shelleys_man_06

You're talking my language 93SilverFD. Lucky for us, the RENESIS has the largest ports of any previous Mazda rotary engine, IIRC. Probably a street port would do fine. I would like to see a bridge port however. Why a T78? I personally love the GT35R, 1.06 A/R. Either way, it's big flow. :D

mighty mouse

I talked with a rep at SFR and asked for some numbers for the turbo system, and this is what he told me. The turbo kit gives 8 a whopping bhp of 360, with 300 to the rear wheels, and 300 ft-lbs of torque @ the flywheel, with 250 to the rear wheels.

winbluerx8sport

DId they say that they had DYNO'D or those are IDEAL or ESTIMATED gains?

8inVegas

i wish those r the numbers

mighty mouse

they said that those are the dyno'd #, and that they should be posting all that information to their web site in a week or so. oh and if your wondering that is with a 7 pound boost.

93silverFD

I have always loved the name "T78"

Its always funny when some kid with his eclipse asks what im running, T78 usually inspires a good size grin. Its funny knowing it would take the exhaust from two of those 4-bangers to get these blades moving.

sausage71

6250$ for a turbo kit aint that a lil pricey i may be wrong but jesus

93silverFD

For a fully tunable package? Not in the slightest.

Look in to a real turbo kit for any car. It's going to cost you big bucks. This includes the fuel, spark, and FI upgrades. Show me any kit under 6k that gives you all that? Hell, hop over to gothamracing.com and price everything you would need for a single turbo kit on a third gen, including an ign. amp, power FC, good Front Mount Intercooler, extreme fuel upgrade kit, and a turbo kit. If you come out under 7k let me know how.

Zaku-8

does 7lbs boost for 120 whp on top of the normal 180 whp sound fishy to anyone?

rotarygod

In a perfect world, you'd exactly double your power for every 14.7 psi of boost that you add. In a perfect world, 7 psi added to a car that makes 180 hp would add up to 265 hp. Now take into account losses through the system from exhaust restriction, added intake heat, etc. Do the math on that one! I'll believe 250 rwhp from an originally 180 rwhp car. If the car made 190, just add 10 more or so to the final number.

GoBerserk

Rotarygod, I appreciate you bringing everyone back to the basics to explain, in a simple way, how boost pressure correlates to power.

But I've been thinking about it and feel like it's not atmospheric pressure that one needs to double (if they wanted to double power), rather it's the pressure (before or after combustion) in the combustion chamber.

Granted you could double the power if the atmosphere double its pressure and you remained naturally aspirated. But we're talking about forced induction and "boost" pressure. Boost pressure is measured (at least in my experience) as the pressure above atmospheric of the intake manifold.

Since the volumetric efficiency of the engine is not 100% in its naturally aspirated state, the pressure you need to double is the MAP pressure. Say the volumetric efficiency is 75%. So the MAP pressure during the intake "sweep" (stroke), (and the pressure of the intake charge would be similar), would be about 11psi.

So to double that pressure (and the power, roughly) you need 22psi. Since boost pressure is just a measure of MAP-14.7, you only need 7.5 lbs of "boost". What some people don't realize is to get an intake pressure of 1atm or 14.7psi you need some sort of forced induction. So even if you have a slight 'vacuum' (say 13 psi) your turbo or supercharger is still the thing that's getting you there.

I think we're saying similar things. The distinction is where we measure the 'boost' and how 'boost' is defined. Let me know what you think of my logic.

Zaku-8

If I'm reading you right, and assuming that the boost of 7.5 lbs is accompanied by increased volumetric efficiency, then 7lbs of marketed boost can be worth a lot more than half of NA power?

On a separate note, 7lbs boost/120whp for the SFR kit doesnt sound quite so unreasonable anymore... because its not just the boost, its the reworked fuel system. If by tuning optimally they manage to find a lot of the emissions and cat-protecting HP that was lost at port, that is.

