Heater Sizing Calculations

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Summary:

Need help sizing a heater for a particular application.
I am trying to determine the size of a single cartridge heater or 2 cartridge heaters to heat a large mass of aluminum to 400F. The aluminum block will be sitting inside an insulated (all sides) enclosure. The following are the specs for this problem.

- Max surface temp of aluminum block is 400F
- Aluminum block is 3.25in tall solid cylinder with 5in DIA
- All surfaces of block are exposed to air (inside enclosure)
- Insulated enclosure is cylindrical shape with ID of 9in and OD of 12in with top and bottom covered (essentially 1.5in of insulation all around)
- Insulation is aerogel with thermal conductivity of 0.0116 BTU/(ft*h*F)
- Outside ambient temp can get as low as 0F
- Heaters must be DC, ideally 12V
- Heaters will be snug fit into aluminum block with thermal paste on all sides

What is the wattage of heater required if I use 2?
 

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  • #2
russ_watters
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Summary:: Need help sizing a heater for a particular application.

I am trying to determine the size of a single cartridge heater or 2 cartridge heaters to heat a large mass of aluminum to 400F. The aluminum block will be sitting inside an insulated (all sides) enclosure. The following are the specs for this problem.

- Max surface temp of aluminum block is 400F
- Aluminum block is 3.25in tall solid cylinder with 5in DIA
- All surfaces of block are exposed to air (inside enclosure)
- Insulated enclosure is cylindrical shape with ID of 9in and OD of 12in with top and bottom covered (essentially 1.5in of insulation all around)
- Insulation is aerogel with thermal conductivity of 0.0116 BTU/(ft*h*F)
- Outside ambient temp can get as low as 0F
- Heaters must be DC, ideally 12V
- Heaters will be snug fit into aluminum block with thermal paste on all sides

What is the wattage of heater required if I use 2?
Welcome to PF.

You seem to have everything you need there and in particular if you understand what the thermal conductivity means it should be a simple calc. Can you take a shot at it? We don't like to just hand out answers here.
 
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  • #3
berkeman
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- Insulation is aerogel with thermal conductivity of 0.0116 BTU/(ft*h*F)
So the Al cylinder is being supported by an aerogel base? How tall?

How is your overall containment structure supported? Is it all aerogel, or is there also a metal support structure to reinforce the aerogel insulation?
- Heaters will be snug fit into aluminum block with thermal paste on all sides
Since it sounds like you are trying to insulate the Al cylinder as much as possible, it seems like the wiring to the metal heater elements will be one of the main heat sinks out of the cylinder. You may want to consider a way to include heating of the wiring to the cylinder so it is not a net heat sink...
 
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  • #4
Tom.G
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I think there is one piece of information missing, or at least not clear.

- Insulation is aerogel with thermal conductivity of 0.0116 BTU/(ft*h*F)

Is that value for the 1.5in of insulation you will be using, or for 1ft thickness, or 1in thickness?
 
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Your problem (as stated) is fairly simple. You really only need to calculate the heat leakage rate at your target differential temperature (and supply that much heat). You should note:

This will produce an 'equilibrium' solution - it does not address how long it takes to get from your initial condition to your target condition - in many practical applications, heater sizing is dominated by 'rate,' rather than equilibrium requirements. Automobile engines are a good example of this issue - try getting to highway speed with an engine capable of producing only as much power as is required to 'cruise' at highway speed.
 
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Thank you all for the replies. I've attached a PDF to show the dimensions and general look of this set up. I am trying not to over complicate it but I really want to understand how heat transfer works in this scenario. I really only need to ensure the surface of the aluminum block is 400F. The following formula is what I'm thinking but there are too many unknown variables including T_air inside the enclosure, T_insidesurface (of enclosure), and T_outsidesurface (of enclosure). I want to avoid making assumptions but maybe that isn't possible.

Watts in = Watt outs

Watts out = Convection from surface of block to air = Convection of air to inside surface of enclosure = Conduction through walls of enclosure = Convection from outside surface of enclosure to air

Do these = signs need to be addition? This is where I get confused. The rate at which the convection occurs off the aluminum block is determined by the internal air temp, which is in turn determined by the surface temp of the inside of the enclosure wall, which is in turn determined by the rate of heat conducting through the walls which is determined by the rate of convection into the ambient air.

