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Unformatted text preview: PROBLEM 14.1 KNOWN: Mixture of O 2 and N 2 with partial pressures in the ratio 0.21 to 0.79. FIND: Mass fraction of each species in the mixture. SCHEMATIC: 2 O N2 p 0.21 p 0.79 = 2 O 32 kg/kmol = M 2 N 28 kg/kmol = M ASSUMPTIONS: (1) Perfect gas behavior. ANALYSIS: From the definition of the mass fraction, i i i i m r r r r = = Hence, with ( 29 i i i i i i i p p p . R T / T T r = = = M M Hence i i i i i p / T m p / T = M M or, cancelling terms and dividing numerator and denominator by the total pressure p, i i i i i x m . x = M M With the mole fractions as 2 2 O O 0.21 x p / p 0.21 0.21 0.79 = = = + 2 2 N N x p / p 0.79, = = find the mass fractions as 2 O 32 0.21 m 0.233 32 0.21 28 0.79 = = + < 2 2 N O m 1 m 0.767. = = < PROBLEM 14.2 KNOWN: Partial pressures and temperature for a mixture of CO 2 and N 2 . FIND: Molar concentration, mass density, mole fraction and mass fraction of each species. SCHEMATIC: 2 A A CO , 44 kg/kmol = M 2 B B N , 28 kg/kmol = M ASSUMPTIONS: (1) Perfect gas behavior. ANALYSIS: From the equation of state for an ideal gas, i i p C . T = Hence, with p A = p B , A B 2 3 1bar C C 8.314 1 m bar/kmol K 298K = = 3 A B C C 0.040 kmol/m . = = < With i i i C , r = M it follows that 3 3 A 44 kg/kmol 0.04 kmol/ m 1.78kg/m r = = < 3 3 B 28 kg/kmol 0.04 kmol/ m 1.13 kg/m . r = = < Also, with i i i i x C / C = find A B x x 0.04/0.0 8 0.5 = = = < and with i i i m / r r = find ( 29 A m 1.78/ 1.78 1.1 3 0.61 = + = < ( 29 B m 1.13/ 1.78 1.1 3 0.39. = + = < PROBLEM 14.3 KNOWN: Mole fraction (or mass fraction) and molecular weight of each species in a mixture of n species. Equal mole fractions (or mass fractions) of O 2 , N 2 and CO 2 in a mixture. FIND: SCHEMATIC: 2 2 2 O N CO x x x 0.333 = = = or 2 2 2 O N CO m m m 0.333 = = = 2 CO 44 = M 2 2 O N 32 , 28 = = M M ASSUMPTIONS: (1) Perfect gas behavior. ANALYSIS: (a) With i i i i i i i i i i i i i i i p / R T p / T m p / R T p / T r r r r = = = = M M and dividing numerator and denominator by the total pressure p, i i i i i i x m . x = M M < Similarly, ( 29 ( 29 i i i i i i i i i i i i i i / T p R T x p R T / T r r r r = = = M M or, dividing numerator and denominator by the total density r i i i i i i m / x . m / = M M < (b) With 2 2 2 2 2 2 O O N N C O CO x x x 32 0.33 3 28 0.3 3 44 0.33 3 34.6 + + = + + = M M M 2 2 2 O N CO m 0.31 , m 0.27 , m 0.42. = = = < With 2 2 2 2 2 CO 2 O O N N CO / m / m / m 0.333/3 2 0.333/2 8 0.333/44 + + = + + M M M 2 2 O m 2.987 10 . = find 2 2 2 O N CO x 0.35 , x 0.40 , x 0.25. = = = < PROBLEM 14.4 KNOWN: Temperature of atmospheric air and water. Percentage by volume of oxygen in the air....
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This note was uploaded on 03/06/2008 for the course ME 343 taught by Professor Shollenberger during the Winter '08 term at Cal Poly.
 Winter '08
 Shollenberger
 Heat Transfer

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