## How to read a turbo compressor map.

To be able to read a compressor map for your specific application will require you to calculate the volume and mass of air moving through your engine.

Hopefully this will also offer some other details.

*Volumetric Flow Equation. *

This equation is for finding the volume of air going into the engine. Lets take an engine with a displacement of 122 cu.in .This is a 4 stroke engine with the intake valves on each cylinder opening once every 2 revolutions. So, for every 2 revolutions the engine takes in 122 cu.in of air. How many pounds of air is that? That depends on the pressure and temperature of the air in the intake manifold. But the volume is always 122 cu.in every 2 rpm.

*volume **of air (cu ft/min)= (engine rpm x engine cid)/(1728 x 2)*

*Ideal Gas Law/Mass Air Flow. *

The Ideal Gas Law relates the air pressure, a handy equation to have. If you know any three of these, you can calculate the fourth. The equation is written:

PV=nRT

P = the absolute pressure (not the gauge pressure),

V = the volume,

n = related to the number of air molecules and is an indication of the mass (or pounds) of air,

R = a constant number.

T = the absolute temperature.

What are absolute temperature and pressure? Do we care? Of course we do!

Absolute pressure is the gauge pressure (measured by a gauge that reads 0 when it is open to the outside air) plus atmospheric pressure. Atmospheric pressure is about 14.7 psi at sea level.

** Example:** A boost gauge reads 0 psi before it is hooked up. After it is hooked up it reads 10 psi under boost. 10 psi is the gauge pressure; the absolute pressure at sea level is 10 + 14.7= 24.7.

When a pressure reading marked “psia” it means “Pounds per Square Inch Absolute” and when the reading is marked “psig” it means “Pounds per Square Inch Gauge” A perfect vacuum is 0 psia, or -14.7 psig.

Absolute temperature is the temperature in degrees F plus 460. This gives degrees Rankine, or deg R. If it is 85 deg F outside, the absolute temperature is 85 + 460 = 545 deg R.

The Ideal Gas Law can be rearranged to calculate any of the variables. For example, if you know the pressure, temperature, and volume of air you can calculate the pounds of air:

**n=PV/ (RT)**

That is useful, since we know the pressure (boost pressure), the volume (which we calculate as shown in the first section (Volumetric Flow Equation), and we can make a good guess on the temperature. So we can figure out how many pounds of air the engine is moving to determine the turbo size needed. The more pounds of air being move the more power will be mademake.

For the purpose of this writing the Ideal Gas Law is rearranged to the two handiest forms, with the required constants:

*To get pounds of air:*

*n(lbs/min)= (P(psia) x V(cu.ft./min) x 29)*

*(10.73 x T(deg R))*

*To get the volume of air: *

*V(cu.ft./min) = n(lbs/min) x 10.73 x T(deg R)*

*(29 x P(psia))*

### Volumetric Efficiency:

In a perfect engine, complete cylinder fill could be achieved under perfect conditions(Theoretically) If there is 15 psi of boost in the intake manifold, we would open the intake valve and get 17 psi in the cylinder before the intake valve closed. Unfortunately, this doesn’t usually happen. With some exhaust remaining in the cylinder and the restriction offered by the intake ports and valves the actual amount of air that flows into the cylinder is somewhat less than ideal. The actual amount of air that does flow, divided by the ideal amount, is called the volumetric efficiency.

For the average basic stock high performance 4 cylinder double overhead cam this number is around 0.85 (or 85%). Things like big valves, big cams, ported heads, custom intakes and intake manifolds, tunnel rams, etc. can get this number closer to 1.0 (or 100%). Due to ram effect some normal aspirated engines/vehicles fitted with tunnel ram can get over 100% at certain rpms.

To determine true actual flow we have to take Volumetric Efficiency or VE into account.

actual air flow = ideal air flow x volumetric efficiency

Let’s use a 2.0 L (122 cu.in) turbo charged engine with an 85% VE @ 8500 RPM as an example.

Remember VE does not change even if you do turbo charge the engine. VE can only change when you modify the intake track such as installing bigger valves and porting the head etc.

*volume, in cu.ft per minute = 8500 x 122 = 290.97 cfm*

*1728 x 2*

This holds true for both an intercooled and a non-intercooled engine. Both intercooled and non-intercooled engines will be moving 290.97 cfm of air into the cylinders at 8500 rpm. As we will see however, the mass of air flowing is not the same.

