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eracers999
04-23-2005, 21:52
I know that pumping compressed air off the turbo heats up the incoming air into the intake.

What about the restriction between the filter and the intake of the turbo, i have 11 inches of restriction ok, water boils at lets just say 212 deg, now for every inch of vaccume, we decrease the boiling point of water, arent we heating up the intake charge even before it gets into the turbo? Wouldent we in fact gain HP off of a less restrictive air filter set up?? I cant seem to come to a answer in my head. Restriction is over 1/2 of the boost psi. You know where im headed with this post. Please elaborate on this subject.

Kent

Billman
04-24-2005, 04:11
Is the filter a stock paper element in your test? I would think that that alone would be worth something. You would be better off just hanging a filter on the turbo inlet.

My tests were done with a 97 Filter Box and a K&N. Removed the lid and gained 25% less restriction. Like many here have stated before, the ribbed elbow is probably the next biggest culprit.

I eliminated that also. My guess would be less resrtiction now.

Pressure raises the boiling point of water(1#=2-3* IIRC). Vacuum lowers it(Never really thought of that). Does make sense though. We're not boiling water here. I'm not too sure how you are relating the two.

Nothing more than the obvious, a pressurized turbo inlet would be ideal, so a vacuum on the inlet to turbo is power lost.

I'm curious to where this thread will lead to.

With a Hot Motor like yours, an Intercooler would be Unbelieveable.

eracers999
04-24-2005, 05:05
Billy

That is kinda why i posted, im not really smart enough to figure it out.
The filter in this test is a flat panel K&N, freshly cleaned and oiled with K&N kit.

I was merging my air conditioning thoughts into this area of the air filter.
The main reasone we vacume the a/c system down is to boil it out to remove moisture from the system. So naturally while i was at work, working on a big truck a/c i had 6.5 air filter in my head and put the two together. Thought ronniejoe could twist it around into shape.

What happens to that air being vacumed hard like that? We got a good idea what happens on the other side of the turbo when it gets pushed hard.
With all the posts on new air filter set ups seemingly making a difference to the good, even tho they are mounted in the engine bay pulling in hot air. Guys are reporting more power and lower egt's. The part im stuck at is, 15 to 20 lbs boost is 15 to 20 lbs boost, the engine is getting the psi, what is changing in air quality with less restriction, pressure being the same, what really are we changing by lowering resriction.
Kent

[ 04-24-2005, 05:21 AM: Message edited by: Kent ]

gmctd
04-24-2005, 05:17
Boiling water is a finite volume in a finite container with an infinite heat source - at least, until you cut the heat source.

Turbo intake is an infinete volume of (hopefully) cool air into the compressor, then into a dynamic pump, of which volume is less than the compressor, resulting in pressure, but never-the-less, has a dynamic flowrate.

Cold air into the compressor will be hot air out.

1psi = 2"hg = 27" H2O

You did not specify H2O or mercury, nor Boost, but a dirty air filter will increase differential across the range.

eracers999
04-24-2005, 05:35
Ok let me be real unscientific here, one wouldent think that in the short distance that the turbo whinds up and feeds the engine psi that it could possibly heat it up to 250 deg so quick but it does, so what happens to the air being vacumed hard?

Kent

Billman
04-24-2005, 05:53
I don't think vacuum can increase temperature anywhere near what pressure does to it.

Reading you edited post, What's changing in less restriction would be less power required to spin compressor? Less backpressure in exhaust?

eracers999
04-24-2005, 07:12
Allright lets go boat prop science. Prop gets turned hard , at the tip it creates a negative pressure causing the water to boil, hence creating cavitation. Lets say the conditions for this discussion are 90deg F and 90% humidity, from the impellers on the turbo to the air filter what happens? On the back side of the impellar tip are we superheating the incoming air, (humid moist air).

Kent

gmctd
04-24-2005, 07:13
The boiling pot is finite, as is the closed ac system.

We're working with a dynamic open-ended system.

Your '94 does not have the 90deg fold-back duct out of the air box, like the '96-up trucks, where a freer filter would alleviate some of that engineered-in restriction.
And, the late 1500 series have an in-line Mass Air Flow sensor to contend with - more restriction.

TBO's '98 K1500 truck will trip the differential gage with the MAF sensor removed, but does not do so when it is installed, yet still makes 15psi easily, in or out.
He uses a K1500 PCM and a K3500 PCM, for the difference.

