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A bunch of engineers are better than one...battery powered MICA/carbon fiber heating element

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JohnnyDC

JohnnyDC

Civil/Environmental
Apr 1, 2015
9
Hello everyone,
I'm fairly new to the forum and I mostly deal with Civil engineering issues as I work with a developer, but I really need help determining something and I'd love to hear your thoughts. I don't want to milk the system, so if anyone has any questions regarding civil engineering I will try to help.
I have a scenario that I need some combined brain power to resolve.
SCENARIO: Extremely safely heat a small (12" x 20" ) thin wafer heating element (MICA or carbon fiber, etc.) with a small sized rechargeable battery source. Must be very safe, no overheating, no explosions, no overheating of battery pack, and as lightweight as possible and I must utilize components already available in bulk production such as slim line 7V batteries, MICA heaters, etc..
-must reach heat setting extremely quickly (would like to hear sample time frames in your suggestions)
-heating element encased on plastic mold, forming thin plastic sheet with small battery pack attached to it
-"low" heat in the 90's F, "high" heat around 105+-F
-Must reach desired heat setting and hold for 20 minutes+-, then will automatically shut off and not be used for many hours
-Need to get numerous uses from one charge
-battery pack needs to be in flat linear arrangement and can be the width of a single battery and as long as 15-20 inches(basically, can't stack the batteries, must align them in a straight line)...and remember, as light as possible
-Does not have to cool quickly
-battery pack will be inches away from the heating element
-will also need to power battery life display, and low,high light
-finished product will be form molded plastic encasing battery (with charging port exposed) and heating element with controls, etc.
-finished product will be somewhat rigid, but will require slight flexing ability but will not be folded or creased, etc, like clothing

Questions (would really like the detail here so I can understand it better):
1.) What is the smallest type and size battery pack that would provide numerous uses (20 minutes of usable heat with 3-6 hours of downtime) on a single charge (7V, 12V??)?
2.) What would discharge current and wattage of this most efficient battery be? Voltage and capacity? Working current and peak current? Anything you can think of that I'm missing...
3.) What would be the most economical flat wafer heating element to use in this application (carbon fiber, Mica, aluminum sheet, etc.)?
4.)What would be the ultimate small rechargeable battery source paired with the ultimate wafer type heating element source that is currently on the market that could easily and safely be paired together and get the requiered output? Looking for point and click ideas that you would suggest if you had to pair these two items together today from readily available extremely economical products.
 
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1> You need to define the power requirements first. Do you know how much power is required to achieve the temperatures you're asking for?
2> Mostly moot, since you need something on the order of >4 hr of continuous battery life, which means its discharge rate can't be any higher than 1/4C
3> Higher voltage and less current is better for higher efficiency of heating
4> Lithium polymer (LiPo) probably, but even conventional lithium batteries like laptop batteries would be reasonable, like: If you can get by with less than 15W, it should last just about 4 hr

> How fast is "quickly?" For some, 1 sec is fast, for others, 1 us is fast
> What are you trying to heat? Does what you're trying to heat need to reach temperature "quickly?"
> How are you controlling heat loss? If you can reduce convective heat loss, you can get by with less power
> What ambient temperature range is required? This has a direct impact on power required to maintain a given temperature

TTFN
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7ofakss

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As IR points out the most important thing is the energy needed to complete the task. Some sort of estimate is needed. For instance, if you're only going to heat up some thin paste you won't need a lot of energy.

Energy = power x time

But if you're heating a pot of water sitting on the heater you might need a huge amount of energy.

Typically you would locate a planar heating element of any kind rated at any usable voltage and mount it in a suitable frame as a mockup of your desired mechanical setup. Then you'd run it using whatever power it needs, say, 120Vac with a Variac. From this setup you can characterize the exact energy you need and the power required for the allotted time frame. From that ALL the battery requirements will fall out of the test.

That done, you'd take the battery situation presented and back into the actual heater element requirements.

