1. To convert a Fahrenheit temperature to an absolute temperature, add 459.69 degrees. (The absolute temperature scale in English units is known as the Rankine scale and is expressed as ^oR.) Therefore our typical value of 59 ^oF is 518.69^oR
   
   
   
2. Now, let's algebraically solve the equation of state, p = R r T, for R.
   R = p/rT.
   
   
3. We are ready to calculate R, the gas constant for air, for typical values of pressure, density, and temperature. Using the values with English units displayed in the Air Properties Definitions slide, substitute into the equation for R, remembering to use the temperature from #1 above.
   R = ( ) / ( )( ). Compute R. R = 11.95 lb-ft²/slug-in²-^oR.
   
   
4. As we go up in altitude, the pressure, density, and temperature decrease. At an altitude of 36,089 ft, p = 472.68 lb/ft², r = 7.06 x 10^-4 slug-ft³, and T = 390 ^oR.
   How many miles high are we? 6.84 miles high.
   
    
5. What is the air temperature in ^oF at this altitude? -70 ^oF
   That's cold! (That is why the walls of an airplane are cold.)
   
   
6. Notice that the units of pressure at the 36,089 ft data set are in lb/ft². To be consistent with our first calculation of R, we must convert to lb/in². Therefore 472.68 lb/ft² = 3.28 lb/in² Comparing this to the typical value of 14.7 lb/in², we notice that we have only about one fifth as much air! That would make it difficult to breathe; therefore, airplanes are pressurized inside.
   
   
7. Calculate R at 36,089 ft (using pressure from #6 above, and r and T in ^oR from #4. R ~ 11.91. Compare this to the R in #3. WOW! R is constant!
   
   
8. Now that we have found (and checked ) R, let's use it to determine the density, r, at an altitude of 65,620 ft where p = 0.80 lb/in² and T = 390^oR.
   Solve the equation of state algebraically for r. r = ( ) / ( )( ). Then substitute in R and the other parameters and compute r.

   r = 1.72 x 10^-4 slug/ft³

Index

Intro

Activity

Worksheet