RTX3080

[chalsall]

Originally Posted by ewmayer

Others have already addressed the incorrect closed-system assumption...

OK, sorry guys. I was trying to have some geeky fun, not cause any agitation.

My model, as stated, was a closed system. Clearly, humans like open systems, for various reasons.

[jwnutter]

Originally Posted by ewmayer

Reminds me, been a while since I did any price-checking on legacy R7s (probably because I am definitively done w/my build-out using same, even if by magic I could score some cheap ones, the electricity bills and you'd-need-to-run-the-AC-even-on-cool-days aspect would make it a no-go) ... usual XFX-R7 link on Amazon shows cheapest used @$1349, but I see a new one one eBay for $999, which is not horrible. And even if no "make offer" option, one can PM the seller to do that ... worst they can do is turn you down.

In my opinion, this is the best resource for current as well as historical hardware prices. If you setup a free account they'll even let you set price alerts which are sent to you via email.

https://pcpartpicker.com/products/video-card/

Originally Posted by chalsall

Please forgive me for this, but I'm "in that mode".

This is similar to the question "Can I cool a room by running a fridge with the door open?

Cold can't be created. Heat can be moved.

Unless there is a route to exhaust heat, a closed system will always get warmer.

Years ago a heat transfer prof. explained this concept to me as follows:

Four Laws of Thermodynamics
Zeroth Law - You have to play the game.
First Law - You can't win.
Second Law - You can't tie.
Third Law - And, you can't stop playing.

This logic has stuck with me for a very long time, much easier to remember than the actual laws.

[jwnutter]

Originally Posted by ewmayer

Others have already addressed the incorrect closed-system assumption, but just by way of Gedankenexperiment let's imagine that the system were partially closed, in the sense that moisture could only enter by way of evaporation from the cooler, but had no way to leave once evaporated. I expect in a place the size of mine, somewhat over 10^4 cft (300 m^3), starting with typical CA summertime ambient air conditions meaning relative humidity ~= 40%, I could evaporate over 5 gal (say, 20 L) of water without the resulting relative humidity getting anywhere near 100% - perhaps someone has a set of suitable tables allowing this to be checked. Using Uncwilly's energy-of-evaporation number of 540 cal/g that translates to ~ 10^7 cal, or ~4*10^7 J. 1 J = 1 W*s, so 4e7 J ~= 11,000 W*h, roughly the same amount of heat put out by my 6-GPU compute setup in ~7 hours. [Any remaining cooling would need the AC, but 7 hours is pretty close to the typical windows-closed-and-AC-on time on a summer day here, i.e. this gets pretty close to obviating the need for AC cooling to offset the heat put out by the rig.]

Now in my above thought experiment, heat can still flow in and out of the space via windows and walls, so at night the cooler outside temps have a chance to remove enough heat to recondense the evaporated water - the only problem is that said condensation will be on every cool surface, collecting the water and returning it to the swamp cooler would be a problem. But aside from that, this describes more or less the same kind of moisture-sealed-in condensation loop used by your refrigerator and AC unit.

Anyhow, it'll be interesting to see how many gallons of H2O the unit I ordered actually manage to evaporate during a warm day, but the math indicates that each gallon evaporated offsets ~1.5 hours of the compute rig's 1.5kW heat production.

Getting back to the thread topic...



Holy crap, up to 600W per card? That would need the FLOPS equivalent of 3 Radeon VIIs running as sclk=3 to achieve parity on a FLOPS/W basis. But it'll be interesting to see the numbers once the 3080/3090 and R7 Pro start shipping and once someone gets an opportunity to run actual GIMPS-useful code on them.

I don't fully understand what you are doing, but a Psychrometric Chart might be of use for these efforts.

https://www.engineeringtoolbox.com/d...art_29inHg.pdf

[ewmayer]

Originally Posted by jwnutter

I don't fully understand what you are doing, but a Psychrometric Chart might be of use for these efforts.

https://www.engineeringtoolbox.com/d...art_29inHg.pdf

Yes - thanks much! Say I keep my place at 75F, I want to know the water content of the air at 100% saturation at the temperature. So we find the point where wet-bulb-temp = dry-buld-temp = 75F, which is at the curved left edge of the crosshatched region, beneath the fourth and final 'T' in 'SATURATION TEMPERATURE'. Following the intersecting horizontal light-blue line to the right edge of the graph, this maps to a humidity ratio - love the units here, metricians will be aghast :) - of 132 grains of H2O per lb of dry air. Per my dictionary:

grain
3 (abbr.: gr.) the smallest unit of weight in the troy and avoirdupois systems, equal to 1/5760 of a pound troy and 1/7000 of a pound avoirdupois (approximately 0.0648 grams). [ORIGIN: because originally the weight was equivalent to that of a grain of wheat.]

So 132 grains ~= 8.6 g of H2O per lb of dry air. Interpolating between the downward-sloping lines of constant specific volume, our 75F 100% saturation point corresponds to a specific volume of ~13.9 ft^3/lb of dry air. There are 35.3 ft^3 per m^3, so this translates to 0.39 m^3/lb = 0.87 m^3/kg, i.e. a dry-air content of 1.15 kg/m^3 = 2.54 lb/m^3. Each lb holds 8.6g of moisture, so we have ~20g H2O per m^3, or a bit over 6kg ~= 6L H2O for 300 m^3 of total interior volume. That strikes me as low, so either my math is wrong or I'm gonna have to hope the evaporated moisture can escape quite readily even with the windows closed. EDIT: The result matches the data in this table at the same EngineeringToolbox site, so 6L of water it is. And that is to get from 0% to 100% relative humidity ... starting at 40% RH translates to around 1 gal of added moisture to reach saturation, thus my 5-gal guesstimate was an order of magnitude too high.