Easiest check is to put your temperature probe before the air filter, then - and you can temporarily remove the CDR tube for this experiment - right at the compressor inlet, for a quick comparo.

At any rate, boiling water removes heat much quicker that when non-boiling, and you're still working with an infinitely renewing - dynamic, not static - cool air mass at the compressor inlet.

On the inlet, 1deg more - or, less - inlet air temp results in only 1deg more - or, less - Boost temp, under the same conditions.
That's the law - base temp effects the final temp - just not as much as you may expect.

Also, by using a larger compressor, where the same Boost can be achieved at lower wheel rpm, Boost temps can be reduced to some extent.
But, that also requires greater power input from the turbine end - more heat, pressure and velocity.
Which can, and does, result in the oft-dreaded, much feared turbo lag.

But, compressing 15psia into 30psia will still result in a major heat increase, little effected - comparitively speaking - by inlet temp, or compressor wheel rpm.

[ 04-24-2005, 07:31 AM: Message edited by: gmctd ]

rjschoolcraft
04-24-2005, 11:11
For an ideal gas (which air is not, but it is close), pressure and temperature are related by the expression PV = mRT. This can be rearranged as T = PV / mR. From this, you can see that as pressure increases, the temperature increases. Conversely, as pressure decreases, temperature decreases.

In the situation that you described in your first post, the boiling is not occurring as the result of increased temperature, but because the boiling point is being lowered to the current temperature of the medium.

Most of the temperature rise across a compressor comes from heat of compression. With 100% adiabatic efficiency, the compressor outlet temperature (T2) is:

T2 = T1 * (P2/P1)^.283

There is some difference of opinion in the literature on the exponent, but that's pretty close. The equation above is a restatement of the first equation that I gave in the first paragraph relating the temperature ratio to pressure ratio. It will give you the temperature rise (T2 - T1) as a result of compression alone. Divide that result by the efficiency to get the actual temperature rise.

Most centrifugal compressors operate between 65% and 80% efficiency, so:

T2actual = T1 + [(T2 - T1) / eff.]

By depressing the inlet (pulling against a restriction) you are increasing the pressure ratio across the compressor for a particular desired level of boost pressure. This means that the compressor has to work harder to achieve your goal. It takes work to pull a vacuum. As a result, a less restrictive inlet does improve performance by reducing the depression at the inlet.

Notice that the ratio equation can be written as:

(T2/T1) = (P2/P1)^.283

In other words, the temperature ratio across the compressor equals the pressure ratio raised to the power of .283. Also note that we're using absolute temperature and pressure here.

Let's look at a desired boost pressure of 15 psi (ignore intercooler and plumbing losses). This would equate to a pressure ratio of about 2. This means that (T2/T1) = 2^.283 for this situation or T2 = T1 * 1.217. This should make it obvious that each degree of inlet temperature (T1) increase results in 1.217 degrees of outlet temperature (T2) increase for a pressure ratio of 2 and 100% adiabatic efficiency. Consider two different inlet conditions: 50F and 100F. On the absolute scale, these are 510R and 560R. At pressure ratio 2, the outlet temperatures (100% adiabatic efficiency) are 620R and 681R (160F and 221F). A 50 degree increase in inlet temperature resulted in a 61 degree increase in outlet temperature. Throw in compressor efficiency of 75% (a middle of the road number) and the outlet temperatures are now 657R and 721R (197F and 261F). Now the outlet temperature increase is 64F for a 50F increase in inlet temperature.

This should help evaluate the trade-offs. If you reduce inlet depression (less restriction), you gain performance improvement. But, if you are now breathing hotter air (underhood vs. outside), you lose some of that gain.

eracers999
04-24-2005, 15:42
Wow, thanks for clearing us up here. There is no way even if i learned all that, that i could possibly remember that kinda stuff. In simple terms Billy was right on.
Appreciate your time!

kent

[ 04-24-2005, 03:55 PM: Message edited by: Kent ]

Billman
04-24-2005, 16:10
You asked for it Kent.

You got RJ's pencilholder in the pocket/glasses held together with tape/ Techno head response.

Thanks again RJ...I think.

rjschoolcraft
04-24-2005, 16:35
Originally posted by Billman:
You got RJ's pencilholder in the pocket/glasses held together with tape/ Techno head response.
I did use a pocket protector for a while... back in the 1980's. I never have needed glasses, though. :D

eracers999
04-24-2005, 19:06
We call em pocket pals over here, laughing...

Kent