Altitude and turbos

[KeltonDSMer]

I have not been around on this forum very long, so I don't know if this was recently dicussed or not.
A turbo running 1bar of boost at sea level is at the same pressure ratio as a turbo at 12psi here. We all know this. Where I get flubbed up is turbo choice. Most upgrades are tested at releatively low altitude and are ment to be used there. Our turbos up here need a higher pressure ratio to make the same boost, but our car's motors flow only about 4/5 of what they do at sea level. This seems like it srews us in two ways. Not only do we have to reach higher pressure ratios for a given airflow, but the turbos can't make that higher pressure ratio at the same flow as the smaller ratio.

Example: A 50 trim compressor can support a 2.5 pressure ratio at 22-23lb/min without crossing the surge line. Assuming a 2psi IC pressure drop in all cases, a 2.5 pressure ratio=20psi at sea level and only 16psi here. Raising the boost to 20psi would create the same added airflow on top of the atmosperic pressure here as it would at sea level, but would effectively raise the pressure ratio to 2.83. We know that the turbo needs to flow at least 22-23lb/min to sustain a 2.5 ratio without going over the surge line; stepping up to a 2.833 ratio would require 27lb/min of airflow. What I'm trying to show is that we need higher pressure ratios here that at sea level to make any given power, but extra flow is required to reach these elevated ratios when our cars already flow much less in the first place....

I've given it some thought and I've come up with the conclusion that a 2.3L stroker up here flows about the same as a 2.0L at sea level at equal pressure ratios. If this were to be true, than spool time and turbo selection would be similar on a sea level 2.0 compared to a stroker up here. All my calculations are accurate, but I haven't taken into account the cange in volumetric efficiency which would definately be present on a stroker motor. I know a few people on this site happen to run strokers, do you guys have some insight on this? Can anyone guess the change in V.E. throughout the RPM range compared to a 2.0 and explain why? Anyone with an opinion or a statement they would like to make should do so indefinately.

I wish I had a buch of cash and resources to figure this out in the real world and not on paper.

[KeltonDSMer]

Can anyone give me a thoughtfull explanation of how/why the volumetric efficiency is different throughout the rev range on a stroker compared to a 2.0.
I only assume it is different because I don't know how to calculate V.E. I just used generic percentages that I found on the internet, which are probaly at sea level anyway. Does it change at diff. altitudes? My airflow calculations for the stroker use the same V.E. values so I'm sure my numbers are skewed even a bit more. Can someone give me some info so I can check this out with more accuracy?:)

[v413nc3]

Given a mass flow rate for air, you must have the density of the air to find the volumetric flow rate. Density is inversely proportional to temperature and can be readily calculated at any temperature. The density of air at 32F and 1 atmosphere is 0.0808 lb/ft^3.

Knowing the density of air at a given temperature, you can use a ratio to determine the density for the incoming air temperature...

t1 over t2 = d2 over d1

Where:
t1 = Temperature of air for a known density (32F @ 0.0808 lb/ft^3)
t2 = Temperature of the intake air measured by the intake air temperature sensor in degrees Rankine
d1 = Density of air for a known temperature (0.0808 lb/ft^3 @ 32F)
d2 = Density of the intake air (lb/ft^3)


Now to explain a few small things, Rankine is a measurement of absolute temp, to convert from Fahrenheit to Rankine you would just add 459.67 to it.

So you want to solve for d2 so that you know the Density of the intake air charge.

After you have d2 you will move on to solve your AVF (Actual Volumetric Flow Rate)

AVF = ft^3/minute

Here you can take your data logger such as DSMLink and get your Mass Flow Rate (MFR) in lb/min

MFR = lb/minute

so ....

AVF = MFR over d2

with me so far? We'll move on...

Knowing the theoretical air flow in cubic-feet per minute (cfm) your car can take in at a given engine speed (rpm) is an absolute must in determining volumetric efficiency. In order to figure out what the theoretical amount of air your engine will take in at a certain rpm, you will need to know... What is the displacement of your engine? For what maximum rpm?

So the equation looks like this...




TAF = (ED)(RPM)(VE) over (ES)C

RPM = maximum rpm chosen
TAF = Theoretical air flow (ft^3/minute)
VE = Volumetric efficiency (100% theoretical)
ED = Engine displacement (in^3)
ES = Engine stroke (2 for a four stroke engine)
C = Conversion factor from in^3 to ft^3
(1728)


Keep in mind this formula is ONLY for fuel injected vehicles.

The volumetric efficiency is simply the actual volumetric flow rate divided by the theoretical volumetric flow rate multiplied by one hundred.



VE = (AVF over TAF times 100)%



As you can see cubic displacement and engine RPM are primary factors in determining the volumetric efficiency of any given engine at given points of your engine speed.

[v413nc3]

Ok, now that I posted all that I will give you the information on my setup (theoretical) for lets say, a 72degree day.

491.67 / 523.67 * .0808

so my d2 is .0747


AVF = 273ft^3/min
(I know the naturally aspirated flow rate for my engine, you can't calculate it off of compressor flow rate.)

148.09 (displacement) * 7500(RPM) * 1.00 (VE 100% theoretical)
/ 2*1728 (conversion from ft^3 to in^3)

TAF = 321.375ft^3


VE = 84.94%

So the VE of my engine is 84.9 or 85% @ 7500RPM

So I'll just do some fast math and show you the numbers for a 2.0 at the same RPM

your CID is 139.28 (compared to a 2.3l at 148.09)
your NA CFM is 226.70 (273.17 on a 2.3)

TAF = 302.25

VE = 75.00%

We just seem to be more efficiant.

[hotrod]

If the intake system is unchanged (including the turbo boost pressure etc.) the VE of a stroker engine will be very similar to the original engine design as long as you are not making a massive change in the engine geometry. The VE numbers will just move to slightly lower rpms.