GoBerserk

The forced induction is what increases the effective volumetric efficiency. But VE is really not what I think I want to be talking about - intake pressure is really the issue. The reason I mentioned VE was to find the theoretical intake pressure of the NA car. It may even be an incorrect usage - I'm not 100% sure.

In any event, The idea is that your power should go up by about the same percetage as intake pressure (assuming you add the correct amount of fuel to complement the new O2 available). At wide-open-throttle intake pressure is not 14.7psi (unless you have some sort of forced induction or fancy intake system). Therefore the pressure you need (to double the power) is twice whatever your NA pressure is. I don't own an 8 (yet - closing on a house first) but get a boost guage, hook it up to a vacuum line, accelerate at WOT and have a passenger (safety, safety) record intake pressure at different RPMs. It will probably read as 'vacuum'. But this is 'vacuum' below ATM. So your absolute pressure is what is the interesting number. If you pull 1 psi 'vacuum' then your MAP is 13.6 psia and you need 27.2 psia (psi absolute) to double power - which is equal to 12.5 psig (psi guage). - this is a different example than I used before-

120whp gain is a lot. Even if they do all the proper A/F management. I know Canzoomer and others have had success with this approach - with gains of around 15% - up to 200whp. They still want a 50% increase above that, and given the example above that would equal 13.6psi x 1.5 - 14.7psi = 5.7 psi boost. Therefore, theoretically, they'll have 300whp and then some. But this is based on a 1psi vacuum at WOT and peak hp RPM. I'm not sure what those numbers are. And of course such a cursory thoeritical analysis like the one RG and I are talking about excludes other factors that will decrease performance.

I wouldn't get my hopes up on any one person's FI theoretical claims on the system they're developing. I am confident that someone will figure it out sooner than later (I believe a couple credible systems are in the works) and will be able to give us all a dyno sheet. 'til that happens all we can do is talk.

rotarygod

When a boost gauge is reading 0 psi, it is actually already at 14.7 psi. This again assumes that it is a perfect day. This vaires a little bit. At sea level everyone is already getting 1 atmosphere of air pressure exerted on them. This is how much the air in the atmosphere above us weighs. The higher you go, the less this number is. If 14.7 psi is 1 atmosphere, then an additional 14.7 psi is twice as much. There is no exponential equation to it in order to increase power. It stays linear. Twice the pressure is twice the air. Again this also assumes that the airflow in cfm is constant. Instead of our gauges reading 14.7 psi all the time, we zero them out and start from there.

If we were running 22 psi, we may or may not double the engine power. We absolutely have to know the flow in cfm. Remember that the turbocharger compresses the air inside of it. Our boost gauge may read a certain psi but this doesn't mean that there is necessarily twice the air there. 14.7 psi is what is needed to theoretically double the horsepower level but this assumes that the turbo is sized properly and flows enough. A turbo that is too small may not have enough flow in cfm to double the engine's capability unless it hits 22 psi. It is really all about how much air the engine itself takes in more than it is about what psi it is at. If the engine takes in twice the air, it can make twice the power. When I do boost pressure conversions, they aren't meant to be taken as absolute. They are really just rough estimates of what should happen if everything was designed properly. In no way are they intended for someone to make claims off of them. Only a dyno can do that.

Richard Paul

iTrader: (5)

Do I feel like doing a lot of typing right now??
Nah, maybe some other time.

shelleys_man_06

I'm going to go with rotarygod in saying that volumetric flow rate is a key parameter. I'm trying to find a link between flow rates and turbocharger sizing. I not only have to look at compressor and turbine maps, but I'm stuck trying to find overall dimensions of the compressor inlet, outlet, and turbine inlet and outlet diameters. I also like to observe the cross sectional area of the toroidal shape a turbocharger makes, or I could just look at the manufacturer's A/R ratio. :D

I believe volumetric flow rate is depedent on the area, from the equation

v[dot]=(AV)

where v[dot] is the volumetric flow rate, dv/dt, A is the cross-sectional area and V is the velocity.

