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-   -   intercooler test (https://www.scoobynet.com/scoobynet-general-1/804721-intercooler-test.html)

typeRv4 26 October 2010 09:55 PM

I think you are confused, thats not how you measure efficiency. In fact im not sure what its measures.

By your logic if ambient temperature were 2 C and the post intercooler temps were 3 C, we could say its gone up by 50% lol

The correct way to measure intercooler thermal efficiency is thus

(Intercooler temp input - Intercooler temp output ) / ( Intercooler temp input - ambient temp) =

(138 - 32.5) / (138 - 21) = 90%

Butty 27 October 2010 08:33 AM

Are those temperature shots taken at idle when stationary?

Tidgy 27 October 2010 08:48 AM

um 50% because the difference of 10 degrees (30-20, i rounded it) is 50% of the initial figure of 20.

the higher the inlet temps go the more power you loose. the temp change shows that inlet temps have gone up by 50%, so power loss, due to temps will be 50% greater than the first pics. But 50% is not 50% power loss, its just 50% additional loss through thermal expansion of air. i can't remember what the figure is for bhp loss for degrees C increase in inlet temps off the top of my head.

Butty 27 October 2010 09:51 AM


Originally Posted by Tidgy (Post 9678466)
um 50% because the difference of 10 degrees (30-20, i rounded it) is 50% of the initial figure of 20.

Nope, the efficiency is based on the turbo out temperature, not just the difference between inlet and charge temps.

Look at the formula TypeRv4 has given.

typeRv4 27 October 2010 07:34 PM


Originally Posted by Butty (Post 9678444)
Are those temperature shots taken at idle when stationary?

Those shots were taken as the car was moving by an assistant.

typeRv4 27 October 2010 07:54 PM

To quote Harvey


Originally Posted by harvey (Post 9402469)
The objective is to get the air entering the inlet manifold at the throttle body as close to ambient temperature as possible.With an efficient front mount that temperature will typically be 6-12 degrees over ambient so let us assume your in car temperature guage is showing 16 degrees then you could expect to see 22-28 degrees on your ACT guage.

In this case if you just measured % gains on the movement of air temps on a good fmic you could say 37-75% - which of course is nonsense.


We normally refer to air charge temperatures as "X" degrees over the ambient.
Rather than worry about efficiency its more important to measure the amount degrees over ambient.

According to the above a good FMIC does 6-12 degrees over ambient @ WOT.

My test shows in this case 11.5 degrees over ambient. So not quite as good as a FMIC but pretty close.

It helps this test was done on the road to get real world airflow thru the tmic and i run a big scoop.

I hope to do some more tests with a watersprayer.

harvey 28 October 2010 08:11 AM

The figures above refer to temperatures over ambient on wide open throttle, either on the road or at the end of a rolling road session where we measure the peak on power.

Tidgy 28 October 2010 08:17 AM

ah sorry i get ya, i think i'm getting my wires crossed lol

typeRv4 11 November 2010 02:24 AM

http://img821.imageshack.us/img821/7673/wstmic2.jpg

Retested the same classic hyperflow top mount with waterspray.

Left is air temp before the airbox, right is exiting the intercooler.

1.7bar peak - md321. Test on exactly the same road.

Temperature increase over ambient 7.1C,

not bad for a top mount ;)

lunar tick 11 November 2010 03:28 PM


Originally Posted by typeRv4 (Post 9706003)
http://img821.imageshack.us/img821/7673/wstmic2.jpg

Temperature increase over ambient 7.1C,

not bad for a top mount ;)

I would be careful Type R4 - all this talk of TMICs doing a decent job wll have the 'FMIC thought police' after you... ;)

lunar tick 11 November 2010 03:34 PM


Originally Posted by Tidgy (Post 9676451)
so in one run it gone up 50%, mmm so much for efficient top mounts lol


:nono:

The % rise in actual temperature should be calculated using the Kelvin scale - ie 11.5/(273 + 21) = 0.039% - not quite 50%!!

Tidgy 11 November 2010 03:45 PM


Originally Posted by lunar tick (Post 9706833)
:nono:

The % rise in actual temperature should be calculated using the Kelvin scale - ie 11.5/(273 + 21) = 0.039% - not quite 50%!!


christ if you realy wanna get that pissy about it.

lets simplify it,

hotter = bad
colder = good

the temps have gone therefore its bad

vindaloo 11 November 2010 06:14 PM


Originally Posted by lunar tick (Post 9706833)
:nono:

The % rise in actual temperature should be calculated using the Kelvin scale - ie 11.5/(273 + 21) = 0.039% - not quite 50%!!

<cough> multiple by 100 </cough>

3.9%


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