I will ignore any additional losses out of the enclosure through wiring etc. I will just add a factor of safety to the overall number generated.

Any direction forward will be much appreciated.
 

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Watts out = Convection from surface of block to air = Convection of air to inside surface of enclosure = Conduction through walls of enclosure = Convection from outside surface of enclosure to air

Do these = signs need to be addition? This is where I get confused.
No, don't add them together. Normally what you do is, write the equation for each individual heat transfer problem (Convection from surface of block to air, Convection of air to inside surface of enclosure, etc.) and then say, at steady-state, they are all equal to the same value. This means that the heat isn't building up anywhere -- the temperatures are no longer changing with time. Steady state.

If you need to find out how the block temperature changes with time (like, what happens when you first turn the heaters on), that is a more complicated problem.
 
  • #8
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I'm assuming that this is an actual design, rather than a homework problem. There is no reason to calculate the heat transfer from the aluminum to the air in the enclosure. It is equivalent to 'measure with a micrometer, mark with chalk, cut with an axe.' If you assume that the air is at 400F and calc heat loss through the enclosure wall, you'll get an answer that is close, and will err on the 'safe' side (over-state the required power slightly). Unless you want to run a numerical model to determine the convection-forced circulation and temperature distribution throughout the enclosure (and on the surface of the slug), you're mistaking precision for accuracy. This is a good example of the difference between Engineering and Physics.
 
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  • #9
Baluncore
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Knowing the mass, thermal capacity and temperature difference, you can work out the energy that must flow into the Al. It will take some time for the Al to reach thermal equilibrium.

The wattage is determined by how quickly you want to heat the Al to the specified temperature.

If the heater wattage is too low, heat loss will exceed heater capacity, before the specified temperature is reached.
 
  • #10
Tom.G
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Once post #4 is cleared up:
https://www.physicsforums.com/posts/6283395

Let's see if we can do a walk-thru of a 'simplified but likely adequate' case. A list of other considerations is included at the end of this post.

'R'= thermal resistance in = (ft2 × Hrs × °F)/BTU (That's the reciprocal of thermal conductance)
  1. If the outer enclosure can be contacted by people, keep the temperature below 115°F. This avoids burns and allows using the approximation of natural convection for heat transfer from the case to ambient.
  2. Natural (convection + radiation) from metal or a pane of glass has an 'R' value of 1.0 to 1.1
  3. The metal skin can be ignored due to its low thermal resistance, and most likely the plastic liner can also be ignored.
  4. This leaves us with the inner surface at 400°F
  5. Assuming a temperature difference of 330°F, use the thickness, surface area, and 'R' value of the aerogel, to calculate the BTUs going thru the inner surface of the box.
  6. All those BTUs will go thru a larger surface area on the outer surface, so calculate the BTUs per ft2 on the outer surface.
  7. Now using those BTUs per ft2 on the outer surface, find the temperature rise from ambient using R=1.1 for surface to air. If this is acceptable, you're done.
  8. If the surface temperature is higher than desired, increase the outer surface area or use better insulation.

Now that you have the BTU loss, divide the BTUs by 3.42 to find minimum heater Watts to maintain temperature. (Remember to allow for heater rating tolerance and low line voltage.)

The items ignored in the above approximation are:
  • Thermal resistance of the plastic liner (as a guess around R=2)
  • Leakage at the seam between the floor and wall of the enclosure
  • If the aerogel is not foamed in place, the seams between the different pieces
  • Conductance of the plastic liner out the bottom of the enclosure
  • Conductance of wires entering the enclosure
  • Leakage around entry point of wires
  • See also the comment by @Dullard re time to reach temperature https://www.physicsforums.com/posts/6283515
Ignoring the plastic liner helps compensate for the rest of these.

Please let us know how this turns out. Oh, and if it's not giving away any secrets, what is the usage?

Cheers,
Tom
 
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  • #11
russ_watters
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I think there is one piece of information missing, or at least not clear.

- Insulation is aerogel with thermal conductivity of 0.0116 BTU/(ft*h*F)

Is that value for the 1.5in of insulation you will be using, or for 1ft thickness, or 1in thickness?
Once post #4 is cleared up:
https://www.physicsforums.com/posts/6283395
Pipe insulation thermal conductivity is sometimes expressed per unit of pipe length. This will be up to the OP to clarify where that number came from and if the source explains or gives any hint about the meaning of the units. But that's my bet.
 
