I ran my PSD engine model with and without the intercooler. In pic #1, I plotted the TOATI=Turbo Outlet Air Temperature Ideal, F, the TOATA=Turbo Outlet Air Temperature Actual, F, and the IMAT=Intake Manifold Air Temperature, F. The TOATI is for an ideal turbo that heats the air the minimum theoretical amount for compressing it to achieve the indicated boost pressures. The TOATA takes into account the TCE=Turbo Compressor Efficiency which for this run was 65% as is shown in the pic #3 data table. The ICEE=Intercooler Exchanger Efficiency is 70 %, and the ICEAT=Intercooler Exchanger Air Temperature, F is 90 F, and with these assumptions the IC cools the charge air from the TOATA temp down to the IMAT value.

Without the IC, TCE=0%, the IMAT temp value is the same as the TOATA temp as no cooling at all takes place. I plotted MAF (lb/min) curves for the flow into the intake manifold W & W/O the IC.

Pic #2 shows the same MAF curves, along with RWHP, W & W/O the IC. The reason the RWHP curves have the strange shape they do is because my model doesn't do the entire HP run at WOT. The throttle position starts at 20% for BP= 0 psi, and ramps up to 100%=WOT at BP=20 psi, and then stays at WOT on up to BP=30 psi.

If you check pic #3, you see the AFIAT=Air Filter Inlet Air Temperature, F is 70 F but due to engine heat soak, the TIAT=Turbo Inlet Air Temperature, F is increased to 80 F. If the IC was bypassed, the IMAT would also be about 80 F, although a little more heat would be picked up in the CAC tube from the turbo directly to the IM. Since the ICEAT=Intercooler Exchanger Air Temperature, F is 90 F due to the waste heat from the A/C condenser flowing past the IC, the IC actually winds up heating the charge air a little at very low boost pressures. As can be seen in pic #1 & #2, the IC makes the biggest improvement at higher boost pressures where the turbo is heating the charge a lot, and then the IC can cool the hot charge air back down towards its heat exchanger temp of 90 F.

I hope these graphs help you explain the benefit of an IC to the folks on your other thread. If the truth be known, the main if not the only reason Ford added the IC in late 1998 was to meet the new NOx spec, because the IC cools the combustion temps and thereby reduces NOx. So at least the EPA specs aren't all bad as we got an IC out of the deal, but I also got a CAT on my early 99 as part of the deal. Guess which one I decided to keep? Just let me know if you need any more analysis, that's what I'm here for.

 

what turbo is this info based after .84, 1.0 or the 1.15 housing? tell me something this graph shows that your turbo is producing absolutley no lag, what RPM does the turbo start producinging and where does it fall off. I am looking at doing a Series twin turbo junk yard set up I am also going to do this setup under the vehicle instead of inside the engine compartment further back on the exhaust so I will not have to have an IC and will keep engine temps down! The reason I want to do this series set up is to get a full range of power from just above idel to 6500 RPM which is NOt obtainable with any one turbo (what I understand. I was thinking of using a quick spooling turbo from some import and then using a deisel turbo but probably need the van 1.0 or even posibly needing to go with a 1.15 is there any vehicels with the 1.15 stock? I think I will want a .6? housing to spool almost imidiatly and am pretty sure the SRT-4 or an STI will have something close to this size. What would you suggest housing sizes to make this realistic as one is starting to fall off the larger is spooled and kicking in?

 

one more thing I want to keep the boost low from between 6-10LBS boost as this is a completely stock LT-1 I am putting this on I am going to step up the injectors a couple of sizes and put a boost-a-pump on it also but that is all I want to have to do. Is this possible? Do the deisel turbos have internal wastegates?

 

The answer is none of the above! All model results I give are always for my early 99 truck, because no one has been either willing or able to furnish me data on the 99.5 engine, but I'm working with Tenn trying to collect enough data to make a computer model for a 99.5 engine. Below is what Banks says are the differences between the early 99 and 99.5 and later turbos.

Ford-Diesel.com: What did Ford change in the turbo on the 1999.5-up Power Strokes?

Gale Banks: There are a few notable differences between the turbo design of the ‘99 (early) and the ‘99.5-up (late) trucks. To begin with, the compressor housing size (A/R ratio) changes from a 1.10 on the early turbo to a 1.00 on the late turbo. The diffuser face inside the compressor housing increases from a diameter of about 5.5" to about 6.0". The compressor outlet diameter increases by about 0.400". These changes account for the difference in physical appearance between the two. In addition to these changes, the compressor wheel itself is of a different design.

