LLC Effects: Some Mysteries Unveiled?

Hello.

in
Click and scroll down a bit (Thanks, leolotus!) you can see that the heat transfer from aluminum to air is better with airflow than without it, but it reaches its maximum value at 6 m/s and does not increase further above that.

Regarding a liquid-cooled system, it should mean that it becomes fully effective with an outside air flow of around 20 km/h.

If the LLK openings are significantly smaller than the frontal surface area of the LLK, its flow rate will only be a fraction of the vehicle's speed, but once this fraction reaches approximately 20 km/h, the LLK is fully effective .

That could explain why our vehicles can manage with such small LLK (Low-Level Cooling) openings, and why enlarging the openings at "normal" speeds doesn't result in a lower LL (Low-Level) temperature... Rainer had, after all, made relevant + unsuccessful attempts .

I wouldn't see it so generally. What's on the page looks like a heatsink for some electronic component with a fixed heat dissipation. However, LLKs (Liquid Cooling Devices) are completely differently shaped, so the flow conditions are also different, and a separate measurement curve would have to be recorded for them. Or someone could just do a FEM simulation. In any case, this can no longer be calculated using the standard cases of heat and mass transfer 1.
The curve will likely have a similar shape, but it can be shifted in (almost) any desired direction during the design of the cooler.

mcgyver2k wrote:

The curve will probably look similar in principle, but it can be shifted in (almost) any desired direction during the design of the cooler.

Okay, what also seems plausible to me is the variable construction depth of the LLK (likely referring to a specific structure) in the direction of the flow.
The deeper the air goes, the more heat it absorbs from the ground before it reaches the last fin for cooling.

"In that case, the only solution, in my opinion, is to increase the airflow speed so that the air doesn't have as much time to absorb heat and so that the last fin can be cooled more effectively. This means a higher minimum airflow speed is required to achieve full liquid cooling performance compared to flat liquid coolers with a large front surface."

But the principle of "heat transfer from aluminum to air" is the same as in the link.
Therefore, I also believe that LLK curves generally look the same, only potentially with different values on the speed axis.

voooooorsicht!!! (Caution!!!)

The open heat sink, or tube, likely has a uniform boundary layer of air that eventually separates (a laminar boundary layer, or something like that).

Regarding the liquid cooling system, I imagine that it might behave differently due to the high flow rates and small fin spacing, which could result in significantly different thermal resistance.

leolotus wrote:

The open heat sink or tube likely has a uniform boundary layer of air that eventually detaches (a laminar boundary layer, or something like that)!!!

Basically, there are only two types: laminar and turbulent (if I remember correctly from my time sailing model airplanes).
However, even under turbulent flow, there is no vacuum, so heat transfer from the aluminum to the air is still possible.

Quote:

Regarding the LLK (likely referring to a specific type of heat exchanger), I imagine that it might behave differently due to the high flow velocities and small fin spacing, which could result in completely different thermal resistance values.

In the diagram, it's not about absolute thermal resistance, but only about the change in some thermal resistance depending on the speed of the airflow.

It is, in my opinion, undeniable that the air flow in a liquid-cooled engine (LLK) is likely slower due to the fin structure compared to a similar LLK without fins.

However, I believe that the effect of any aerodynamic device initially increases with increasing airspeed, but reaches its maximum at relatively low speeds and remains constant above that -> basically, as shown in the diagram from your link, only with higher numbers on the m/s axis (e.g., double the speeds; in that case, it wouldn't matter whether the aerodynamic device is exposed to an airflow of 50 or 150 km/h, as long as it's above 50 km/h).

laminar / turbulent, that's exactly what I meant.

As long as a laminar (i.e., uniform) flow exists on the surface, the diagram mentioned above is valid. A boundary layer forms, which acts as an insulator. However, in the liquid coolant, there is definitely turbulent flow, meaning that the layers of air are mixed all the way to the metal surface.

Hi,
If you want to know the exact details, it's complicated.
Heat transfer is greater in turbulent flow than in laminar flow because the turbulence constantly brings cooler air closer to the warm surface. In laminar flow, heat transfer primarily occurs through conduction, which is less efficient in air.
The electronic cooler cannot really be compared to the liquid-cooled intercooler. In electronics with the large fins, there is still relatively cool air in between. For plate heat exchangers, what Ulf said in his previous posts applies to the air on the cooling side. However, it's important not to forget that the heat transfer on the charge air side does not improve with increasing speed. Therefore, I can imagine that the power output increases only slightly from a certain speed onwards, even if the fins are very deep and close together.

christians wrote:

However, it must not be forgotten that the heat transfer on the charge air side does not improve with increasing vehicle speed. Therefore, I can imagine that the transmission power increases only slightly from a certain speed onwards, even if the fins are very deep and closely spaced.

Yep, I think the intercooler fins are probably designed to minimize airflow resistance in the intercooler system.
-> In this case, "electronic" conditions (with laminar flow) will prevail, meaning that there will be no further increase in speed above 6 m/s. And that 6 m/s speed is likely to be reached even at idle.
And since the cooling capacity inside does not increase, each cm³ of intake air has less energy extracted from it, the faster it flows through the intercooler -> at maximum power (Pmax), the "internal" cooling effect of the intercooler is at its worst.

