Possible Electric build with silicone heat mats

[Weezy]

Not sure where you are coming up with your efficiency factor or what you are basing it on, but heating in general with electricity tends to be highly efficient compared to other heat sources. That is because it has a low energy content (3,412 BTU) per unit of measure (kilowatt) versus a fuel source such as propane which has 92,000 BTU per gallon. So while a high efficiency rating sounds good, it's not the whole picture.

You're misunderstanding the efficiency we're talking about. You're thinking about how efficient electric elements are at converting electric energy to thermal energy. The efficiency is how effective the element is at transferring that energy to what you want heated. In this case it's the transfer of energy from the heat pad, through the metal pot, to the wort inside. There are energy losses as the heat energy much move through and between each material (generally referred to as losses 'to the environment'). An immersed water heater element is 90%+ efficient because it is directly heating the wort and is completely immersed in what we want heated. Here's a silly example to explain the differences. Say you have a small, electric, ceramic, space heater with a fan in it. You use it to heat a small bathroom. It works well enough since the room is small. It's working at, lets guess 75% efficiency. Now put that heater into a cardboard box inside that room and see how well it warms the room, through the box. That box is going to get really hot and it will warm the room somewhat, but the efficiency of this system is going to be much, much lower than the unhindered space heater. Maybe the efficiency is now 20%.

 

[dyqik]

You're misunderstanding the efficiency we're talking about. You're thinking about how efficient electric elements are at converting electric energy to thermal energy. The efficiency is how effective the element is at transferring that energy to what you want heated. In this case it's the transfer of energy from the heat pad, through the metal pot, to the wort inside. There are energy losses as the heat energy much move through and between each material (generally referred to as losses 'to the environment'). An immersed water heater element is 90%+ efficient because it is directly heating the wort and is completely immersed in what we want heated. Here's a silly example to explain the differences. Say you have a small, electric, ceramic, space heater with a fan in it. You use it to heat a small bathroom. It works well enough since the room is small. It's working at, lets guess 75% efficiency. Now put that heater into a cardboard box inside that room and see how well it warms the room, through the box. That box is going to get really hot and it will warm the room somewhat, but the efficiency of this system is going to be much, much lower than the unhindered space heater. Maybe the efficiency is now 20%.

Actually, it'll be exactly as efficient in the steady state. Where do you think the rest of the energy is going? All of the net energy going into the cardboard box must end up as heat in the room.

Space heaters in closed rooms operate at 100% efficiency. Every bit of energy that enters the room and doesn't leave as electrical energy is converted to heat in the room.

The heater issue the cardboard box will reach equilibrium more slowly, but it will be 100% efficient.

The efficiency losses in immersed elements are due to the base of the element being heated and losing heat energy that doesn't go via the wort. If you could put the element on a cable into the wort and immerse the base, it would be practically 100% efficient (only losses would be resistance in the cable before it enters the wort). For silicone heating mats, the losses are due to heat being lost from the outside of the mats directly to the air and table. The overall kettle efficiency in both cases needs to also include the heat losses from the walls and top of the kettle as distinct from the heat energy that goes into increasing the temperature of the wort.

 

[trboyden]

No, I understood the efficiency you meant, which is why I questioned the 80%. If an immersed element is 90%, how can an external one be 80% that has to heat both the pot and the water. That doesn't pass the smell test. I also question the results. The amount of energy needed to boil water/wort is measured in Joules. For 5 gallons, you'd need 38029761.11 Joules to bring the 5 gallons to a boil. If the OP really is only using 2200 watts of energy from his mats, there is no way that could be done in less than 288 minutes (Joules required/2200 watts = seconds / 60 = minutes). And that's assuming 100% thermal transfer to the water/wort. So 360% efficiency? No. There is missing variable(s) that is skewing the results.

