Gene would be the best choice, but he doesn't have his PSD any more, more's the pity. Whomever might be our guinea pig, they should have access to a dyno, a heavy trailer, and AE (or similar) like you mentioned. But really, CFM should be doing this themselves.

-- OR --

They have done it and the numbers flat out stink.

 

I just KNOW theyve done something and they're ashamed of it...what company wouldnt have done their homework...
Its probably too late to cancel manufacturing of it or something...

 

The site says specs will be posted Feb 7th. I wonder what they will have to say then. Probably some BS numbers.

 

Man, some of you guys are real negative nancy's. I'd at least give the company a chance to post some info... The pic on their website is just a CAD drawing at this point, not the real product, so they might not actually have any flow data yet. Who knows? A lot can be done with computer modeling.

If it's like the rest of their products, I would expect it to flow better than the stock piece. How much more, I've got no idea.

 

Yea, I guess youre right. Ill give them a chance...for now!

 

There is just not the same bottle neck as in the 5.9's, 6.0's and 6.4 trucks... There is just not room for improvement...

 

I'll keep my pig-tail Nancy wig on, thanks Jeremy!!!

 

On a bench, it might flow a little more than the stocker...maybe just a little. Enough to actually see a benefit in application on the truck, honestly, probably not, especially at the level that a relatively stock truck would be with a stock or drop-in turbo. I do think there are better places to address airflow restriction.

BTW, I don't mean to be knocking any of you guys with my "negative nancy" comment, so please don't take offense to it.

 

I don't think the heat ENERGY that's transferred between "the hot side and the cold side" is the dominant mechanism, but even if it was Aluminum doesn't provide any thermal insulation to block such a transfer!

You need to employ the concept of "heat capacity" to calculate/estimate the TIME rate at which the heat ENERGY that's transferred between two objects increases the temperature of the cooler object. For example when you touch a hot stove it takes only an instant to transfer enough heat ENERGY from the stove to the skin on your finger to increase the skin's temperature to a high enough value to cause a third degree burn. That's because the HC=Heat Capacity of the stove is almost infinitely higher than the HC of the skin on your finger.

In the "steady state" the spider's temperature is about the same temperature as the turbo's compressor housing because the spider is in direct "thermal contact" with the compressor housing. After pulling a long grade in the summer this temperature can exceed 250F so that the entire spider becomes the hot stove in the above example and the HC of the spider is also almost infinitely higher than the HC of the air passing through it!

I'll do some calculations just to see if I can come up with any additional insight into this question of how much the MAT increases due to the intake air passing through a hot spider, but the bottom line is that someone would need to measure MAT for both spiders under various identical ambient conditions to get the exact answers.

Each air molecule has an average kinetic ENERGY that's given by KEavg=(3/2)kT ft-lb, where k=Boltzmann constant=5.65731x10^-24 ft-lb/R, and T=absolute temperature R, R=Rankine, R=F+459.67. If you let N=number of air molecules in a given volume, then the total average kinetic ENERGY of all the N molecules in the volume V is given by KE=(3/2)kTN ft-lb.

If you let Wam=average weight of a dry air molecule lb, and then divide both sides of the equation KE=(3/2)kTN by Wam and move N and T to the left hand side you get {KE}/{(N)(Wam)(T)}={(3/2)(k)}/{Wam}, which has the units of ft-lb/lb-R, and this defines the HCA=Heat Capacity Air which can be written as, HCA={KE}/{(N)(Wam)(T)} ft-lb/lb-R, or as HCA={(3/2)(k)}/{Wam} ft-lb/lb-R, and both of these HCA formulas specify the number of ft-lb of KE that's required to increase the temperature of 1 lb of air by 1 degree R=F+459.67.

In round numbers dry air consists of 23% O2 molecules and 77% N2 molecules. When you consider the trace amounts of the other molecules in air the exact average weight of a dry air molecule is Wam=1.060347x10^-25 lb. The total weight of N air molecules in a volume V is given by Wat=(N)(Wam) lb.

Using the second version of the HCA formula gives, HCA={(3/2)(k)}/{Wam}={(3/2)(5.65731x10^-24)}/{1.060347x10^-25}=80.0 ft-lb/lb-R. So you need to transfer exactly 80 ft-lb of KE into each 1 lb of air to increase its absolute air temperature by 1 degree R=F+459.67.

Using the first version of the HCA formula and solving for KE gives, KE={(HCA)(N)(Wam)(T)} ft-lb, and using HCA=80.0 ft-lb/lb-R and Wat=(N)(Wam) lb in this equation gives, KE={(80)(Wat)(T)} ft-lb as the relationship between the absolute temperature T in R=Rankine, R=F+459.67 of a total weight of air Wat lb, and the total amount of KE ft-lb that's contained in that weight of air. If you solve for T you get T={KE}/{(80)(Wat)} which gives the absolute temperature of a given weight of air that contains a given amount of KE that's been stored in the air due to the motion of the air molecules.

So we need to calculate Was=the weight of air that's contained in the volume of the spider through which the air flows towards the engine, the KEb=the KE that this Was contains "before" entering the spider, the KEa=the KE that this Was contains "after" leaving the spider, and then Tb={KEb}/{(80)(Was)} and Ta={KEa}/{(80)(Was)}, and Ta-Tb then gives the increase in absolute air temperature due to the air flowing through the hot spider, and this can be converted to F to give the final answer!

Since all this brain work has given me a headache I'll pick up at this point in my next installment. I need someone to post the spider dimensions or better yet if you've got a spare spider duct tape the engine side closed and fill it with water and then pour that into a measuring cup to get the volume. I also need an estimate of the cross-sectional area A in^2 through the air flows.

 

Subscribing, just cause I saw my man Gene post up some formulas, and I want to get smarter.

 

I have NO idea what I just read. I could read it ten times and STILL have no idea... Just waiting for the yes if flows more or no it doesn't flow more..

 

Well go touch a hot stove for a split second and then read it again! Never trust anyone's "yes" or "no" answer without at least making an attempt to understand for yourself if it's correct or not! In other words don't be like Congress and write blank checks just because someone says "yes" a blank check will "fix" the problem!

Right now I'm working on the "heat transfer" issue, and then I'll take a stab at the "flows more" issue. However both issues are actually interrelated because if the "heat transfer" of the improved spider vs the OEM spider is less the MAT is reduced for a given VAF ft^3/min and this in turn increases the AD=Air Density lb/ft^3 and therefore the MAF={(AD)(VAF)} lb/min into the cylinders.

On the other hand if the improved spider vs the OEM spider has less PD=Pressure Difference psi for a given VAF ft^3/min this increases the MAP psi which in turn increases the AD lb/ft^3 and therefore the MAF={(AD)(VAF)} lb/min into the cylinders. So the improved spider can potentially increase the MAF by deceasing the MAT and/or by increasing the MAP.

To make more HP you need to increase the MAF into the cylinders and then add more fuel. If you just increase the MAF into the cylinders without adding more fuel you might still get a tad more HP due to the enhanced combustion of the same amount of fuel, and you'll also get lower EGT because the additional air mass filling the cylinder gives more total "heat capacity" for absorbing heat ENERGY from the fuel at a lower CAT=Cylinder Air Temperature.

 

Geeze apparently I'm blind, I missed this thread completely :

 

Subscribing for the same reasons Brandon did!! Had to re-read Eugene's posts a couple of times, but it's setting in a little.

 

Do you have any pics of cutting these apart? I'm interested in what they look like. Is the hot side ruined when you do this or could you essentially separate the 2 and make your own like they're coming out with?