Higher temperatures =/= more heat into your room; it's all about power consumption and heat transfer

[Agall]

A topic that I've seen come up that I've had to address, an understandable misinterpretation, is that higher temperatures of components mean more heat into your room. This is far enough removed from a proper understanding of thermodynamics enough for me to make a post I can reference back to as a detailed explanation of why.

 

TLDR: The primary controlling variable for energy/heat into your room from your computer is its power consumption. More energy in = more energy out. The variable that controls this is energy in, which is compensated for by either changing fan speeds or allowing the difference in temperature of the components and your environment to change. Higher temperatures under load are simply the way a heat source compensates for an insufficient proportional increase in the other variables, like airflow or heat capacity/surface area/heat transfer coefficients, etc.

What can you do to change this? Lower your power consumption. A way to mitigate its effect to the user can also be to transport that heat more effectively away from the user (shoutout to whole room watercooling, RIP).

 

Now to show this qualitatively with some simple equations and definitions to understand:

 

 

 

Temperature- the average random molecular kinetic energy of a substance. This is a localized approximation of kinetic energy by its nature and not indicative of kinetic energy of a whole system.

 

A fundamentally important thing to understand is that computers are practically space heaters, where most of the electrical energy into a computer is converted to waste heat.

 

BTU/hr being proportional to Watts (3.41 conversion) since Watts are just Joules/sec.

 

dT (difference in temperature) can be defined as (Tsource – Tsink), being the difference in temperature between the heat source and heatsink (CPU and your room, as examples). Heat transfer cannot occur without a difference in temperature. If dT is 0, then no heat transfer is occurring between those materials. The higher the difference, the more heat transfer can occur. This is why allowing your temperatures to go up at a lower fan speed can reach an equilibrium, Tsource will keep going up until it does or it throttles back power consumption, therefore reducing the heat transfer requirement.

 

More unit definition/clarifications:

 

'U,A' in the top equation being complex variables, since it’ll include the heat transfer coefficient of every layer or be an average across multiple layers, where this in some equations is defined as the ‘material constant’ since its largely dependent on material properties and are relatively constant. The simplest explanation of these variables is they’re proportional to your heat sink, like a 120mm AIO versus an NH-D15, where an NH-D15 will have a higher UA.  A way of visualizing this is with a cross section of heat transfer layers of a microprocessor.

 

 

Notice how we’re already at 3 layers of materials without adding thermal paste or the heat sink, which the heatsink will have multiple layers between it alone. Each of these materials will have their own U (heat transfer coefficient), where generally you’re most limited by the number of layers and the lowest heat transfer coefficient. The overall value of these heat transfer coefficients when factoring in their thickness and individual heat transfer coefficients will determine U. This is the major reason why 3D v-cache operates so hot and requires lower maximums, since the extra copper and cache layers require a proportionally lower temperature at the point of measurement to maintain a safe die temperature.

 

'A' (cross sectional area) is the third dimension of the heat transfer profile, being the contactable surface area to transfer heat from a heat source to a heat sink. Things like Heat Flux come into play with this value, but its outside the scope of this discussion.

 

Now finally to the second equation…

 

'M' (mass flow rate) is the volumetric flow of the fluid (whether that’s water or air) that’s acting as the heat transfer medium. This would be the fluid in a CPU AIO or the fluid in the heatpipes of a tower cooler, and then again for the air that the fans move across the radiator or heatsink fins.

 

'C' (specific heat capacity) is practically speaking a measurement of density, where it’s the amount of energy per mass per degree a material can absorb. The higher this value, the more energy that material can absorb without changing temperature independently of its total mass. This is a material property, water having one of the highest, especially as a relatively safe heat transfer medium. A quantifiable measurement for this being that water can absorb ~4.5x more energy than aluminum of the same mass per degree change of temperature.