I'm not sure what the Turbine A/R is on my early 99 because I can't get a good look at that end, but the compressor clearly has 1.10 A/R stamped on it, which agrees with the Banks input.

My graphs give a steady state output of MAF and RWHP, for each value of BP at 2800 RPM! Turbo lag is the time delay between applying the throttle, and building up to a given steady state boost level at a given RPM, and my model doesn't include this time dynamic aspect.

I've given some additional information in the pic below. The graph of RWHP vs BP, is a plot of the RWHP values at 2800 RPM in the column shown in pink vs the BP (0 to 30 psi) in the first pink column. The ER=Equivalence Ratio, % is like a throttle position, and for example, if you pushed the throttle to say 60%, you'd have to wait for the turbo lag until you got to the 10 psi BP value shown to the right of 60% ER in the pic. Then if a load dyno was properly adjusted, you could achieve a steady state situation where you were measuring the RWHP at 60 % throttle, at 10 psi boost, and at 2800 RPM.

If you want some more detail, ER=AFRS/AFR, where AFRS=Air Fuel Ratio Stoichiometric, which as shown in the data table for this run is 14.7, and AFR=Air Fuel Ratio, which is defined as AFR=(MAF)/(MFF) where MAF=Mass Air Flow, lb/min and MFF=Mass Fuel Flow, lb/min.

Note, with my current configuration, my truck at WOT maxes out at about 22 psi boost and 256 RWHP. I used to run a K&N cone air filter and a Banks big head wastegate actuator with a 70-HP chip, and I could hit 27+ psi boost, and measured 267 RWHP on several dyno runs. That turbo blew at 50 K miles, and now I run a stock wastegate actuator, an AIS, a DP 40 tow, and make less boost and HP, but hopefully things will hold together longer.

I like the idea of not having the hot turbos under the hood which heat soaks the entire engine compartment. But I don't like your plan to go without an IC. The green curve shows the temp of the air coming out of an ideal turbo, and the red curve shows the temp coming out of an actual turbo, and that's with a turbo inlet temp of 80 F. Compressing air with an ideal turbo makes it hot, and compressing it with a real world inefficient turbo makes it even hotter. You need an IC to cool it back down, or all that work you do on your twin turbos will be in vain, and you'll wind up with less MAF and less HP than you can get using an upgraded under the hood turbo cooled by an IC. That's not just an opinion, it's just the way the physics work out when you compress air.

 

Damn, You're good.

 

It's as simple as cooler air is more dense, while hotter air is less dense. Denser air means "more" air which means more power (long as you've got fuel to burn).

Ernest's model is much, much nicer, though. Pretty cool stuff, Gene.

 

the research I did yestarday there is a loss of 1psi from the rear of the car to the front of the car and 100* drop in tempearture. this is why I was not planning on useing the IC along with the cost factor and the the question of where to mount it on the frame rail.

 

Oops, I forgot to qualify this statement by saying that this would be a good approximation for the IMAT W/O an IC at 0 psi boost when the turbo isn't heating the air so that the TOATA is about the same as the TIAT. As soon as you start increasing boost, the turbo starts heating the air, and then TOATA is greater than TIAT by an amount that depends on boost, and that's when the IC can start being effective in cooling the charge air.

 

Compare the red and blue curves at 30 psi boost, and you'll see about a 250 F temp difference between the turbo outlet temp and the inlet manifold temp, that means the IC cooled the charge air by 250 F, from about 425 F going into the IC, down to about 175 F coming out of the IC. At 425 F the air density, AD, is 0.0449 lb/ft3, and at 175 F it's 0.0626 lb/ft3, so having an IC increases the AD by 40%, and that means for a given CFM the MAF increases by 40%, and that means you can add 40% more fuel while keeping the AFR constant, and that means having an IC gives the potential to generate 40% more HP for a given CFM.

Please note that this 40% increase in MAF applies for the CFM flow through the turbo that's producing the 30 psi of boost. As I'll explain later, this 40% higher MAF that's flowing into the engine due to the use of an IC is exactly the same MAF value that's flowing into the turbo inlet, which is exactly the came MAF value that's flowing into the air filter inlet! The use of an IC doesn't magically produce more lb/min of air flow into the engine, but it does reduce the CFM that a turbo needs to flow in order to get the given MAF into the engine.

