Thermodynamics anyone?
Moderator: slothrob
13 posts
• Page 1 of 1
Thermodynamics anyone?
Hello fellow forumers!
I was just wondering if there are any engineers in our midst that know how to calculate heat output and heat loss for specific heat sources and specific vessels respectively. I imagine there are some calibration methods that can be used to calculate these figures based on measured temperatures at specific time intervals. What are the most common units used to measure these values? If you have any advice, please steer me in the right direction. Thanks!
Jeff
I was just wondering if there are any engineers in our midst that know how to calculate heat output and heat loss for specific heat sources and specific vessels respectively. I imagine there are some calibration methods that can be used to calculate these figures based on measured temperatures at specific time intervals. What are the most common units used to measure these values? If you have any advice, please steer me in the right direction. Thanks!
Jeff

jeff  Imperial Stout
 Posts: 1459
 Joined: Sun Jan 09, 2000 9:16 pm
 Location: Hollywood, SC
So you want to calculate the heat capacity(kJ/
Cheers,
Jeremy
"To see a World in a Grain of Sand
And a Heaven in a Wild Flower,"
Hold a beer in the palm of your hand
And be glad it's happy hour.
Adapted from W. Blake "Auguries of Innocence"
Jeremy
"To see a World in a Grain of Sand
And a Heaven in a Wild Flower,"
Hold a beer in the palm of your hand
And be glad it's happy hour.
Adapted from W. Blake "Auguries of Innocence"
 JSGilbert
 Posts: 9
 Joined: Mon Aug 02, 2004 1:45 pm
 Location: Denver, CO
Precisely.
That is the basic idea. I have read some scattered information on calibration procedures in using a measured volume of water. Heat source output being determined by measuring the length of time to raise the given voume of water 10

jeff  Imperial Stout
 Posts: 1459
 Joined: Sun Jan 09, 2000 9:16 pm
 Location: Hollywood, SC
Jeff,
Yes, your methods appear to be valid. I agree with not worrying about the absolute output of your heat source, it makes everything much easier.
What units will you be using? Metric or English?
You'll need the specific heat of water (4.18 J/gK), but I'm sure you already knew that.
Yes, your methods appear to be valid. I agree with not worrying about the absolute output of your heat source, it makes everything much easier.
What units will you be using? Metric or English?
You'll need the specific heat of water (4.18 J/gK), but I'm sure you already knew that.
Cheers,
Jeremy
"To see a World in a Grain of Sand
And a Heaven in a Wild Flower,"
Hold a beer in the palm of your hand
And be glad it's happy hour.
Adapted from W. Blake "Auguries of Innocence"
Jeremy
"To see a World in a Grain of Sand
And a Heaven in a Wild Flower,"
Hold a beer in the palm of your hand
And be glad it's happy hour.
Adapted from W. Blake "Auguries of Innocence"
 JSGilbert
 Posts: 9
 Joined: Mon Aug 02, 2004 1:45 pm
 Location: Denver, CO
Glad to hear I am on the right track!
I am probably more accustomed to English units, but I can go either way. Do you know the forumulae that I need to use to perform the tests? I imagine the variables include:
It is probably a good task for a computer to perform so I plan on plugging all this stuff into a spreadsheet or program so I only need to worry about it once. Again, your kind assistance is appreciated.
Jeff
 Thermal mass of the vessel and the water
Environment temperature
Heating and cooling times
Various vessel and water temperature changes
It is probably a good task for a computer to perform so I plan on plugging all this stuff into a spreadsheet or program so I only need to worry about it once. Again, your kind assistance is appreciated.
Jeff

jeff  Imperial Stout
 Posts: 1459
 Joined: Sun Jan 09, 2000 9:16 pm
 Location: Hollywood, SC
Thermodynamics
Your methods seem okay, but because I don't know exactly what your setup is, nor the final process you will do, I want to make two points.
1) However you go about measuring the output of the burner, make sure that the water does not undergo a phase change. Assuming you are only recording temperature you cannot measure the energy required to change phase. Bottom line, don't heat ice, don't make steam.
2) Keep in mind that if you have your system outside that it will probably cool faster in the winter. Also, if your vessel is made from metal if you put it on a metal framework or table, their may be heat lost to heat up the frame or the table. This amount may be trivial though.
Tomorrow sometime, if I can remember, I will post the relevant equations. I will take a little time to reaquaint myself with English system.
Nate
1) However you go about measuring the output of the burner, make sure that the water does not undergo a phase change. Assuming you are only recording temperature you cannot measure the energy required to change phase. Bottom line, don't heat ice, don't make steam.
