Air Permeability Changes from LLM: How Much?

Hello,

To avoid completely overwhelming the LLK puzzle thread, here's a new (and slightly different) topic.

I am trying to understand the impact of a Large Language Model (LLM), including its potential for a new or expanded research article, in quantitative terms.

The starting point is that the turbocharger, as the first stage of density increase, heats the air it draws in by approximately 75°C for each bar of pressure increase, which I was able to calculate based on some formulas and characteristics of the air itself (I hope this is correct so far).

The liquid cooler reduces the temperature of the heated air, bringing it closer to ambient temperature, which further increases the air density. However, I haven't found any information regarding the quantitative impact of this effect.

Therefore, my question is:
For example, if air at an outside temperature of 0°C and a charge pressure of 1 bar, with a theoretical temperature of 75.3°C and a doubled density of 2.58 kg/m³, passes through the low-temperature coolant (LLK) and the LLK cools it down:
By how much does the air density increase again per Kelvin (or per 10 Kelvin) ?

It's possible that I'm just not understanding something, but at this point, I keep getting stuck...

WARNING: Ulf.

You need to specify exactly WHERE you have the 1 bar of boost pressure.

The older TD engines used to regulate the turbocharger's boost pressure. Therefore, you had to rev the engine higher after installing an intercooler. Otherwise, the pressure at the cylinder head would actually be lower than it was before the intercooler was installed!

"Our TDI sensors measure the pressure at the intake, regardless of how it's affected by factors like heating/expansion or cooling by the coolant, which causes contraction/tightening/pressure loss. The control system doesn't take these factors into account."

If your intercooler suddenly became "better," your turbo would immediately need to draw in significantly more air (i.e., spin faster). Because if the air behind the intercooler becomes denser, its pressure decreases (since the volume of the space, like the pipes, remains the same).

Was that understandable? (Sorry, I can't express it better - maybe I'll try to explain it better later tonight.)

- I can't imagine 75 K/bar (that seems a bit too much).

- The invoice is missing the heat output from the turbine (which causes the air to expand). This creates a pressure increase without an increase in mass.

m;

Hello Ulf,

As a first approximation, you can consider the air to be an ideal gas. The air density is proportional to the absolute temperature at -273.15 Celsius. This results in an increase in density when the temperature decreases from 75 to 0, approximately by a factor of 75/273, which is about 0.274.

Best regards,
Christian.

dieselmartin wrote:

You need to specify exactly WHERE you have the 1 bar of boost pressure... . . . Our TDI sensors measure the pressure at the intake, and

regardless of how the previous process (warming -> expansion / cooling by coolant -> contraction / tighter / pressure loss!!) occurs,

Quote:

the control system doesn't care. Yep, I meant like with the TDI.

If your intercooler suddenly starts working "better," your turbo would immediately need to draw in significantly MORE air (i.e., spin faster). Because if the air behind the intercooler becomes denser, its pressure decreases (because the space, i.e., the pipes, etc., remains the same).
Was that understandable? (Sorry that I can't express it better - maybe I'll do some calculations tonight).

Quote:

It's already okay - I've already written something similar when I "decoded" "LLK"

.

- I can't imagine the 75 K/bar (that seems a bit too much).

- The invoice is missing the heat output from the turbo (which causes the air to expand. This creates a pressure increase without an increase in mass.

So, the 75 K/bar value, which is a theoretical figure, comes from the assumed equation "compression work = heating energy."

Unraveled:

Volume * Pressure increase = Mass * Specific heat capacity * Temperature increase.

Volume is replaced by mass / density, which results in:

Pressure increase * volume / density = mass * specific heat capacity * temperature increase.

Mass cancels out.
Pressure increase / Density = Specific heat capacity * Temperature increase.

Rearranged: }Temperature increase = Pressure increase / (Density * specific heat capacity)}

Using the values for air (1.29 kg/m³ and 1.01 kJ/(kg*K), this results in the aforementioned 75.3 K/bar.


The order of magnitude seems correct to me.

In a diagram showing the performance of an industrial turbocharger with intercooling (TDI/AFD), the intake air temperature before the intercooler increases with the boost pressure, starting at 50°C (idle) and reaching approximately 127°C at 1900 rpm (above this, it remains constant).
So, here it's approximately 77°C / bar, likely including the heating caused by the charger.