 

Done with explaining the equations, now for some actual qualitative analysis. This involves assuming static quantities for certain variables to create proportions between either the one's on topic or to demonstrate a relationship between those variables. In this case, we'll look at Q, M, and Tsource:

 

Take the equations and strip them down to a proportion between Q (heat transfer rate), dT (difference in heatsink to heatsource) and M (mass flow rate)these are the variables we can control in our computer. The rest we can assume to be relatively constant for the purposes of this discussion. Even within dT, we can assume that Tsink is relatively constant, since this would ultimately be the temperature of your room/environment. Mass flow rate is practically speaking the speed of your fans, and we’ll eliminate the variable of a watercooling pump and its speed from this to make it simpler. This leaves specifically Q (power consumption of the PC, Tsource (component temperature like CPU/GPU), and M (speed of cooling fans).

 

While your computer is ‘idle’ in a relatively low power state, lets say 50W, this would require less Q (heat transfer rate) since there’s less heat being generated per second (Watts), therefore the Tsource (your CPU/GPU) doesn’t have to be as high to remove said heat, creating a proportionally lower dT (assuming constant fan speed).

While your computer is under load, lets say 500W, the heat transfer rate will go up, requiring one of the equation’s variables to go up proportionally. This requires a change to either dT or M (mass flow rate), usually both. This results in a higher Tsource to create a higher dT and higher fan speeds to increase the mass flow rate, since its unlikely one would be able to compensate for 10x the amount of heat simply through an increase in dT (while maintaining operating limitations like 105C).

 

 

Some caveats:

Yes, a silicon semiconductor at a lower temperature will therefore have a lower resistance therefore being more efficient, but that’s an insignificant contribution to overall power consumption in the operating range of computers.

 

The average power consumption, heat soak characteristics, and overall heat capacity of a computer does contribute. This is applicable to full eATX towers versus SFX systems, where a 500W draw SFX system will take far less time to heat soak than an eATX system which might have 2x the steel/aluminum and 10x the internal air volume. This will affect the time it takes for the whole system to reach equilibrium.

 

Anyone else with knowledge of thermodynamics who would like to contribute to this discussion or if you find anything I can improve, I’ll edit the OP accordingly with citation. 

 

Feel free to ask questions and for clarifications, and I'll try as best as I can to not require a whole essay to explain. Most of the information I would need to reference would be in OP.

 

An analogy for this using fluid dynamics is to take a tank of water with an inlet and outlet. If the flow into the tank exceeds the flow out of the tank, then the tank level will rise. However, the tank level might rise high enough for it to increase the head pressure on the outlet, therefore increasing the outlet flow, which may match the inlet flow rate once the tank reaches a certain level creating an equilibrium. Water in this analogy is equal to kinetic energy, where the inlet flow are the components generating heat, flow out being the heat transfer out via fans and the tank volume being the overall heat capacity of the system which includes the air inside, heatsinks, etc. Temperature would operate similarly to head pressure and therefore tank level.

 

In this analogy as well, if there's no outlet, then the tank will overflow or become pressurized, just like a computer in a sealed environment will eventually shutdown because it'll overheat, regardless of how much it tries to thermal throttle with no energy out. So if there's energy going in, then there must be energy going out equal to what's going in, otherwise the kinetic energy will continue to rise. To full circle the analogy, neglecting water's surface tension, the tank would be perforated and slowly leaking water, just like how anything with a higher temperature than its surroundings would slowly leak energy to its surroundings and eventually reach equilibrium unless it has an active energy input.

 

To flesh the analogy out more, my 'mythbusting' is that tank level = flow out of the tank; which isn't true, since its not considering flow in and volume of the tank. To put the importance of the tank's volume into perspective, a 1" diameter tube with 5" height of water creates the same head pressure as a 5 gallon drum with 5" height of water, just like how a GT1030 at 75W can operate at the same temperature as a RTX 4090 at 600W. They're both creating the same dT, but at drastically different wattages/volume.

[NorKris]

its kinda an easy observation (to be fair i dont know if all CPU (or GPU) designs is like this) 

but having a cpu run a render (cinebench..or what ever) with A cooler vs a B cooler will make the watt go up on Cooler B if its not good enough. and more watt is more heat  

[Agall]

24 minutes ago, NorKris said:

its kinda an easy observation (to be fair i dont know if all CPU (or GPU) designs is like this) 

but having a cpu run a render (cinebench..or what ever) with A cooler vs a B cooler will make the watt go up on Cooler B if its not good enough. and more watt is more heat  

Like in one of my caveats, you can get low power consumption at a lower temperature for the same score in such benchmarks because of the better heatsinks+lower temperature, but that's an efficiency argument and not an argument based on temperature. Temperature is secondary in that discussion to the primary discussion of efficiency. 

