I believe you're there. If you try to measure the vacuum anywhere except where there is TOTAL air flow, you're not measuring the full effect of the system.

Full air flow to the turbo is anywhere from the internal filter opening right at the neck, all the way to the turbo inlet where the housing opens back up for the impeller (wheel).

 

BTW, thanks Pete!! Just typing out my thought process helped me see it!!

 

I know what you mean, Joe, and you're welcome.

I tend to think in "pictures", too. I guess that's why I had so much trouble with the calculus and differential equation courses that were forced down my throat in school. It's just hard for me to visualize some of those "higher math functions", but fuel pressures, exhaust temperatures, air flows, chemical reactions, combustion, etc... those things I can get a pretty clear picture of easily enough.

 

I struggled with Calc, too. I guess for the same reasons. Your explanation was perfect, though. Hopefully others benefited from my "impairment".

EDIT: BTW, I saw where you were going and was thinking about integration when I read your example...

 

Well, all I can say is that it's a good thing you said "almost", and then backed that up by misspelling "bet", cause I'd sure like to have a 450 HP truck to play with! I calculated a slight pressure drop using "Bernoulli's Principle" for flow through a straight pipe, and it introduces at most a slight error at a 500 cfm flow, but since the error is proportional to the square of the flow velocity its negligible at lower cfm flows! If Tenn only had an air box like me, he could measure just upstream and downstream of the element using 2 fittings, and essentially eliminate all errors.

My calculation assumes a straight pipe, and I'd need to know if Tenn's using the OEM CCV coupler, but even then he'd need to have his fitting right where that narrow segment is to cause a problem. Tenn, did you use this coupler Click for full size image and if so where did you drill for the fitting. Based on your pic, it looks like you used a piece of plain (white) PVC pipe, which my calc's indicate isn't a problem!

Also, your IR pointer will work pretty good for making temp measurements, because the thermal mass of the objects of interest is such that they maintain their temp for many minutes at idle with the hood up.

The largest source of error is the uncertainty in the cfm vs H20 restriction of the filter element, and to do the most accurate test, the element should be new and clean, and have some bench test data, which is why I offered to chip in and get a new DONALDSON B085011 element, since there appears to be test data for it. Of course Tenn could always get a leaf blower and make up his own test rig.

If anyone's interested, or I'll probably get around to doing it anyway, there's a couple of ways to calculate the effect of the slight pressure drop in a moving air flow, and they both give a worst case of a little over 1" H20, say 1.5", and that's at 500 cfm. If you halve the cfm to 250, you get a X4 reduction to a little over 0.25" H20.

 

Ok, I just got this far so here's the short version (for me) off the top of my head. Your last statement refers to acceleration, and "Bernoulli's Principle" only applies in a steady state along with a number of other restrictions that are never completely satisfied in a real world application, but the first part regarding a conservation of energy is one way to approach the problem.

When a mass of air is stationary, all its molecules are in random thermal motion with a velocity of around 950 mph, and they exert a standard static pressure of 14.7 psi by bouncing off of everything they come in contact with. Neglecting gravitational potential energy which doesn't change because the flow tube is horizontal, the total energy of the air mass is the 1/2 m v^2 kinetic energies of all the molecules in the air mass.

Now if this mass of air is flowing with a steady state flow velocity of V= 60 mph, which is about the velocity in a 4" tube at 500 cfm, all the molecules are moving along the tube at 60 mph, and also buzzing around in a random way as well. Imagine a swarm of bees. Since a 1/2 m V^2 energy has been added to all the molecules, each of their 1/2mv^2 energies must be reduced so as to maintain the total energy in the air mass at its original valve. This reduction means that the molecules aren't buzzing around in a random way and hitting the wall of the tube (where the static port is) with quite as much force as before, and this results in a reduction of the static pressure by an amount that's given by (60/950)^2(14.7)=0.0586 psi which is 1.6" H20.

Strictly speaking, the above calculation is for a flow tube with NO filter element on its end, like Tenn said he'll do some day when the dust settles down some. In that kind of test, the above calculation compares still air in the engine compartment which has a static pressure of 14.7000 psi, and concludes that the pressure sensed by a static port inside the tube with an air flow velocity of 60 mph past it is reduced to 14.6414 psi, which is a reduction of 0.0586 psi which is 1.6" H20. Also, for a 250 cfm flow, V=30 mph, and the reduction in static pressure is only 0.4" H20!