Keith Cress
kcress -
 
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IRStuff/ITsmoked/MacGyverS2000,
Thanks so much for your responses. Sorry for the delay, was out in the field yesterday.
Here is a little more information. We are trying to make a temporary warming tray for some small molds we are making that will also need to be manipulated by hand at a certain temp. These small pieces sit in this tray that radiates 100 degreesF while being treated for another step.
The tray itself is going to be a very thin tray. Basically just enough hard plastic to cover the thin sheet heating element. The heating element will be about 12 inches by 20 inches, and plastic will be molded around the wafer heating element. It essentially looks like a school lunch tray...very thin. The battery pack has to be small and also incased in the mold with only access points being the charging point (and we would be able to see battery power reading and temp reading displays that will be built into the tray. So the battery pack would be built in, not accessible, and on one side of the tray molded in.
So, radiant heat of 100 degrees only passes through a very thin sheet of plastic, which serves as the top of the tray. Doesnt matter if the molded pieces actually stay at 100 degrees, as long as the radiant heat is 100 degrees.
**Also, we only need the heat for 20 minutes. The heat will automatically turn off after 20 minutes and it won't be used again for about 6 hours...so not really a long time to heat the tray which is why we were thinking a typical 7V could work. Heat needs to be at 100 within a few minutes.
There is no heat loss control...there may be 2-3 small pieces on the tray or there may be 15.
We will have a temp thermostat withing the heating element controls that will hold the temp of the heating element to produce approx 100 degrees radiant heat. Also, heat can fluctuate 5+- degrees, so temp doesn't have to be exact.
Basically, +-100 degrees from a very thin sheet heating element encased in a molded plastic tray, need the heat for 20 minutes, will be off for 4-6 hours at a time, would like to get numerous uses out of a single charge, and we would like to make the tray from a custom battery pack that will also be encased in the mold (for example, a 7V battery pack that the cells are in a straight line instead of stacked, so it will be a long low profile battery pack), and likely a thin MICA heating element.
What I don't know is how much power is required to run a small MICA heating element to heat it to 100 for 20 minutes, what the discharge current and wattage should be, voltage, working current/peak current, etc..
I know it's probably something that's better figured out in trial and error, but a good starting point is what we're looking for. In order to make a custom battery, we have to provide the information mentioned above. Thanks so much for all of your help.
 
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Oh, and this is done via battery pack because it will sometimes be done offsite with no other access to power.
 
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Here is an example of a battery pack and heating element that would work well with regards to size and possibly heating capabilities:
Here are the stats on the battery packs for the coat application:
Capacity: 1700mAh (1.7Ah)
Volts: 3.7V (3.7 Volt)
Type: Li-ion
2 of these run coat heaters for Columbia Jackets

Here is an example of the heat element and battery with charger pack:
 
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It looks like Columbia recalled most of those jackets and batteries, and I can't even find anything their site relating to that product, but that's neither here nor there.

The Columbia battery has a total energy 6.37 Whr, while the laptop battery I cited is 97 Whr, so roughly 15x more energy.

OK, so what's the ambient temperature? Assuming that it's about 70F, you'd need to sustain about 26W of heater power to stay at temperature, and that's assuming the bottom is insulated so that all the heat goes through the top. Otherwise, you'll need to double the power. So those Columbia batteries are way undersized for this application. If the bottom is insulated, the laptop battery I cited would last for at least eleven 20-min heating cycles. You could possibly get more, since you're not necessarily restricted to the voltage limits that a laptop has to have.

With the Dell battery, we can assume a minimum of say, 9V, at which you'd still want the same 26W, so that requires 3 ohms of total heater resistance. Minco has a bunch of foil heaters. Whatever size and resistance combination results in something on the order of 2.5 ohms that would leave some space for the power control.

TTFN
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7ofakss

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Note: If you plan on using a laptop battery of the more recent ilk (LiPoly, et. al.), plan on having a battery controller circuit. They will not provide you juice until it starts talking to one. Gone are the days of grabbing an old laptop battery and connecting directly to the +/- posts.