It's a thermal physics equation. http://www.physicsclassroom.com/class/thermalP/Lesson-2/Measuring-the-Quantity-of-Heat

 

[Weezy]

Ugh. Yes the cardboard box example was bad. A straw man argument doesn't change the fact that external elements experience energy loss to the environment. Period. An electric hot plate, heating a can of soup, sends more heat into the room than into the pot of soup. Period. How well an external element is attached to what you want heated is critical in how efficient the thermal energy transfer will be. These silicone pads obviously experience energy loss to the air....I'd bet $100 that they're too hot to touch on the outside when running, yes? That's energy loss to the environment. That's energy not going to heating the pot or wort. That's a reduction in the efficiency of the system.

 

[dyqik]

The only real difference between the silicone pad case and the immersed element case is in the difference in temperatures between the outside of the pot in the immersion element case and the outside of the heating mats in the silicone pad case, and the relative heat transfer coefficients to the air for the two materials. Everything else is the approximately the same.

If the silicone pads have a relatively high heat transfer coefficient to the wort (they are glued to the pot over a large surface area, and a recirculation pump is used to stir the wort), then they could approach the efficiency of an immersed element - the temperature of the pads would be approximately the same as that of the wort. The losses from the pot in the immersion element case would then be about the same as the losses from the pads in the silicone heating pad case. 80% efficiency doesn't seem unreasonable at all to me, and 90% or higher would also be plausible in still air (which would lower the heat transfer coefficient to the air).

The post above shows a 70 C change in 60 minutes for 25l of water. This is a total energy of 4200 J/litre/C x 25 litres x 70C = 7.35 MJ supplied to the wort in 3600 seconds, for an average power of 2041 W supplied to the wort. With an average power supplied of 2200W, this is an efficiency of 93%. This seems very good, but not implausible, and certainly not unphysical. I'd like to be sure that the volumes and temperatures were properly calibrated, have an accurate specific heat and know that the power draw was constant before being more precise than estimating that at 90-95% efficiency.

For 5 gallons, you'd need 38029761.11 Joules to bring the 5 gallons to a boil.

I don't know where this figure of 38 MJ required to boil 5 gallons comes from. It's wrong by a factor of 4 or so.

 

[augiedoggy]

The only real difference between the silicone pad case and the immersed element case is in the difference in temperatures between the outside of the pot in the immersion element case and the outside of the heating mats in the silicone pad case, and the relative heat transfer coefficients to the air for the two materials. Everything else is the approximately the same.

If the silicone pads have a relatively high heat transfer coefficient to the wort (they are glued to the pot over a large surface area, and a recirculation pump is used to stir the wort), then they could approach the efficiency of an immersed element - the temperature of the pads would be approximately the same as that of the wort. The losses from the pot in the immersion element case would then be about the same as the losses from the pads in the silicone heating pad case. 80% efficiency doesn't seem unreasonable at all to me, and 90% or higher would also be plausible in still air (which would lower the heat transfer coefficient to the air).

The post above shows a 70 C change in 60 minutes for 25l of water. This is a total energy of 4200 J/litre/C x 25 litres x 70C = 7.35 MJ supplied to the wort in 3600 seconds, for an average power of 2041 W supplied to the wort. With an average power supplied of 2200W, this is an efficiency of 93%. This seems very good, but not implausible, and certainly not unphysical. I'd like to be sure that the volumes and temperatures were properly calibrated, have an accurate specific heat and know that the power draw was constant before being more precise than estimating that at 90-95% efficiency.


I don't know where this figure of 38 MJ required to boil 5 gallons comes from. It's wrong by a factor of 4 or so.

Not to interrupt your scientific argument but
In reality, isnt half the heating pad or element surface exposed to air and therefore being wasted by transferring /releasing half the heat energy to the air outside the pot?

How can this possibly be compared to the efficiency of an immersion element where 100% of the element surface area and heat is in direct contact with the liquid being heated? id say 96% or more of the heat generated is directed to the liquid vs 50% in the heating pad and thats before you take into effect the cooling properties of the ambient air against the pad that the element must overcome?

 

[dyqik]

Isnt half the heating pad or element surface exposed to air and therefore being wasted by transferring /releasing half the heat energy to the air outside the pot?

How can this possibly be compared to the efficiency of an immersion element where 100% of the element surface area and heat is in direct contact with the liquid being heated?