2. Am I familiar with the dry adiabatic ascent of air in the atmosphere (due to thermal convection)?
For every 100 meters of altitude, the air temperature decreases by 1°C (rule of thumb 1).
The barometric altitude is 8 meters per millibar at sea level (rule of thumb).

If you combine both rules of thumb, you can even reach approximately 80 K/bar "in nature's laboratory."

chris11 wrote:

As a first approximation, you can consider the air as an ideal gas. The air density is proportional to the absolute temperature at -273.15 Celsius. This results in an increase in density when the temperature decreases from 75 to 0, which is approximately 75/273 = 0.274.

Aha

This means that as a general rule, one can assume:

An intercooler that reduces the air temperature by 75 K replaces 0.27 bar of boost pressure, or, translated into more easily understandable values:
A cooling reduction of 10K is equivalent to a 37 mbar reduction in boost pressure...??

Hello,

Okay, based on the available data, we can only calculate this with a very rough approximation. More generally, we're talking about flows here, so not volumes, but flow rates, etc.

For accurate calculations, a much larger amount of data is required. Even for calculations involving an idealized process, this is the case. For a real-world scenario, even more data is needed, such as pressure and heat loss in the pipelines, etc.

Essentially, these are actual facts upon which one can build.

Since the temperature of the intake air after the turbocharger is not only determined by the compression within the turbocharger but also by the heat from the exhaust gases acting on the turbocharger itself, you need a fixed, measured value. Otherwise, you'll just be going in circles. The pressure of the EGR system should also be taken into account.

Here's an example calculation, assuming ideal conditions: You determine the temperature of the air after the charger.
To do this, one determines the pressure. This should allow you to calculate the density and speed of the air in the system (assuming air is an ideal gas, which is only an approximation).

Then, it is also possible to calculate the air mass flow rate (or volumetric flow rate) in the system.

Then, the outside temperature and the vehicle's speed are needed to calculate the mass flow rate of the outside air and, consequently, the energy flow.

Then, by comparing energy values, one can determine the temperature, pressure, and other parameters within the system.

Okay, something like this should work. I unfortunately don't have a book handy to check it. Of course, you still need to extract values from some tables and diagrams, especially if you don't assume air is an ideal gas and you take humidity into account.
Then, there are probably still some intermediate steps required.
It depends on how precise the values need to be.
With that, you would have a static value. However, by using functions instead of fixed values, you can obtain meaningful results.
Or one can simply settle for the key aspects of an idealized process.

I believe the 'Industrie TDI' refers to an engine where the external mass airflow is not variable.
Given that the cooling performance of a liquid-cooled system in a car naturally increases with speed, I would be cautious about making a direct comparison.

Greetings.

With an ambient temperature of 0°C, it will be difficult to bring the temperature of the charging circuit below approximately 20°C, as a temperature difference is always needed for heat exchange. Given the large flow rate and the relatively small heat exchange surface area, this seems reasonable. Otherwise, when working with the ideal gas law, the effect of pressure is negligible for air at low pressures!
Best regards, Schnappi.

Roadagain wrote:

Hello, Therefore, based on the available data, this can only be calculated with a very rough approximation. Bitte gib mir den deutschen Text, den du übersetzt haben möchtest. To perform an accurate calculation, you also need a lot more data. Even just to calculate with an idealized process. For a real-world scenario, you need even more, such as pressure and heat loss in the pipes, etc.

Hi,

I understand that with such relatively complex processes, you only get truly precise results when you consider "a thousand and one parameters."

On the other hand, IMO can often establish useful heuristics, even for complex situations, by only considering the most important influencing factors (and thus reducing the computational effort to a fraction ).

For example, during the heating of the air between the compressor and the low-pressure cooler, heat transfer through the compressor itself likely plays a minor role, so the rule of thumb "75K per bar" is probably sufficient for the level of discussion we're having.

And I wanted to have a comparable rule of thumb for the change in air density due to cooling at constant pressure.
(I am aware that a real air conditioning system can never cool the air down to the exact temperature of the outside air.)

If you can confirm (or correct) that the rule of thumb "10 kPa of cooling corresponds to approximately 37 mbar of boost pressure" in terms of air density, that would be helpful.

Then I wonder why anyone bothers with the effort of cooling the LL (likely referring to a liquid-cooled component).

