that the eff. is a constant, I think it is linear and related to the output of the burner at that time.

I was talking to a friend today who has a degree in physics... although I have no interest in LP boils, I may start working on a jacketed kettle that will trap more of the burners heat against the side of the kettle. Increasing the heating surface area and increasing the contact time of the heat and the kettle bottom and sides. I am interested in what the gains would be over the typical turkey fryer stand.

Anyhoo, I am learning a lot here.
 
When I used to use gas, I fantasized about taking a SS tube (keg spear) and welding it to the center of the keg to allow heat to flow up the inside of the wort, as well as around the sides. More surface area, and especially access to the center of the wort, would result in much faster boils I think.

Then I stumbled onto Pol's electric keggle thread...
 
PP the physics major actually recommended the tube through the center too... accompanied by a a gas (heat) retaining jacket around the outside of the kettle. Increase contact surface area and time.

I may try to get my hands on some sheet metal and fabricate something on my old turkey fryer kettle to see some raw data. I know there will be in increase in performance, just not sure to the degree, and Id like to know!
 
What kind of boil? I'd say it all depends.

In your first post in the second video:
That's about the boil I shoot for.

It is not violent, not roiling, not exciting, but definitely rolling.
I want to see the fluid rolling.

The volume of steam generated is not terribly relevant. You will get some steam, but I should think that the wilder the boil the more dark your wort will get. and as a side effect the more evaporation for which you will have to compensate.
You might want the darker color.
 
1 BTU to get 1lb of water to rise in temp 1F

13 gallons water is
108 pounds of water

From 60F to 210F:
16,200 BTUs required to heat in an hour
If you have 32.4K BTUs it will take you 30 minutes to accomplish this.

970 BTUs per POUND to create water vapor.
Creating 1.5 gallons of vapor per hour (boil off)
12.5 POUNDS of vapor per hour
12.5 x 970 BTUs per POUND per hour
12,125 BTUs to create 1.5 gallons of boil off per hour

1kW equals 3412 BTUs
12,125 BTUs equals 3.5kW
3.5kW equals 3500W

3.5kWh at $.12 per kWh
$.42 to boil off 1.5 gallons in one hour (unless you are in Hawaii)

Or NY mine is around $.23 . The electricity alone is $.11 and then for each KWH I get hit with a delivery charge of roughly the same amount .. And they are getting a 4% rate hike this year.Wish I had NG at the house
 
OK, how about this:

Boiling is the visible product of a phase change from a liquid to a gas. Boiling occurs when a fluid is at its http://en.wikipedia.org/wiki/Boiling_point and can therefore exist in both the liquid and gas phases. A mass existing entirely in the liquid phase at its saturation temperature is called a saturated liquid and has a quality of 0. For this same Control Mass existing entirely as a vapor at the saturation temperature, It is called a saturated vapor(or saturated steam) and has a quality of 1.

-Quality is a term used to express the fraction of a mass in the vapor phase and is found by the M[/SIZE]vapor/Mtotal. Quality is only used to describe a fluid at the saturation temperature
-lbm: pound mass, the amount of mass that exerts 1 pound of force on earth

Back to Boiling.
For water, The amount of energy need to convert 1lbm saturated liquid to 1 lbm saturated steam is 970 BTU/lbm at 14.7psia. this quantity is known as laten Heat of Vaporization

to approximate our brew pot, we will assume it is a control mass and that there is no heat transfer into or out of the system. to keep it simple, we will assume the pot contains 1lbm of water.

The specific heat of water is 1BTU*(ºF*lbm)^-1.

So, in an engineering problem, the units always cancel out leaving the desired quantity.

Say we want to heat the water to 212ºF from 62ºF, so ΔT is 150º

150ºF*1lbm*1BTU*(ºF*lbm)^-1=150BTU ºF and lbm cancel so the answer is in BTU

Since we are now at our saturation temperature, say we want to boil 10% of our water off, what we are saying from a thermodynamic perspective is that we want to convert 10% of our water into steam, or change the quality from 0 to .1

Since the water is at its saturation temperature, any additional energy transferred into it will cause a phase change, but no temperature increase until it is at a quality of 1.

So, back again to the example. We now want to boil off .1 lbm of water, we use the previously stated so

.1lbm*970BTU/lbm=97BTU where lbm cancels.

So thats easy enough. we just "boiled" .1lbm of water. Now, this process could have happened quickly over the period of a couple seconds or very slowly over an hour or something, but it still could be considered boiling. If we were to take the steam that was created during this process and condense it back into a liquid, that 97 BTU's would be released into the condenser, similar to Monti's calendria.

Some other things to keep in mind:
The properties of water are not constant, You need a certain number of properties to define the state of a fluid at any state, If the pressure was changed, all the values in the problem would change.
A BTU is a unit of energy, A Watt is a unit of energy flow, not a unit of energy. So saying a burner is 100,000 BTU is technically incorrect, a better statement would be 100,000 BTU/hr. Granted, it can usually be assumed that burner measurements are on a per hour basis, when talking about a quantified heat transfer, it would generally be better to state terms properly.

