Volumetric efficency= (v. e.) and choice of turbo

VOLUMETRIC EFFICENCY (V. E.) and the CHOICE OF the TURBO



Hi to all dear friends some time ago on the forum I sent out a post pretty similar to this section, off-limits, even if
it was a little more "raw" of this to ****llo math let's say that from then I am a little also evolved arrhythmically and
being that even more, I can access most of this section I have decided to rewrite two lines on a topic maybe a little
known and little understood or even, dare I say, understood by few and unknown by many that I hope remains available
for all the other people who want it is time to learn new things...

starting from the assumption that I am not a professor, much less an engineer,I do not pretend or presume to
insegrare anything to anyone,in fact if someone wants to correct concepts believes are wrong are quite ready for it to rectify as
written here but still given that I believe that in life there is always more to learn and learn new
I want to put at the disposal of the concepts,and formulas the basic
that to some person may also seem interesting and useful...
I start with the specify that all of the calculations and equations you are using IMPERIAL units,
what are the English units...then later you can convert in our the METRIC system by using equivalences.
to start with, we provide the equivalences between imperial and metric measurements and the explanation
of the abbreviations used in the text:
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(Cubic inch = inch cube) = 16,39cm3; 1cm3=0,061 which

Psi (pounds on the square inch = pounds on the square inch = 0,068 bar 1 bar=14.7 psi

°F (degree Fahrenheit = degree Fahrenheit) = 32+1,8*x°C; °C=(°F-32)/1,8; for example, 0°C=32°F and 100°C=232°F.

R (degree Rankine = degree Rankine) = °F+460 = °C+273,15; it is the absolute scale
(i.e. which has the source at a temperature of absolute zero)
used in some English-speaking countries, instead of the Kelvin scale. For example: O°C=273,15 K=32°F=492R

Cu ft (cubic feet = cubic feet) = 28.317 cm3 = 28,317 l

Cfm (cubic feet per minute = cubic feet per minute) = 28.317cm3/min = 28,317 l/min

Lbs (pounds = lb) = 0,454 kg; 1kg=2.2 lbs

Mph (miles per hour = miles per hour (mph) = 1,6093 km/h, 1km/h=0,62 mph

Psig = psi gauge (psi its measured by the gauge)

Psia = psi absolute (psi absolute, which is obtained by adding the atmospheric pressure equals 1 bar=14.7 psi,
to the overpressure provided by the turbine)

R universal constant of perfect gases) = 8,315 J/(mol*K)

Tin = temperature input
Tout = temperature output
Pin = inlet pressure
Pout = outlet pressure


this not so brief post tries to explain in a simple way(for how short you can summarize)
using the example of an engine Lampredi, in the specific FIAT 1582CC 8Valvole turbizzato

the meaning of:
• volumetric efficiency,
• mass flow rate (quantity, not always constant) ,
• volumetric flow rate (volume constant),
• importance of the intercooler, or how it affects the cooling of the incoming air in the engine and
how will it affect the performance.

a little bit of the essential formulas....
to convert from cubic centimeters to WHICH:
WHERE= c.c. * 0,061
WHERE=1600cc * 0,061 =97,6 WHICH

The equation of the volumetric flow rate of the engine:
This equation serves to find the air volume (volumetric flow rate)
our engine ingurgita every two revolutions of the crankshaft,
(2 in as much as our engine is 4-stroke and the intake valve opens every 2 turns)

The Volume of air (cfm) = [(engine RPM)*(engine displacement)]/(1728*2)
(“1728” is a constant,
the “2” is used to take account of the fact that the air intake is only every two revolutions of the engine
given that we are dealing with a 4-stroke engine)

Once you have calculated the volume, which is always constant, we must calcolarci the mass flow rate,
that is, the LB/MIN (MASS),which may vary according to the pressure and temperature of the air.

