![]() | ||
|
![]() |
#1 |
Cooling Neophyte
Join Date: Jun 2004
Location: USA
Posts: 11
|
![]()
I am toying with the idea of moving to pelt power for my socket A system.
I am wondering how hard is it to get the grease off? Is there is an easy to remove subsitute for dieletic grease if not? I ask this because I only want to use it during bursts of overclocking fun. I am gonna slap the pelt onto some old socket 7 stuff first to hone my anti-condensation skills. How much power do I need to run this thing in the first place, at 12 volts? Its the dangerden 226w model. I plan on going with the ghetto method. (Article on this site) using a parrell PSU setup, so that is why I asked. |
![]() |
![]() |
![]() |
#2 |
Cooling Neophyte
Join Date: Jun 2004
Location: USA
Posts: 11
|
![]()
I see how it is. Make me find it myself eh?
Anyhow, here is the answer to my question (for insulation removal anyways) http://www.phase-change.com/articles...on/Image19.jpg Electronics cleaner, nuff said. Now, can someone please tell me how much amps I am going to need at 12v for the DD 226w? |
![]() |
![]() |
![]() |
#3 |
Cooling Savant
Join Date: Mar 2003
Location: MO
Posts: 781
|
![]()
Got a link to the pelt? Helping is fun, but don't make us work too hard to help ya.
Electronic cleaners, yeah. Or any xylene based cleaner. |
![]() |
![]() |
![]() |
#4 |
Cooling Neophyte
Join Date: Jun 2004
Location: USA
Posts: 11
|
![]() |
![]() |
![]() |
![]() |
#5 | |
Cooling Savant
Join Date: May 2003
Location: Okotoks, A.B. Canada
Posts: 726
|
![]() Quote:
12V x ?Amps= 226Watts ?Amps = 226Watts/ 12V =18.8333Amps edit: Now looking at the link Vmax = 15.2 Volts same formula at 15.2Volts = 14.875 Watts so if the Amps is constant and you lower the voltage...to 12 volts you would be usig 178.5 Watts.. i think If i'm wrong.. someone will correct me for shure...
__________________
"Great spirits have always encountered violent opposition from mediocre minds" - (Einstein) |
|
![]() |
![]() |
![]() |
#6 |
Cooling Savant
Join Date: Mar 2003
Location: MO
Posts: 781
|
![]()
You're wrong.
![]() The Qmax of 226W is not the amount of power it will use, but the maximum amount of heat it will removed from the cold side. The 226W is removed using a current of Imax=24 amps which takes Vmax=15.2 volts to produce. So, at 12V you're appyling 79% of Vmax and need a a supply that can deliver at least 79% of Imax, or about 19 amps plus a margin of error. |
![]() |
![]() |
![]() |
#7 |
Cooling Neophyte
Join Date: Jun 2004
Location: USA
Posts: 11
|
![]()
I get it now! See, I need stuff put in examples to understand it.
So basically its 79% of the amps and volts. Gotcha. (Percent of the spec'd rating according to the percent of a change. 15 vs 12) 1. So I just take for instance, a 24v pelt @ 20 amps and run it at 12v, then its roughly 12v @ 10amps. Does the watt rating go in half too then? Say from 200w to 100w? Last edited by Heavytank2; 07-17-2004 at 12:52 PM. |
![]() |
![]() |
![]() |
#8 |
Cooling Savant
Join Date: Mar 2003
Location: MO
Posts: 781
|
![]()
Heat removed isn't linear, it'll be roughly Q=2*Qmax*I%-Qmax*(I%)^2. So at 79% of Imax of the original 226W pelt, you'll get 2*226*.79-226*(.79)^2=216 watts of cooling.
If you make a graph of that, you can see why it's best to choose a large pelt and undervolt. |
![]() |
![]() |
![]() |
#9 |
Thermophile
Join Date: May 2001
Location: UK
Posts: 1,064
|
![]()
Current drawn also varies somewhat with heat load - the more heatload you apply the greater the current drawn.
__________________
Once upon a time, in a land far far away... |
![]() |
![]() |
![]() |
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests) | |
|
|