[NorKris]

15 minutes ago, Agall said:

Like in one of my caveats, you can get low power consumption at a lower temperature for the same score in such benchmarks because of the better heatsinks+lower temperature, but that's an efficiency argument and not an argument based on temperature. Temperature is secondary in that discussion to the primary discussion of efficiency. 

okey what makes the watt go up?

[37748585862723896]

I think you can skip the part of heat transfer, it makes things simplier, it eventually flows evenly

 

And title is confusing to me

Temperature is kinetic energy, so also more heat energy

[Agall]

1 minute ago, NorKris said:

okey what makes the watt go up?

The component in question request more power for various reasons, either through increasing voltage and/or current.

 

Example being my rig, which I've been playing with lower draw configurations due to the triple digit ambient temperatures in my area. I've dialed back the power consumption nearly 200W, measured at the wall and validated in software.

[NorKris]

5 minutes ago, Agall said:

The component in question request more power for various reasons, either through increasing voltage and/or current.

 

so not cuz it got 8-11c hotter?

[Agall]

1 hour ago, NorKris said:

so not cuz it got 8-11c hotter?

If you're referring to a scenario where you run cinebench r23 and your CPU temperature jumps 8-11C, that's because your processor is generating more heat due to an increase in voltage/current. Its likely your fan speeds have increased, but you'll inevitably see an increase in the Tsource, being the CPU, since the fan's effective mass flow rate would otherwise have to be 1:1 proportional with the increase in power consumption. 

 

Let's do a quantitative representation. Let's say my 7950x3D idles at 30W. I run cinebench R23 and it jumps to 150W since the CPU is now increasing its voltage/current draw to satisfy the computational demand. Lets say its idling at 50C and jumps to 90C with an ambient of 30C. That's a 5x increase in wattage draw with a 3x increase in temperature difference. Therefore, if my fan speeds don't increase the mass flow rate by 1.67x (its multiplicative, not additive), then I'll see either an increase in temperature (spoilers, you won't, since that's the thermal limit for the 7950x3D) or it'll reduce its power draw to compensate. Caveat here being that this is a simplified version of the math, which can get very deep into calculus with a multitude more variables otherwise (including heat soak of the various heatsinks in the system), but its a solid representation of the concept. Equation used to represent this is Q=MCdT shown above. It would be even more accurate if we referenced to Kelvin, but that's beyond the scope.

 

If your assumption is that the CPU temperature increasing is an indication of a higher wattage draw, then that's potentially true, but not always, since there's more than one variable in that equation. CPU temperature could go up simply because there's a lack of airflow, maybe from an incidental restriction in airflow (someone blocking it) or a fan has stopped working. The reduction in mass flow rate in this case would force an increase in dT, therefore in Tsource, to get the same Q (heat transfer rate).

 

In a perfect world for computers, the Tsource doesn't change at all, because either you're providing proportionally higher airflow or changing the material properties of the heatsink for the increase in wattage draw, the latter of which isn't feasible for most people (an example would be a material that phase chases when reaching a certain temperature and pressure. Like a scenario where you're running a saturated fluid that boils at a higher temperature, like alcohol).

 

I've theorized using water at saturation, however you'd have to run quite the vacuum in the loop and you'd also have to get the coolant temperature to 60C. So the actual CPU temperature would be in the 80-90C, if not higher to do so.

 

Water - Saturation Pressure vs. Temperature (engineeringtoolbox.com)

 

 

[Agall]

1 hour ago, 37748585862723896 said:

I think you can skip the part of heat transfer, it makes things simplier, it eventually flows evenly

 

And title is confusing to me

Temperature is kinetic energy, so also more heat energy

Temperature is localized.

 

A passively cooled 5W tablet CPU could be running at 90C versus a 500W RTX 4090 at 60C. Obviously the RTX 4090 is going to output more thermal energy, because its consuming more.