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I'm confused by this units question; conductivity is Btu/hr-ft-F (as the OP showed) or W/m-K; if you want resistance use th/k where th= thickness to get hr-ft^2-F per Btu.

Or you can think of it this way

##k=\frac {\text{heat flux}} {\text{temp gradient}} = \frac {\frac{Btu}{hr-ft^2}} {\frac {degF}{ft}} = \frac{Btu-ft}{hr-ft^{2}-degF} = \frac{Btu} {hr-ft-degF}##
 
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  • #13
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Thanks everyone for the input. This has really clarified it for me.

The purpose of this design is a highly sensitive oven for gas chromatography. The surface temperature of the aluminum block is meant to maintain temperature to very tight tolerances, around +/- 0.03C, even when ambient temperature changes abruptly. The aluminum block in the diagram is very simplified to what will actually be there. There will also be some 1/16" stainless steel tubing running in and out of the enclosure adding to heat loss.

There are many variables that also play into the issue such as maximum temp of block that is allowable due to explosion-proof standards in North America, voltage size of heaters due to limitations of control board, etc. My approach to the problem was to gain understanding on how to conduct a study on heat flow in and out of an oven using a conductive heater so I could design better and actually size my heaters with more confidence.

Lastly to clarify, the value for the Aerogel came from Engineering Toolbox website. It's value is 0.02 W/mK in metric units.
https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html.

Thanks again to all who commented. I may not be able to respond much after this. I am swamped with work (including this) but am very appreciative of all the smart people out there!
 
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  • #14
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I'm confused by this units question; conductivity is Btu/hr-ft-F (as the OP showed) or W/m-K; if you want resistance use th/k where th= thickness to get hr-ft^2-F per Btu.

Or you can think of it this way

##k=\frac {\text{heat flux}} {\text{temp gradient}} = \frac {\frac{Btu}{hr-ft^2}} {\frac {degF}{ft}} = \frac{Btu-ft}{hr-ft^{2}-degF} = \frac{Btu} {hr-ft-degF}##
Yes.

For me the units issue for thermal conductivity is best reconciled by thinking of the "length" as "surface area /thickness". Otherwise I always get confused....

Incidently I get ~<40W for the heat requirement steady state worst case.
 
  • #15
jrmichler
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The surface temperature of the aluminum block is meant to maintain temperature to very tight tolerances, around +/- 0.03C, even when ambient temperature changes abruptly.
This is an extremely challenging specification to meet. If your temperature control is a sensor in the block, it will not sense changes in ambient temperature until after the block starts to change temperature. That is a lag in the control system, plus there is another lag from the heater to the sensor. Control systems lags cause temperature oscillations unless the control system gain is low. Low system gain causes a lag in response to ambient temperature changes. There are limits to how accurately a system with lags can be controlled.

I suggest reducing the need to respond to ambient temperature changes by reducing the total heat loss. Use a Dewar flask instead of aerogel insulation. Dewar flasks are made in many sizes and shapes, so you should be able to find one that meets your needs. I see that Amazon sells a large selection of Dewar flasks.

Further reduce the response to ambient temperature variation by a thick insulated lid. Make the lid of two pieces of insulation with an aluminum plate in the middle. Put a heater and temperature sensor in the plate, and control its temperature to the same temperature as the aluminum block.

Minimize the heat gain through wires and tubes to the block. Make wires the smallest practical diameter, use the smallest tubing with the thinnest wall practical, and coil them through the insulated lid to get longer length.

Now that the heat loss is as low as practical, calculate the steady state heat loss. Determine the longest practical time for heating up (over a weekend would be ideal), and calculate the heat rate required. Size your heating element to be just barely large enough to meet the total heat requirement. Now add a temperature sensor and a feedback controller.

Hint 1: Controlling temperature to +/- 0.03 C requires a temperature sensor accurate and repeatable to at least +/- 0.01 C, and ideally accurate and repeatable to +/- 0.003 C.

Hint 2: Keep in mind that the resolution of a digital display does not tell you the accuracy of the sensor and associated electronics.

Hint 3: Accuracy and repeatability are two different things. They are specified separately.
 
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