The law of ideal gas / the mass flow rate of air:
The law of perfect gases is the equation is very convenient.
It relates the pressure, temperature, volume and mass (i.e. pounds) of air.
Knowing three of these quantities, we can calculate the fourth. The equation is:

P*V = n*R*T
where:
P is the absolute pressure (NOT the pressure measured by the pressure gauge),
V is the volume,
n is the number of moles of air (which is the unit of mass in this case),
R is a constant
T is the absolute temperature.

What are the temperature and the pressure absolute?
absolute pressure(psia)= pressure gauge(psig) + atmospheric pressure
The absolute pressure is the pressure in a duct is measured by a pressure gauge that marks 0 when this is atmospheric)
added to the atmospheric pressure. The atmospheric pressure at the ****llo of the sea is approximately 14.7 psi.=100kpa=1bar

Example:
"turbo pressure 1.5 bar + atmospheric pressure where you live 0,97 bar =2,47 bar absolute"

The scale of the pressure gauge can be in psia or psig. The “a” stands for “absolute”, “g” for “gauge” (that measured by the gauge).
The perfect vacuum is 0psia, or -14,7 psig.

The absolute temperature is the temperature expressed in degrees Fahrenheit + 460.
This gives degrees Rankine, marked with R. If the outside temperature is 80°F, the absolute temperature is 80+460=540R.

The Law of perfect gases can be rewritten to calculate any of the variables.
For example, knowing pressure, temperature and volume of air, we can calculate the pounds (mass) of air:
n = (P*V)/(R*T)

These we need from the moment that we know the pressure (or the pressure),
the volume, and we can make a good estimate of the temperature.
In this way, we can calculate how many pounds of air the engine is processing.
And the greater the mass of air processed, the greater the power that can be obtained.

Reverse formulas:
To calculate the mass of air:
n[lbs/min] = {P[psia]*V[cfm]*29}/{10,73*T[R]}

To calculate the volume of air:
V[cfm] = {n[lbs/min]*10,73*T[R}]/{29*P[psia]}

(as before, the “29” is the constant of equivalence, 10,73, instead, serves to convert the moles of air in pounds of air)

volumetric efficiency of the engine or volumetric efficensy:
Expressed as a percentage, indicates the actual amount of air that can get into the cylinders,
and changes in function of the conformity of the engine, 2valvole/4 valves/5valvole,
ducts, pipes, and various restrictions,to develop the engine head in an effective way,
tends to increase the volumetric efficiency

For an engine LAMPREDI, this value is estimated to 0.85 (85%).
Larger valves, cam shafts, machining of the head, turbocharger dynamics, etc.,
closer this value to 1 (100%).
when we calculate the amount of air that enters the engine,
we have to multiply the amount of air that is ideal for the efficiency to obtain the real value.

Amount of air = (amount of air in LB/MIN)*(volumetric efficiency in %)

Example: let's calculate the pounds of air flowing into the engines of two cars different
a equipped with intercooling and without intercooling.

We assume a volumetric efficiency of 0.85 for both engines.
Suppose we perform the calculation at 5000RPM. What is the volume of air drawn from both?

Volume [cfm] = (5000*97)/(1728*2) = 140,34 cfm (3974l/min)

This value is valid for both engines, with intercooler and the one that does not.

At 5000RPM each process 140,34 cfm of air per minute.

We take the model without the intercooler:
The temperature of the air in the intake manifold is about 212°F (100°C).
The pressure provided by the compressor is 19psi (1.3 bar).
That air mass is developing the engine?
Absolute temperature = 212°F+460 = 672R
Absolute pressure = 19psig+14,7 = 33,7 psia
n [lbs/min] = (33,7 psia*140,34 cfm*29)/(10,73*672R) = 19,02 pounds of air per minute (ideal)
the air volume actual = (lbs/min ideal)*(volumetric efficiency) = 19,02*0,85 = 16,167 lbs/min (7,35 kg of air)

We now turn to the model with the intercooler:
In this case, the temperature of the air in the intake manifold is only 77°F (25°C).
This engine is supercharged with 17psi (1,15 bar).
Absolute temperature = 77°F+460 = 537R
Absolute pressure = 17psig+14,7=31,7 psia
n [lbs/min] = (31,7 psia*140,34 cfm*29)/(10,73*537R) = 22,39 pounds of air per minute (ideal)
the air volume actual = (lbs/min ideal)*(volumetric efficiency) = 22,39*0,85 = 19,032 lbs/min (8,65 kg of air)

What denotes a first look at these two examples?
our engine equipped with intercooler, while having pressures turbo low
processes a greater mass of air, as cooler temperatures help them breathe in more oxygen..
and more oxygen means more power.