To see what effect an intercooler can have we take the same 122 cu.in engine and calculate the lbs of air/minute for both conditions.

Suppose the temperature in the intake manifold for the non intercooled engine is about 250 deg F. The engine is running 15 psig boost. What is the mass of air the engine is using?

*Absolute temperature = 250 deg F + 460 = 710 deg R*

*Absolute pressure = 15 psig + 14.7 = 29.7 psia*

n(lbs/min)= 29.7 psia x 290.97 cfm x 29 = 32.9 lbs of air per minute (ideal)

10.73 x 710 deg R

lbsair per minute actual = lbs/min ideal x vol. eff.

= 32.9 x 0.85

= 27.97 lbs air/minute

Now let’s take the same engine and put an intercooler on it with the manifold temperature now at 125 deg F and 15 psi boost.

*Absolute temperature = 125 deg F + 460 = 585 deg R*

*Absolute pressure = 17 psig + 14.7 = 31.7 psia*

*n(lbs/min)= 29.7 psia x 290.97 cfm x 29 = 39.93 lbs of air per minute (ideal)*

*10.73 x 585 deg R*

*lbsair per minute actual = lbs/min ideal x vol. eff.*

*= 39.3 x 0.85*

*= 33.41 lbs air/minute*

Note that the inter-cooled engine got 5.44 lbs of air/min more than the non inter-cooled engine.

Let’s increase VE by installing bigger valves and port the cylinder head. Bench numbers for the head increased form 265 cfm ? 300 cfm thus meaning that an overall increase of 12% was achieved.

*New VE.*

*= 85 x 1.12 (12%)*

*= 95.2*

Therefore

*lbs air per minute actual = lbs/min ideal x vol. eff.*

*= 39.3 x 0.952*

*= 37.41 lbs air/minute*

By porting the head and installing bigger valves we now have increased the air intake of the intercooled engine with and additional 4 lb/minute

*Turbo Charging.*

Turbo ChargerThe compressor is the part of the turbocharger that compresses air and pumps it into the intake manifold. Air molecules get sucked into the rapidly spinning compressor blades and get flung out to the outside edge. When this happens, the air molecules get stacked up and forced together. This increases their pressure.

It takes power to do this. This power comes from the exhaust side of the turbo, called the Turbine. Not all of the power that comes from the turbine goes into building pressure. Some of the power is used up in heating the air. This is because we cannot build a perfect machine. If we could, all of the power would go into building pressure. Instead, because of the design of the compressor, the air molecules get “beat up”, and this results in heat. Just like rubbing your hands together will warm your hands due to the friction between your hands, the friction between the compressor and the air and between the air molecules themselves will heat up the air.

Compressor efficiency is determined by dividing the amount of power that goes into building pressure by the total power put into the compressor.

For example, a 70% efficient compressor means that 70% of the power put into the compressor is used in building air pressure. The other 30% of the power is used heating up the air. That is why we like high efficiency compressors because more of the power is being used on building pressure and less is used heating up the air. Turbos, Paxtons, and Vortechs are all centrifugal superchargers. They are called this because the centrifugal force of flinging the air molecules from the center of the housing to the outside edge is what builds air pressure. The maximum efficiency of these kinds of superchargers is usually between 70% and 80%. Roots blowers like the 6-71, work differently with a 40% lower efficiency. When you try to build lots of boost with a blower you have to put in a lot of power and more than half of that power is heating up the air instead of raising pressure.

*How Hot is the Air Coming out of the Compressor? *

The equation used to calculate the discharge temperature is:

*Tout = Tin + Tin x [-1+(Pout/Pin)0.263]*

*efficiency*

Example: the inlet temperature is 70 deg F, the suction pressure is -0.5 psig (a slight vacuum), the discharge pressure is 15 psig, and the efficiency is 72%. What is the discharge temperature?

*Tin= 70 deg F + 460 = 530 deg R*

*Pin= -0.5 psig + 14.7 = 14.2 psia*

*Pout= 15 psig + 14.7 = 29.7 psia*

*Pout/Pin = 29.7/14.2 = 2.092 (this is the compression ratio)*

*Tout = 530 + 530 x (-1+2.0920.263 ) = 687.53 deg R – 460 = 227.53 deg F*

*0.72*

At 15 psi boost with and an inlet temperature to the compressor of 70 deg F the theoretical outlet temperature is **227.53 deg F.**

Compressors do not always operate at the same discharge pressure. The discharge pressure that the compressor produces depends on the volumetric flow into it (not the pounds of air, but the CFM of air), and the rpm that it is turning. The performance of a compressor can be shown on a graph by a series of curves. Below is a GT 2876, GT3076, GT3571 and GT3776 compressor maps from Garrett.

