stichl let me help you out, the last 3 points are generally the true the rest makes no sense.
1 st we need to consider the values that are to be plotted on the comprssor map, they are given for a reason.
along the y axis we have the pressure ratio, this is the absolute boost pressure (boost plus atmospheric)/boost pressure. The boost pressure is something we decide in advance for this exercise lets call it 14.7psi (1 Bar), this gives a Pr=2.
We then need to consider what is happening on the x axis. this is the Engine's air flow requirements it is not the turbo's air flow output, this disqualifies the made up number in the middle of your post. Here we are considering 2 situations one is a std head and one a mildly gas flowed head.
so the air flow along the x axis is calculated by
(L x rpm x VE x Pr)/ 5660 with the result in CFM cubic feet per minute. It convienient to tell at this point you that this graph will be corrected for 20deg C, at about 900 ft above sea level, as air desity reduces with increasing temperature and altitiude.
so for engine 1 standard
L = 2 liters
rpm = 7000 (max rpm but 7k for simplicity)
VE = 90% (for 4 valve)
Pr = 2 (worked out above)
= (2 x 7000 x 90 x 2)/5660
= 445.2 CFM
In Lb/min multiply by .07
= 31.164 lb/min
Revise the VE for engine 2 with head work, lets say a 95%VE, this maybe a bit tight but a full race engine only has 105% VE.
CFM = 469.9
or
32.897 lb/min (an increase over std)
We then plot these 2 points on the graph (see my photo attached very hi-tec
)
http://img.photobucket.com/albums/v74/sediciRich/IMG_0428.jpgAs, was done before on the old forum we need to plot some things to estimate whether we will meeet the left hand surge line, here we would estimate that the car would reach full boost by half rpm (it will do this sooner in reality) thus half max air flow.
For std unit at 15.82lb/min
Mod 16.449 lb/min both at Pr 2.
Finally we can add another estimation plot to say at 20% air flow we would expect to be at Pr =1 eg no boost.
std 6.233 lb/min
mod 6.579 lb/min
Looking on the graph you can see a number of things, the std max air flow leaves the turbo in the 72% range at Pr =2, while the modified unit requires more air which asks more of the turbo at a point where its efficiency is dropping off I estimate this part of the graph at 67% although there is no value on the graph. So this disproves the bit about increasing VE extending turbo efficiency or words to that effect. This plot avoids the max efficiency island at 1bar, saying that for mid to high rpm this turbo perfomrs well at .6 bar boost. Essentially the max efficiency could be utilised by having a variable boost up the rev range allowing a plot to run through the max eff island, up to say the 1.4 bar on the std engine - this needs electronic control. However using 1.4 bar on the modified unit will have the turbo spinning at what is the highest plotted speed on the graph (the more horizontal lines)potentially 135,000 rpm - this is not good for bearing life.
We can go on from here to calculate the difference in temperature from these 2 max air flows at Pr=2 (std and mod)
First we can calculate the temperature rise (theoretical) using
Tr= (F x (At + 273) x 100)/E
Where Tr is temp rise
F is a correction factor for Pr=2 which is 0.217
At + 273 is the inlet temp in deg Kelvin.
E is the efficiency at the max rpm on our map.
Std unit
= (0.217 x (20 + 273) x 100)/72
Tr std = 88.3 degrees C
For the mod unit estmate on map 67%
Tr mod = 94.3 degrees C.
Discharge from the compressor will be the increase + the ambient (20 degrees C as the map correction)
Dt std = 108.3 deg C
Dt mod = 114.9 deg C
from this we can work out how much more air we are flowing towards the engine, this is important because our pressure ratio of 2 does not mean we are pumping 2 the mass of air, because it heats up its desity is reduced despite have a pressure of 2 atm.
New desity formula
Dr = 293 x Pr / Dt + 273
293 is inlet temp in deg K
Dr std = 1.53
Dr mod = 1.51
this says the std unit will creat 1 bar boost at 7k rpm creating 53 % more air mass, with the modified unit turbo pumping less mass 53%. Incidentally its easy to estimate bhp if std was 200 then with 50 % more air power will be 300. This assumes many things A/f ratio. After intercooling the actual pressure will drop, thus ultimately in the calculations we have to decide where we take our pressure ratio from pressure after the intercooler will be slightly lower then at the turbo outlet.
So what has this shown, well it shows a car with better VE demands more air over its std counter part; although at the expense of compressor efficiency. But it also shows that there isn't great deal of difference at the end in air flow terms from the turbo for a small increase in VE. But as I said in an earlier post the improved VE assists with cylinder events. Allowing more of the air flow to be inducted in each inlet cycle, this ultimately along with a correct AF ratio increase power. But there are so many other factors apart from the turbo, its easier to think of NA tuning to see factors (not directly appliacble to forced induction) which improve power, the main one is VE, and the quality of combustion a function of many things including ignition timing. The turbo compressor is only a part in the chain and like Stichl quoted at the end the turbine has a part in the chain of events.
Just and end note - I have not pasted this of any site, some people will no doubt be thinking of insults to reply to this; Please don't. Its taken me a while to type this, I do so to help other enthusiasts and not to be slated. Its also taken me a long time to learn and understand these topics, so I no longer rely on books for explainations, but I do for the equations. I hope this is of use to someone, as you can follow the same work as a GUIDE to other turbos if you can get maps for them.
Hopefully there aren't any glaring errors.
Rich