PROBLEM 14.6 KNOWN: Pressure and temperature of hydrogen stored in a spherical steel tank of prescribed diameter and thickness. FIND: (a) Initial rate of hydrogen mass loss from the tank, (b) Initial rate of pressure drop in the tank. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional species diffusion in a stationary medium, () Uniform total molar concentration, C, () No chemical reactions. ANALYSIS: (a) From Table 14.1 or CA,o CA,L CA,o NA,r = = Rm,dif 1/4 DAB 1/r 1/r i o ( ) ( π )( ) 4π 0. 10 1 m /s1.5 kmol/m N 1 A,r = = 7.5 10 kmol/s ( 1/0.05 m1/0.05 m) 1 1 na,r ANA,r kg/kmol 7.5 10 = = kmol/s= 14.7 10 kg/s. M (b) Applying a species balance to a control volume about the hydrogen, Hence M& A,st = M& A,out =na,r ( ρ ) π ρ π π d AV D d A D dpa D A dp M& M A A,st = = = = dt 6 dt 6RAT dt 6RT dt ( )( ) dpa 6R T 60.0814 m bar/kmol K 00 K n 1 A,r 14.7 10 kg/s dt = D = π M A π 0.1m kg/kmol ( ) dp A =.50 10 7 bar/s. dt COMMENTS: If the spherical shell is appoximated as a plane wall, N a,x = D AB (C A,o ) πd /L = 7.07 10-1 kmol/s. This result is 4% lower than that associated with the spherical shell calculation.
PROBLEM 14.1 KNOWN: Oxygen pressures on opposite sides of a rubber membrane. FIND: (a) Molar diffusion flux of O, (b) Molar concentrations of O outside the rubber. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conditions, () Stationary medium of uniform total molar concentration, C = C A + C B, () Perfect gas behavior. PROPERTIES: Table A-8, Oxygen-rubber (98 K): D AB = 0.1 10-9 m /s; Table A-10, Oxygenrubber (98 K): S =.1 10 - kmol/m bar. ANALYSIS: (a) For the assumed conditions From Eq. 14., Hence ( ) ( ) dca CA 0 CA L NA,x = J A,x = DAB = D AB. dx L CA( 0) = SpA,1 = 6.4 10 kmol/m CA( L) = SpA, =.1 10 lmol/m. N 9 A,x = 0.1 10 m /s ( ) 6.4 10.1 10 kmol/m 0.0005 m 9 NA,x = 1.1 10 kmol/s m. (b) From the perfect gas law pa,1 bar C A,1 = = = 0.0807 kmol/m RT 0.0814 m bar/kmol K98 K ( ) CA, = 0.5CA,1 = 0.0404 kmol/m. COMMENTS: Recognize that the molar concentrations outside the membrane differ from those within the membrane; that is, C A,1 C A (0) and C A, C A (L).
PROBLEM 14. KNOWN: Column containing liquid phase of water (A) evaporates into the air (B) flowing over the mouth of the column. FIND: Evaporation rate of water (kg/h m ) using the known value of the binary diffusion coefficient for the water vapor - air mixture. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, one-dimensional diffusion in the column, () Constant properties, () Uniform temperature and pressure throughout the column, (4) Water vapor exhibits ideal gas behavior, and (5) Negligible water vapor in the chamber air. PROPERTIES: Table A-6, water (T = 0 K): p sat = 0.105 bar; Table A-8, water vapor-air (0.5 atm, 0 K): Since D AB ~ p -1 T / find D AB = 0 6 10 4 m /. / s 1.00 / 0.5 0 / 98 = 1157. 10 4 m 1 6 1 6 / s ANALYSIS: Equimolar counter diffusion occurs in the vertical column as water vapor, evaporating at the liquid-vapor interface (x = 0), diffuses up the column through air out into the chamber. From Eq. 14.7, the molar flow rate per unit area is C D N = AB ln 1 x A,L A,x L 1 xa,0 where C is the mixture concentration determined from the ideal gas law as p 0.5 atm C = = = 0.00997 kmol/m RuT 8.05 10 m atm/kmol K 0 K where R u = 8.05 10 m atm/kmol K. The mole fractions at x = 0 and x = L are x A,L = 0 (no water vapor in air above column) xa,0 = pa / p= 0105. / 0. 5 = 0. 41 where p A is the saturation pressure for water at T = 0 K. Substituting numerical values 0. 00997 kmol / m 1157. 10 4 m / s 11 06 N A,x = ln 0.150 m 11 0.416 N = A,x. 964 10 6 kmol / m s or, on a mass basis, m A,x = N A,x MA m = A,x 964. 10 6 kmol / m s 600 s / h 18 kg / kmol m A,x = 0. 57 kg / m h