2) Keep in mind that if you have your system outside that it will probably cool faster in the winter. Also, if your vessel is made from metal if you put it on a metal framework or table, their may be heat lost to heat up the frame or the table. This amount may be trivial though.
Tomorrow sometime, if I can remember, I will post the relevant equations. I will take a little time to reaquaint myself with English system.
Nate
 Dr Strangebrew
 Pale Ale
 Posts: 77
 Joined: Tue May 13, 2003 7:01 pm
 Location: Lincoln, NE, US
Temperature range for calibration.
Thanks for the reply!
I was figuring a calibration range of 6070
1) However you go about measuring the output of the burner, make sure that the water does not undergo a phase change. Assuming you are only recording temperature you cannot measure the energy required to change phase. Bottom line, don't heat ice, don't make steam.
I was figuring a calibration range of 6070

jeff  Imperial Stout
 Posts: 1459
 Joined: Sun Jan 09, 2000 9:16 pm
 Location: Hollywood, SC
More info please
Sorry it took so long to get a response... I forgot. This morning I began composing a response. The first part is rather straight forward. The second part has some nuances that I think need to be appreciated. I would like to know more about what you are trying to do. Furthermore, some of the equations would look better in an equation editor in Microsoft Word. Would this paste easily to the forum? Or maybe I can email you with a response? Note, for the benefit of others in the forum I would still paste the same thing on the forum, the equations just wouldn't be as pretty they would be from notepad.
Thanks,
Nate
Thanks,
Nate
 Dr Strangebrew
 Pale Ale
 Posts: 77
 Joined: Tue May 13, 2003 7:01 pm
 Location: Lincoln, NE, US
My intentions
I am currently working on a program that will help me take the guesswork out of brewing session timing. It is pretty easy to come up with rest temperatures and rest times but there is a lot of waiting and heating that goes on in between those phases. I would be nice to be able to get a good idea of how long I have to wait for my kettle to heat to mashin temperature, or how long it will take to mashout. I would like to answer questions such as:

jeff  Imperial Stout
 Posts: 1459
 Joined: Sun Jan 09, 2000 9:16 pm
 Location: Hollywood, SC
Okay, here we go...
I am more certain about the validity of the first part than I am for the second part.
:= is notation for, "by definition is equal to"
Q := amount of heat (in Joules, heat is defined as the energy transferred because of a change in temperature)
m := mass (in kilograms)
c := specific heat (in this case water; therfore, c = 4.19*10^3 J/(kg*C)
dT := change in temperature (in C)
The equation that you need is then:
(1) Q = m*c*dT (where * denotes multiplication)
If you assume constant burner output then the amount of heat transferred per unit time is simply Q divided by time. With the provisos we have already discussed this should take care of the output of your burner.
Now on to heat loss in the mash tun. As a mathematical treatment of the heat loss of your mash tun I will use Newton's law of cooling. This model well describes coffee cooling on a counter or a pie heating up in an oven. It states that the rate of change of temperature of an object with respect to time is proportional to the difference in temperature of the object and its enviroment. So:
dT := change in temperature of the object
dt := change in time
k := some positive number that can only be determined through experiment (the proportionality constant from above)
T := current temperature of the object
R := temperature of the surroundings
A := initial temperature of the object
and the resulting differential equation is:
(2) dT/dt = k*(TR)
which has the following solution
(3) T=R+A*exp[k*t]
Let me explain. "dT/dt" is a differential operator. Mathematically it is called the first time derivative of temperature. Or, stated with real words, it is the time rate of change of the temperature of the object. "exp" is the exponential function. Another way to write the last equation is:
(4) T=R+A*2.718^(k*t)
or, if you are familiar with the NUMBER e,
(5) T=R+A*e^(k*t)
Without taking measurements this equation cannot be solved. As I said above, k must empirically be determined. I have heard of other versions of this law. One involves crosssectional area, but as far as I know k, or some other unknown parameter, is still present. Here is what I recommend to determine k.
Using algebra, equation (5) can be rewritten as:
(6) ln[(TR)/A]/t = k (where ln is the natural logirithm)
I would be great if you have software where you could plot (5) to see how it behaves. If the temperature difference between mash tun and enviroment is large then the temperature of the tun will decrease exponentially. As the mash tun approaches enviroment temperature, its temperature decreases much, much slower.
An example of this is your wort chiller. Have you noticed how quickly your wort cools to 100 degress F? The last 25 degrees take much., much longer than the first 112. My model agrees (overall) with wort chilling at least.
