Modern Engineering Thermodynamics

416 CHAPTER 12: Mixtures of Gases and Vapors
EXAMPLE 12.4
Though oxygen is necessary to sustain life, breathing oxygen at elevated pressure has toxic effects. It causes changes in lung
tissue and affects the liver and brain. Acute oxygen poisoning at high pressures can cause convulsions that can lead to death
(even at atmospheric pressure, pure oxygen can be breathed safely for only two hours). Oxygen poisoning at elevated environmental
pressure can be avoided by maintaining the oxygen partial pressure equal to that of atmospheric air at standard
temperature and pressure (STP). Also, since atmospheric nitrogen is very soluble in blood and body tissue, rapid depressurization
causes nitrogen bubbles to form in the blood and tissue (nitrogen embolisms) producing a condition commonly
called the bends.
Divers going to great depths in the sea are able to circumvent this problem somewhat by breathing a compressed heliumoxygen
mixture in which the oxygen partial pressure is adjusted so that it is always equal to its value in atmospheric air at
STP. Helium, being much less soluble in body tissue than nitrogen, decreases the time required for depressurization when
the diver returns to the surface.
The engineering problem that we must solve is stated as follows:
a. For a deep water diver, determine the proper helium-oxygen breathing mixture composition for a dive to 100. m below
the surface of the water, where the pressure is 1.08 MN/m 2 . Give your answer in mole, volume, and mass fractions.
b. Determine the effective gas constant, the specific heats, and the specific heat ratio for this mixture.
Assume helium and oxygen behave as ideal gases with constant specific heats.
Solution
a. From Example 12.3 and Eq. (12.23), we find that the partial pressure of oxygen in air at STP is
p O2 = χ O2
p m = 0:2095ð0:1013 MN/m 2 Þ
= 0:0212 MN/m 2
Therefore, at a total pressure in 100. m of water of 1.08 MN/m 2 , this same partial pressure requires a mole and volume
fraction of oxygen of only
χ O2
= ψ O2
= π O2 = p O2 /p m = 0:0212/1:08 = 0:0196
and, from Eq. (12.10), the helium mole and volume fractions are
χ He = ψ He = 1 − χ O2
= 1 − 0:0196 = 0:980
The equivalent mass fractions are given by Eq. (12.13), where the mixture equivalent molecular mass can be computed
from Eq. (12.11), as
then,
and
M m = χ O2
M O2 + χ He M He = 0:0196ð32:00Þ + 0:980ð4:003Þ = 4:55 kg=kgmole
w O2 = χ O2
ðM O2 /M m Þ = 0:0196ð32:00/4:55Þ = 0:138
w He = 1 − w O2 = 0:862
Therefore, the required oxygen concentration is only 1.96% on a volume or molar basis, but it is 13.8% on a mass or
weight basis.
b. The mixture equivalent gas constant can be computed from Eq. (12.15) as
R m = R = 8:3143 kJ/ ð kgmole .KÞ
= 1:86 kJ/ðkg .KÞ
M m 4:55 kg/kgmole
and the mixture specific heats can be determined from the equations in Table 12.5 and Table C.13b as
and
c vm = w O2 c vO2 + w He c vHe = 0:138ð0:657Þ + 0:862ð3:123Þ = 2:78 kJ=ðkg .KÞ
c pm = w O2 c pO2 + w He c pHe = 0:138ð0:917Þ + 0:862ð5:200Þ = 4:61 kJ=ðkg .KÞ