I could pull out an equation and plug some numbers into it, which I've done, and get the exact same result, except it might go up or down by a 0.1" H20 or so depending on the exact ambient air temp that's used. I tend to like the swarm of bees explanation better, because it better shows the physics of what's really behind the static pressure reduction. Imagine a swarm of bees buzzing around inside the engine compartment with random flight speeds of 950 mph causing a static pressure due to their bouncing off everything. When they start getting sucked into Tenn's 4" inlet tube, their random flight speed decreases ever so slightly to keep their total energy constant, and the static pressure they exert bouncing off the walls of the tube as they flow along it is just a tad lower than when they were flying around freely in the engine compartment.

Note it's a lc v vs an uc V, when I previewed this the FTE font didn't seem to make this as clear as in my email window!

 

It is a used Napa unit, about 4k miles on it. I have a brand new Fleetfilter which I believe is a Donaldson unit which I can install. I used a piece of PVC which is drilled/tapped dead center.

I disagree that it would have the same affect under pressure as that is what happens under pressure when the whoosh happens and I get a positive/pressure reading on the gauge. More testing needed here.

This project has taken on a life of it's own, so I want to get it right. So I will be deliberate. Which is another word for slow. Life and work can get in the way.

And now UT is playing football. I will be further distracted.

 

I think this "I used a piece of PVC which is drilled/tapped dead center" is the best spot to measure just the effect from the filter.

"I disagree that it would have the same affect under pressure", I didn't mean to blow a leaf blower into your filter, I should've said a leaf vacuum, but I was just kidding anyway. Those flow test rigs are VERY expensive, like 100K$, from what I've been able to find out.

 

I was referring to the venturi effect that was brought up earlier where it was posted that it would do this under pressure too. From what I have seen of this gauge and my initial readings, that is not going to happen, but I will continue to test.

 

The end of the filter where Tenn took his measurments IS the correct place to take a static pressure measurment. There are three seperate types of pressure at work here: Static, Velocity, and Total pressure. You can only directly measure static and total pressures and you can get velocitity pressure from subtracting static from total pressures. This is what a pitot tube does. Btw, in Genes other thread with the pics of the guy's TAG measuremnet device, that is a crude pitot tube. Velocity pressure is used to calculate the actual air velocity in the duct. Ok, back to static pressure, the jar analogy is correct. If half the jar is at -10" H20, the whole jar is at -10". What Tenn's measurements mean, is the filter media and size is not creating a large pressure drop. If you were to measure the static pressure on the other side of the filter, it would be the same. Now, the trouble is turbulence and there will be a lot of it on that other side of the filter. The filter neck itself most likely causes a largest pressure drop than the filter media itself. The problem is, we don't know where or how the numbers were arrived at in Gene's chart. Oh yeah, this is what I do for a living. (no, not measuring airflow in the 6637).

 

If it were me, I'd just use a thermistor of the type used in gasser MAF sensors and see if there is a detectable flow difference with & without the TAG. You could do it with different types of filters as well, I guess. That in conjunction with the vacuum measured at the intake opening would tell the story. Seems to me that this is getting over complicated...

 

Yeah it is. The truth is, there is not a good way to measure the airflow of an intake with any degree of acuracy or reapeatability while it is on the truck. Horsepower would probably be the best way to measure differences in intakes. Meanwhile, I'll keep my 6637.

 

I'm not trying to be a wise *** here, but just adding some more food for thought! I stand by my assertion of unequal air flow in a cylindrical filter that's open at one end and closed at the other. My pic is an electrical equivalent circuit of such a filter. Electron flow=air molecule flow, voltage=pressure, and the resistors=restriction. If Rp=0, then all 8 resistors have the same voltage across them, and each flows the same current. The Rp represents the longitudinal delta P for each unit length inside the filter, an I'll agree that's a much smaller value than the Re which is the restriction for each unit length of the element, but Rp is not 0! Instead of a 1 ft long filter, consider one that's 10 ft long or even 100 ft long, and it might be easier to see that the first 1 ft of length flows more cfm than the last 1 ft does.

