making heat with a peltier

bowman1964 · 07-12-2003, 06:26 AM

ok guys i need some imput on this.

i have a 172 watt pelt that i use to make heat for a load block for testing evaperators.
now i know right now i have 13.4v x 6.0 amps =80.4watts now add the heat made during this process you get 13.4v x 6.0a x 1.4 now this = 112.56 watts. but it has a water cooled cold side to give more heat to the block but it is too hard to caculate the total heat output excaty.

now i have 2 brand new 130 watt peltiers laying around also.
what would be the best way to make large amounts of heat and me able to get a precise heat output.

i thought of using the 2 130's in a side by side plate and just let the cold side reach 0 dt and then i can just caculate volts x amps.

or what if i run them stacked would that be better.
stacking them i think? will work with lets say 13.4v x 6.0a x 1.4 + the 13.4 x 6.0a of the second one giving me a totel of 192.96 watts I THINK?

any ways any ideas?

ok tell me some ideas on how to set these up to get a fixed heat output .

thanks

Since87 · 07-12-2003, 07:10 AM

You need power resistors or something designed to be a heating element. It's always going to be a pain figuring out how much power is coming out of a pelt setup.

Keep it simple.

bigben2k · 07-12-2003, 08:03 AM

I have to agree with Since87: that 1.4 factor could be off by 0.1, which means that you've got a 7% margin of error right there (all figures theoretical).

To measure the cold side, you'd need a thermal probe at the inlet and outlet, and if you measure those temps to within 0.1 deg C, which will be expensive, you'll still have a +/- 0.2 error margin, on a measurement that should turn out to be about 0.6 degrees, so an added ~10% margin of error on the total (all figures are approximate).

The heater is much simpler: measure the current and voltage, within 1% (or better), and you've got yourself a power measurement that's within 2%.

bowman1964 · 07-12-2003, 09:00 AM

ok what if i tried to use these resistors, they will gererate a specific heat.but i am not sure if they will be enough

bigben2k · 07-12-2003, 09:12 AM

A much better choice.

Dan from www.dansdata.com uses a similar resistor, for his testbed. Now all you have to do is get a multimeter that's fairly accurate, and measure the voltage and current, to calculate the actual power.

Oh yeah, you also have to insulate the heck out of the assembly, because you want to make sure that all the heat goes to the waterblock, and nothing else.

bowman1964 · 07-12-2003, 09:19 AM

well these power resistors will pull 2.69amp's each at 13.59v

bigben2k · 07-12-2003, 09:34 AM

Quote:

Originally posted by bowman1964
well these power resistors will pull 2.69amp's each at 13.59v

P=VI, so 13.59 * 2.69 = 36.56 Watts.

redleader · 07-15-2003, 05:47 PM

Heres what you do for DIRT CHEAP heating. Forget waterheating elements. Goto wallmart and buy a rice cooker. I got one they sell for 14 bucks IIRC. Inside was a 330w, 120v AC heating element enclosed in side an Aluminium hot plate. You could vary the power, easily measure dissipation and heat water without risking electrocuting yourself

bigben2k · 09-02-2003, 02:58 PM

Redleader, what size is it, once you're ripped it out?

07-12-2003, 06:26 AM	#1
bowman1964 Cooling Savant Join Date: Jul 2002 Location: virginia usa Posts: 126	making heat with a peltier ok guys i need some imput on this. i have a 172 watt pelt that i use to make heat for a load block for testing evaperators. now i know right now i have 13.4v x 6.0 amps =80.4watts now add the heat made during this process you get 13.4v x 6.0a x 1.4 now this = 112.56 watts. but it has a water cooled cold side to give more heat to the block but it is too hard to caculate the total heat output excaty. now i have 2 brand new 130 watt peltiers laying around also. what would be the best way to make large amounts of heat and me able to get a precise heat output. i thought of using the 2 130's in a side by side plate and just let the cold side reach 0 dt and then i can just caculate volts x amps. or what if i run them stacked would that be better. stacking them i think? will work with lets say 13.4v x 6.0a x 1.4 + the 13.4 x 6.0a of the second one giving me a totel of 192.96 watts I THINK? any ways any ideas? ok tell me some ideas on how to set these up to get a fixed heat output . thanks __________________ xp2800barton@2712 226x12 dual bank 2x256 twinmos pc3200 8RDA epox full mods.vdd,vdimm,vcore. custom r502 cooled unit.CPU running @-18C GPU -21C Chipset -20c only one like it.in one case. custom r502 cooling unit 3D2001SE 21315 3DMARK03 6659

07-12-2003, 08:03 AM	#3
bigben2k Responsible for 2% of all the posts here. Join Date: May 2002 Location: Texas, U.S.A. Posts: 8,302	I have to agree with Since87: that 1.4 factor could be off by 0.1, which means that you've got a 7% margin of error right there (all figures theoretical). To measure the cold side, you'd need a thermal probe at the inlet and outlet, and if you measure those temps to within 0.1 deg C, which will be expensive, you'll still have a +/- 0.2 error margin, on a measurement that should turn out to be about 0.6 degrees, so an added ~10% margin of error on the total (all figures are approximate). The heater is much simpler: measure the current and voltage, within 1% (or better), and you've got yourself a power measurement that's within 2%. __________________ WBTA (Water Block Testing Alliance) Nordic Hardware WC 101

07-12-2003, 09:12 AM	#5
bigben2k Responsible for 2% of all the posts here. Join Date: May 2002 Location: Texas, U.S.A. Posts: 8,302	A much better choice. Dan from www.dansdata.com uses a similar resistor, for his testbed. Now all you have to do is get a multimeter that's fairly accurate, and measure the voltage and current, to calculate the actual power. Oh yeah, you also have to insulate the heck out of the assembly, because you want to make sure that all the heat goes to the waterblock, and nothing else. __________________ WBTA (Water Block Testing Alliance) Nordic Hardware WC 101

07-12-2003, 09:19 AM	#6
bowman1964 Cooling Savant Join Date: Jul 2002 Location: virginia usa Posts: 126	well these power resistors will pull 2.69amp's each at 13.59v __________________ xp2800barton@2712 226x12 dual bank 2x256 twinmos pc3200 8RDA epox full mods.vdd,vdimm,vcore. custom r502 cooled unit.CPU running @-18C GPU -21C Chipset -20c only one like it.in one case. custom r502 cooling unit 3D2001SE 21315 3DMARK03 6659

09-02-2003, 02:58 PM	#9
bigben2k Responsible for 2% of all the posts here. Join Date: May 2002 Location: Texas, U.S.A. Posts: 8,302	Redleader, what size is it, once you're ripped it out? __________________ WBTA (Water Block Testing Alliance) Nordic Hardware WC 101

07-12-2003, 07:10 AM	#2
Since87 Pro/Guru - Uber Mod Join Date: Sep 2002 Location: Indiana Posts: 834	You need power resistors or something designed to be a heating element. It's always going to be a pain figuring out how much power is coming out of a pelt setup. Keep it simple.

07-15-2003, 05:47 PM	#8
redleader Thermophile Join Date: Jun 2001 Location: The deserts of Tucson, Az Posts: 1,264	Heres what you do for DIRT CHEAP heating. Forget waterheating elements. Goto wallmart and buy a rice cooker. I got one they sell for 14 bucks IIRC. Inside was a 330w, 120v AC heating element enclosed in side an Aluminium hot plate. You could vary the power, easily measure dissipation and heat water without risking electrocuting yourself

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