I brewed only using BIAB, besides some canned kits at the beginins. Every body that does BIAB say that this method, using the entire volume of water for mashing is better for keeping a good constant temperature during the mash. OK, seems logic, but the theory says that this way you will rather loss more heat than using a smaller volume of water.
I am refering to this formula Q=U*A*ΔT, saying that the heat loss of an area of size A is determined by the U heat transfer coeficient and the difference in temperature between inside and out. Is there anyting that i am missing?
Post #2 made 3 years ago
That formula doesn't apply to heat transfer through radiation.
All hot objects lose energy by radiation. [EDIT: or conduction or convection]
All hot objects lose energy by radiation. [EDIT: or conduction or convection]
Last edited by Mad_Scientist on 31 Jan 2015, 03:03, edited 1 time in total.
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Post #3 made 3 years ago
Bionut, if you have studied BTU, there is a 1 Degree F/Pound of water, number.
The Larger the Volume, the larger the Btu's stored.
A kettle will let the Btu's lost/gained from the temperature difference of the OUTSIDE vs Inside, and the temperature lost/gained can be changed from the insulation, around the Kettle.
The larger the Volume also increases the Thermal Mass, or the Btu's being stored.
Depending on the Kettle being used, and the insulation around the kettle, the Btu Loss is then determined.
What I am getting to is, a smaller volume has low thermal mass due to it's size, and therefore depending on conditions will lose just as many BTU's per time as any other Volume. But the smaller volume will run out of heat sooner, than a larger volume.
The Larger the Volume, the larger the Btu's stored.
A kettle will let the Btu's lost/gained from the temperature difference of the OUTSIDE vs Inside, and the temperature lost/gained can be changed from the insulation, around the Kettle.
The larger the Volume also increases the Thermal Mass, or the Btu's being stored.
Depending on the Kettle being used, and the insulation around the kettle, the Btu Loss is then determined.
What I am getting to is, a smaller volume has low thermal mass due to it's size, and therefore depending on conditions will lose just as many BTU's per time as any other Volume. But the smaller volume will run out of heat sooner, than a larger volume.
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Post #4 made 3 years ago
Sounds ok to me, but i need some sort of equation to prove that. I am arguing with a friend about this, he is a 2 vessel brewer.
Post #5 made 3 years ago
Bionut, this may be over the top, but, your looking foe an entire course on Thermodynamics.
But, here is a link to a general Discussion(with Formulas) about heat transfer.
http://physics.bu.edu/~duffy/py105/note ... nsfer.html
Good Luck!!!
But, here is a link to a general Discussion(with Formulas) about heat transfer.
http://physics.bu.edu/~duffy/py105/note ... nsfer.html
Good Luck!!!
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Post #6 made 3 years ago
No formula yet, but found this;
http://coolcosmos.ipac.caltech.edu/cosm ... nsfer.html
http://coolcosmos.ipac.caltech.edu/cosm ... nsfer.html
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Post #7 made 3 years ago
Thanks, the radiaton heat transfer formula doesn't use the mass. I let it be, he thinks that with biab you could only loose heat. I know that it isn't true and continue with my biab schedule :D
Post #8 made 3 years ago
bionut there are some absolutely excellent and spot on (totally correct) answers above. Instead of you proving to your friend why basic physics is correct, maybe get your friend prove to us why basic physics is wrong? It is, after all, a very basic physics question.
For example, which of the following would reach 45C first?
1. A 50 ton/tonne block of concrete at 50 C with an ambient temperature of 45 C.
2. A 50 gram block of concrete at 50 C with an ambient temperature of 45 C.
Forget the formula and just ask your friend that.
PP
P.S. Also, your formula in post #1 does not explain what Q,U A and T mean. Regardless, the posts above are definitely correct. Good question though .
For example, which of the following would reach 45C first?
1. A 50 ton/tonne block of concrete at 50 C with an ambient temperature of 45 C.
2. A 50 gram block of concrete at 50 C with an ambient temperature of 45 C.
Forget the formula and just ask your friend that.
