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# McMath

```McMath
by Dave Lawson,  1997

I know there is some interest in learning about turbos
and intercoolers and the effect they have on our cars,
so I spent some time "doin' the math", which I hope will
build a technical foundation in which we can discuss
various upgrades.

A while back I set off to educate myself on what my car
is doing, so I could make informed decisions on expanding
the performance envelope. A few months ago I did a post,
turbo.edu which listed a number of sights where those with
inquiring minds could go and learn about turbos and such.

Here I have literally gone through an example of 1 operating
condition of an MC engine, explaining the various inputs
which enter into turbocharger and intercooler analysis, along
with handing out the actual math to do the calculations.

If your are interested in why your car is slower driving
through the arizona desert in august or why your car will
drive differently cruzzin' up pikes peak to meet with
Ms. Mouton, get out your calculators. If a technical
analysis bores you, hit the D key. Enjoy.

Take our much discussed 2.2l MC engine, the amount of air
the engine consumes can be calculated using the following
equation:

w * D
Fi =  -------
(1728 * 2)

where
Fi = ideal air flow [ft^3/min]
w  = engine speed [rpm]
D  = engine displacement [in^3]

Letting D = 135.8 and w = 6000 the ideal air flow of
the MC engine is:

Fi = 235.8 ft^3/min

The problem here is our engine isn't 100% efficient, so
we need to factor in the volumetric efficiency and calculate
the actual air flow:

Fa = Fi * VE

Assuming the MC is 95% efficient, we get

Fa = 224.0

I know this VE might be a bit high for the MC, but have never heard
an actual number mentioned.

Remember the ideal gas law which Bob Meyers described a
few days ago. Look closely and you will see some of what
Bob was describing in the math below.

PV = nRT

where
P = absolute pressure [psig]
V = volume
n = number of moles
R = constant = not used directly in this example
T = absolute temperature [ degrees Rankin]
= [deg F] + 460

Now lets throw some other conditions into the example

Ta = ambient air temp [deg F] = 80
a nice summer day

Pb = boost pressure [psi] = 13.2 ==> .9bar
close to those IA stage II mods floating around

Pi = turbo inlet pressure [psi] = 12.2
let altitude = 5000ft (because thats where I live), this effects
the turbo inlet pressure, which sets Pi = 12.2 psi instead of
the 14.7 psi for sea level

Ei = intercooler efficiency = .7
Another number whose actual value can be debated, remember it is
0 when sitting still and is less efficient when it gets the heat
soak treatment

Now lets calculate the ideal temperature gain which occurs when
the turbo compressor does its job and, ta-da, compresses the air

Tgi = (Pr^0.283 * (Ta + 460)) - (Ta + 460)

where
Pr = pressure ratio
= (Pi + Pb) / Pi
= 2.29 <=== note this is above the upper limit which QSHIPQ
has mentioned previously, 2.0, and shows why you need
think in terms of pressure ratio, especially those
living @ altitude, or when you are out touring in
the hills

Using the above described conditions we get

Tgi = 142.6 [deg F]

Now we need to include the turbo efficiency...

Et = turbo charger efficiency = 65%
This number comes from those rare compressor maps which KKK
doesn't release. It's based in the design of the turbo and
includes both the hot and cold side characteristics. This actual
number is based on the pressure ratio and the actual air flow
calculated above. See, those maps really do have a use.

Like the engine,  the turbo isn't 100% efficient and this
effects the temperature gain.

Tg = Tgi / Et

= 219.5 deg F

So now the actual turbo outlet temp is

To = Ta + Tg

= 299.5 deg F

Now lets calculate the air density ratio based on the compressed
air

Rd = ((Ta+460)/(To+460)) * Pr
= 1.63

Now let this air flow into the intercooler and calculate how much
the intercooler reduces the air temperature:

Tir = Tg * Ei
= 153.6 deg F

Now calculate the actual air temp at the intercooler outlet
or intake manifold inlet:

Tm = To - Tir
= 145.9 deg F

So the net rise in the air temp after being compressed and
cooled is

Tr = Tm - Ta
= 65.9 deg F

With this information we can calculate the intake air tract
overall efficiency

Eintake = (Ta+460)/(Tm+460)
= 0.89

And the net air density ratio after intercooling

Rm = Pr * Eintake
= 2.04

And now we can calculate the effective air flow of our
turbocharged, intercooled MC engines

Fe = Fa * Rm
= 457.0 ft^3/min

By using the turbocharger/intercooler we have more than doubled the
air flow our engine would produce if it was normally aspirated.
Remember this example is for 1 engine speed at 1 ambient air temp
and 1 boost pressure using 1 point off the K26 turbo compressor map.
I leave it as an excercise to the reader to calculate the results for
cold days, an engine coming off idle(why does my engine come alive @
3000 rpm?), driving at altitude, a more efficient intercooler, etc.
I have developed a spreadsheet to do all these calcs and even keep
tabs on the competition. How much air does that 944 turbo S consume
or what is the air flow for those 12 sec talons?

We can take this further and start calculating the fuel
required to mix with this compressed air or how the pressure
drop across an intercooler effects the system, but... lets save
some subjects to discuss next week.

I do hope that owners of the 1.8L turbos have been following
the discussions, as these concepts are just as applicable. I
haven't yet ran the numbers for these cars, but I would be
interested in the results if others have. I would also like
to hear about the design of the 1.8T systems, intercooler sizes,
diameters of the pipes, measured boost pressures and air temps,
etc.

-
Dave Lawson  dbldmnd@compuserve.com
83 ur-q

```