 

The title is alluding to the specific scenario where people seem to desire the lowest temperatures possible for the sake of 'heat into their room'. It's something I've seen/discussed a few times. Temperature within itself isn't a simple concept, since its nuanced and non linear, since unless we're talking in Kelvins, there's not a proportional relationship to kinetic energy.

[NorKris]

8 minutes ago, Agall said:

If you're referring to a scenario where you run cinebench r23 and your CPU temperature jumps 8-11C, that's because your processor is generating more heat due to an increase in voltage/current.

 

 

controlled and same for every run

 

9 minutes ago, Agall said:

Its likely your fan speeds have increased

 

controlled and same for every run

 

For every run with an Air cooler the watt was 15-25w higher than on water. test was done over 12 times each

[Agall]

2 minutes ago, NorKris said:

controlled and same for every run

 

controlled and same for every run

 

For every run with an Air cooler the watt was 15-25w higher than on water. test was done over 12 times each

Those are different heatsinks designs entirely, even if an air cooler is really just watercooling but smol. Watercooling does a far better job at transporting heat away from the heat generating component while also having a higher heat capacity.

 

You're still likely making an argument for efficiency, not for thermal output being proportional to temperature. As I've mentioned before, silicon microprocessors do run more efficiently at a lower temperature, simply due to them requiring less voltage to switch transistors when running at a lower resistance. Silicon's resistance goes up as it increases in temperature. That's a fundamentally different argument than the myth I'm dispelling, which is why I even mentioned it in my OP.

[NorKris]

4 minutes ago, Agall said:

 

You're still likely making an argument for efficiency, not for thermal output being proportional to temperature. As I've mentioned before, silicon microprocessors do run more efficiently at a lower temperature, simply due to them requiring less voltage to switch transistors when running at a lower resistance. Silicon's resistance goes up as it increases in temperature. That's a fundamentally different argument than the myth I'm dispelling, which is why I even mentioned it in my OP.

aah okey then i agree with you. even tho the statement of the thread is not 100% true when u can see more heat output due to "thermal run away" 

 

Statement A;  My pc heats up my room cuz its 80c when gaming, i want to lower the temp  --- your right with ur topic

 

Statment B;  My CPU is pulling 300w and is at 100c can i cool it down better and get the watt down --- here the topic misses abit  

[thrasher_565]

1 hour ago, NorKris said:

okey what makes the watt go up?

the cpu see oh its cooler so up the wats to stay at its limit. getting a bostbin.

but amd will give you more proframce the cooler well intel wont. there videos on how temp and under volting works.

[Agall]

43 minutes ago, NorKris said:

aah okey then i agree with you. even tho the statement of the thread is not 100% true when u can see more heat output due to "thermal run away" 

 

Statement A;  My pc heats up my room cuz its 80c when gaming, i want to lower the temp  --- your right with ur topic

 

Statment B;  My CPU is pulling 300w and is at 100c can i cool it down better and get the watt down --- here the topic misses abit  

The answer to statement B is to lower the power consumption which I mention in my TLDR: "What can you do to change this? Lower your power consumption."

 

Especially true for Intel CPUs, where a 125W TDP limit can net similar performance. Same goes for the RTX 4090, something I've been doing recently with how hot its been. I've been running a stock clock for both Diablo 4 and Warframe with 90% TDP. Its cut the power draw by nearly 200W and took my office from 'still too hot' to cycling my AC unit to keep it from being too cold. I'm also testing an 8c/16t configuration, but that only amounted to like 20W less.

 

There's a balancing point here with the ability for my AC unit to remove energy and the nature of my office that's reached a tipping point making that singular change. I don't think it's simply 200W less output specifically makes it magically better, but there's a lot of variables within that which dip into chaos theory.

 

First summer with the RTX 4090 obviously, and its noticeable  Its no longer my '2 birds 1 stone' space heater like it was during winter.

[NorKris]

7 hours ago, Agall said:

The answer to statement B is to lower the power consumption which I mention in my TLDR: "What can you do to change this? Lower your power consumption."

