i was working off 8v at 11amps 30*dT figures from the TEC performance graph, gave me 528 watts power usage.
so going by your numbers little more power usage but better dT. Sh#t COP but i already knew that :P
so 600+ watts, hehe, i'll take that if it can give me that delta
EDIT: Just realized why my Qmax was off, was working off a 66w/TEC calc from a previous voltage/amps combo. good thing i had a bit up my sleeve
Last edited by Liam_G; 05-19-2012 at 11:23 PM.
my app will calculate pretty much the same as the graphs you are referring to when everything is equal
See this
http://www.tecpeltier.com/viewtopic.php?f=4&t=41
based on the graphs it will draw more than 11 amps.
ALSO im assuming the hot side is 30c not 26.86c which affects Everything
At the end of the day they are all just estimations
COP is only poor because you've got a large delta if you had a delta of 0 then COP would be awesome.
you can't have a large delta and high COP
yeah i went back and had a look at the graph, realized i had used 66wQmax/TEC from a previous calc, its more like 55-60w Qmax/TEC as your program shows... edited my last post with as much.
now here is a question for you. when i look at the performance graph i select my desired dT range and then my volts and i calc my amps from where they intersect, and then my qmax from the other graph with my amps. so somewhere in there you are getting a different result, so what do you target in the volts/amps graph. i was looking at the 30*dT line for 11amps so what does your program look at?
It looks like your program is looking at the 0*dT line on the volts/amps graph. is that what i should be doing?
doesn't that amps figure drop because i am effectively cooling the TEC, otherwise why do they display different dT lines on that graph?
Well like i said im actually calculating something slightly different because the hot side of the TEC is different.
and you can't tell me your putting in accurate numbers based on the graph like i said with my eyes 8 volts for a delta of 30 uses more than 11 amps
And 66 Qmax that can't be accurate too. But here is the big thing assuming 66 is right then the TEC's can move 396watts to a delta of 30 but your not applying 396 your applying 350. Not applying the 396watts will affect the electricity used. So your numbers can't be accurate.
But as i said we are all making a huge number of assumptions so reality will be different so dont worry about it to much
720W Qmax + block/rad(200w cpu block/600w 480 rad*2) = 2120w cooling power - 600w load = 1520w (2:1 ratio) cooling power left - 250w cpu = 1270w cooling power left for holding lower temps or cooling more parts @ least another 600w load(2 high end gpus maybe).Tecs on htpc will allow on the fly cooling when needed instead of a 24/7 setup that will always run.
no, no, no, 66w qmax / tec was a mistake on my behalf, i had used 66w/TEC which was actually a result from a previous TEC calculation i had done, not this one. your 55-60w Qmax/TEC is correct....
so am i right in thinking that if i know my radiators can cool a certain heat load to give me a certain delta i should be looking at that delta T line on the volts/amps graph??? or should i always look at the 0*deltaT line on that graph. because that fits what your program is showing, 8v, 80amps. 80/6 = ~13.3amps which is where the 0*dT line intersects with 8v..
and yeah i understand a lot of this is assumed knowledge, i don;t really know how to accurately find the heat load my cpu puts out so can't really get an accurate result, so thats why i overshoot it a bit, to give me room to move.
@ Hell Hound
your not getting this are you.
this:
720W Qmax + block/rad(200w cpu block/600w 480 rad*2) = 2120w cooling power - 600w load = 1520w (2:1 ratio) cooling power left - 250w cpu = 1270w cooling power left for holding lower temps or cooling more parts @ least another 600w load(2 high end gpus maybe).Tecs on htpc will allow on the fly cooling when needed instead of a 24/7 setup that will always run.
makes no ing sense
720watts Qmax is the total of your cooling power from the TECs, this is where your getting confused. For your TECs to be able to cool 720watts of heat you need to cool your TECs with your radiator. So, for your TEC to cool 720watts Qmax, your TEC consumes 600 odd watts of power. What this means is you need a radiator that can cool the 720watts that the TEC transfers from cold side to hot side, plus the heat load created by the TEC from consuming 600watts of power. So what we do is add the Qmax to the power consumed, so you get 1320watts... this figure 1320watts is what your radiator needs to be able to cool down and not have the water inside the radiator exceed a 10* delta from ambient temperature. If your radiator can't handle this heat load you will have a higher water temp in your radiator. The higher the temp the water in your radiator the more of the delta T you gained by adding in TEC's is being chewed up, which results in virtually no difference in temp on the cold side of your TEC chiller under ambient, effectively you have a deltaT of 0*degrees, which means you just consumed 600watts of power for zip, nada, nothing.... good luck to you chap! either you do what i;m doing and learn about TEC's and how they actually work, or just stay away from them.
