Predicting gauge pressure, perfect Model A

L

Larry Brumfield

BANNED
Quite often people try to predict gauge pressure on compression tests by multipying the atmospheric pressure times the compression ratio. If I've heard it once I've heard it 40 times, "If your Model A is exactly like it was when they rolled off the line, your pressure at sea level ought to be about 62 pounds." They determine this by multiplying the 4.22 compression ratio x 14.7 p.s.i. (atmospheric pressure at sea level)

Well that's close but no cigar!

To be more accurate you must take into account the temperature rise caused by compression when the test is being performed plus any possible heat loss to the cylinder. To compensate, the following adjustment is necessary.

At cranking speed, the absolute pressure is a function of the 1.25 power of the compression ratio. For example, 4.22^1.25 x atmospheric pressure = the absolute pressure.

Now the 1.25 is not accurate in all cases and may vary according to some from 1.2 to 1.3, since the amount of heat will vary. But it does provide useful results for purposes of comparison. Also, the atmospheric pressure can be determined wherever you are by finding out the pressure in your area and multiplying by .491. For example, 29.92 in. Hg. x .491 = 14.69 or 14.7 p.s.i @ sea level. You can find out the pressure on any given day by looking at your local weather channel.

So to predict gauge pressure on a perfect, stock Model A engine at sea level:

4.22^1.25 = 6.048401

6.048401 x 14.7 = 89 p.s.i.

89 - 14.7 = 74.3 p.s.i. gauge pressure


Larry B.
 
" You can find out the pressure on any given day by looking at your local weather channel."

Larry,

This is not correct. Here at Longmont,CO.,my pressure is about 25 inches. During the weather report, the indicated pressure is always around 30 or so inches as it is corrected to the equivalent pressure at Sealevel. Weather stations though out the US always correct the pressure to the "equivalent value at Sealevel".

There is a fairly complicated formula relating air pressure to altitude since the relationship is non-linear.

Marc
 
Since I'm down south in the flat country fairly close to the sea I never thought about the weather channels elsewhere. However, if you know the inches from your own personal mercury barometer then it is just a matter of multiplying by .491. A simple relationship exists between pressure measurements in p.s.i. and in inches Hg.. One cu. in. of mercury weighs .491 pound.

Marc wrote: "There is a fairly complicated formula relating air pressure to altitude since the relationship is non-linear."

Nothing complicated about it. You can figure the pressure if you know the altitude whether you have a barometer or not. Atmospheric pressure at sea level is approximately 14.7 p.s.i. As altitude increases, the atmospheric pressure decreases appoximately 1 (one) p.s.i. for every 2,343 feet above sea level. So if you were at 5,280 ft. for example, the atmospheric pressure would be approximately 12.44 p.s.i. (not quite as accurate as with a barometer reading multiplied by .491 but close).

Moreover, all this is not perfect without the use of sophisticated mesuring devices. Pressure varies at any given point in the atmosphere. On a standard day, at sea level, pressure will be 29.92 in. Hg. On nonstandard days, pressure at sea level will vary considerably above or below this figure.

Larry B.
 
"As altitude increases, the atmospheric pressure decreases appoximately 1 (one) p.s.i. for every 2,343 feet above sea level. "

Larry,

Pretty good formula. However, above 5000 ft the pressure starts to drop off even faster. Think about a large "blob" of air and the heavy stuff is at the bottom.

Marc
 
True but not near as much as you think. At 10,000 feet above sea level, standard pressure is 20.58 in. Hg. or 10.10 p.s.i.

If you figured it the altitude way you would get 10.42 p.s.i. That's still pretty close.

10,000 divided by 2,343 = 4.27

14.69 - 4.27 = 10.42 p.s.i.


Larry B.
 
You must log in or register to reply here.
Back
Top