[Bmwguy11]

Originally Posted by Overboost

Obviously it doesn't work like this in real life because we would never be able to keep wastegate duty cycle and turbo rpm or environmental conditions constant. If we could then the heat energy given off by the turbos will be constant and the increase in temps would be also be somewhat constant since change in heat capacity is not that much at a delta of a few degrees. Take everything i say in the context it was given please.

Friction and inertia etc would come in to play in this but again i said keeping all things CONSTANT

But I just showed you actual numbers keeping all other things "constant" that disproved what you said. I'm just trying to correct misinformation. 8 degree temp difference pre-turbo, "all other things constant", will not equate to 8 degree temp difference at turbo outlet. That's not how thermodynamics works. You implied you took physics courses, which is what you were using for the basis of your knowledge in the post...

[vasillalov]

Overboost:

Excellent comments!

[happy4nk8er]

A typical flat plate air to air heat exchanger has a 60% or thereabouts transfer efficiency.

[Bmwguy11]

Originally Posted by Overboost

Intercoolers cool by conduction and convection which change with the environment so there is no way that the percentage efficiency can always be the same. As i said before heat capacity of any form of matter usually changes with temperature. Cooler air providing greater efficiency than hotter air because it has a scientific greater specific latent heat capacity. You are right though. Hotter air does have a greater delta because the heat capacity is lower which means it also loses heat faster but that is not the point of since one can argue that the hotter air will raise the intercooler temperature faster also reducing its efficiency. Intercooler efficiency simply cannot be constant.

Intercooler efficiency in the same ambient air temp is typically constant until the point of heat soak. Its efficiency is not really affected by the temperature of the air being pushed into it, unless it is nearing the threshold. In fact, ambient air temp has a greater effect on IC efficiency than anything.

[Overboost]

Originally Posted by Bmwguy11

But I just showed you actual numbers keeping all other things "constant" that disproved what you said. I'm just trying to correct misinformation. 8 degree temp difference pre-turbo, "all other things constant", will not equate to 8 degree temp difference at turbo outlet. That's not how thermodynamics works. You implied you took physics courses, which is what you were using for the basis of your knowledge in the post...

Sorry I went through the link you posted and didnt see any number keeping the conditions i stated constant. I saw the formula we use PV=nRT as its written in my text books but that has more to do with pressure and volume than it than it does heat transfer. There were no actual calculation on that link. If you can point them out to me I'd be happy to admit I am wrong. We are all learning here and i'm no nasa scientist.

[vasillalov]

Originally Posted by happy4nk8er

A typical flat plate air to air heat exchanger has a 60% or thereabouts transfer efficiency.

Any reputable source as a proof for that? ... I am just wondering how on earth some FMIC vendors here claim 80%+ efficiency if what you say is true?

..on the flip side: if the true efficiency is about 60%, then this means that Cold Air Intakes are even more important!

[Overboost]

Originally Posted by Bmwguy11

Intercooler efficiency in the same ambient air temp is typically constant until the point of heat soak. Its efficiency is not really affected by the temperature of the air being pushed into it, unless it is nearing the threshold.

Agreed, but thats given that you can keep the metal in the intercooler at the same temperature. I'm not disproving anything you say but since you understand a bit of what I am saying you know that in real world situations it becomes a bit more complicated to calculate than the simple way i tried to state it so most on the forum can understand.

[Overboost]

The other thing Bmwguy11 that I would have to ask you is. What would you say the point of heat soak is? For me heat is always being transferred to the intercooler so in essence it is always soaking up heat. As long as air temperature is above intercooler temperature, the direction of heat will always be towards the metal in the intercooler.

[Bmwguy11]

Originally Posted by Overboost

Sorry I went through the link you posted and didnt see any number keeping the conditions i stated constant. I saw the formula we use PV=nRT as its written in my text books but that has more to do with pressure and volume than it than it does heat transfer. There were no actual calculation on that link. If you can point them out to me I'd be happy to admit I am wrong. We are all learning here and i'm no nasa scientist.

Here's another good resource for formulas.

http://www.stealth316.com/2-adiabat1.htm

[Bmwguy11]

Originally Posted by vasillalov

Any reputable source as a proof for that? ... I am just wondering how on earth some FMIC vendors here claim 80%+ efficiency if what you say is true?

..on the flip side: if the true efficiency is about 60%, then this means that Cold Air Intakes are even more important!

Typical FMIC's these days are around 70-80%.

[Bmwguy11]

Originally Posted by Overboost

The other thing Bmwguy11 that I would have to ask you is. What would you say the point of heat soak is? For me heat is always being transferred to the intercooler so in essence it is always soaking up heat. As long as air temperature is above intercooler temperature, the direction of heat will always be towards the metal in the intercooler.