PROBLEM 14. KNOWN: Radius of coal particles burning in oxygen atmosphere of prescribed pressure and temperature. FIND: (a) Radial distributions of O and CO, (b) O molar consumption rate. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conditions, () Uniform total molar concentration, () No homogeneous chemical reactions, (4) Coal is pure carbon, (5) Surface reaction rate is infinite (hence concentration of O at surface, C A, is zero), (6) Constant D AB, (7) Perfect gas behavior. PROPERTIES: Table A-8, CO O ; D AB (7 K) = 0.14 10-4 m /s; D AB (1450 K) = D AB (7 K) (1450/7) / = 1.71 10-4 m /s. ANALYSIS: (a) For the assumed conditions, Eq. 14.5 reduces to d dc r A 0 dr = dr r ( dc A /dr) = C1 or CA = ( C 1/r) + C. From the boundary conditions: CA( ) = C C = C ( ) CA ro = 0 0 = C 1/ro + C C1 = Cr o. Hence, recognizing that C = C A + C B, CA = C C( r o/r) = C1 ( r o/r) CB = C CA = C( r o/r. ) (b) The conditions correspond to equimolar, counter diffusion ( ) NA = N B, with dxa dca Cr N o A,r = N A,r4πr = CDAB4πr = DAB4πr = 4πDABr + 4πD = ABCr o. dr dr r With find p 1atm C= = = 8.405 10 kmol/m R T 8.05 10 m atm/kmol K 1450 K ( ) N 4 A,r = 1.71 10 m /s 4π 8.405 10 kmol/m 10 m 8 NA,r = 1.81 10 kmol/s.
PROBLEM 14.4 KNOWN: Radius of a spherical organism and molar concentration of oxygen at surface. Diffusion and reaction rate coefficients. FIND: (a) Radial distribution of O concentration, (b) Rate of O consumption, (c) Molar concentration at r = 0. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, one-dimensional diffusion, () Stationary medium, () Uniform total molar concentration, (4) Constant properties (k 0, D AB ). ANALYSIS: (a) For the prescribed conditions and assumptions, Eq. (14.40) reduces to DAB d dca r k 0 = 0 r dr dr dca k0 r r = + C1 dr DAB k 0 r C C 1 A = C 6D AB r + With the requirement that C A (r) remain finite at r = 0, C 1 = 0. With C A (r o ) = C A,o k 0r C o = CA,o 6DAB ( )( ) CA = CA,o k 0/6DAB ro r Because C A cannot be less than zero at any location within the organism, the right-hand side of the foregoing equation must always exceed zero, thereby placing limits on the value of C A,o. The smallest possible value of C A,o is determined from the requirement that C A (0) 0, in which case C A,o k0r o /6DAB ( ) (b) Since oxygen consumption occurs at a uniform volumetric rate of k 0, the total respiration rate is R = k 0, or ( ) π o 0 R = 4/ r k Continued..
PROBLEM 14.4 (Cont.) (c) With r = 0, ( ) = CA 0 CA,o k0r o /6DAB C 5 4 4 8 A 0 = 5 10 kmol / m 1. 10 kmol / s m 10 m / 6 10 m / s ( ) ( ) 5 A ( ) C 0 = 10 kmol/m COMMENTS: (1) The minimum value of C A,o for which a physically realistic solution is possible is C 5 A,o = k0 r o /6DAB = 10 kmol/m. () The total respiration rate may also be obtained by applying Fick s law at r = r o, in which case A ( o ) AB ( π o ) A r= r ( )( ) ( ) o AB π o o AB o π o 0 R = N r =+ D 4 r dc /dr = D 4 R k /6D r = 4/ r k. The result agrees with that of part (b).