No, the heat energy losses from each side of the heating pad are not equal, because the heat transfer coefficients are different. Heat transfer by conduction to a well stirred high heat capacity medium like wort is much much more rapid than heat transfer to a low heat capacity medium like air that is stirred only by convection.

 

[augiedoggy]

Eh? How does the size of the unit of measure affect the efficiency? That's just nonsense. Efficiency is heat energy into wort/energy used. Nothing more or less.

Even when calculating cost, the size of the unit doesn't matter, because units are divisible in both cases. The cost per unit is a factor, but not the size of the unit.

From what I understand size can have a lot to do with efficiency.... how efficient would it be to heat your house with a small undersized furnace vs the correctly sized one?

 

[augiedoggy]

No, the heat energy losses from each side of the heating pad are not equal, because the heat transfer coefficients are different. Heat transfer by conduction to a well stirred high heat capacity medium like wort is much much more rapid than heat transfer to a low heat capacity medium like air that is stirred only by convection.

If that were soo true than insulating a kettle would have little effect on efficiency... all one has to do is wrap a towel around a heating kettle to see it does in fact have a huge impact.... The same principal with a cooler of ice... what am I missing here? what your saying does make sense but I think how much efficiency/ energy is lost is still greater than I believe you are insinuating.

The liquid itself is a better conductor and if all the energy is with in it...

plus how much energy is lost by just the poor surface tranfer from the rough silicone surface on the mat and the surface of the pot? we all know how much difference even thermal paste makes in transferring energy from one medium to another? (yes because of your point about air being a poor medium)

 

[dyqik]

if that were soo true than insulating a kettle would have little effect on efficiency... all one has to do is wrap a towel around a heating kettle to see it does in fact have a huge impact....

The same principal with a cooler of ice... what am I missing here? what your saying does make sense but I think how much efficiency/ energy is lost is still greater than I believe you are insinuating.

The liquid itself is a better conductor and if all the energy is with in it...

plus how much energy is lost by just the poor surface tranfer from the rough silicone surface on the mat and the surface of the pot? we all know how much difference even thermal paste makes in transferring energy from one medium to another? (yes because of your point about air being a poor medium)

Sorry, I have to completely edit my post because of your changed questions.

The "half the power going each way" thing is just plain wrong as a starting point. It would be right for a heating pad with similar volumes of wort or similar volumes of air on each side, but that isn't the case here at all.

Taking still water and air on either side of a heated plate as the starting point, the thermal conductivity of water is 20 times higher than that of air. The heat lost to the water will be at least 20 times greater than that to air, and actually much more because water also has a much higher heat capacity. You know this - if you drop a piece of hot metal into water it will cool much much more rapidly than if you leave it hanging by a thread in mid air. Even if you have only small pot of water and a much bigger volume of air so that they have equal heat capacities.

You need to do the calculations properly, and you'll need the equations here. Note that the heat transfer to the wort is given by the "Forced convection, external flow, vertical plane" equation, while I think the heat transfer to still air is governed by the "Internal flow, laminar flow" equation, modified for an approximately flat plate, and by something like "External flow, vertical plane" for the blown air case. The Prandtl and Rayleigh numbers will be different for air and water as their densities, viscosities, speed, thermal conductivities, etc. are all different so the heat transfer coefficients will be different.

In particular, the Prandtl numbers of air and water are different by a factor of 10 (and the difference will be higher for a more viscous wort). These are raised to something like the 2nd power in the empirical equations for the heat transfer coefficient. There are other differences in other parts of the equations as well.

 

[trboyden]

Doh! My bad. I'll be the first to admit I made a math error. My calculator was set to a 40 gallon pot size and I didn't change that value. So re-checking my calculations based on the formula at the Physics Classroom, the real math is...

1. Calculate the delta between start and goal temp
   • Start temp = 20*C
   • Goal temp = 100*C
   • Delta = 80*C
2. Calculate the volume in Grams
   • Grams = Gallons * a conversion factor of 3785.4118
   • Total Grams for 5 gallons = 18927.059
3. Multiply Delta * Volume * Energy Constant (4.186 Joules)
   • Total energy required to boil = 6338293.518 Joules
4. Divide Total Joules Required by Energy Applied (2200 Watts)
   • 6338293.518 / 2200 = 2881.042508 seconds
   • Total seconds / 60 = 48.01737514 minutes

So 48 minutes based on 100% thermal transfer and no losses.

