Problem 3.45 - o e g z R T = 1 ( ) Evaluating: R 8314.3 N m...

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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: RT d ρ dz ρ g = Therefore: d ρ ρ g RT dz = Integrating this expression: ρ o ρ ρ 1 ρ d 0 z z g RT d = ln ρ ρ o gz RT = or ρ ρ
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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 = 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 = 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: 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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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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Problem 3.45 - o e g z R T = 1 ( ) Evaluating: R 8314.3 N m...

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