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Problem 3.45

# Problem 3.45 - o e g z ⋅ R T ⋅ − = 1 Evaluating R...

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Problem 3.45 [Difficulty: 3] Given: Martian atmosphere behaves as an idel gas, constant temperature M m 32.0 = T 200 K = g 3.92 m s 2 = ρ o 0.015 kg m 3 = Find: Density at z = 20 km Plot the ratio of density to sea level density versus altitude, compare to that of earth. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dz ρ g = (Hydrostatic Pressure - z is positive upwards) p ρ R T = (Ideal Gas Equation of State) R R u M m = (Definition of Gas Constant) Assumptions: (1) Static fluid (2) Constant gravitational acceleration (3) Ideal gas behavior Taking the differential of the equation of state (constant temperature): dp R T d ρ = Substituting into the hydrostatic pressure equation: R T d ρ dz ρ g = Therefore: d ρ ρ g R T dz = Integrating this expression: ρ o ρ ρ 1 ρ d 0 z z g R T d = ln ρ ρ o g z R T = or ρ ρ o e g z

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Unformatted text preview: o e g z ⋅ R T ⋅ − = 1 ( ) Evaluating: R 8314.3 N m ⋅ kg mol ⋅ K ⋅ ⋅ 1 32.0 × kg mol ⋅ kg ⋅ = R 259.822 N m ⋅ kg K ⋅ ⋅ = ρ 0.015 kg m 3 ⋅ e 3.92 m s 2 ⋅ 20 × 10 3 × m ⋅ 1 259.822 × kg K ⋅ N m ⋅ ⋅ 1 200 × 1 K ⋅ N s 2 ⋅ kg m ⋅ × ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − × = ρ 3.32 10 3 − × kg m 3 = For the Martian atmosphere, let x g R T ⋅ = x 3.92 m s 2 ⋅ 1 259.822 × kg K ⋅ N m ⋅ ⋅ 1 200 × 1 K ⋅ N s 2 ⋅ kg m ⋅ × = x 0.07544 1 km ⋅ = Therefore: ρ ρ o e x − z ⋅ = These data are plotted along with the data for Earth's atmosphere from Table A.3. 0.2 0.4 0.6 0.8 1 5 10 15 20 Earth Mars Density Ratios of Earth and Mars versus Elevation Density Ratio (-) Elevation z (km)...
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Problem 3.45 - o e g z ⋅ R T ⋅ − = 1 Evaluating R...

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