Also, we can use the volumetric flow rate as part of my favorite, and heavily abused Ideal Gas Law to find mass flow rate, for which shaft power is dependent on. However, isn't inlet pressure just as important for determining how much power we can get from the system? Mass flow rates are dependent on pressure, so it would be important to keep in mind as well. Systems, work producing and consuming, can vary in input/output based on ambient pressure and temperature.

There is more to this when it comes to the design of turbocharger systems. I just wanted to flex my thermodynamic muscle once more for old times sake.

rotarygod

I'm sure it can get really technical but most people won't understand that. Explaining things in simple terms makes accuracy difficult to convey sometimes.

93silverFD

Jesus guys, don't make me break out my physics books.

shelleys_man_06

The higher I get in my courses, the more complicated posts I'll put.

GeorgeH

iTrader: (2)

Then, when you graduate, you'll get some experiece and your posts will get simpler again. :p

93silverFD

lol

Lets see if I can complicate this anymore. I think what you all are trying to do is come up with a definite number increase in hp according to a specific increase in the manifold absolute pressure (MAP). This is physically impossible. If you could pump a pure gaseous oxygen containing mixture (how’s that for a oxymoron) in to the combustion chamber and completely react it with the fuel, I could give you the volumetric outcome and pressure resulting there of, of the products given the exact volume within which the reaction was contained. I could then extrapolate the work being done as a result of exerted force. You could then estimate with excellent accuracy, the percent change in applied force, at the flywheel of course. It doesn't work like that though. Your "air" is pumped in at an un-constant temperature, fuel is horridly inconsistent, and NOx's are always going to be produced in the combustion chamber as a result of the heat and spark. You can’t accurately predict the amount of energy lost due to heat absorption, nor predict that it is in some way linear. I do agree though, that you can give a pretty good estimate knowing the type of power produced per/pound of pressure increase (std. temp and pressure, one atm @ 25C if I remember correctly) in an engine of similar size, compression, and flow symmetry. Exact figures however, are impossible to determine. It is my experience though, that certain snails produce more power at a given boost pressure than others. This can be attributed to engine rpm, compressor efficency, charge temp, and probably a million other factors.

I will tell you this though, I cant see an RX8 with 7.5 psig pushing 300 at the wheels. Maybe the compression will be the wild card, but I don't see a difference of 1 compression point making that big of a difference (as opposed to the pre FD 9:1s).

I guess we will all have to wait and see. My bet though, is that the numbers are grossly overstated.

93silverFD

I would also like to reiterate what RG already touched on. Its all based on CFM, you can have all the pressure in the world built up in your manifold, if its not efficiently making it in to the combustion chamber, than it means nothing. Looking at the intake ports on the Renny though, I’m not sure this is going to be a problem. I'm not sure about the IM though, that would require "Scientific Analysis".

Richard Paul

iTrader: (5)

That's pretty good silver. You guys don't want to hear this but the rotary is not an efficent thing. In fact it is about 30% less then a piston engine. You can tell this from the MPG.
A good piston engine gets about 10 HP per Lbs of air per min. The rotary will give you 7. Don't be offended, there has been almost no development done on the rotary when compared to the piston engine. Think about it ONE little company has done this.

Given all that I'll tell you once more it's the DENSITY ratio not the PRESSURE ratio that you need. To keep me from having to type it all again you can find all the formula in the first 5 pages of my thread "axial flow supercharger".

This is in there someplace but here it is again. Attached.

93silverFD

I think that’s what I was getting at but I got ahead of my fingers somewhere in there. Density is mass per unit of volume (mass/vol.) Since mass is affected by temperature there is no way to predict charge density and the resulting molar amounts of product without having experimental output and temperature results for the specific turbo. Even assuming you knew the exact composition of all reactants involved.

93silverFD

Sure would be nice to have a Density gauge.