TwinTurbo.NET: Nissan 300ZX forum - Greg - based on what I have been reading --the work
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Subject Greg - based on what I have been reading --the work
     
Posted by Slowninja(GT28RS Inconel Z) on October 20, 2005 at 2:26 AM
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In Reply To Seb's car is not behaving like it has a bigger displacement posted by Greg D. (Specialty-Z) on October 19, 2005 at 11:08 PM
     
Message that you are talking about will increase the Volumetric efficiency of the motor closer to 100% from where it was stock and this will allow more air to enter with each stroke. With more air and more fuel --that equals more power at the same PSI levels. That is a simple concept that everyone understand and buys into.

One of the most important parts of turbo efficiency that has been discussed here but not focused on his absolute temperature. If the absolute temperature is lower - and there is more air - the turbo can make more power - more power that it is rated for - at the same PSI levels.
"
Example
Time for an example. Lets calculate the pounds of air flowing into an engine for two different cars, an intercooled '87 and a nonintercooled '85. For both cars we will use a volumetric efficiency of 0.85. For both cars the engine is turning at 5000 rpm. What is the volume of air it is using?

volume, in cu.ft per minute = 5000 x 231 = 334.2 cfm
1728 x 2

This holds true for both cars, both intercooled and nonintercooled will be moving 334.2 cfm of air into the cylinders at 5000 rpm. As we will see however, the mass of air flowing is not the same.

Suppose the car an '85, so it isn't intercooled. The temperature in the intake manifold is about 250 deg F. The car is running 19 psi boost. What is the mass of air the engine is using?

Absolute temperature = 250 deg F + 460 = 710 deg R

Absolute pressure = 19 psig + 14.7 = 33.7 psia
****
n (lbs/min)= 33.7 psia x 334.2 cfm x 29 = 42.9 lbs of air per minute (ideal)
10.73 x 710 deg R ****

lbs air per minute actual = lbs/min ideal x vol. eff.
= 42.9 x 0.85
= 36.4 lbs air/minute

What if the car is an '87, it IS intercooled, so the temperature in the intake manifold is only 130 deg F. This car is running 17 psi boost.

Absolute temperature = 130 deg F + 460 = 590 deg R
Absolute pressure = 17 psig + 14.7 = 31.7 psia
****
n(lbs/min)= 31.7 psia x 334.2 cfm x 29 = 48.5 lbs of air per minute (ideal)
10.73 x 590 deg R ****

lbs air per minute actual = 48.5 x 0.85 = 41.3 lbs air/minute
"
Absolute Pressure

benefits of larger intercoolers

Benefits of H20 injection

You can see that the determining factor really here is the absolute temperature in the formula. This is controlled by large intercoolers / Larger piping / Water injection.

With the Massive Ash-spec intercoolers and increased piping and water injection - new limits should be able to be reached without passing the surge / choke line.

My setup is different than Seb's car. I have less backpressure with a larger exhaust - I have dual fuel pumps and 850cc injectors (wont run out of fuel) - and with the Zemulator the ignition timing can be adjusted throughout the RPM band - better and increased cooling capacity - and water injection. These things WERE NOT DONE on Seb's car. I know that for a fact. Therefore, my car will not be a fair comparison to Seb's car.

On a side note -

I have never heard of tuning a car to the rated turbo rwhp limit. I have heard of tuning a car and staying under the rated surge / choke line. But I do agree with you about safety. But my example would be that you can change your tires every 7,500 miles instead of 12,000 --but why would you if you still have meat on the tires?

The safety of the tuning is the key - Low EGT's , standard A/F curve, and proper ignition timing.

The compressor map issue is the key to the whole question of what the turbo can / cannot do.

Ofcourse - I am not on the same level as you on knowledge --I am only regurgitating what I've read from what I've seen from researching the issue and talking with folks like yourself.

Idiots guide for people like me on how a turbo works


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