For example, if you start with 1 bar of boost pressure (= 75K temperature increase) and assume a possible cooling of 50K, you could achieve the same effect in terms of air density by running 0.18 bar more boost pressure - minus the pressure loss caused by the intercooler itself (are there any empirical values or rules of thumb for this?).

Quote:

Regarding the industrial TDI engine, I believe it's a type of engine where the external mass airflow is not variable. Since the engine oil cooler in a car logically increases its cooling performance the faster you drive, I would be cautious about making a direct comparison.

Probably not; the peripheral descriptions, such as the location of individual components in the engine compartment, suggest that it's intended for installation in a vehicle.
The LL temperature chart does not provide any information regarding the vehicle's speed.

Then I wonder why anyone bothers with the effort of LL cooling:

For example, if you start with 1 bar of boost pressure (= 75K temperature increase) and expect a cooling effect of 50K, you could achieve the same effect in terms of air density by running 0.18 bar more boost pressure - minus the pressure loss caused by the intercooler itself (are there any empirical values or rules of thumb for this?).



Hello Ulf,

Perhaps the key to understanding LLK (Low-Temperature Combustion) also lies in the compression temperature within the cylinders. If the incoming air is 50K warmer, how much warmer is the air at maximum compression?
It is possible that this will then have negative effects on the district processes, or on the efficiency and exhaust emissions.

Greetings.

Eike.

Hello,

Before I can say anything specific about the rule of thumb, I still need to get a book.
This week, I'm going to the bookstore anyway, so I'll take a look.

And the reason is that the air-tight seal decreases with warming. Therefore, less air enters the cylinders. Consequently, less fuel can be burned. Thus, less energy can be released while maintaining the same desired lambda value. Therefore, the engine's efficiency decreases.
When you cool the air, you get more air (especially more oxygen), which allows you to burn more fuel and release more energy per combustion. The result is more power, more torque, and better efficiency.

Greetings.

Roadagain wrote:

And the point is that the air tightness decreases when heated. Therefore, less air enters the cylinders.

Understood.
The question was, why would someone use a liquid coolant (LLK) as an additional component to increase density, instead of simply adjusting the boost pressure with the already existing hardware to a level that allows the air to enter the combustion chambers with the desired density, even without cooling.

Eike brought up an interesting point, which I unfortunately can't comment on.

The question was, why would someone use a liquid coolant (LLK) as an additional component to increase density, instead of simply adjusting the boost pressure with the already existing hardware to a level that allows the air to enter the combustion chambers with the desired density, even without cooling.



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- ignition pressure
- thermal load
- compressed drive power
- thermal efficiency
- ...

CU Gremlin.

Gremlin wrote:

ulf wrote: The question was, why would someone use a low-temperature coolant (LTC) as an additional component to increase density, instead of simply adjusting the boost pressure with the already existing hardware to a level that allows the air to enter the combustion chambers with the desired density, even without cooling. - firing pressure. - thermal load - compressed drive power - thermal efficiency - ...

Ah, so it's probably due to solid, technical reasons related to the engine.
While I may not be able to fully develop the answer based on the keywords provided (and may even need some help to understand it), my question has been answered - thank you!

eike wrote:

It is possible that this will then have negative effects on the district processes, or on the efficiency and exhaust emissions. Greetings. Eike

Yes, the amount of work required for compression increases as the intake temperature rises. For example, the (idealized) Carnot process operates with isothermal compression.

The warmer the air, the lower the density, which means less oxygen, and therefore a lower efficiency.

I already said that. So, the warmer the air, the less clean air you can pump in, because the pressure is already higher at higher temperatures.

Greetings.

dieselmartin wrote:

The older TD models used to regulate the turbocharger's boost pressure, which is why you had to increase the RPM if you installed an intercooler. Otherwise, the amount of (pressure) reaching the cylinder head would be less than it was before the intercooler installation.

Hi Martin,

Contrary to my previous assumption, my intercooler also has internal cooling fins (which can be seen when the hose is removed). While these fins can dissipate a relatively large amount of heat from the intake air, they also create significant turbulence, which increases the flow resistance and thus leads to a loss of boost pressure due to the intercooler.

Do you or someone else know how much pressure a normally sized LLK (Linear Low-Density Polyethylene) consumes at approximately Pmax?