So now the results makes more sense to me. At 120 minutes he had a 40% efficiency. That would make more sense given the external heat pads, the absorption of the pot, and the losses to the air.

That's something worth noting on my Excel calculator. The fields in yellow require manual entry. So if you use the unit convertors at the bottom, make sure you type those values into the relevant fields. When I changed the volume value from 40 gallons to 5 gallons, I didn't do that.

-Tim

The post above shows a 70 C change in 60 minutes for 25l of water. This is a total energy of 4200 J/litre/C x 25 litres x 70C = 7.35 MJ supplied to the wort in 3600 seconds, for an average power of 2041 W supplied to the wort. With an average power supplied of 2200W, this is an efficiency of 93%. This seems very good, but not implausible, and certainly not unphysical. I'd like to be sure that the volumes and temperatures were properly calibrated, have an accurate specific heat and know that the power draw was constant before being more precise than estimating that at 90-95% efficiency.


I don't know where this figure of 38 MJ required to boil 5 gallons comes from. It's wrong by a factor of 4 or so.

 

[dyqik]

He has something like 90-95% efficiency for 20C to 90C, which seems about right, given the relative thermal conductivities. At higher wort temperatures, the heat loss to the air increases so those losses will go up as the temperature rises. Above 90C, the evaporative losses become more and more important, and as boiling starts, you start losing significant amounts of heat to evaporation. At boiling, you lose 2700W for every gallon boiled off per hour. To get over that last hump from about 90C to boiling, in a reasonable time, you need quite a bit more power than you do to get from 20C to 90C.

Augie, to answer your other question about thermal transfer from silicone to steel, the thermal conductivity of thermal interface silicone is about 1.2 W/m/K, about 3 times that of water. The thermal conductivity of glues e.g. epoxies is a bit better still, and the layer will be very thin if applied correctly. As long as a bit more than about 50% of the surface of the silicone is making contact with the glue or the steel, the thermal conductivity is limited by that of the water, not the glue or silicone.

 

[trboyden]

Not sure why you keep cherry picking numbers. What does it matter what the efficiency is for 20C to 90C? The goal is 100C. The overall efficiency of hitting the goal is what matters, not some cherry picked number to make the efficiency look better than it was. If his efficiency went down considerably over 90C, which in all indications based on his photos of temperature shows, than he's not going to have 90%+ efficiency. 194*F is not boiling.

He has something like 90-95% efficiency for 20C to 90C, which seems about right, given the relative thermal conductivities. At higher wort temperatures, the heat loss to the air increases so those losses will go up as the temperature rises. Above 90C, the evaporative losses become more and more important, and as boiling starts, you start losing significant amounts of heat to evaporation. At boiling, you lose 2700W for every gallon boiled off per hour. To get over that last hump from about 90C to boiling, in a reasonable time, you need quite a bit more power than you do to get from 20C to 90C.

Augie, to answer your other question about thermal transfer from silicone to steel, the thermal conductivity of thermal interface silicone is about 1.2 W/m/K, about 3 times that of water. The thermal conductivity of glues e.g. epoxies is a bit better still, and the layer will be very thin if applied correctly. As long as a bit more than about 50% of the surface of the silicone is making contact with the glue or the steel, the thermal conductivity is limited by that of the water, not the glue or silicone.

 

[dyqik]

Not sure why you keep cherry picking numbers. What does it matter what the efficiency is for 20C to 90C? The goal is 100C. The overall efficiency of hitting the goal is what matters, not some cherry picked number to make the efficiency look better than it was. If his efficiency went down considerably over 90C, which in all indications based on his photos of temperature shows, than he's not going to have 90%+ efficiency. 194*F is not boiling.