On typical modern design 4 valve 4 stroke engines, the VE max will occur at torque peak rpm and is typically about 95%. On most engines VE will fall into the low 90% to 85% range at max usable rpm. On engines that have significant ram effects designed into the intake manifold runners, these numbers can go above 100% at some points in the rpm range.

The drop in VE at high engine rpm is a function of both absolute manifold pressure (boost+altitude) and the air charge temperature as both are necessary to determine the true air charge density. The actual mass flow into the cylinder (the mass volumetric effeciency) is directly effected by the true manifold air charge density and the design of the engine.

If you have access to MAF numbers you can assume that everything that goes past the MAF must go into the engine, so you can figure VE and VE changes simply using the MAF numbers. If all your interested in is relative changes, than oldMAF/newMAF is good enough. If you want to see the limiting CFM a given engine can process, then you need the swept volume of the engine added to the equation. If you are trying to compute expected mass flow from that limiting volumetric flow, you must include the local atmospheric pressure to make sense of the MAF numbers.

When the swept volume of the engine is computed as in the above post it is a simple function of bore, stroke and rpm. That CFM value needs to be converted to a mass flow based on the temperature AND the absolute pressure of the air at the MAF sensor. You cannot use a constant of 0.808 lb/ft cubed @ 32 deg unless you are at sea level. What you really need to convert the volumetric max air flow of the engine to a mass flow number is the density ratio between the air at the MAF sensor and the air in the intake manifold, and a predicted VE for the engine at that rpm.

In the Ingersol Rand Compressed Air and Gas Data book (which uses a value of 0.08071 lb/ft^3 @ 14.696 psia) they list the following standard atmospheric pressures at altitudes.

sealevel 14.696
5000 ft 12.23
6000 ft 11.78

As you can see, here in the Denver area our base pressure is from 2.4 - 2.9 psi lower than the standard sealevel pressure the 0.0808 constant was figured on. Our typical weight of a cubic ft of air will be about .83 - .86 of the sealevel value.

For a given altitude/boost profile you can ignore the pressure only if it does not change. It will not change unless you change the manifold pressure by cranking up the boost, changing the altitude, or change the turbo IC design which effects the ability to maintain both boost at high rpm AND the absolute manifold air pressure at high rpm.

The second effect that determines the engines operational VE is a function of the head, intake valve, cam timing and bore stroke configuration of the particular engine.

At the end of the intake stroke just as the intake valve closes the actual pressure in the cylinder is very near the true manifold absolute pressure. Frequently due to inertial or accoustic ram effects it is 2-5 psi higher than the manifold pressure. Most of the VE effects are actually due to temperature changes in the intake charge as it is drawn into the cylinder. Something like 30% of the heat gain is due to conduction of heat from the intake runners and hot intake valve and valve seat. About 22% or so of the heat gain is due to the mechanical work performed on the air charge as it is sucked into the cylinder. At peak engine rpm and maximum piston velocity the air flow through the intake valve approaches the speed of sound. This causes significant heating of the air and is a significant factor in the reduction in VE at max rpm.

A stroker motor will have higher piston speeds, and will reach max torque usually at a lower rpm than the orginal design (if you have not changed valves, cams etc.). Due to the higher piston speeds at high rpm's it will also tend to fall off in VE a little sooner than the original because of the higher intake velocities.

Most engines hit max power near the point mean intake mixture speed reaches 220-250 ft/sec depending on cam timing and other variables.

Larry

[KeltonDSMer]

Hey, thanks man; that is some really good stuff you posted. The equations you gave me are basically the same as the equatiuons I got from a great turbo matching article in SCC. I have to guess on lots of stuff because I'm 15 and haven't found the right DSM for me yet, so I have no real life experience/info nesessary to do some of these calculations accurately.

Oh, and I'd like you all to help me make a decision of what mod route I should take my car on. I'll have only 6-8 grand to spend on the car and mods for the first 6 or so months, but that is about when I get my licence anyway; wow that sucks shit. What are the opinions on buying a stock 1g awd with a broken t-belt that f***ed up the head and bent some valves or something. I bet I could get a real good price; say, can the bottom end get messed up from the valves hitting the pistons? I was thinking of replacing the old 14b with a used 16g right off the bat and a 3" cat back, intake pipe, boost control and a fuel pump. I'll definately clean the IC and upgrade all IC piping too. I'll also buy some data logging equipment and an AFC, along w/ an EGT gauge. What do you think is the best way to read O2 voltage, I think air/fuel gauges are stupid w/o numbers because optimium O2 values change for different driving conditions. If I were to use an atmospheric dump off the IWG, would that bypass sufficient exhaust at full boost to get away w/ a 2.5" pressbent downpipe and a 2.5" test pipe, especially if the turbo had the 34mm flapper? I'll replace things like the Ex. mani when they need to be with better aftermarket parts too. Boulder cops are pretty ignorant, do you think I'll be fine with a dump tube if I set a fairly high boost level and don't drive like a cock sucker around town?

Sorry about that crap, I'm really tired. I just want some discussion of first mods on a 1G up at 6000 ft. Thanks for the help.

Oh, and tomarrow I'm going to compare some detailed flow calculations across the rev range for a stroker here an a 2.0 at sea level.

[KeltonDSMer]

Wow, that was a really great post hotrod. I'm glad that 12 psi is pretty close to the actuall atmospheric pressure here, (5500ft), because that's what I've been using this whole time to do airflow tables. Everday I want a stroker more and more; I can't afford to be driving around at 8000RPM because I will have no extra money. Theoreticly, you could use a gigantic compressor on a stroker motor, even with only 12psi atmospheric pressure. Cool.

Everday I get more into this stuff, when will it end?