Standard pressure and temperature

[Mike Arman]

For aviation purposes, a "standard day" is defined as 29.92 inches of
mercury at 59 degrees F, equivalent to 1013.5 (I have also seen 1013.2)
millibars of pressure at 15 degrees C, this being the "average pressure and
temperature in the temperate latitudes" (real precise definition, right?).
Note that this is AT SEA LEVEL.

Standard temps are 50 degrees at 2,500 feet, 41 degrees at 5,000 feet, 32
degrees at 7,500 feet and 23 degrees at 10,000 feet.

Normal pressure drop is about one inch of mercury per 1,000 feet of
altitude. Normal temperature drop is about 3.5 degrees F (2 degrees C) per
thousand feet of altitude.

YT&BPMV

(Your temperature and barometric pressure may vary . . . )


Turbocharged and supercharged engines have what is called a "critical"
altitude - it is the altitude at which the turbocharger or supercharger can
no longer maintain sea level pressure in the inlet manifold. This makes our
calculations more difficult because we are starting at sea level, and
running increased pressure. 1.5 bar is 30 plus 15 inches of pressure, or a
total of 45 inches of manifold pressure (30 being normal atmospheric.)
(Note also that the effective positive pressure in the intercooler and
plumbing is only 15 psi - there's already 30 psi on the outside, even parked.)

If we are driving around on a road 7,500 feet above sea level, we can
expect the normal atmospheric pressure to be 30 inches minus 7.5 inches, or
22.5 inches of mercury. If we want to recover performance equal to 1.5 bar
referenced to starting at sea level, we'll need to see 2 bar on the gauge
at 7,500 feet - 45 inches is twice the outside air pressure at 7,500 feet,
whereas it is only 1.5 times the sea level pressure.

To make things more confusing, if the temperature is hotter than standard,
the air will be less dense than standard, and reverse. (This is called
density altitude - pressure altitude corrected for non-standard temperatures).

This can make a big difference - at 100 degrees F and 10,000 feet (altitude
of Leadville, CO, for instance), the density altitude is now equivalent to
about 15,000 feet, and the outside air pressure is only 15 inches of
mercury, which means we need THREE bar showing on the boost gauge to get
sea level air pressure (1 bar) plus .5 bar boost (total 1.5 bar) in the
manifold and thus sea level performance.

For the sake of our collective sanity, let's establish a standardized
barometric pressure, temperature and altitiude of the road above sea level
for our discussions of boost! Apples to apples, right?

Best Regards,

Mike Arman