Ideas for battery powered heater wanted.

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ptribbey
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Ideas for battery powered heater wanted.

Post by ptribbey » Fri Jun 30, 2006 12:25 pm

I am working on a project where i need to heat 200cc of water from about 90f(+1 -5F) to 99f in less than an hour. One time use. It should be able to be run off one 9v or one d or c cell or possibly a couple aa cells.
I have some ideas of my own but I always enjoy other peoples take on a problem.
Thanks
Paul

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MrAl
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Post by MrAl » Fri Jun 30, 2006 1:10 pm

Hi there,

You can use a power resistor or a goup of smaller resistors as
the heating element, so you can adjust resistance for whatever
source you want to use.
You can also use an inverted light bulb with most of it under
the water and the base above water level.
LEDs vs Bulbs, LEDs are winning.

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Bob Scott
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Post by Bob Scott » Fri Jun 30, 2006 2:58 pm

Mr. Tribbey,

I was just thinking that I'd have to break out my old grade 9 school physics text in order to refamiliarize myself with exactly how much energy is needed to heat the water that much, using both CGS and FPS sytems 'cause you're mixing farenheit degrees and cc's. Hmm calories to joules to volt-amp-hrs. I have an idea:

Pour the 200cc of liquid into a small sealed container. Now if you hold a 9V battery in your right hand and hold the sealed container of liquid under your left armpit for half an hour, the liquid should rise to 98.6F, which when rounded off to 2 significant figures is pretty close to 99F.

What do you think?

Bob :cool:

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jwax
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Post by jwax » Fri Jun 30, 2006 6:52 pm

Put a pinch of table salt in the water, drop in a 9 volt battery and a thermometer, and remove when the temp hits 99 F!
I like Bobs' armpit solution too! :grin:

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Chris Smith
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Post by Chris Smith » Fri Jun 30, 2006 7:20 pm

You can do the simple math.

Will how many batteries at this wattage/ calories/ Joule do the job?

Batteries have how many amp hours, wattage is volts and amps, heat is efficiency plus loss.

1 cal = Energy needed to raise the temperature of 1 gram of water 1 degree Celsius

http://www.ac.wwu.edu/~vawter/PhysicsNe ... fHeat.html

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jollyrgr
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Post by jollyrgr » Fri Jun 30, 2006 7:33 pm

Some figures to work out....

200cc of water is 6.76 ounces or 0.053 gallons US.

One BTU will raise one pound of water by one degree F. This is 251.9 calories or 1055 joules.

Water weight 8.5 pound per gallon.

You have 0.4505 pounds of water.

You want to raise it by 9 degrees. (90F to 99F)

For a pound of water this would take 9 BTUs. For your slightly less than half a pound it will take 4 BTUs (some round off dropped). This is 4220 joules.

One joule is 2.778 x 10 E-4 watt-hours.

This means you need 1.17 watts for this to occur in one hour. Slightly more than that for less time.

Suggestion. Get a light bulb like those used in the low voltage outdoor lights. These come in 4W, 7W and larger settings. The equivalent would be a 194 automotive bulb found for cornering lights and insturment panels. I doubt that a 9 volt battery would last long enough to do the job. I'm guessing these have a mAH (watt) power rating of about 100mAH. This is just a guess. You'd probably need something more like 8 D cell batteries. Another choice would be to use a rechargable battery pack like those found on cordless drills.

If anyone would be so kind as to check my calculations, I would not mind.
No trees were harmed in the creation of this message. But billions of electrons, photons, and electromagnetic waves were terribly inconvenienced!

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Chris Smith
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Post by Chris Smith » Fri Jun 30, 2006 8:18 pm

A 9 volt is 250ma or less.

Mine are 225. [rated]

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MrAl
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Post by MrAl » Sat Jul 01, 2006 4:57 am

Hello there,

My 9v alkalines are 500mAh each.
Maybe a 50 ohm resistor or five 10 ohms in series.