 

Especially true for Intel CPUs, where a 125W TDP limit can net similar performance. Same goes for the RTX 4090, something I've been doing recently with how hot its been. I've been running a stock clock for both Diablo 4 and Warframe with 90% TDP. Its cut the power draw by nearly 200W and took my office from 'still too hot' to cycling my AC unit to keep it from being too cold. I'm also testing an 8c/16t configuration, but that only amounted to like 20W less.

 

There's a balancing point here with the ability for my AC unit to remove energy and the nature of my office that's reached a tipping point making that singular change. I don't think it's simply 200W less output specifically makes it magically better, but there's a lot of variables within that which dip into chaos theory.

 

First summer with the RTX 4090 obviously, and its noticeable  Its no longer my '2 birds 1 stone' space heater like it was during winter.

maybe you missunderstood. but the idea with B  was to only change cooler 

[Agall]

8 hours ago, NorKris said:

maybe you missunderstood. but the idea with B  was to only change cooler 

There's something to be said for targeting points of efficiency, something the 3D v-cache CPUs are forced to do. The 7950x I have as an example is set to a 105W TDP which draws a maximum of 120W. It gets less performance than the unlocked 250W setting, but if a fan fails, I won't have to worry about stability, even though its in a temperature controlled environment (its a server in a server room). 

 

Regarding efficiency from lower temperatures, I've only seen single digit percentage reductions in power consumption, something along the lines of 4% reduction in power consumption per 10C drop. Getting that may cause a proportional increase in power draw from other components though, like increasing fan speeds or pump speed or dialing up the AC, so its not a simple equation.

 

One could simply have a more effective cooling system that doesn't proportionally increase the power consumption, being an AIO will draw more power than an air cooler simply due to the additional electrical component of the pump.

[NorKris]

8 minutes ago, Agall said:

 

Regarding efficiency from lower temperatures, I've only seen single digit percentage reductions in power consumption, something along the lines of 4% reduction in power consumption per 10C drop. Getting that may cause a proportional increase in power draw from other components though, like increasing fan speeds or pump speed or dialing up the AC, so its not a simple equation.

 

yes i agree  

 

8 minutes ago, Agall said:

One could simply have a more effective cooling system that doesn't proportionally increase the power consumption, being an AIO will draw more power than an air cooler simply due to the additional electrical component of the pump.

yes but at the thermal throttle-ish level the Air cooled did draw more power. but generally yes. 

you bring up AMD alot,  and i would clarify that i did not test this on an AMD system. it was on an 7940x  

[Agall]

11 minutes ago, NorKris said:

yes i agree  

 

yes but at the thermal throttle-ish level the Air cooled did draw more power. but generally yes. 

you bring up AMD alot,  and i would clarify that i did not test this on an AMD system. it was on an 7940x  

 

 

Last HEDT system I owned was a 4930k  I don't consider 3950x, 7950x and 7950x3D 'HEDT'. I could see this chip having some weird power behavior with insufficient cooling. That chip was cooled by a 240mm AIO at the time and I never tried air cooling on it.

[RevGAM]

20 hours ago, Agall said:

Heat transfer cannot occur without a difference in temperature. If dT is 0, then no heat transfer is occurring between those materials

Correct me if I'm wrong but that suggests a steady state in which SOMETHING is keeping the temperature stable, which would be the components. Thus, since a temperature of,  say,  +15 over ambient cannot maintain itself without a source, that means that the source (eg CPU, GPU) will, in reality, still be trickling heat into the cooler, albeit at a much slower pace. Correct?

[NorKris]

1 hour ago, Agall said:

 

Last HEDT system I owned was a 4930k  I don't consider 3950x, 7950x and 7950x3D 'HEDT'. I could see this chip having some weird power behavior with insufficient cooling. That chip was cooled by a 240mm AIO at the time and I never tried air cooling on it.

the behavior seems to be normal 

[Agall]

1 hour ago, RevGAM said:

Correct me if I'm wrong but that suggests a steady state in which SOMETHING is keeping the temperature stable, which would be the components. Thus, since a temperature of,  say,  +15 over ambient cannot maintain itself without a source, that means that the source (eg CPU, GPU) will, in reality, still be trickling heat into the cooler, albeit at a much slower pace. Correct?