At this point I am not sure how useful this theoretical discussion is. Consider (6), if you know T, R, A, and t you could find k. This would require a trial run. Say, get it all set up and begin to work through a batch. Measure the initial temperature of object. Measure enviroment temperature. After 10 min exposure to enviroment, with no applied heat, again measure the temperature of the object. Plug all that into (6) and you could find k, assuming that 10 min was enough to notice an appriciable temperature decline. You may have to try different time intervals.
Unfortunately, k will probably also vary with the mash volume and water:grist ratio. Hopefully, this will be trivial. I submit, without proof, or even a well contructed arguement that k won't vary with the enviroment. As I understand the equation, k is a parameter dependent only on the object, in this case mash tun.
The easiest equation to work with would be (4) or (5), provided that k varies little with mash composition and volume. In other words, one way to do this would be to assume that k won't vary with enviroment temperature or mash volume. This hopefully could give you the temperature of your mash tun as a function of time.
If k does vary significantly with mash volume, I suspect it surely will if you are doing a halfbatch or doublebatch, by keeping good records you could come up with a function for k based entirely upon observations of mash volume and water:grist ratio.
Keep in mind that you may have to make adjustments if the enviroment temperature varies during the mash. If you mash for 6090 min there probably won't be a problem, but there may be days...
Suppose that we are looking for how to set your burner so that your mash tun stays at an even, or close to even temperature. I foresee a couple ways to do this. One would be to apply constant heat, exactly balanced so that what thermal energy leaving the mash tun is exactly balanced by what is coming coming in from the burner. This would require knowing the heat output of several burner settings.
A better way would be to apply heat intermittently from the burner using the setting investigated in the first method. This seems like the better option. Using the results from the above equations, including the results of the first method, you could determine an appropriate time interval to apply the bursts.
Frankly, I don't know to what extent the above equations will be helpful. I haven't formally established the validity of Newton's law of cooling. Even if it is valid, I suspect that a successful system will require excellent record keeping, especially regarding parameters that affect k. Furthermore, burner output is expected to be constant on windy days it may not be.
I hope this helped,
Nate
:= is notation for, "by definition is equal to"
Q := amount of heat (in Joules, heat is defined as the energy transferred because of a change in temperature)
m := mass (in kilograms)
c := specific heat (in this case water; therfore, c = 4.19*10^3 J/(kg*C)
dT := change in temperature (in C)
The equation that you need is then:
(1) Q = m*c*dT (where * denotes multiplication)
If you assume constant burner output then the amount of heat transferred per unit time is simply Q divided by time. With the provisos we have already discussed this should take care of the output of your burner.
Now on to heat loss in the mash tun. As a mathematical treatment of the heat loss of your mash tun I will use Newton's law of cooling. This model well describes coffee cooling on a counter or a pie heating up in an oven. It states that the rate of change of temperature of an object with respect to time is proportional to the difference in temperature of the object and its enviroment. So:
dT := change in temperature of the object
dt := change in time
k := some positive number that can only be determined through experiment (the proportionality constant from above)
T := current temperature of the object
R := temperature of the surroundings
A := initial temperature of the object
and the resulting differential equation is:
(2) dT/dt = k*(TR)
which has the following solution
(3) T=R+A*exp[k*t]
Let me explain. "dT/dt" is a differential operator. Mathematically it is called the first time derivative of temperature. Or, stated with real words, it is the time rate of change of the temperature of the object. "exp" is the exponential function. Another way to write the last equation is:
(4) T=R+A*2.718^(k*t)
or, if you are familiar with the NUMBER e,
(5) T=R+A*e^(k*t)
Without taking measurements this equation cannot be solved. As I said above, k must empirically be determined. I have heard of other versions of this law. One involves crosssectional area, but as far as I know k, or some other unknown parameter, is still present. Here is what I recommend to determine k.
Using algebra, equation (5) can be rewritten as:
(6) ln[(TR)/A]/t = k (where ln is the natural logirithm)
I would be great if you have software where you could plot (5) to see how it behaves. If the temperature difference between mash tun and enviroment is large then the temperature of the tun will decrease exponentially. As the mash tun approaches enviroment temperature, its temperature decreases much, much slower.
An example of this is your wort chiller. Have you noticed how quickly your wort cools to 100 degress F? The last 25 degrees take much., much longer than the first 112. My model agrees (overall) with wort chilling at least.
At this point I am not sure how useful this theoretical discussion is. Consider (6), if you know T, R, A, and t you could find k. This would require a trial run. Say, get it all set up and begin to work through a batch. Measure the initial temperature of object. Measure enviroment temperature. After 10 min exposure to enviroment, with no applied heat, again measure the temperature of the object. Plug all that into (6) and you could find k, assuming that 10 min was enough to notice an appriciable temperature decline. You may have to try different time intervals.