EDIT: you need to understand some fundamentals of TEC's, they don't "cool" heat as such, they transfer heat from one side of the TEC to the other, you, ie your radiators needs to cool the heat.
Last edited by Liam_G; 05-20-2012 at 12:52 AM.
wattage used will depend on ambient temp @ time of use.Rad cooling power will be determined by fan speed+temp @ time of usage.So total cooling power minus watts used @ time of use determines delta.With a max of 2.1kw of cooling power and a average use of 1.3kw -1.6kw should net good temps no matter the weather that day.Being in a rack setup really changes temp depending on load on receiver/amp @ time of use.Upgrading to faster htpc within thermal spec of cooling setup should be easy.
720W Qmax + block/rad(200w cpu block/600w 480 rad*2) = 2120w
This just makes no sense ?
If your Qmax is 720watt then this is the max load you can cool . It doesn't matter if you have a trillion Radiators cooling the hot side of the TEC's.
The reality is there's heaps of uncounted for losses ie thermal resistance of the hot and cold side blocks blah ... so you'll never get my delta of 34 or yours of 30.
But you'll still get kick ass cooling
he makes less and less sense, if he ever made any, i have given up after trying to interpret his last post.
yeah there is lots of stuff we can't take into account, i will just run it and see what i get when i have built it, which will be a while off yet. and i'm confident i will get kickass cooling too
Last edited by Liam_G; 05-20-2012 at 02:17 AM.
Thread build Tec/fan controller = Ex:set temp and tec will use the wattage (no more than) needed to achieve temp,fans spin @ rpm set for what ever temp.If goal is 10c above ambient loaded,then today was 27c but next day could be 32c or 22c the fan speed and tec watt usage would go up/down with out having to touch a thing.Ok now have to determine how much cooling power needed to cool max used wattage @ max ambient temp to find goal temp.I'm using tec to have a controlled temp environment.So a max temp stays the same but the watts used does not instead of lowest temp possible @ max clk.I want stay under max spec temp and keep good power usage/cooling ratio.So say the chip spec is 64c max I would want 50c 100% loaded no matter ambient temps,How I get there is the fun part.
Rad cools tec >tec cools water>water cools component
Water temp is what I will be adjusting to achieve component temp
Qmax 2:1 = 360w MAX load traveling threw water
So blocks on components don't take away from load traveling threw water.
actually this is't right either
720watts Qmax is the total of your cooling power from the TECs
Assuming you are applying max amps and volts and keeping the hot side at 26.85c
This is where your getting confused. For your TECs to be able to cool 720watts of heat you need to cool your TECs with your radiator.
Qmax is the max amount of heat the TEC's can move but this is NOT the amount of electricty used to move 720watts from the cold side to the hot side. at full power it's going to use 1512watts
So, for your TEC to cool 720watts Qmax, your TEC consumes 600 odd watts of power. What this means is you need a radiator that can cool the 720watts that the TEC transfers from cold side to hot side, plus the heat load created by the TEC from consuming 600watts of power.
So what we do is add the Qmax to the power consumed, so you get 1320watts... this figure 1320watts is what your radiator needs to be able to cool down
So, for your TEC to cool 720watts of heat, your TEC consumes 1512 odd watts of power. What this means is you need a radiator that can cool the 720watts that the TEC transfers from cold side to hot side, plus the heat load created by the TEC from consuming 1512watts of power. So what we do is add the 720watt (the load your cooling) to the power consumed, so you get 2232watts... this figure is what your radiator needs to be able to cool down
This is where i point out that no ones cpu is generating 720watts of heat so the above is not reality but i just went with it anyway
To me it sounds like EVERYONE should read this
Everything you've wanted to know about TEC's
http://www.tecpeltier.com/viewtopic....7782d7d2b72050
i'm not sure we are talking about the same thing...... he wanted to use 48, yes 48 TEC's at low volts and amps to cool 720watts. at those low amps and volts i approximated a 600watt power usage. so i added the 600watts to the 720 watts to get 1320watts. i'm not talking about the tec i mentioned, he had some other tec entirely.
temps go up and down w/ just a rad.This way temps won't move but power usage will.plan was to use a rack in a closed closet w/ plexi window.
From earlier post:
Rad cools tec >tec cools water>water cools component
Water temp is what I will be adjusting to achieve component temp
Qmax 2:1 = 360w MAX load traveling threw water
So blocks on components don't take away from load traveling threw water ?
tec cal.jpg
This is wrong.. amperage will decrease as dt increase at a set voltage..
At the proposed heat load the tecs will be running at 11 amps with a dt of 30 if the voltage is set to 8v...
NOT 13 amps which is the amperage at dt=0
Muffy you need to fix your program.
chart.png
Last edited by mindchill; 05-20-2012 at 08:16 AM.
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