Yes, the energy input to the system would have to have a magnitude on average, otherwise the heat transfer would eventually equalize until dT=0 and therefore Q = 0. In the context of computers, their existence while powered on is a constant positive heat source, though variable in its output based on load. 

 

Q (heat transfer rate) in thermodynamics is dependent on context, since plenty of equations can be manipulated to create functions which isolate it. In this context, it serves as a measure of energy into the system and the energy out of the system. If those aren't equal, then temperature within that system will rise.

 

The best way to understand thermodynamics in my opinion is to look at the context and units of an equation. Q as an example is a measurement of energy over time, which can be either as a source of energy or the transfer of it to another material. Same goes for the other variables; if the units don't work out in the end, you did something wrong.

 

An analogy for this using fluid dynamics is take a tank of water with an inlet and outlet. If the flow into the tank exceeds the flow out of the tank, then the tank level will rise. However, the tank level might rise high enough for it to increase the head pressure on the outlet, therefore increasing the flow, which may match the inlet flow rate, therefore creating an equilibrium and a constant tank level. Water in this analogy is equal to kinetic energy, where the flow in is the heat generating components, flow out being the heat transfer out and the tank volume being the heat capacity of the system. Temperature would operate similarly to head pressure and therefore tank level.

 

In this analogy as well, if there's no outlet, then the tank will overflow or become pressurized, just like a computer in a sealed environment will eventually shutdown because it'll overheat, regardless of how much it tries to thermal throttle with no energy out.

 

To flesh the analogy out more, my 'mythbusting' is that tank level = flow out of the tank; which isn't true, since its not considering flow in, flow out, and volume of the tank.

[RevGAM]

1 hour ago, Agall said:

Yes, the energy input to the system would have to have a magnitude on average, otherwise the heat transfer would eventually equalize until dT=0 and therefore Q = 0. In the context of computers, their existence while powered on is a constant positive heat source, though variable in its output based on load. 

 

Q (heat transfer rate) in thermodynamics is dependent on context, since plenty of equations can be manipulated to create functions which isolate it. In this context, it serves as a measure of energy into the system and the energy out of the system. If those aren't equal, then temperature within that system will rise.

 

The best way to understand thermodynamics in my opinion is to look at the context and units of an equation. Q as an example is a measurement of energy over time, which can be either as a source of energy or the transfer of it to another material. Same goes for the other variables; if the units don't work out in the end, you did something wrong.

 

An analogy for this using fluid dynamics is take a tank of water with an inlet and outlet. If the flow into the tank exceeds the flow out of the tank, then the tank level will rise. However, the tank level might rise high enough for it to increase the head pressure on the outlet, therefore increasing the flow, which may match the inlet flow rate, therefore creating an equilibrium and a constant tank level. Water in this analogy is equal to kinetic energy, where the flow in is the heat generating components, flow out being the heat transfer out and the tank volume being the heat capacity of the system. Temperature would operate similarly to head pressure and therefore tank level.

 

In this analogy as well, if there's no outlet, then the tank will overflow or become pressurized, just like a computer in a sealed environment will eventually shutdown because it'll overheat, regardless of how much it tries to thermal throttle with no energy out.

 

To flesh the analogy out more, my 'mythbusting' is that tank level = flow out of the tank; which isn't true, since its not considering flow in, flow out, and volume of the tank.

Excellent job explaining this and busting that myth. Hopefully, it'll help a lot of people, even if the technical aspect is too much. 

[Agall]

22 minutes ago, RevGAM said:

Excellent job explaining this and busting that myth. Hopefully, it'll help a lot of people, even if the technical aspect is too much. 

Thank you, I added my analogy to the OP with some slight modifications. I think its amusing how I noticed LTT's new video where Linus finally tests his pool cooling idea. In the equations, that's practically speaking giving a 1000x increase in the 'UA' variables in the equation where cooling even 500W is a joke with that amount of heat capacity. Even with that amount of water and flow available, its still going to have to be hotter than the pool to actually provide cooling.

[Agall]

Modified title and header to be less provocative and more informational.

[RevGAM]

17 hours ago, Agall said:

Modified title and header to be less provocative and more informational.

What, you're not going to go Mr. Beast on us???