# How to make a low voltage heating element?

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by cgk009, Dec 13, 2010.

1. ### cgk009New Member

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I am tasked with a project to create a heating element using low voltage, such as batteries, to make a heating element to approx 100 degrees that will last for approx 30 minutes. It has to be approximately the size of a deck of cards.
Any suggestions would be helpful especially how to use a standard resistor to make a consistent 100 degrees.

Thanks

2. ### crutschowWell-Known MemberMost Helpful Member

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The heat generated by a resistor is simply V²/R = W so it's just a matter of substituting the battery voltage to determine the resistor value you need for the desired watts. The resistor could be a discrete power resistor or just some high resistance wire, such as made from iron or nichrome.

Knowing the watts you need is the problem. You may have to experimentally heat the device with a power supply and resistor, and measure the temperature, to know how much power is required.

To regulate the heat you could use a simple snap action thermostat designed for the desired temperature to turn the resistive element on and off. For close control you could use a thermistor to trigger a comparator on and off which, in turn, applies power to the resistor.

3. ### #12New Member

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When I needed a high temperature test chamber, I used a hollow ceramic wirewound resistor. Plenty of room inside to place the D.U.T. (device under test) and a thermocouple. That's the easy part. Making the temperature "consistent" is more difficult. How consistent? Within a degree? A tenth of a degree? F, C, or K? While I'm at it, insulation is a good thing. A tiny amount of power can make terribly high temperatures if there is no escape route.

Be more specific and you will get better answers.

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5. ### cgk009New Member

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Thank you for the responses so far.

I need it to be 100 degrees Fahrenheit plus/minus 3 degrees.

Last for approx 30 minutes.

Use 9volt batteries unless you advise a different battery.

What type of resistor?

Thank you

6. ### Mr RBWell-Known Member

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What is the load? How much heat will be drawn off?

100'F is only about 37'C so you can use an electronic temp sensor like a LM335 and a comparator with a transistor as the heating element.

The main problem will be the load and how much power it requires.

7. ### BoncukNew Member

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What is the medium to be heated and what is the volume?

8. ### cgk009New Member

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I need it to be 100 degrees Fahrenheit plus/minus 3 degrees.

Last for approx 30 minutes.

Use 9volt batteries unless you advise a different battery.

What type of resistor?

It will be heating 2 liters of water through small plastic tubing over 30 minutes.

Size: is approx the size of a deck of cards with heating element and batteries.

Thank you

9. ### #12New Member

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First calculation shows you will need 38.68 watt-hours to heat 2Kg of water from 70F to 100F (not including loss of heat into the surrounding air).
A nine volt battery is advertised at 440ma-hr and this would convert to 3.96 Watt-hours if the battery voltage did not become lower as the battery discharges.
Therefore, you will completely empty more than 10 batteries on this job.

Do you still want to use 9 volt batteries?

10. ### cgk009New Member

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I do not have to use 9volt. I need to use something small but open to any suggestions.

Thank you for the response.

11. ### RMMMNew Member

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Lithium Ion batteries.

Even D-cell NiMH or NiCad will help will do better than a 9v. 9v "transistor" batteries are designed for VERY low draw applications. They should not be used as the main "load" supply.

They are good to switch a transistor on and off for a few years, but any "decent" load will kill them quickly.

12. ### #12New Member

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Still sneaking up on this...

I am guessing 2 liters of water won't fit in a deck of cards, so you must have the water flowing. Do you have 30 minutes to get to 100 degrees? No. You want to get there quickly and "maintain" the temperature. Right?

A 12 volt lead acid battery with about 4 to 5 amp-hours in it would carry enough energy and weigh about 2 kilograms (same as the water). To heat the water in 3 minutes would require 61 amps of current. I don't know if a lead acid battery will discharge at 12 or 15 times its C rating and stay healthy. (A little help here, fellows.)

Anyway, the smallest battery that will work is about 2 liters, and I think we just violated the "small" requirement.

13. ### cgk009New Member

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the heating element and batteries needs to be card size. I will be running 2L of water through the card size element through small tubing.