I'm really not cherry picking 90C, I'm referring to actual numbers from the experiment above, which are also relevant to brewing. I chose 90C because it gave a longer time to average over than e.g. 75C. Those numbers give the actual efficiency for heating strike or sparge water, or maintaining mash temperatures. And it's much higher than the 50% that a number of posters naively guessed would be the case at those temperatures. If you'd prefer to stop at 75C for sparge or strike water, then you'd get similar efficiencies for those processes.

The reason for calculating the heat transfer efficiency in this range is that you don't have to factor in any significant changes in heat losses from the wort. Above this range as you approach boiling, the heat transfer efficiency from heating pad to steel to wort won't change very much. However, above about 90-95C, the heat losses from the wort change rapidly, regardless of the heating method, because the losses increase due to evaporation. So I'm stopping at 90C because the losses from the wort become seriously nonlinear somewhere between 90C and 100C. Immersion elements won't get anywhere near 95% efficiency in that range, if you measure the heat into the wort just by the temperature change. Certainly my propane burner and 9000 BTU/hr gas stove burner lose a lot of efficiency there, with the final 5 C to boiling taking much longer to achieve than the 5 C before that, and if you look at the "my element won't reach boil" thread, you'll see that a 1500W immersion element can have zero efficiency by this measure. You have to be very careful by what you mean by efficiency in this temperature range, and any differences could be down to the geometry of the pot and other things that affect the losses.

If you really wanted to do a good job on this, you should use similar figures for a 2000-2500 W immersion element all the way to boiling to compare to these figures to see how much of the increase in time to boiling is attributable to the nature of the element. I'll bet that a 2000W immersion element isn't that much faster, if at all. I should get my hot-rod components this week, so I could do that test if you'd like.

 

[Weezy]

This is dumb. im ready to pull out the thermogoddammics book and do some math but I have better things to do. Dygik, by inspection, his efficiency can't be as close to a fully immersed heating element as you're suggesting. There is going to to be double digit loss to the air and poor conductivity to the pot.

 

[dyqik]

This is dumb. im ready to pull out the thermogoddammics book and do some math but I have better things to do. Dygik, by inspection, his efficiency can't be as close to a fully immersed heating element as you're suggesting. There is going to to be double digit loss to the air and poor conductivity to the pot.

I'd love to know why both my estimates and the experiment above are wrong. Please do the maths.

BTW, it's 1h20m or 80 minutes to boiling in the reported results, not 120 minutes. That gives a total efficiency of 77% for the entire heating process to boiling based on wort temperature (although if you look at the video, he's actually drawing 2240W at that point in time).

 

[trboyden]

You are correct, I misread that. Still, 48 / 80 = 60% but I digress at this point. We don't know the final time because he didn't snap a photo of when it hit 97 - 99*C, if at all, which is the range I usually get a roiling boil in. Could have taken another half an hour or more for all we know. Time is an important factor in the efficiency calculation. So we have an important missing data point and as a result, no one here can be fully factual. at best there is a fudge factor due to the missing info.

Regardless of the math/science, I do think the experiment was a cool one and props to the OP. Do I think it's practical as a regular brew rig? no. That thing would be a bitch to clean and the pads cost too much compared to other heating options. Still an interesting project and I definitely see the uses in an apartment setting with electrical restrictions. I think a heat stick would be more practical in those circumstances though.

I'd love to know why both my estimates and the experiment above are wrong. Please do the maths.

BTW, it's 1h20m or 80 minutes to boiling in the reported results, not 120 minutes. That gives a total efficiency of 77% for the entire heating process to boiling based on wort temperature (although if you look at the video, he's actually drawing 2240W at that point in time).

 

[mysobry]

As anticipated as follow up of my tests I'm trying to insulate the pot in order to increase efficiency and heat transfer

I have selected a very special material with incredible insulation power in a small thick : Aerogel

I got some foils 5mm from UK company

The blanket seems easy to cut ad adapt but is very powdery so don't know how to wrap and insulate

Any suggestions ?

BR
Davide

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[Matreco]

I Davide
I'm thinking building a similar mash tun inspired by yours. Are you satisfied with the results? Is it feasible to apply the same solution based on silicone pads only to a dedicated 20 or 30 gallon mash tun? What do you think?

Thanks
Hugo