PP
P.S. Also, your formula in post #1 does not explain what Q,U A and T mean. Regardless, the posts above are definitely correct. Good question though .
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Re: Re:
Post #9 made 3 years ago
He says that mass doesn't appear in that formula and so i doesn't matter. Even that the 50 tone block will loose more heat due to the big surface area (A in the formula)PistolPatch wrote:bionut there are some absolutely excellent and spot on (totally correct) answers above. Instead of you proving to your friend why basic physics is correct, maybe get your friend prove to us why basic physics is wrong? It is, after all, a very basic physics question.
For example, which of the following would reach 45C first?
1. A 50 ton/tonne block of concrete at 50 C with an ambient temperature of 45 C.
2. A 50 gram block of concrete at 50 C with an ambient temperature of 45 C.
Forget the formula and just ask your friend that.
PP
P.S. Also, your formula in post #1 does not explain what Q,U A and T mean. Regardless, the posts above are definitely correct. Good question though .
Q is the heat loss, U is the heat transfer coeficient of a material, A the area and T is the temperature difference.
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Last edited by bionut on 01 Feb 2015, 00:51, edited 1 time in total.
Post #10 made 3 years ago
Bionut, area is important for heat loose, True.
The Mass depends on Volume and Density.
The mass is what Contains Heat.
The Larger the Mass the greater the heat Capacity.
The higher the Density the Smaller the Volume, and the Smaller volume gives less area.
Larger things of the same mass lose heat faster than Small things, TRUE
BUT, smaller things contain Geometrically less heat, and have Geometrically smaller area.
The heat loss per time, are the Same for large and small objects of the same density, and same environmental conditions.
Since the small object starts with less Total heat, the small object runs out of heat sooner.
Example:
Say I have a block of "something" 1 cubit by 1 cubit by 1 cubit or 1 cubic cubit, and it contains 1 Q heat quantity.
It has an area of 1 square Cubit per side or 6 square cubits total.
I also have a block of the same "something" that is 10 cubits by 10 cubit by 10 cubits. It is 1000 times larger than the small block and contains 1000 Q Heat Quantities.
The area is 100 square cubits per side and has 6 sides for a total of 600 Square Cubits
The big block has 1000 times the volume, but only 600 times the Surface area.
The Big block will lose heat at a rate 0.6(600/1000) compared to the small Block (6/1).
I hope this Confuses, more than Enlightens.
The Mass depends on Volume and Density.
The mass is what Contains Heat.
The Larger the Mass the greater the heat Capacity.
The higher the Density the Smaller the Volume, and the Smaller volume gives less area.
Larger things of the same mass lose heat faster than Small things, TRUE
BUT, smaller things contain Geometrically less heat, and have Geometrically smaller area.
The heat loss per time, are the Same for large and small objects of the same density, and same environmental conditions.
Since the small object starts with less Total heat, the small object runs out of heat sooner.
Example:
Say I have a block of "something" 1 cubit by 1 cubit by 1 cubit or 1 cubic cubit, and it contains 1 Q heat quantity.
It has an area of 1 square Cubit per side or 6 square cubits total.
I also have a block of the same "something" that is 10 cubits by 10 cubit by 10 cubits. It is 1000 times larger than the small block and contains 1000 Q Heat Quantities.
The area is 100 square cubits per side and has 6 sides for a total of 600 Square Cubits
The big block has 1000 times the volume, but only 600 times the Surface area.
The Big block will lose heat at a rate 0.6(600/1000) compared to the small Block (6/1).
I hope this Confuses, more than Enlightens.
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Post #11 made 3 years ago
That is what i understand from that also, but i can't get him tu understand this logic. I am happy with my setup and i am sure that i don't lose more than 12°C per hour.
Post #12 made 3 years ago
Google "squarecube law", pretty straightforward scientific explanation for this.

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