Dan - Owner
Footwell%20Animation%20Tiny.gif
 
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This is all great information, especially the Minco size chart. I hope everyone had a good weekend and I'm just getting back into the swing of things. I really wish I could make the calculations you guys are able to make, but then that would mean I actually had a high level of understanding of watt/ohm/voltage, etc.. and unfortunately I don't. I can give an example of the heater element (very similar to the Minco), and I can also safely assume we won't be able to use a laptop battery because of the manufacturing. The battery will likely be manufactured through a custom application overseas and we'll just give them our requirements. Maybe that will help with some unanswered questions I've neglected to provide the initial information on. Another key feature to these parts will be the need to mass produce once a prototype has been completed. The custom battery packs can be produced very cheap, as well as the foil heaters overseas. The battery pack needs to be narrow and long...like a roll of C batteries in a line and it can be 10-12 inches long.
5-10, 20-min heating cycles would be sufficient. The main goal is to keep the battery pack small and narrow, with just enough battery power to safely heat for 5-10+- cycles. Some laptop batteries fit the shape, but it cost to much for our purpose because it has to many additional features that make it a "laptop battery".
Here is a simple two terminal Li-ion 1800mAh 11/1V battery pack that I found that looks like (shape/dimensions) could fit the application:

Heating element:

Basically, to get a prototype I need a long narrow battery pack, which can be customized, that will power the flat sheet heating element, which can also be customized. Where I am having great difficulty is matching a good pair to fit my needs. Would that battery mentioned above do the trick if paired with one of those heating elements mentioned above?
 
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We would need to consult with that manufacturer to coordinate the effort to get the raw battery plus the necessary electronics for a safe charge, and I believe many of these custom battery manufacturers are able to do that. We'll also have to do the same for the heating element, although most of the manufacturers I've found are able to include some type of very basic built in thermostat to limit heat output to +-5degrees from required temp.
The other items we will need to coordinate are the simple battery power display, the "on/off", "low", "high" heat functions, and the autoshutoff electronics. All of which will be in close proximity to people working on these trays, so all must be very safe and pass all property safety guidelines.
 
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What we wish and the laws of physics frequently collide and conflict. Your new cited battery has a total capacity of 1.8Ahr*11.1V = 20Whr

My estimate for your heating requirements was 26W. 20Whr/26W = 46.1 min, which is barely two cycles; For 10 cycles, you will need 5 of these battery packs, which is at least as much volume as the Dell battery, which has 96Whr capacity. Regardless of what you want, the end result is that you will need the volume of at least a laptop battery, since wolumetric energy density for a given chemistry is pretty much fixed.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com:
 
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IRstuff, what is the conversion you use to arrive at 46.1 min from Whr/W (20/26)? Please excuse my ignorance in this arena.
 
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19.998 Whr/26 W = 0.786 hr = 46.1 min

Whr = watt*hour --> energy


TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529


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We could work with 63Whr and shorten the run cycle to 15 minutes to get more cycles. check this one out, 4.4A, 14.4V, 8 cell:
Our biggest problem is getting it narrow and short enough to fit within the tray. A short battery pack without all the external attachments to fit a laptop would definitely work.
 
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IRStuff,
The manufacturer sent over the exact specs and we feel it will work well as far as physical dimensions. Also, it's around 74wh which we can make work. There is a question that I have regarding the stas shown below:
14.8V 5000MAH Li-ion battery pack
Max continuous current: 3-3.5A
Does the 3-3.5A max continuous current affect our particular scenario?
 
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Generally, you need to be below about 1/5 of the ampacity in A*hr to reliably achieve the rated capacity, i.e., 0.5A load. I would derate the battery capacity by at least 20% to provide margin.

Note also, most discharge curves are constructed for constant current loads. Your heater is not constant current; as the battery voltage drops, the load current will drop correspondingly. This means that there might be some duration where the heater is not fully powered, but still heating to some degree, past the "official" point of "discharge."

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com:
 
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