The bottom of the graph shows the lbs/min of air that the compressor is moving, corrected to a standard temperature and pressure. The standard industry practice is to put this part of the graph in actual volumetric flow (such as ACFM) since the compression is constant for a given volumetric flow and compressor speed, NOT for a given mass flow. We have to figure out the pounds of air moving and correct it from the actual inlet temperature and pressure to their standard temperature and pressure.

The left side of the graph shows the outlet pressure to inlet pressure ratio.

There are two different sets of curves in the graph; efficiency curves and rpm curves. The area where there are lines drawn is the operating envelope. It is best to operate the compressor within its envelope. It will still run if you go to the right of the envelope, just not well. To the left of the envelope, where it is marked “surge limit”, the flow through the compressor is unstable and will go up and down and backwards unpredictably. This is surging. Do not pick a turbo that will operate in this area! It can be very damaging.

Pick a turbo that is close to the peak turbo efficiency at the engine’s torque peak while still maintaining at least 60% efficiency at the maximum rpm of the engine.

*How to read the graph.*

Figure out the pounds of air that you are moving through the engine. In our example, we were passing 33.41 lbs/min of air, at inlet conditions of -0.5 psig and 70 deg F. Now correct that flow to the standard temperature and pressure.

*Corrected flow = actual flow x (Tin/545)0.5*

*(Pin/13.949)*

Note that 13.949 has been used because everything is measured in psia instead of in inches of mercury

*13.949 psia = 28.4 inches mercury absolute.*

*29.92 inches mercury is atmospheric pressure at sea level, so 29.92 – 28.4 = 1.52 inches mercury vacuum.*

That is the standard suction pressure.

Standard temperature is 545 deg R, or 545 – 460 = 85 deg F.

So we are correcting the flow from 70 deg F and -0.5 psig to 85 deg F and -0.75 psig (or 13.949 psia, or 0.75 psi vacuum, or 1.5 inches mercury vacuum, or however you want to look at it.)

Again, temperature and pressure have to be absolute.

*Tin = 70 + 460 = 530 deg R*

*Pin = -0.5 + 14.7 = 14.2 psia*

*Corrected flow = 33.41 x (530/545)0.5 = 32.46 lb/min*

*(14.2/13.949)*

Mark that point on the bottom of the graph, and draw a straight line upward from that point.

The next step is to figure out the compression ratio, using absolute pressures. Using our example, we had 15 psi boost in the intake manifold. Let’s suppose the pressure drop from the turbo outlet to the manifold is 3 psi; so the actual compressor outlet pressure is 3+15=18 psig. The air pressure is 0 psig, but since the turbo is sucking air to itself the pressure at the inlet is lower than that.

Let’s say it is -0.5 psig at the inlet. Then the compression ratio, Pout/Pin is :

*Pout/Pin = (18 + 14.7) = 2.30*

*(-0.5 + 14.7)*

Now we plot 2.3 on the left side of the graph and draw a line horizontally from that point. Where the two lines meet is where the turbo will operate at 15 psi of boost.

Look at the efficiency curves, which look like circles. For the GT2876 it is inside the 74% curve, 77% for the GT3076, 77% for the GT3571 and 78% for the GT3776.

The other curves are rpm curves for the turbo. Our point on the rpm curve for the GT2876 is at 101500 rpm, for the GT 3076 at 101000, for the GT 3571 at 115500 and for the GT3776 at 98,000 rpm to get the pressure up to 15 psig from -0.5 psig. The Turbine has to provide enough power to spin it that fast.

Change any of these numbers, and the point at which the compressor runs at changes. More engine rpm means more air flow, so the operating point moves to the right. Colder intake temperatures means more pounds of air which moves our point to the right. Raising the boost probably means more air into the cylinders, but also the compression ratio goes up so our point definitely moves up and should move right. And so on.

If you are trying to choose between 2 turbos, pick the one with the better efficiency where most of your driving is done.

Note: This document is not produced to recommend a specific turbo, it is produce to help the reader understand the basic principals required in choosing a turbo for his/her application.