14. ### MikeMlWell-Known MemberMost Helpful Member

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And you have negotiated a reprieve from the Laws of Thermodynamics????

15. ### #12New Member

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Simply repeating that you want it small will not make that possible on this planet at this time. You will either have to change your requirements or go into to the future and get a really tiny matter to energy converter. I prefer that you change the requirements, but the other option would make life easier for me.

16. ### MosaicWell-Known Member

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Some calcs involved.

A battery wont do it. U need some serious fuel. U must construct a fuel cell. Paraffin wax packs 42 kJ per gram at .8g/mL. A pack of cards is about 71mL or 56g of wax.

The Specific Heat Cap of water is 4.2 KJ/Kg/deg C. A litre of water is 1 kG so u need 8.4 KJ per deg C. If your ambient is 25 deg C , the delta Temp is 37-25 = 12 Deg C.
Which is 8.4 * 12 = 101 KJ of energy required. Estimating heat transfer at 70% we have 100/70 *101 = 144 KJ required.

To get 144 KJ u need to burn 144/42 = 3.5g of wax. this uses3.5/56 = about 6.25% of the pack of cards. Double this to provide for the flame & air intake leaves u with about 87% of the space for the valve and valve controls. Although the valve is a 12V device , it will pull with a fresh 9VDC alkaline source.

The 9VDC batt occupies about 18 mL of space.
The valve is 3.2cm x 1.9cm x 6.2 cm = 38 mL volume

So far we have 9 ml for the wax/flame, 18mL for the Duracell 9V, 38mL for the valve.
That's about 65mL. Add in another 2mL for an SMD LM358 dual opamp, thermistor, flywheel diode, & MOSFET driver and a couple resistors and you have your control for the valve.

Total 67ml, or about 94% the size of a pack of cards. I'd use the remainingg space to apply thermal insulation to keep the heat transfer efficiency up.

Setup device so that multiple wicks from the wax heat the flow valve. Heat the aluminium part of the valve that houses the flow channels. U can use the tiny NPT fittings supplied. The valve is good to 122 Deg F

Use a valve like this one:
EBC Electronic Boost Control Solenoid Kit DIYAutoTune.com

It is a 3 port valve, so u can throttle the valve using an opamp to achieve a flow of 37 deg C. Basically, set it up so that if the outlet water is under the tgt temp, u throttle the valve towards recirculating the water to the source.

Of course, the valve is entirely off (recirc) during the first heating phase until the entire 2L source has reached perhaps 36 Deg C. Then the valve is opened fully to flow to the target. Remember the fuel is sized to run out when the desired temp is achieved. So no chance of overheating.

http://www.sciencebyjones.com/heat_of_combustion.htm

Last edited: Dec 14, 2010
17. ### #12New Member

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Very interesting! Apparently electronics people rarely think of burning something to get heat. Can you elaborate on how the burning wax heats a valve that is rated at a maximum of 122 degrees F in a chamber the size of a deck of cards, and how the flame heats the 2 liters of water at the same time?

18. ### MosaicWell-Known Member

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Certainly!

In order to flow water there must be a pump.

The pump pushes water through the 3 way valve which when closed recirculates the water back to the 2L reservoir. The flame is heating the aluminum valve and that heat is transferred to the flowing water and cooling the valve at the same time.

When the flowing water achieves , say, 36 to 38 deg C. the op amp comparator/mosfet opens the valve outlet and closes the recirc. feed. Thus the water is now fed at the correct temp to the tgt destination.

In terms of the chamber, I have allocated 4.5 ml for flame size/air intake and that exceeds a normal candle flame. For a faster heat generation camphor is an alternative as a fuel source but the calcs would have to be done for the energy density of camphor.

Because the system is designed to circulate water with the valve unpowered it is relatively failsafe. If the pump fails and the candle is burning the valve may overheat.

Last edited: Dec 14, 2010
19. ### #12New Member

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Excellent. I hope the original poster can use the idea.

20. ### DerStrom8Super ModeratorMost Helpful Member

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I'm assuming you don't want to use a wall wart?

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