The point of heat soak is the point at which the efficiency of the intercooler drops below what it would "normally" work at given the same ambient air temp. In other words, it's the point at which the intercooler no longer removes as much heat from the air passing through it as it should. Heat soaking an IC on a moving vehicle is not a common thing unless you're running a hard track day maybe...

I'm trying to find the article/thread on s2ki.com that had the data back from when I was a moderator there... I apologize I haven't been able to get it quicker. :/

[Overboost]

Originally Posted by Bmwguy11

Here's another good resource for formulas.

http://www.stealth316.com/2-adiabat1.htm

All good reads. Thanks for the information. Both links are great but do not disprove what I said. In the last link they basically show that temperature increase is exponential in regard to turbo output. Both link vary different factors than what I proposed. First link varied volume and pressure, second link varied turbine output. I varied only temperature in my hypothetical situation keeping all other forces constant. In essence keeping the turbo and motor in a hypothetical steady state. If you have anymore links to post please do. The more information we can get the better it will be for the community.

[happy4nk8er]

Originally Posted by vasillalov

Any reputable source as a proof for that? ... I am just wondering how on earth some FMIC vendors here claim 80%+ efficiency if what you say is true?

..on the flip side: if the true efficiency is about 60%, then this means that Cold Air Intakes are even more important!

I design air handling equipment with flat plate air to air energy recovery systems. These systems are based on airflow velocities of 750 fpm to 1500 fpm thru the heat exchanger. I have not calculated the velocity thru our IC's but no matter how much I massage the design I can't get above 65% efficiency. So I am estimating that our IC's are probably near that. 80% efficient I don't buy that.

[Joshboody]

Originally Posted by Overboost

The other thing I forgot to mention although i can't write is as a fact is that. Wastegate duty cycles could be lower for DCI simply because there is an increase in temperature. We know that and increase in temperature for any given fluid in a container or closed environment increases the pressure. Since we target absolute pressure and not air mas this means that the intake that ingests hotter air can essentially hit the boost target at a lower turbine rpm because hotter air is aiding in the pressure increase.

I can add this to the first post if the community agrees.

compressor moves air volume, NOT mass (volume and pressure have direct relationship). In actuality I would think higher temps yield higher WGDC due to decreasing compressor efficiency with density (not sure though)... something to do with momentum, inertia, centripetal forces. I bet you will see higher WGDC in the summer then winter at same psi.

WGDC is actually quite complicated and does NOT have a direct relationship to flapper position due to exhaust pressures.

EDIT: i have made a point recently to always log WGDC when datalogging.... good thing to keep track of.

[Overboost]

Originally Posted by Joshboody

compressor moves air volume, NOT mass. In actuality I would think higher temps yield higher WGDC due to decreasing compressor efficiency with density (not sure though)... something to do with momentum, inertia, centripetal forces. I bet you will see higher WGDC in the summer then winter at same psi.

WGDC is actually quite complicated and does have a direct relationship to flapper position due to exhaust pressures.

Everything has a mass. Density is equal to the Mass divide by the volume. I was basically saying we do not know how much air we are moving by weight only how much space it takes up (volume). If we used the links that Bmwguy posted, PV=nRT then we can see the direct relation between Pressure (manifold or charge pipepressure) and Temperature is direct in relation to each other. That is an increase in temps equals an increase in pressure at any point in the system.

Volume then become Mass divide by density.

[BoostinBimmer]

Originally Posted by Overboost

Intercoolers cool by conduction and convection which change with the environment so there is no way that the percentage efficiency can always be the same. As i said before heat capacity of any form of matter usually changes with temperature. Cooler air providing greater efficiency than hotter air because it has a scientific greater specific latent heat capacity. You are right though. Hotter air does have a greater delta because the heat capacity is lower which means it also loses heat faster but that is not the point of since one can argue that the hotter air will raise the intercooler temperature faster also reducing its efficiency. Intercooler efficiency simply cannot be constant.

Actually the efficiency does stay the same because for fins (which an intercooler is essentially composed of) the efficiency is measured by the equation: dQfin/dQfin,max. Which is the Actual heat transfer rate through convection and conduction/ The Ideal heat transfer rate if the fin were at base temperature

knowing this, as well as the equation for dQfin,max= hAfin(Tb-Tamb).

Where h is a constant, Afin is the surface area of the fin and Tb= base temp and Tamb= ambient temperature.

Also, the equation for dQfin= Sqrt(hPKAc)*(Tb-Tamb). where h and K are constants, Ac is the cross-sectional area of the fin, and P is the fin perimeter.