I came up with this saying to emphasize why you don't want to ingest hot under the hood air with an open element air filter, and I like the way it sounds, so I'll use it again because it's even more appropriate for your planned twin turbo setup. Installing twin turbos without using an IC would be penny wise but "pound of air" foolish!

It's interesting to compare the AD of the 70 F air coming into the air filter from an ambient air inlet which is 0.0750 lb/ft3 with the AD of 0.0626 lb/ft3 for the 175 F air coming out of the IC. That's a 20% decease in AD using a stock IC. That's why some upgrade their IC in an attempt to get some of this 20% back. However, my "pound of air" foolish saying applies if you upgrade the IC, and then needlessly ingest hot under the hood air using an open element air filter.

Here's the correct way to look at this whole business of air flow. The CFM flowing into the air filter inlet, along with the density of this air flow (which is determined by its temp), determines the MAF that's flowing into the air filter inlet. The conservation of mass dictates that this same value of MAF that's flowing into the air filter inlet, flows through the turbo, the IC, the IM, and into the engine. To say otherwise implies that some where in the air flow path air molecules are either escaping (which is possible with a boost leak) or that air molecules are being created, which isn't possible. This is why you want to minimize the air temp going into the air filter inlet by using a cold air intake, and not an open element filter. Once the MAF flowing into the air filter inlet is established by a given CFM and inlet air temp, that's all there is, and no amount of cooling by an IC is going to increase the MAF into the engine to a higher value than is flowing into the air filter inlet to begin with.

When I comment on the adverse effect of sucking in hot air with an open element air filter, many people on this site claim that the air temp going into the filter isn't that important because the IC can "make up the difference" by cooling the air back down again. But this is like saying that the IC injects additional air molecules into the charge air stream as it passes through the IC. The MAF (lb/min) going into the air filter inlet (which for a given CFM depends on inlet air temp) is all that you'll ever get flowing into the engine no matter how efficient the IC is for cooling the charge air.

So in summary, the correct way to look at the benefit of an IC is that it decreases the CFM requirement to produce a given MAF. Since a turbo is a constant volume compressor, it produces a given CFM for a given exhaust drive pressure on the turbine. Using an IC makes the best use of the given CFM from the turbo for producing a given MAF into the engine, but the MAF into the engine is always exactly the same value as the MAF into the air filter inlet. If this isn't clear, I can also explain it using my swarm of bees analogy from my thread on the cooling system!

 

no I completely understand it, I think! What you are saying is that you are still getting the same amount of air coming out of the compressor side as what is going into the hot side, the turbo just makes it move faster and the IC just compacts it so it is more efficiant. the more efficiant the more power. Is this correct?

 

alright gene and or anyother turbo guru! I am coming up with a a 1.41 pressure ratio since I want to run 6lbs of boost then add it back to the 14.7 and divide by 14.7 should get this number correct? this should be my Y-axis of the compressor map. the next step gets complicated to me and I would think most, I may be off on my calculation here which is my major question. alright with an 85% correction of efficiancy I came up with 46.83lbs/min. 6000rpm X 346CID = 2,076,000/3,456=600.69 rounded this is air flow of motor at 6KRPM correct? alright air pressure at sea level is 14.7+my 6lbs of boost = 20.7 X 600.69 X 29= 360,594.21 over 10.73 X rankin+temp at IM alright this is how I figured that i read that without an IC it is 250* or 130* for IC application well all the info I have read with my turbo set remote mount you drop 100* from the rear to the IM so it would be 150*+460 * for rankin = 610* alright back to the formula 10.73 X 610= 6545.3. am I correct in formula so far? 360,594.21/6,545.3= 55.09lbs/min X 85% efficantcy = 46.83lbs/minute = X-axis is this correct or did I mess something up here for I am having a real hard time finding a turbo compressor map that this will fall in at least a 70% efficancy on and only 2 or 3 at most that even hit the map at all............HELP please

 

also does anyone know the formula to convert volumetric flow M^3/minute back to lbs/min?