There will no doubt be a small amount of lost heat while
the water is heating up. What wasnt mentioned yet
was the ambient temperature, which could make a
huge difference in the amount of power needed.
If you are doing this in sub zero temperatures you'll
need much more power to accomplish the same thing.
LEDs vs Bulbs, LEDs are winning.

Will
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Post by Will » Sat Jul 01, 2006 10:05 am

I think Jolly Rog's calc nailed it fairly precisely (my calc said 1.16 Watt-hour) so if you could get 250 mA out of a 9.0 battery then you would do it in little over a half hour. The problem is (I have no experience of this) that if you were taking 250 mA out of a 9.0 volt battery you might be expending too high a proportion of the available energy on the internal resistance of the battery ? Chris says his batteries are 250 mA but I think he means 250 mAh. You could get a lantern battery (6 volt about 5 inches by 3 inches by 3 inches but these are about five bucks or something ?
BB

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ptribbey
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Post by ptribbey » Sat Jul 01, 2006 10:53 am

Thank you all for your ideas.

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jwax
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Post by jwax » Sat Jul 01, 2006 11:35 am

Curious, ptribbey- What ya building? :?

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Chris Smith
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Post by Chris Smith » Sat Jul 01, 2006 1:04 pm

Yes 225 ma/h

My batteries are rated at 500ma/h, but the voltage drops very fast at this rate.

When using it at 225 ma/h, the 9 volt rating is fairly constant and lasts many more hours at this given.

At present I get a full 20 hours with the power off, and Im into changing the 9 volt batteries into a six pack of AA batteries for almost five times the term. I need longer protection.

Five times the draw will suck the battery down more than 7 times faster.

You need to find a battery and use the standard chart for joules, resistance and time to figure out over time if this battery totally will deliver the calories for your needs.

Depending on total draw and time, will affect all batteries, but their values are calculative using the joule.

http://www.duracell.com/oem/Pdf/MN1604.pdf

If you need a good source of 9 volt batteries [MN 1604] to experiment with, I have several hundred pulls out of service.

A lot of them are in peak condition with a full 9 volt rating, while many are lesser, and others are depleated.

You can start with a 9 volt value using one of these units, use a given load, a short exact time, and start your value charts.

Instead of wasting $5.00 a pop on new ones for your experiment, I can send you a load of them with a 9 volts value [and if necessary 8 volts] with shipping rates at far less than the price of just one?

Even if you purchase them in bulk like I do, the cost goes up exponentially. PM me for a price if you want some used ones?

[real cheap]

k7elp60
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Post by k7elp60 » Sat Jul 01, 2006 5:42 pm

Perhaps a C size alkaline and a resistor may work, a the normal C size alkaline has a capacity of 7AH(amp-hours. So a 1.5 ohm resistor would
produce 1 watt/hour.

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ptribbey
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Post by ptribbey » Mon Jul 03, 2006 10:21 am

jwax, its a medical device for a physician friend. I dont know all of the details.
Chris, Yes, I would like to buy some 1604's from you. do you subscribe to paypal?
Again thanks for all your input.
This is what Ive learned so far:
The specific heat of water is 1 btu per pound for change in degrees f.
300 cc of water weighs .63 pounds.
9 degree rise times .63 pounds equals 5.67btu.
5.67 btu equals 1.66 watt hours.
do-able with a 250 mah 9v battery and nichrome wire.
I bought some 36 g nichrome to play with from ebay today.
say i form up 36 ohms worth of wire and connect to a fresh 1604. It will draw about 250 mA.
How do I calculate the useful time from the battery? If it is rated for 250mAh, is the useful time 1 hour? just checking.
Thanks
Paul

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Post by dyarker » Mon Jul 03, 2006 12:26 pm

20 to 30 minutes ... maybe

Divideing 250mAHr by 250mA does equal 1Hr, but batteries are commonly rated based on a 20Hr discharge time. Under heavy load losses to battery internal resistance increases, and over-heating caused by the resistance shortens the battery life even more.

see spec for 9V battery at http://www.panasonic.com/industrial/bat ... _Sheet.pdf
Dale Y

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