Part 2

THE COMPRESSOR AND ITS EFFICIENCY:
The compressor is the part of the turbocharger that compresses the air and pump it into the intake manifold.
This increases the pressure.
To make this happen, there is a need for power. The power comes from the hot side of the turbo, called the turbine.
Not all of the power comes from the turbine is used to create overpressure.
A part of the power is “needed” to heat the air.
It's the same thing when sfreghiamo the palms of the hands between them.
The friction between essegenera heat, as well as the friction between the compressor and the air,
and the friction between the air molecules generates heat.
By dividing the amount of power that is used to create overpressure
for the amount of power that is transmitted to the compressor, you get the performance (efficiency) of the compressor.
this concept can also be applied to displacement compressors, but for them to force things or how it is conformed
the compressor itself its efficiency will be very low which means precisely with high ****lli overpressure
the temperature of the air is scalderranno much at this point that not even the intercooler is very efficient, able to remove
effectively heat from the air.

For example, if a compressor has an efficiency of 70% means that 70% of the power transmitted to the compressor
it serves to create pressure. The remaining 30% of power warms the air.
And this is the reason why we have to prefer compressors with a high performance;
the more power is used to create excess pressure, the less he is wasted to heat the air.
If the temperature rises much increasing pressure, it may happen that you find yourself
even with less air entering the engine, and this causes a loss of power.

HOW HOT IS The AIR coming FROM the COMPRESSOR?:
That's a good question. The equation to use to calculate the temperature output is:

Tout = Tin+{Tin*[(Pout/Pin)0,263-1]/(yield)}

Example: the temperature input is 70°F (21°C), the inlet pressure is between-0.5 psig (a slight depression),
the output pressure is 19psig, and the yield is 72%. What is the temperature out?

Tin = 70°F+460=530R
Pin = 14,7-(-0,5)=15,2 psia
Pout = 19psig+14,7=33,7 psia
Pout/Pin = 33,7/15,2 = 2,21 (this is the PRESSURE RATIO)

Tout = 530+[530*(2,210,263-1)/0,72] = 700R-460 = 240°F = 115°C

Therefore, the temperature of the theoretical output is 240°F. it Would obviously be preferable to have an intercooler
that cool the air before it enters the engine!
The compressors do not always work with the same pressure output.
This in fact depends on the volumetric flow rate that enters
(not the pounds of air but the CFM of air, the volume then) and the rotation speed of the engine.
The performance of a compressor can be summarised on a chart by a series of curves.

Here's how to read the chart that you will find in the annex:

Calcolatevi the pounds of air that your engine processes,
have to do it for each rpm you interested,if you want to get a final graph the most detailed as possible,
the theory says that, we need to find a turbo with the best performance as close as possible to the point of torque maximum
of your engine (remembering that each rpm has a range of different air)

example:
we hold them to be good to have a motor with maximum torque at 5000 rpm
1. Since we are analyzing an engine 1600cc, let's convert this value to "where"
1582cc * 0,061 which =96,502 (97cui)

2. find the volume of air:
The Volume of air (cfm) = [(5000rpm)*(97cui)]/(1728*2)=140,34 cfm (3974l/min)

3. we calculate the LB/MIN. (mass flow rate):
n = (P*V)/(R*T)

