Re: B747 technical questions

Date:         05 Aug 97 17:45:51 
From:         Bob Standaert <standaert@chemvx.chem.tamu.edu>
Organization: Texas A&M University
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shumaker@eisner.decus.org wrote:
> In article <airliners.1997.1633@ohare.Chicago.COM>, "Michel Gammon" <jmgammon@sympatico.ca> writes:
> > shumaker@eisner.decus.org wrote in article <airliners.1997.1560@ohare.Chicago.COM>...

snip

> > Water is a liquid and therefore will not be subjected to the same rules and
> > also is incompressible.  Water vapour is microscopic liquid water droplets,
> > and steam is water in the gaseous phase.

No.  Water vapor is a gas.  Steam -- water vapor -- water in the gaseous
(vapor) phase.

> There is a significant volume increase when liquid-phase water changes
> into gaseous-phase water, and the phase change begins to occur over the
> range of temperatures which tires encounter.

True.  A kilogram of water at 0 deg C would occupy a volume of 1.0 L as
a liquid, 1.1 L as a solid, or 1250 L as a (hypothetical) gas at 1 atm.
pressure (14.7 psi).

However, I have trouble believing that the pressure fluctuation caused
by the condensation or freezing of water vapor is a serious issue.  For
the sake of illustration, assume the volume of a large aircraft tire is
1000 L; whether this is right or wrong doesn't affect the conclusion.
Imagine the tire is pressurized to 150 psi with sultry summer air at IAH
(40 deg C, 100% relative humidity), and that no liquid is introduced.
I'll treat the Houston air as an ideal gas (hardly an assumption anyone
who has breathed it can accept, but bear with me). The vapor pressure of
water at 40 deg C is 55 mm Hg (1.07 psi), and the total pressure of 150
psi would reflect the sum of the partial pressures:  149 psi from dry
air and 1 psi from water vapor.  The mass of the contained gaseous water
would be approximately 50 grams (equivalent to 50 mL of liquid, or about
5 teaspoons).  If the tire were then cooled to -50 deg C, where the
vapor pressure of water is 0.03 mm Hg (0.0006 psi), essentially all of
the water would condense and freeze.  The pressure in the tire would
drop to about 106 psi, but the condensation of water would contribute
only about 1 psi to the drop.  It would be as if the tire had initially
been inflated to 149 psi instead of 150.

On the other hand, if you started out with some liquid sloshing around
inside the tire and heated it up, you could generate a significant
pressure increase over what you'd get otherwise (the water will
contribute a partial pressure equal to its vapor pressure, which
increases sharply with temperature).  If we heat the tire (no extra
liquid inside) from 40 to 100 deg C, its pressure will increase from 150
to 179 psi; if there were extra liquid at the outset, the pressure could
go as high as 192 psi (water has a vapor pressure of 14.7 psi at 100
deg). You would need to vaporize 0.6 L of excess liquid to see the full
effect.  If we heated the tire (no extra liquid) from 40 to 150 deg C,
the pressure would increase to 203 psi.  With excess liquid water at the
outset, the pressure could go as high as 271 psi because water has a
vapor pressure of 69 psi at 150 deg C.  You would need to vaporize about
2.5 L of liquid to see the full effect.  Less liquid, less effect; no
liquid, no effect.

As long as you don't start out with liquid water inside the tire, the
water won't have much effect on pressure.  If small amounts of
condensation are a concern, I expect it is because of the potential for
freeze/thaw damage or corrosion rather than pressure changes.  Since it
would be just as easy if not easier to use dry air as nitrogen, and air
is a little cheaper, I expect that the chemical inertness of N2 is a
significant factor.

Regards,

Bob