Absolute pressure = 15 psig + 14.7 = 29.7 psia

To determine true actual flow we have to take Volumetric Efficiency or VE into account.Turbo Compressor Map How to read a turbo compressor map.

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.

10.73 x 585 deg R

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 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.

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.

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:

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.

Pin = -0.5 + 14.7 = 14.2 psia

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.

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?

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

How to read the graph.

(14.2/13.949)

Pout= 15 psig + 14.7 = 29.7 psia

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.

1728 x 2