Question 2.SP.5: If an artificial atmosphere consists of 20% oxygen and 80% n......

Table A.5:

R \text { (oxygen })=1554 \mathrm{ft}^{2} /\left(\sec ^{2} \cdot{ }^{\circ} \mathrm{R}\right),

\begin{aligned}& R \text { (nitrogen) }=1773 \mathrm{ft}^{2} /\left(\sec ^{2} \cdot{ }^{\circ} \mathrm{R}\right)\end{aligned}

Eq. (2.5): 100 \% \mathrm{O}_{2}: \quad \dot{\gamma}=\frac{32.2(14.7 \times 144)}{1554(460+60)}=0.0843 \mathrm{lb} / \mathrm{ft}^{3}

Eq. (2.5): 100 \% \mathrm{~N}_{2}: \quad \gamma=\frac{32.2(14.7 \times 144)}{1773(520)}=0.0739 \mathrm{lb} / \mathrm{ft}^{3}

(a) Each \mathrm{ft}^{3} of mixture contains 0.2 \mathrm{ft}^{3} of \mathrm{O}_{2} and 0.8 \mathrm{ft}^{3} of \mathrm{N}_{2}.

So for 20 %  \mathrm{O}_{2}, \quad \gamma=0.20(0.0843)=0.01687 \mathrm{lb} / \mathrm{ft}^{3} \quad

From Eq. (2.5), for 20 %  \mathrm{O}_{2}, \quad p=\frac{\gamma R T}{g}=\frac{0.01687(1554) 520}{32.2}

=423 \mathrm{lb} / \mathrm{ft}^{2} \mathrm{abs}=2: 94 \mathrm{psia}

Note that this = 20%(14.7 psia).

(b) For 80 \% \mathrm{~N}_{2}, \quad \gamma=0.80(0.0739)=0.0591 \mathrm{lb} / \mathrm{ft}^{3}.

\text { Mixture: } \gamma=0.01687+0.0591=0.0760 \mathrm{lb} / \mathrm{ft}^{3} \quad

\gamma=\frac{g p}{R T}        (2.5)