So when you divide dQfin/dQfin,max the temperature components (Tb-Tamb) cancel eachother out:

so, Fin (intercooler) efficiency = Sqrt(hPKAc)/hAfin

Which is not temperature dependent.

This IS different than the effectiveness of the intercooler which is temperature dependent often referred to as the efficiency and follows the equation:
Eff= T2-T3/T2-T4

where T2= Turbo outlet temp
T3= Intercooler exit temp
T4= Temperature of the intercooler (which depends on heatsoak and
ambient temperature)

[Bmwguy11]

Originally Posted by Overboost

All good reads. Thanks for the information. Both links are great but do not disprove what I said. In the last link they basically show that temperature increase is exponential in regard to turbo output. Both link vary different factors than what I proposed. First link varied volume and pressure, second link varied turbine output. I varied only temperature in my hypothetical situation keeping all other forces constant. In essence keeping the turbo and motor in a hypothetical steady state. If you have anymore links to post please do. The more information we can get the better it will be for the community.

I guess if that link can come up with something that does account for a few more variables, isn't that a more accurate a representation, while still keeping other harder-to-nail-down variables "equal"? As we remove more and more variables, the less and less it applies to real-world scenario. My .02.

[Overboost]

Originally Posted by BoostinBimmer

Actually the efficiency does stay the same because for fins (which an intercooler is essentially composed of) the efficiency is measured by the equation: dQfin/dQfin,max. Which is the Actual heat transfer rate through convection and conduction/ The Ideal heat transfer rate if the fin were at base temperature

knowing this, as well as the equation for dQfin,max= hAfin(Tb-Tamb).

Where h is a constant, Afin is the surface area of the fin and Tb= base temp and Tamb= ambient temperature.

Also, the equation for dQfin= Sqrt(hPKAc)*(Tb-Tamb). where h and K are constants, Ac is the cross-sectional area of the fin, and P is the fin perimeter.

So when you divide dQfin/dQfin,max the temperature components (Tb-Tamb) cancel eachother out:

so, Fin (intercooler) efficiency = Sqrt(hPKAc)/hAfin

Which is not temperature dependent.

This IS different than the effectiveness of the intercooler which is temperature dependent often referred to as the efficiency and follows the equation:
Eff= T2-T3/T2-T4

where T2= Turbo outlet temp
T3= Intercooler exit temp
T4= Temperature of the intercooler (which depends on heatsoak and
ambient temperature)

From you formula i agree that fin efficiency stays the same and also agree that as a whole intercooler efficiency does not stay the same since it depends on heat soak. The fins are very thin and transfer heat from the air to the bar or plate of the intercooler to the fins that are exposed to ambient air correct me if i'm wrong. What I think i should have said is the rate at which the fins transfer heat to the bar or plate of the intercooler changes as the heat of the whole intercooler changes.

[vasillalov]

Originally Posted by happy4nk8er

I design air handling equipment with flat plate air to air energy recovery systems. These systems are based on airflow velocities of 750 fpm to 1500 fpm thru the heat exchanger. I have not calculated the velocity thru our IC's but no matter how much I massage the design I can't get above 65% efficiency. So I am estimating that our IC's are probably near that. 80% efficient I don't buy that.

Thank you for contributing!

[Overboost]

Originally Posted by Bmwguy11

I guess if that link can come up with something that does account for a few more variables, isn't that a more accurate a representation, while still keeping other harder-to-nail-down variables "equal"? As we remove more and more variables, the less and less it applies to real-world scenario. My .02.

Think of it as my scenario being instantaneous. If you plot each instantaneous point changing one variable at a time then you derive a dynamic curve. Again thanks for the information.

[BoostinBimmer]

Originally Posted by Overboost

From you formula i agree that fin efficiency stays the same and also agree that as a whole intercooler efficiency does not stay the same since it depends on heat soak. The fins are very thin and transfer heat from the air to the bar or plate of the intercooler to the fins that are exposed to ambient air correct me if i'm wrong. What I think i should have said is the rate at which the fins transfer heat to the bar or plate of the intercooler changes as the heat of the whole intercooler changes.

+1 this is due to the temperature gradient as the driving force. As the two temperatures come closer to being equal, the rate decreases.

[GeorgiaTech335coupe]

Quick correction to original post - latent heat is used the change state without temperature change. Sensible heat is what you are looking for. This is the amount of energy needed to raise/lower temperature.

A rough equation, assuming adiabatic compression (no heat loss/gain through transmission in the turbo), to get the temp rise from the compression of the air is T2=T1*Rc^(k-1/k). Temp is in kelvin. Rc is ratio of compression in absolute (~2 for 15 psi boost), k = 1.4 for ideal gas. This equates to a factor of ~1.22 temperature rise in Kelvin.