But let's go back to the external cooling air side:

Even though the electronic diagram cannot be directly transferred due to the other flow conditions and geometry of the heat exchanger, I still believe that the basic principle of the dependence of the cooling performance on the flow rate will be the same:
So, above a flow velocity of v=0, there is a decreasing external thermal resistance (K/Watt) up to a certain flow velocity, and with further increases in flow velocity, the thermal resistance no longer decreases.

The question is, what is the exact value of this "specific flow rate"?

Or do you fundamentally see things differently at this level?

Hello,

in
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Regarding heat transfer, the explanation about the laminar boundary layer is generally correct. However, with a higher airflow rate, the air passing through becomes less heated, which creates a larger temperature difference across the liquid cooling system. This, in turn, leads to an improved heat transfer rate.

It's also important to remember that the heatsink itself can only dissipate a certain amount of heat over a given time. Anything beyond that cannot be removed by the air. Therefore, this aspect also leads to a limit, no matter how high that limit may be with liquid cooling.

voooooorsicht!!! (or voooooorsicht!!! - caution!!!)

The open heat sink, or tube, likely has a uniform boundary layer of air that eventually separates (a laminar boundary layer, or something like that).

I imagine that with the liquid cooling system, the behavior could be quite different due to the high flow velocities and the small fin spacing, which could result in completely different thermal resistance values.}

The closer the rib spacing, the easier a laminar boundary layer forms. An increase in flow velocity counteracts this. The longer the area over which the flow passes, the more a laminar boundary layer develops. However, this is counteracted constructively through the use of deflectors.

The flow conditions are described by the Reynolds number. Up to approximately 2500, the flow is laminar. The Reynolds number is calculated as: density * velocity * diameter / kinematic viscosity.

@joegolf: The heat flow is determined by the formula k*A*delta T.
So, you can change the amount of heat transferred by altering the temperature difference.

schnappi wrote:

... @joegolf: The heat flow is determined by the formula k*A*delta T. So, you can change the amount of heat being transferred by altering the temperature difference.

Yes, but definitely not below the temperature of the air flowing through it! So, the delta T defines the limit within the cooler itself.

schnappi wrote:

@joegolf: The heat flow is determined by the formula k*A*delta T. So, you can change the amount of heat being transferred by altering the temperature difference!

Hm, I'll interpret that as...
A = effective area for heat transfer.
delta T = temperature difference
k = Thermal conductivity (inverse of the thermal resistance) of the building envelope between the interior and exterior air.

Then, k changes with the flow velocity at the lower limit of the curve (see above).
A is constructively defined.
The delta T can probably not be freely chosen, but is determined by the instantaneous charging pressure and the outside temperature .


By the way, I once tried to calculate the increase in intake air temperature caused by the work of compression.
Result: Each increase in barometric pressure heats the surrounding air by a significant 75.3 Kelvin.

It also fits well with a diagram in a document about industrial TDI: the temperature before the turbocharger } increases from 50 to approximately 127 °C between 1000 and 1900 rpm and then remains constant above that level (like the boost pressure).
-> 1 bar pressure results in a temperature increase of 77°C, regardless of the mass flow rate.

@joegolf: The heat flow is determined by the formula k*A*delta T.
So, you can change the amount of heat being transferred by altering the temperature difference!
Hm, I'll interpret that as...
A = effective area for heat transfer.
delta T = temperature difference
k = Thermal conductivity (inverse of the thermal resistance) of the building envelope between the interior and exterior air.

Then, k changes with the flow velocity at the LLK (see above).
A is constructively defined.
The delta T can probably not be freely chosen, but is determined by the instantaneous charging pressure and the outside temperature .


By the way, I once tried to calculate the increase in intake air temperature caused by the work of compression.
Result: Each increase in barometric pressure heats the surrounding air by a significant 75.3 Kelvin.

It also fits well with a diagram in a document about industrial TDI: the temperature before the turbocharger increases from 50 to approximately 127 °C between 1000 and 1900 rpm and then remains constant above that level (like the boost pressure).
-> 1 bar results in a temperature increase of 77°C here, regardless of the mass flow rate}.

k is the heat transfer coefficient! k = 1/(1/αᵢ + s/λ + 1/αₐ)
Here, 'i' represents the inner side, 'a' represents the outer side, 'alpha' is the heat transfer coefficient, 's' is the wall thickness, and 'lambda' is the thermal conductivity of the cooler material.

alpha, in turn, is a function of the Nusselt number, which provides information about the flow conditions (e.g., laminar boundary layer thickness, etc.). Explaining everything in detail would take too long!

schnappi wrote:

k is the heat transfer coefficient! k = 1/(1/alpha i + s/lambda + 1 / alpha a) Here, 'i' represents the inner side, 'a' represents the outer side, 'alpha' is the heat transfer coefficient, 's' is the wall thickness, and 'lambda' is the thermal conductivity of the cooler material. alpha, in turn, is a function of the Nusselt number, which provides information about the flow conditions (e.g., laminar boundary layer thickness, etc.). It would take too long to explain everything in detail!

Hmm, so it seems that ultimately, these are parameters that a regular driver can only influence "indirectly" through their driving style or speed.

Or, as a "non-standard" driver, by modifying/replacing the engine control unit (ECU) .

What would be the actual benefit of additionally spraying the LLK (presumably a type of material) with water droplets?
Has anyone ever tested this before?

I would like to repurpose a portion of the unconnected SRA and then, for a while, pour distilled water into the washing water pre-tank!