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

If an artificial atmosphere consists of 20% oxygen and 80% nitrogen by-volume, at 14.7 psia and 60°F, what are (a) the specific weight and partial pressure of the oxygen and (b) the specific weight of the mixture?

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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)

TABLE A.5   Physical properties of common gases at standard sea-level atmospheric pressure  { }^a   .
Gas Chemical formula Molar mass ,   \mathbf{ M} Density   \rho   , Absolute viscosity, {}^ b Gas constant,    \boldsymbol{R} Specific heat,   c_p \quad c_v Specific heat ratio, k=c_p / c_v
  \textbf { at } 68^{\circ} \mathbf{F}  slug/ slug-mol    \textbf { slug/ft }{ }^3 10^{-6} \mathbf{lb} \cdot \mathbf{sec}^2 \mathbf{ft}^2   \begin{aligned}& \mathbf{f t} \cdot \mathbf{l b} /\left(\text { slug } \cdot{ }^{\circ} \mathbf{R}\right) \\& =\mathbf{f t}^2 /\left(\sec ^2 \cdot{ }^{\circ} \mathbf{R}\right)\end{aligned} \begin{aligned}& \mathbf{f t} \cdot \mathbf{l b} /\left(\text { slug } \cdot{ }^{\circ} \mathbf{R}\right) \\& =\mathbf{f t}^2 /\left(\sec ^2 \cdot{ }^{\circ} \mathbf{R}\right)\end{aligned}
Air carbon 28.96 0.00231 0.376 1,715 6,000 4,285 1.40
 dioxide carbon CO_2 44.01 0.00354 0.310 1,123 5,132 4,009 1.28
quad monoxide CO 28.01 0.00226 0.380 1,778 6,218 4,440 1.40
Helium He 4.003 0.000323 0.411 12,420 31,230 18,810 1.66
Hydrogen H_2 2.016 0.000162 0.189 24,680 86,390 61,710 1.40
Methane   CH_4 16.04 0.00129 0.280 3,100 13,400 10,300 1.30
Nitrogen   N_2 28.02 0.00226 0.368 1,773 6,210 4,437 1.40
Oxygen O_2 32.00 0.00258 0.418 1,554 5,437 3,883 1.40
Water vapor   H_2O 18.02 0.00145 0.212 2,760 11,110 8,350 1.33
  \textbf { at } 20^{\circ} \mathbf{C} kg / kg-mol   \mathbf{kg} / \mathbf{m}^3   10^{-6} \mathbf{~N} \cdot \mathbf{s} / \mathbf{m}^2    \begin{aligned}& \mathbf{N} \cdot \mathbf{m} /(\mathbf{kg} \cdot \mathbf{K}) \\& =\mathbf{m}^2 /\left(\mathbf{s}^2 \cdot \mathbf{K}\right)\end{aligned}   \begin{aligned} & \mathbf{N} \cdot \mathbf{m} /(\mathbf{k g} \cdot \mathbf{K}) \\ & =\mathbf{m}^2 /\left(\mathbf{s}^2 \cdot \mathbf{K}\right) \end{aligned}
Air carbon 28.96 1.205 18.0 287 1003 716 1.40
 dioxide carbon CO_2 44.01 1.84 14.8 188 858 670 1.28
quad monoxide CO 28.01 1.16 18.2 297 1040 743 1.40
Helium He 4.003 0.166 19.7 2077 5220 3143 1.66
Hydrogen H_2 2.016 0.0839 9.0 4120 14450 10330 1.40
Methane   CH_4 16.04 0.668 13.4 520 2250 1730 1.30
Nitrogen   N_2 28.02 1.16 17.6 297 1040 743 1.40
Oxygen O_2 32.00 1.33 20.0 260 909 649 1.40
Water vapor   H_2O 18.02 0.747 10.1 462 1862 1400 1.33

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