# Thread: First time jetta owner modifications

1. Originally Posted by MechEngg
tl;dr cliffs:
I'm a ****ing boss, all up in your ****ty cheerios.

2. Originally Posted by MechEngg
Let me do my best!

Assuming you are running 20psi boost in the intake manifold, atmospheric conditions are sea level so ambient air pressure is normally 14.7psi. Let us judge the atmospheric temperatures of 10C, so we can assume the CAI inlet temp to be 10C and we can take the short ram air intake temp to be 20C

As you know turbines and compressors are never 100% efficient. Looking off our K03s compressor map let's assume we are running at a ~68% efficiency range at 20psi at our current rpms.

Now let's break this down, there is a heating stage which is where heat is added to the air from the compressor, and then there is a cooling stage where the intercooler does its best at reducing the heat before the air is pulled into the cylinders.

For stage 1, the heat addition from the compressor.
T2 = T1 x PR^0.283
Then add inefficiencies of the turbo
T2 = T1 + ((T1xPR^0.283) - T1)/Eff)

T2 for 10C = 283 + ((283*2.36^0.283) - 283)/0.68) = 283 + 114 = 397
T2 for 20C = 298 + ((293*2.36^0.283) - 293)/0.68) = 293 + 118 = 411

So a 10C difference in air intake temperature actually leads to a 14C increase in temperature after the turbo but before the intercooler.

For stage 2, the heat reduction from the intercooler.
We all know our intercoolers suck, and honestly i don't know the efficiency of them so let's assume a fairly standard sucky 70%.

Tdrop = Eff*(Ti-Ta), but for this we must remember that the ambient air temp is the same for both scenario's

T3 for 10C = T2-(Eff*(Ti-Ta)) = 397-(0.70*(397-283)) = 397-80 = 317
T3 for 20C = T2-(Eff*(Ti-Ta)) = 411-(0.70*(411-283)) = 411-90 = 321

So overall an intake temperature of 10C will lead to a final temperature difference of 4C. Is this small temperature difference really worth the losses that you pick up in the CAI? I think that we need to actually do the flow modelling and calculate out the pressure restrictions in the CAI to truely see but as it is brutally obvious, a more efficient intercooler core will give you far better gains than trying to install a CAI over a SRI because as the efficiency of the intercooler grows closer to 100%, the closer the final temperatures get to becoming the same value.

Does this help to solve everybodies questions??
I don't believe in mathematics, this information is irrelevant.

3. Originally Posted by GLI-R
Castrol is the best Oil i have ran in my engines

F*uck Castrol? F*ck Your engine then
My engine doesn't run as well with castrol. Mobil 1 has always been good with it

4. Originally Posted by MechEngg
Let me do my best!

Assuming you are running 20psi boost in the intake manifold, atmospheric conditions are sea level so ambient air pressure is normally 14.7psi. Let us judge the atmospheric temperatures of 10C, so we can assume the CAI inlet temp to be 10C and we can take the short ram air intake temp to be 20C

As you know turbines and compressors are never 100% efficient. Looking off our K03s compressor map let's assume we are running at a ~68% efficiency range at 20psi at our current rpms.

Now let's break this down, there is a heating stage which is where heat is added to the air from the compressor, and then there is a cooling stage where the intercooler does its best at reducing the heat before the air is pulled into the cylinders.

For stage 1, the heat addition from the compressor.
T2 = T1 x PR^0.283
Then add inefficiencies of the turbo
T2 = T1 + ((T1xPR^0.283) - T1)/Eff)

T2 for 10C = 283 + ((283*2.36^0.283) - 283)/0.68) = 283 + 114 = 397
T2 for 20C = 298 + ((293*2.36^0.283) - 293)/0.68) = 293 + 118 = 411

So a 10C difference in air intake temperature actually leads to a 14C increase in temperature after the turbo but before the intercooler.

For stage 2, the heat reduction from the intercooler.
We all know our intercoolers suck, and honestly i don't know the efficiency of them so let's assume a fairly standard sucky 70%.

Tdrop = Eff*(Ti-Ta), but for this we must remember that the ambient air temp is the same for both scenario's

T3 for 10C = T2-(Eff*(Ti-Ta)) = 397-(0.70*(397-283)) = 397-80 = 317
T3 for 20C = T2-(Eff*(Ti-Ta)) = 411-(0.70*(411-283)) = 411-90 = 321

So overall an intake temperature of 10C will lead to a final temperature difference of 4C. Is this small temperature difference really worth the losses that you pick up in the CAI? I think that we need to actually do the flow modelling and calculate out the pressure restrictions in the CAI to truely see but as it is brutally obvious, a more efficient intercooler core will give you far better gains than trying to install a CAI over a SRI because as the efficiency of the intercooler grows closer to 100%, the closer the final temperatures get to becoming the same value.

Does this help to solve everybodies questions??
I took your formulations into excel, so we could have a more graphic idea... check it out

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