pyro
03-15-2017, 09:42 PM
I boil on a 14 gauge SS pan, and have had an interest on how much improvement I would see if I went to a ~20 gauge pan. A new pan is fairly expensive so I wanted to understand if I would see a significant improvement. Therefore I started researching some thermal dynamics basics and would like to share what I learned as I haven't seen it presented before. This thread is for the engineering nerds out there, but I hope everyone can learn something as well.
Our goal is to get energy transferred from the fire to sap which will convert the liquid to steam. To do this we need a lot of energy. But its not just energy alone we are concerned with. We don't want to boil all day and night, so we want to do this as fast as possible which is why we need to talk about the rate of this heat transfer called the "heat transfer rate" or H. Heat transfer is (Eq 2) H = T/R.
We can model the heat transfer rate of two different temperatures across a medium by equating it to a basic electrical circuit with two voltages across a resistance. One temperature is the fire temperature say 1000 F. The other side is boiling sap at a constant of 212 F (I know what your thinking, but its not significant). To transfer energy from one side to another it must go across the pan which has some unknown thermal resistance. Again equate heat transfer to current in an electric circuit. In design phase, our goal is to reduce the resistance to increase the heat flow (or current in electrical analogy).
16082
Thermal resistance is defined as (Eq 3) R = L/kA. The thermal resistance is made up of three variables. To reduce the thermal resistance we have only three options:
1. Increase area A
2. Increase thermal conductivity k (by switching materials)
3. Decrease pan thickness L
Substituting Equation (3) into Equation (2) we get Equation (4) which shows the total energy transfer of the system. It proves that decreasing the pan thickness L by 50% is equivalent to doubling the surface area A! It is this that is why the pan thickness is significantly important.
It also shows that if we found a different material such as copper it could increase k and significantly improve energy transfer. For example, copper has a thermal conductivity (=1/R) of about 15-30x more than 304 stainless steel. That means the thermal resistance is 15-30x less and would improve thermal energy transfer 15-30x!
Well not so fast. why are the few people with copper pans not raving about their 15x improvement? Why are the manufacturers not using them? I suspect there is something I am missing but I can't find what it is. We can find the losses in the system, but I still don't think it could explain where the 15-30x improvement is lost. Maybe someone has ideas?
Let's investigate losses anyways since it will help settle the sap boil depth question. The most significant loss is due to sap depth. The sides lose heat to ambient air. We can model this using the same thermal resistance. However the temperature delta is not as large as the bottom of the pan since the outside is not 1000 but ambient outside air. As an example, I have compared 1" to 3" boiling depth at 1000 degree F fire temperature, and 73 degree F ambient air on the side of the pan. I made up 73 degrees F (room temp) since the evaporator heats up the surrounding area and I have never measured it directly. These value are percentage of energy lost to the sides. Less is better.
For a 2x3 flat pan:
1": 5%
3": 15%
For a 2x4 flat pan:
1": 4%
3": 13%
For a 2x6 flat pan:
1": 4%
3": 12%
So based on all that math that I believe, I still sometimes convince myself I see a better boil with 3" deep, rather than 1". But that likely is more due to the correlation between the fact that I never have 1" of sap in when the fire is ripping.
I appreciate any comments, especially if you can prove or disprove using thermal equations.
Our goal is to get energy transferred from the fire to sap which will convert the liquid to steam. To do this we need a lot of energy. But its not just energy alone we are concerned with. We don't want to boil all day and night, so we want to do this as fast as possible which is why we need to talk about the rate of this heat transfer called the "heat transfer rate" or H. Heat transfer is (Eq 2) H = T/R.
We can model the heat transfer rate of two different temperatures across a medium by equating it to a basic electrical circuit with two voltages across a resistance. One temperature is the fire temperature say 1000 F. The other side is boiling sap at a constant of 212 F (I know what your thinking, but its not significant). To transfer energy from one side to another it must go across the pan which has some unknown thermal resistance. Again equate heat transfer to current in an electric circuit. In design phase, our goal is to reduce the resistance to increase the heat flow (or current in electrical analogy).
16082
Thermal resistance is defined as (Eq 3) R = L/kA. The thermal resistance is made up of three variables. To reduce the thermal resistance we have only three options:
1. Increase area A
2. Increase thermal conductivity k (by switching materials)
3. Decrease pan thickness L
Substituting Equation (3) into Equation (2) we get Equation (4) which shows the total energy transfer of the system. It proves that decreasing the pan thickness L by 50% is equivalent to doubling the surface area A! It is this that is why the pan thickness is significantly important.
It also shows that if we found a different material such as copper it could increase k and significantly improve energy transfer. For example, copper has a thermal conductivity (=1/R) of about 15-30x more than 304 stainless steel. That means the thermal resistance is 15-30x less and would improve thermal energy transfer 15-30x!
Well not so fast. why are the few people with copper pans not raving about their 15x improvement? Why are the manufacturers not using them? I suspect there is something I am missing but I can't find what it is. We can find the losses in the system, but I still don't think it could explain where the 15-30x improvement is lost. Maybe someone has ideas?
Let's investigate losses anyways since it will help settle the sap boil depth question. The most significant loss is due to sap depth. The sides lose heat to ambient air. We can model this using the same thermal resistance. However the temperature delta is not as large as the bottom of the pan since the outside is not 1000 but ambient outside air. As an example, I have compared 1" to 3" boiling depth at 1000 degree F fire temperature, and 73 degree F ambient air on the side of the pan. I made up 73 degrees F (room temp) since the evaporator heats up the surrounding area and I have never measured it directly. These value are percentage of energy lost to the sides. Less is better.
For a 2x3 flat pan:
1": 5%
3": 15%
For a 2x4 flat pan:
1": 4%
3": 13%
For a 2x6 flat pan:
1": 4%
3": 12%
So based on all that math that I believe, I still sometimes convince myself I see a better boil with 3" deep, rather than 1". But that likely is more due to the correlation between the fact that I never have 1" of sap in when the fire is ripping.
I appreciate any comments, especially if you can prove or disprove using thermal equations.