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Thread: First time jetta owner modifications

  1. Member
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    08-19-2012 12:14 AM #36
    Quote Originally Posted by MechEngg View Post
    tl;dr cliffs:
    I'm a ****ing boss, all up in your ****ty cheerios.

  2. Member oakdub's Avatar
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    08-19-2012 12:21 AM #37
    Quote Originally Posted by MechEngg View Post
    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. Member rbuicki's Avatar
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    08-19-2012 04:14 AM #38
    Quote Originally Posted by GLI-R View Post
    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. Member GLI-R's Avatar
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    08-19-2012 03:33 PM #39
    Quote Originally Posted by MechEngg View Post
    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

    -2011 MK4.5 1.8T VW Motorsport GLI R Proyect. Codename ".:Rolling Thunder"

    -"The ammount of Love, Money and HorsePower We Can Have Is Never Enough" -Sensei Mawy Luna

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