Unfortunately, k will probably also vary with the mash volume and water:grist ratio. Hopefully, this will be trivial. I submit, without proof, or even a well contructed arguement that k won't vary with the enviroment. As I understand the equation, k is a parameter dependent only on the object, in this case mash tun.
The easiest equation to work with would be (4) or (5), provided that k varies little with mash composition and volume. In other words, one way to do this would be to assume that k won't vary with enviroment temperature or mash volume. This hopefully could give you the temperature of your mash tun as a function of time.
If k does vary significantly with mash volume, I suspect it surely will if you are doing a halfbatch or doublebatch, by keeping good records you could come up with a function for k based entirely upon observations of mash volume and water:grist ratio.
Keep in mind that you may have to make adjustments if the enviroment temperature varies during the mash. If you mash for 6090 min there probably won't be a problem, but there may be days...
Suppose that we are looking for how to set your burner so that your mash tun stays at an even, or close to even temperature. I foresee a couple ways to do this. One would be to apply constant heat, exactly balanced so that what thermal energy leaving the mash tun is exactly balanced by what is coming coming in from the burner. This would require knowing the heat output of several burner settings.
A better way would be to apply heat intermittently from the burner using the setting investigated in the first method. This seems like the better option. Using the results from the above equations, including the results of the first method, you could determine an appropriate time interval to apply the bursts.
Frankly, I don't know to what extent the above equations will be helpful. I haven't formally established the validity of Newton's law of cooling. Even if it is valid, I suspect that a successful system will require excellent record keeping, especially regarding parameters that affect k. Furthermore, burner output is expected to be constant on windy days it may not be.
I hope this helped,
Nate
 Dr Strangebrew
 Pale Ale
 Posts: 77
 Joined: Tue May 13, 2003 7:01 pm
 Location: Lincoln, NE, US
WOW!! Many thanks o wise master of thermodynamic wisdom!
Thank you for spending the time to help me with this problem! I will get to work plugging in the equations; it will be very interesting to see how they predict real world conditions. It seems to me that these methods can be extended to determine wort chiller efficiency and the effects tubing and plumbing have on temperature too.
Next I will move on to figuring evaporation rate and the specific enthalpy of liquid water. You don't know anything about that do you?
Thanks again!
Jeff
Next I will move on to figuring evaporation rate and the specific enthalpy of liquid water. You don't know anything about that do you?
Thanks again!
Jeff

jeff  Imperial Stout
 Posts: 1459
 Joined: Sun Jan 09, 2000 9:16 pm
 Location: Hollywood, SC
Re: WOW!! Many thanks o wise master of thermodynamic wisdom!
You need three things.
1) the Q=m*Cp*(T1To) equation above
and the specific heat (cp) for the metal and the water and the grains.
Here they are:
stainless steel 0.12 Btu/lbm/F
grains 0.4 Btu/lbm/F (estimated)
water 1 Btu/lbm/F
Calculate the Q (heat) for each one by setting the ending temp (t1) and starting temp (To). The sum of the heats is the output of your burner
edit: by output, I meant 'effective output' i.e. how much heat it adds
1) the Q=m*Cp*(T1To) equation above
and the specific heat (cp) for the metal and the water and the grains.
Here they are:
stainless steel 0.12 Btu/lbm/F
grains 0.4 Btu/lbm/F (estimated)
water 1 Btu/lbm/F
Calculate the Q (heat) for each one by setting the ending temp (t1) and starting temp (To). The sum of the heats is the output of your burner
edit: by output, I meant 'effective output' i.e. how much heat it adds
 TomMeier
 Posts: 2
 Joined: Thu Aug 19, 2004 4:48 pm
A couple of pointers.
Just to make sure that you stay on the right track and don't get frustrated with your process, make sure that you use Kelvin as your temperature units (273.15K = 0C) or else your calculations will not work. Also enthalpy is based on two things, temperature and pressure. Since the system is open (heat and mass are transferred) you may have to look at the entropy as well. There are tables for the known values for water, air, and common refrigerants at certain temperatures and pressures, but I have not encountered any for beer, plus the specific gravity of the wort may change those values.
Overall the process seems pretty sound but I only took a couple minutes to look over it.
I definitely recommend using metric units, its easier that way.
Have fun with the application of the 2nd Law!
kru
Overall the process seems pretty sound but I only took a couple minutes to look over it.
I definitely recommend using metric units, its easier that way.
Have fun with the application of the 2nd Law!
kru
 krussader
 Light Lager
 Posts: 11
 Joined: Fri Apr 25, 2008 9:59 pm
13 posts
• Page 1 of 1
Who is online
Users browsing this forum: No registered users and 2 guests