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Equation of State Worksheet
- 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 __________oR.
(Note: Zero degrees Rankine (or Kelvin in metric units) is defined
as absolute zero which is the hypothetical temperature at
which all molecular activity ceases.)
- Now, let's algebraically solve the equation of state, p = R r
T, for R.
R = ___________.
- 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. _____________________.
- 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? _______________.
- What is the air temperature in oF at this altitude?
That's cold! (That is why the walls of an airplane are cold.)
- 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 = _______________ 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
- Calculate R at 36,089 ft (using pressure from #6 above, and r
and T in oR from #4.
R = _______________. Compare this to the R in #3. WOW! R is
- 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 = 390oR.
Solve the equation of state algebraically for r. r = ( ) / ( )( ).
Then substitute in R and the other parameters and compute r.
r = ______________slug/ft3.