[jwnutter]

Originally Posted by ewmayer

Yes - thanks much! Say I keep my place at 75F, I want to know the water content of the air at 100% saturation at the temperature. So we find the point where wet-bulb-temp = dry-buld-temp = 75F, which is at the curved left edge of the crosshatched region, beneath the fourth and final 'T' in 'SATURATION TEMPERATURE'. Following the intersecting horizontal light-blue line to the right edge of the graph, this maps to a humidity ratio - love the units here, metricians will be aghast :) - of 132 grains of H2O per lb of dry air. Per my dictionary:

grain
3 (abbr.: gr.) the smallest unit of weight in the troy and avoirdupois systems, equal to 1/5760 of a pound troy and 1/7000 of a pound avoirdupois (approximately 0.0648 grams). [ORIGIN: because originally the weight was equivalent to that of a grain of wheat.]

So 132 grains ~= 8.6 g of H2O per lb of dry air. Interpolating between the downward-sloping lines of constant specific volume, our 75F 100% saturation point corresponds to a specific volume of ~13.9 ft^3/lb of dry air. There are 35.3 ft^3 per m^3, so this translates to 0.39 m^3/lb = 0.87 m^3/kg, i.e. a dry-air content of 1.15 kg/m^3 = 2.54 lb/m^3. Each lb holds 8.6g of moisture, so we have ~20g H2O per m^3, or a bit over 6kg ~= 6L H2O for 300 m^3 of total interior volume. That strikes me as low, so either my math is wrong or I'm gonna have to hope the evaporated moisture can escape quite readily even with the windows closed. EDIT: The result matches the data in this table at the same EngineeringToolbox site, so 6L of water it is. And that is to get from 0% to 100% relative humidity ... starting at 40% RH translates to around 1 gal of added moisture to reach saturation, thus my 5-gal guesstimate was an order of magnitude too high.

You got it - engineeringtoolbox.com is a great resource. Yes, these are some of my favorite units.

I should note that I believe the chart I linked in the previous post was for an atmospheric pressure of 29.92 in-Hg. So, depending on your elevation above/below sea level your specific psychrometric values could differ from the chart slightly. That said, there are a number of online calculators that will spit out the exact values from the chart without having to interpolate.

As an example (though I haven't tested the results from this site): http://www.sugartech.co.za/psychro/index.php

Here's a resource for how to read a Psychrometric Chart - but it sounds like you have the idea already: https://www.youtube.com/watch?v=s7J6R9wECh8

[jwnutter]

Originally Posted by Cheetahgod

If you put that air conditioning unit you bought outside then you can cool outside of your house creating a temperature differential causing the heat to leave your house.

This is actually a very true statement. It would just require a very large refrigeration unit.

[retina]

Originally Posted by jwnutter

This is actually a very true statement. It would just require a very large refrigeration unit.

So basically, put your entire house inside a refrigerator. Sure, that should be easy.

So why don't we put the entire planet inside a refrigerator? Hey look, I just solved global warming!

[jwnutter]

Originally Posted by retina

So basically, put your entire house inside a refrigerator. Sure, that should be easy.

So why don't we put the entire planet inside a refrigerator? Hey look, I just solved global warming!

That is incorrect. He stated that one could satisfy the equation Q=U*A*LMTD by conditioning the air outside of the home. This is very possible. Not practical, but very very possible. I've been in subzero freezers the size of many sporting fields. You could fit a small neighborhood inside of these facilities.

[jwnutter]

You guys should become friends with your local heat transfer engineer. The modeling these guys do on various heat and mass transfer systems is impressive, and at times in very unique situations. I recall talking to a heat transfer colleague about his use of molten salt to cool a reactor. Very interesting topics about how these system react under extreme temperatures and pressures and the impact this has on the object that contains the system (reactor, HX, tank, etc.).

[jwnutter]

https://www.pcgamer.com/nvidia-amper...ector-adapter/

A new 12-pin PCIe power adapter purportedly set to be used with the Nvidia Ampere generation graphics cards has been confirmed by a power supply manufacturer. The new connector takes a 2x8-pin power direct from the PSU and shrinks it down into a single 12-pin connector.

What wasn't clear from the initial rumours, and not made obvious until Andreas Schilling over at HardwareLuxx received a Seasonic-made connector, is the scale of this new 12-pin configuration. It's considerably smaller than a 2x8-pin connection, and nearly occupies the same footprint as a single 8-pin connector.

A traditional 8-pin PCIe connector (or 6+2-pin) features three +12V pins and five ground pins, and is able to deliver 150 watts per connector to a graphics card.

HardwareLuxx reports that the new 12-pin connector comes with six +12V pins, four ground, and two Sense pins (used to indicate what configuration the connector is being used in, in order to adjust power delivery accordingly (as is the case with 8-pin and 6-pin connectors).

As listed on the exterior of the box, the Seasonic power connector recommends a PSU capacity of 850W or more for use with the cable—and one would assume the graphics card to go with it.

[jwnutter]

Forgot to add this image.