4. we develop the equation:
• P we know the absolute pressure of the turbo because the set we
• V we know the volume because we have previously calculated
• R we know it because it is a constant of the perfect gases(10,73)
• T we lack to know the absolute temperature in the manifold, but if we have no way to measure it with instruments,
we can more or less define it to be approaching with a formula,
remember that the absolute temperature is calculated by adding "°F+460"..let's see

5. calculation of absolute temperature:
Tout = Tin+{Tin*[(Pout/Pin)0,263-1]/(yield)}
Example:
• the temperatures of the external input is 50°F (10°C),
• the inlet pressure upstream of the turbo is to 0.5 psig (a slight depression),
• the output pressure is 21,75 psig,(1,50 bar)
• and the efficiency of the turbo is equal to 77%.(found written on the inside of the chart garrett)
• What is the temperature out?
Tin = 50°F+460=510R
Pin = 14,06 psi-(or 0.5 psig)=14,56 psia (depression + atmospheric pressure)
Pout = 21,75 psig+14,06 psi=35,81 psia (manifold pressure + atmospheric pressure)
Pout/Pin = 35,81/14,56 = 2,46 (this is the PRESSURE RATIO)

course:
Tout = 510+[510*(2,460,263-1)/0,77] = 686,92 R-460 = 227°F = 108°C(temperature outside the compressor)

however, having an intercooler, the temperatures further decrease,
it also depends on what type of intercooler you use, whether air/air or air/water..
we suppose we have one air/water, where we can lower the temperature by up to 80% (yield of the ic)

6. the performance of the equation for the calculation of the LB/MIN:
n [lbs/min] = (35,81 psia*140,34 cfm*29)/(10,73*529R) = 25,68 pounds of air per minute (ideal)

7. now calcoliamoci the amount mass effective:
the air volume actual = (lbs/min ideal)*(volumetric efficiency) = 25,72*0,85 = of 21.88 lbs/min (9,95 kg of air)

8. result:
LB/MIN/5000RPM =21,88
PRESSURE RATIO =2,46

now just cross these data on the MAP FLOW, or COMPRESSOR MAP (as in the picture) you have chosen to see in that area
performance of the turbo are, and figure out if that turbo can do to your case..

Conclusions:
From this post, it denotes, that the theories are not the only calculations that are incomprehensible and meaningless written on a white sheet of paper,
or big word to fill the mouth, but (obviously) if used in the correct way, by filling in the equations with the data
as close as possible to reality, "the apple doesn't fall far from the tree".There is, however, to say that
it only takes a small change of a given, to change completely the final result,
and busting the reading of the chart..
Basically, more higher the number of spins that you get from your engine, the more are the lb/min that they swallow,
and the more power you get.
Lower the temperature of air circulating in the engine, the more I ingurgiterà the engine, resulting in greater power,in
as of course you will be able to enter more fuel,
if the outside temperature is higher, it will change the temperature in the manifold, as a consequence, you get fewer powers,
if you get it wrong the value of the VE of the engine at the point of maximum peak torque, you will get a given falsified..
all this to say that the variables in play here are many, but not having equipment,
it room test the engine as in F1, we could not have perfect data, but how much less,
you will be able to obtain information as less truthful...

Good job to all.

Munro

for the sake of completeness of info I would add the link to garrett that in addition to the above formulae and simplified that I use it when I choose a turbo there is a whole series of useful explanations on how to read a compressor map.
it's in English, but it is all quite understandable.

Good evening,

I'm imbattendo now in the editing and reading
the volumetric efficiency of the engine, and I found myself in front of
to a doubt.

I found in the map of my car with the curves of the mass air flow sensor, and knowing the
in the diesel engine works with excess air I tried to increase
the curves, the result is that the car was much more responsive, but unfortunately, as
that warmed it began to have issues with the RPM fluctuating (800/ 2000 RPM)
by the tool you gave me, however, that the volumetric efficiency was increased by 91% (EGR closed)
to 220%.

it is possible that this issue was due to a catkin excessive volumetric efficiency?

Do you think it would be enough to get the E. V. a to 100% to get the maximum from the mass air flow sensor?

Greetings.