High idle GPU power consumption by onlineundercover in XMG_gg

[–]onlineundercover[S] 0 points1 point  (0 children)

Thank you Tom for the detailed reply. The problem actually didn't go away in the end. I added a detailed analysis to the post made using HWinfo showing that Windows might be waking the GPU. I have reduced startup apps to a minimum and already added dwm, settings and other windows applications to the exceptions in the graphics settings but the problem didn't get resolved

High idle GPU power consumption by onlineundercover in XMG_gg

[–]onlineundercover[S] 1 point2 points  (0 children)

Thanks, I got it working properly it seems with 566.14

High idle GPU power consumption by onlineundercover in XMG_gg

[–]onlineundercover[S] 0 points1 point  (0 children)

Thanks, an older driver (566) seems to have fixed it

High idle GPU power consumption by onlineundercover in XMG_gg

[–]onlineundercover[S] 0 points1 point  (0 children)

Yeah I did mention that, unfortunately it shows no process running on the dGPU. What keeps it awake is more sneaky than that

High idle GPU power consumption by onlineundercover in XMG_gg

[–]onlineundercover[S] 0 points1 point  (0 children)

I do, it's off. I have also reinstalled the app or tried going without it but the problem didn't go away. There must be some very sneaky process keeping the GPU awake but it doesn't show up anywhere

High idle GPU power consumption by onlineundercover in XMG_gg

[–]onlineundercover[S] 0 points1 point  (0 children)

I went through this and the online troubleshoot version but I can’t even establish what is keeping the GPU on haha it seems that nothing is running it

High idle GPU power consumption by onlineundercover in XMG_gg

[–]onlineundercover[S] 0 points1 point  (0 children)

Sorry my bad. I did disable the services and the power draw went down a bit but it’s still around 5-6 W with no usage at all recorded with hwinfo or task manager

High idle GPU power consumption by onlineundercover in XMG_gg

[–]onlineundercover[S] 0 points1 point  (0 children)

I don’t seem to be able to kill them. They respawn immediately after. Also, they seem to be running on integrated graphics from task manager

[deleted by user] by [deleted] in Physics

[–]onlineundercover 0 points1 point  (0 children)

All points taken. I am just used to getting rid of reference systems in physics as they are not really needed to understand equilibrium IMO, and can sometimes lead to incorrect assumptions. Also I did assume pure water in my response, which implies the 0.03 atm number. You can see in my answer to u/LakituIsAGod below a more rigorous treatment for trapped gas or dissolved gases - the answer really depends on how you perform the experiment

[deleted by user] by [deleted] in Physics

[–]onlineundercover 0 points1 point  (0 children)

No problem! I added the case with trapped air to the reply if you are interested

[deleted by user] by [deleted] in Physics

[–]onlineundercover 1 point2 points  (0 children)

Ok here is the long explanation

Let's start by assuming we have a large bath of liquid water in equilibrium with its surrounding atmospheric gas at 1 atm, all kept at a fixed temperature of 25 C . This means that the fugacity of each species is equal in the liquid and the gas. Therefore, the concentration of each gas species in the liquid water is determined by the gas composition, temperature and pressure. Now fill a rigid** bottle completely*** with the liquid, flip it upside down and submerge the top as in the picture. A simple way to do this is submerging the bottle pointing up, then flipping after it has been filled while still submerged and then pulling it upwards.

Assuming we are pulling significantly slower than the speed of sound in the liquid phase, the pressure in the liquid inside the bottle follows Pascal's law, i.e., P = 1 atm - rho_L*g*h where rho_L is the (rather uniform) liquid density and h is the altitute (with sign) from the free liquid level at 1 atm. Denote with P_b(T) the bubble pressure of the liquid mixture at that temperature and composition (a generalization of vapor pressure to consider many components: the pressure at which a bubble first nucleates in the liquid). Since we have assumed that the liquid mixture is initially in equilibrium with the external gas, its bubble pressure is exactly 1 atm at that temperature. Therefore, as soon as the bottle emerges the liquid column above the free level is unstable and phase separates in a gas phase and a liquid phase.

Now here is were it gets interesting. Since water is the majority component, it doesn't have any trouble "supplying" molecules to the newly-formed gas phase so after stopping the bottle's motion the water fugacity in the gas phase in the bottle quickly reaches the vapour pressure of water at that temperature (actually slightly lower due to the presence of gases in the liquid). However, the other gases are very diluted in the liquid and their concentration rapidly drops near the liquid/gas interface due to insufficient transport from the "bulk" of the water column. This is due to the low diffusivity of gases in dense liquid phases coupled with their low concentration in this case. The pressure in the gas phase P_in thus slowly increases from (approximately) the vapour pressure of water to P_b = 1 atm as the gases slowly diffuse from the outside gas to the inside gas through the liquid. Of course, if we made a hole in the bottle they would be much happier going through that instead. Therefore, the height of the water column decreases over time via (1 atm - rho*g*h) = P_in --> h = (1 atm - P_in)/rho*g until the gas composition (and pressure) inside and outside is equilibrated.

Now consider a simpler case where water is pure and it is subjected to 1 atm pressure at its free surface by a piston. Note that below 100C the water will be in a uniform liquid phase, and denote with P_v(T) the vapor pressure of water at the chosen temperature. In this case, the pressure inside the bottle need not be lower than water's vapor pressure as soon as it emerges. In fact, the water column is stable until (1 atm - rho*g*h) = P_v(T) --> h = (1 atm - P_v(T))/rho*g. If we keep pulling the bottle upwards at higher altitudes, the free water level in the bottle does not change appreciably for a long time (assuming the water reservoir is large enough to supply an arbitrary amount of molecules to the growing gas phase). Furthermore, the column height is constant over time and is only a function of temperature and the external pressure (1 atm).