So back to the original question:
Boiling is a process, Not a state point, so saying something is boiling doesn't mean much at all actually other than its undergoing a phase change.
 
OK, how about this:

Boiling is the visible product of a phase change from a liquid to a gas. Boiling occurs when a fluid is at its http://en.wikipedia.org/wiki/Boiling_point and can therefore exist in both the liquid and gas phases. A mass existing entirely in the liquid phase at its saturation temperature is called a saturated liquid and has a quality of 0. For this same Control Mass existing entirely as a vapor at the saturation temperature, It is called a saturated vapor(or saturated steam) and has a quality of 1.

-Quality is a term used to express the fraction of a mass in the vapor phase and is found by the M[/SIZE]vapor/Mtotal. Quality is only used to describe a fluid at the saturation temperature
-lbm: pound mass, the amount of mass that exerts 1 pound of force on earth

Back to Boiling.
For water, The amount of energy need to convert 1lbm saturated liquid to 1 lbm saturated steam is 970 BTU/lbm at 14.7psia. this quantity is known as laten Heat of Vaporization

to approximate our brew pot, we will assume it is a control mass and that there is no heat transfer into or out of the system. to keep it simple, we will assume the pot contains 1lbm of water.

The specific heat of water is 1BTU*(ºF*lbm)^-1.

So, in an engineering problem, the units always cancel out leaving the desired quantity.

Say we want to heat the water to 212ºF from 62ºF, so ΔT is 150º

150ºF*1lbm*1BTU*(ºF*lbm)^-1=150BTU ºF and lbm cancel so the answer is in BTU

Since we are now at our saturation temperature, say we want to boil 10% of our water off, what we are saying from a thermodynamic perspective is that we want to convert 10% of our water into steam, or change the quality from 0 to .1

Since the water is at its saturation temperature, any additional energy transferred into it will cause a phase change, but no temperature increase until it is at a quality of 1.

So, back again to the example. We now want to boil off .1 lbm of water, we use the previously stated so

.1lbm*970BTU/lbm=97BTU where lbm cancels.

So thats easy enough. we just "boiled" .1lbm of water. Now, this process could have happened quickly over the period of a couple seconds or very slowly over an hour or something, but it still could be considered boiling. If we were to take the steam that was created during this process and condense it back into a liquid, that 97 BTU's would be released into the condenser, similar to Monti's calendria.

Some other things to keep in mind:
The properties of water are not constant, You need a certain number of properties to define the state of a fluid at any state, If the pressure was changed, all the values in the problem would change.
A BTU is a unit of energy, A Watt is a unit of energy flow, not a unit of energy. So saying a burner is 100,000 BTU is technically incorrect, a better statement would be 100,000 BTU/hr. Granted, it can usually be assumed that burner measurements are on a per hour basis, when talking about a quantified heat transfer, it would generally be better to state terms properly.

So back to the original question:
Boiling is a process, Not a state point, so saying something is boiling doesn't mean much at all actually other than its undergoing a phase change.

You just laid out my thoughts that lead me to post this thread. Thank You.

I could never have put your post together so that others would totally understand what I was asking and thinking about. That is why I posted the videos of a boil with less vigor.

To give an example of what you just said. Sunday, I placed 1.5 gallons of water on the stove and brought it up to 201-203* and held it there for 1 hour. I had it sitting on a 1250 watt burner set just past half way. I lost just about half a gallon. At first I thought holy crap, that's impossible. Then I did the math and calculated the entire time frame and it is just about right. I never boiled a single drop.

So, now what? The boil helps out in a few ways like driving out DMS. Do we really have to have a rolling boil at all, that is if one is willing to just wait longer?
 
I have found that the eff. of a gas burner increases as you reduce the gas flow rate, so if you used 1 pound of LP for the hour boil, that is 22,000 BTUs


This is nice to hear. I say that cause living in my dream world, I would like to get real close to not needed to run the boiler any higher then the 2.5 PSI mark. I am a tight bastard.
 
What kind of boil? I'd say it all depends.

In your first post in the second video:
That's about the boil I shoot for.

It is not violent, not roiling, not exciting, but definitely rolling.
I want to see the fluid rolling.

The volume of steam generated is not terribly relevant. You will get some steam, but I should think that the wilder the boil the more dark your wort will get. and as a side effect the more evaporation for which you will have to compensate.
You might want the darker color.


It was my understanding that, beer wont get any more dark if the sugar isn't caramelized? I believe CodeRage did a test on that with the electric elements. IIRC, it was determined that the electric elements don't make the beer any darker. Therefor no scorching is happening. In my steam setup I am trying to refine, I will not be able to scorch sugar since it scorches at about 300* and my steam will be running at about 225*. As to the evap.....DMS is driven off in the evap process. So in your lighter beers you may want to step up the vigor a bit???
 
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