If you put a new one of these filters in my engine compartment, but didn't hook it up, it would still get uniformly dirty pretty quickly, just like everything else does in there. This is yet another reason for putting the filter element in a box, and running an intake to cool, fresh, relatively clean air from ambient, because that way the element only has to deal with the dirt load in the air that's actually needed by the engine.

If a pool filter is vertical, and the end cap is at the bottom, the gravity pressure gradient might cause more flow at the dead end than at the open end.

 

I just Googled "Venturi effect", and guess what I got back on about page 4 or so? An authoritative sound summary blurb that linked to this very thread. OMG, you really do have to be careful when using the internet to get reference information!

I've done the "you can breath in outer space" bit, and the "swarm of bees" bit, here's an interesting proposition to ponder regarding just what Bernoulli’s principle really means! More food for thought.

An airplane’s barometric altimeter operates by measuring the pressure at a static port on the surface of its fuselage. The static port is typically oriented at about a 90 deg angle to the airflow. Doesn't this sound similar to the air flow past the static port fitting that Tenn is using to measure the pressure in his turbo inlet tube?

In accordance with Bernoulli’s principle, shouldn't this high velocity air flow past the static port of an airplane’s barometric altimeter be associated with a lower pressure there? Doesn't this imply that this lower pressure at the static port would cause errors in the altimeter as a function of airspeed? If these kinds of errors do exist, wouldn't it take a fancy computer to calculate the corrections as a function of air speed and provide real time inputs to the altimeter?

How were people such as myself able to fly for so many years using a plain old simple minded (before computers were invented) barometric static port altimeter without killing ourselves? Was the altimeter somehow ignoring Bernoulli’s principle, and giving the correct altitude independent of airspeed, or were we just lucky?

 

I see I've had no answers to my airplane question. I'm not the one who started us down the road of Bernoulli’s principle and the Venturi effect, but now that I'm headed in that direction I'm coming up with several enigmas! BTW, I'm not asking the following questions to try and trick anyone. I think I have an explanation for some of these, that's at least internally consistent with my views of physics, but if I believe some of the inputs I've gathered from the internet on these types of questions, my opinion might well be a minority one. So please join in the fun, and no equations or fancy calculations are needed, just word answers are fine. Higher, lower, the same, somewhere in between, it all depends on ..., etc.. will get the ball rolling.

I gave a 1.5" H20 or so number at 60 mph using bees, that's why I'm using a 60 mph truck and wind speed in the following thought experiments. Unless we can talk Tenn into actually drilling the hole and doing these experiments, we'll just have food for thought answers to go with the food for thought questions.

1) Tenn drills a hole in the roof of his truck and mounts a static port there. He then travels down a flat road with no head, tail, or cross wind, at 60 mph. He then takes his gauge which has the hose attached to the vacuum port and the pressure port vented to the cab, and see's that it reads 0, as it should with both ports reading the pressure in the cab. Now he hooks the hose to the static port in the roof, and what does he read on the gauge?

2) Tenn is at a dead stop, and along comes a perfectly horizontal straight line head wind of 60 mph. This allows Tenn to repeat the above experiment, except now the 60 mph air flow over his roof is due to a wind instead of his motion through still air. What result does he get this time?

3) Both Bernoulli’s principle and the Venturi effect have been used to explain how planes can fly. The typical explanation is that as the mass of air molecules encounters the leading edge of a curved wing they part company. Some take the longer path over the curved top of the wing while others take the shorter path under the flatter bottom of the wing, and they all meet up again at the trailing edge of the wing. Since the ones going over the top traveled a longer distance in the same time, they must've gone faster, and by flowing faster over the top than under the bottom they create lift because faster moving air has a lower pressure than slower moving air. Is this a correct explanation to explain how an airplane wing generates lift? Can any explanation that involves air pushing up from the underside of the wing possibly be correct for explaining how a plane flies at altitude?

Tenn, please note I've actually been hard at work on the real problem at hand, and I'll post 6 pics of some computer outputs and a write up later tonight or maybe tomorrow at the latest.