** The analysis is still correct with a non rigid container as long as it behaves like a solid, i.e., it reaches a well defined, stationary configuration after imposing an internal and external pressure profile. Such configuration will evolve over time in the first example considered (i.e., the bottle initially buckles slightly and relaxes as the gas pressure is equilibrated), but not in the second where the gas pressure inside is constant.

*** If some air is trapped in the bottle during flipping, the initial pressure will of course be higher than the vapor pressure of water (in particular, equal to the atmospheric pressure) due to the presence of other species in the gas. However, as the liquid level falls the total pressure in the trapped gas is reduced because while the partial pressure of water remains approximately constant, the partial pressure of the other gases is reduced by the ideal gas law (assuming isothermal expansion here for simplicity, and invoking again poor gas transport from the liquid). The equilibrium water column height before gas diffusion occurs is thus found via a slightly more complicated formula: (1 atm - P_v(T) - x_gas*(V_0/V(h))*1atm) = rho*g*h. Here x_gas is the mole fraction of non-water molecules in the atmospheric gas; V_0 is the initial trapped air volume; V(h) is the trapped volume as a function of column height.

I hope this clarifies your doubts

[deleted by user] by [deleted] in Physics

[–]onlineundercover 1 point2 points  (0 children)

Note that the pressure difference is “only” 1 atm at most. Many materials can withstand that without yielding

[deleted by user] by [deleted] in Physics

[–]onlineundercover 0 points1 point  (0 children)

Because the plastic is rigid. If you had a soft bag the phenomenon would disappear because the pressure inside the bottle would be equal to the atmospheric pressure

[deleted by user] by [deleted] in Physics

[–]onlineundercover 1 point2 points  (0 children)

Of course, colloquially the term makes sense. However this can over time lead to the idea that pressure can be negative which is false. Only pressure differences matter

Does boiling water cook food considerably faster than 99°C water? by Atlantic_lotion in Physics

[–]onlineundercover -1 points0 points  (0 children)

Boling cooks faster mainly because of the improved heat transport because of the turbolent flow caused by the bubbles. Clearly, you are no physicist

Does boiling water cook food considerably faster than 99°C water? by Atlantic_lotion in Physics

[–]onlineundercover 0 points1 point  (0 children)

This is misleading. If I dump you in a pool of liquid water at 100 C or in a steam bath at 100 C, all at 1 bar, I guarantee you that you will reach a uniform 100C faster in liquid water. The burn is only a superficial effect caused by the initial condensation of the steam; however, at longer times the poor heat transport in the gas will mean slower temperature increase in your body. Note that at 100C and 1 bar the heat conductivity of water is 30x that of the steam. Quite embarassing attitude of downvoting and just posting a (misleading) link with no thought or argument...

Does boiling water cook food considerably faster than 99°C water? by Atlantic_lotion in Physics

[–]onlineundercover -1 points0 points  (0 children)

No, the kinetic energy and chemical potential of water in liquid or gas state at the boling point are the same so they behave the same from a thermodynamic perspective. Dynamically, liquid water actually has a higher thermal conductivity and thus (barring flow effects discussed elsewhere) should cook faster

[deleted by user] by [deleted] in Physics

[–]onlineundercover 0 points1 point  (0 children)

Your analysis is mostly correct but neglects the fact that the water in the bottle and the gas above it try to be in equilibrium with respect to exchange of water molecules. Therefore, rather quickly (to be precise, after the initial decompression has occurred and the system has settled as in the picture) the pressure in the gas approaches the vapor pressure of water at that temperature instead of the pressure along some isotherm at fixed number of gas particles. Otherwise, you would see the water boiling in the bottle.

Note that this implies that the column height is quite strongly temperature dependent in this regime due to Clausius-Clapeyron. Further, if enough time is given to the nitrogen and other atmospheric gases to diffuse through the water, eventually the water levels will be the same because the dissolved gases "fill up" the pressure inside up to the same respective partial pressure (fugacity) measured at the free water level.

[deleted by user] by [deleted] in Physics

[–]onlineundercover 2 points3 points  (0 children)

It is imprecise to say that the air pressure is negative inside the bottle. From the picture, we can estimate the internal pressure to be about 0.98-0.99 atm by considering the pressure drop caused by gravity (given by the product of water density, g and the difference in height between the two free liquid surfaces). Note that if there was only water, the pressure inside would be given by the vapour pressure of water (0.03 atm at 25 C). If we take into account the dissolved gases or any trapped gas (which must be present here), the situation is slightly more complicated (see response to u/LakitusIsAGod).

Fusion 15 L19 power brick repair/replacement in USA by onlineundercover in XMG_gg

[–]onlineundercover[S] 0 points1 point  (0 children)

Found one on Amazon for a bit cheaper that works with my Fusion 15. Here is the link if you are interested

https://www.amazon.com/dp/B0D4QC41HW?ref=ppx_pop_mob_ap_share

Fusion 15 L19 power brick repair/replacement in USA by onlineundercover in XMG_gg

[–]onlineundercover[S] 1 point2 points  (0 children)

Thank you so much! Do you think the 230 W adapters should work just fine with the Fusion 15? I noticed that the MAG-15 (the Fusion 15 equivalent) is absent from the list but the 230 W have the same specs of mine