 

I'm not a turbo guru, but I'll gladly give you some inputs based my limited knowledge of the subject. First since you're dealing with a gasser and not a 7.3L diesel you'll probably get more appropriate inputs on a forum for your particular application. If you look at my graphs you'll see that the difference between having an IC and not having one varies from no difference at 0 psi boost to a large difference at 30 psi boost. When I saw your mention of "twin turbos" I thought of the typical diesel applications where twins are used to get BP's of 50 psi and higher, and then you definitely need a good heavy duty IC to get rid of the compression heat.

If you're only interested in running a BP of 6 psi on a gasser, an IC probably isn't mandatory, but I'm not sure the performance increase from only 6 psi boost without an IC is worth all the trouble of installing a turbo in the first place. It would give about a 20.7/ 14.7= 1.4 pressure ratio as you said, which would provide about a 40% increase in manifold pressure. But if you look at the red and blue curves in my graph, this 40% increase in pressure for a 6 psi boost without an IC is accompanied by about a 100F increase in manifold air temp, but only about a 20F increase for a turbo with an IC.

In the first case your manifold air density is about 0.0736 lb/ft3 at 80F without a turbo, and about 0.0621(1.4)=0.0869 lb/ft3 at 180F with a 6 psi turbo without an IC. So at 6 psi your X1.4 increase in pressure only gives a (0.0869/0.0736)=1.18 or 18% increase in density due to the 100F increase in temp without an IC. This does give the potential for an 18% increase in HP, but it doesn't seem to me like that's enough to justify the cost of a turbo.

If you added an IC to the above picture, you'd only have about a 20F increase, and this gives a 0.071(1.4)=0.0994 lb/ft3 at 100F with a 6 psi turbo with an IC, and a (0.0994/0.0736)=1.35 or a 35% increase in density due to the 20F increase in temp with an IC. So at a 6 psi boost, your manifold air density increases by 35% with an IC, but only by 18% without an IC.

The formula you ask for is MAF(lb/min) = VAF(ft3/min) * AD(lb/ft3). Also, AD=(1.5018)(AP)/(ATK), where AD=Air Density, lb/ft3, AP=Air Pressure, psi, and ATK=Air Temperature Kelvin, K, where ATK=(5/9)(ATF-32)+273.15, where ATF=Air Temperature Fahrenheit, F. You can also work in Rankin, R, but for all my formulas you need to convert F to K, and use the K values in the formula.

Some compressor maps give MAF or lb/min on the X axis, and others give VAF or CFM. It's best to use VAF, because the turbo is a constant volume or CFM compressor. In other words, a turbo tends to flow the same CFM for different inlet air temps, but the AD and therefore the MAF varies with inlet temp. I just posted a new input on this general subject here... https://www.ford-trucks.com/forums/655390-engine-performance-vs-increases-in-air-filter-inlet-temp.html that might be of interest.

 

I provided a new graph (pic#1) of Volume Air Flow VAF (CFM) and Mass Air Flow MAF (lb/min) for a 30 ft-length of pipe for several diameters. The flow scales as proportional to length, so for example to get the restriction for a 10 ft-length at a given CFM, divide by 3.

The MAF=VAF*AD, and the MAF scale on the right hand side is for a 14.7 psi AP and a 70F AT, which gives an AD=0.0750 lb/ft3. If you're flowing hot air at a higher pressure from a turbo like discussed in my previous post, for AD=0.0869 lb/ft3 at 180F with a 6 psi turbo without an IC, the MAF is increased by a X (0.0869/0.0750)=1.16 or about 16% from what's shown in the graph. For the case of AD=0.0994 lb/ft3 at 100F with a 6 psi turbo with an IC, the MAF is increased by a X (0.0994/0.0750)=1.33 or about 33% from what's shown in the graph. So you can use this graph with the different scale factors to estimate your MAF with and without an IC.

The graph in pic #2 gives the AFV=Air Flow Velocity, in MPH on the right hand scale. For a given restriction, the VAF scales as (Dia)^2, and the CSA also scales as (Dia)^2. Therefore the AFV through all the different dia pipes is given by the same gold dotted curve. For example, look at a restriction of 10" H20, and you see that the gold curve indicates an AFV of about 100 MPH for all pipe diameters. For example, that means at a 10" H20 restriction a 2.5" dia pipe (red curve) flows 300 CFM with an AFV of 100 MPH, a 3.0" dia pipe (blue curve) flows about 425 CFM with an AFV of 100 MPH, a 3.5" pipe about 575 CFM, and the 4.0" pipe about 750 CFM at a 100 MPH AFV.