# Gas Absorption.docx

Description
10.2-1 Gas Solubility in Aqueous Solution. At 303 K the concentration of CO2 in water is 0.90 x Variables that can be set 10-4 kg CO2/kg water. Using the Henry’s law constant from Appendix A.3, what partial pressure of CO2 must be kept in the ags to keep the CO2 from vaporizing from the SOLUTION: aqueous solution? F=C–P+2 F=3–2+2 F=3 Given: T = 303 K Variables that can be set: xA = 0.9 x 10-4 kg CO2 /kg H2O Required: PA of CO2 Solution: 10.3-2 1. total pressure 2. temperature 3. m
Categories
Published

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
10.2-1 Gas Solubility in Aqueous Solution. At 303 K the concentration of CO 2  in water is 0.90 x 10 -4  kg CO 2 /kg water. Using the Henry’s law constant from Appendix A.3, what partial pressure of CO 2  must be kept in the ags to keep the CO 2  from vaporizing from the aqueous solution? Given: T = 303 K x  A = 0.9 x 10 -4  kg CO 2  /kg H 2 O Required: P  A  of CO 2  Solution: From A.3- 18 for Henry’s law constant (Geankoplis p. 884)  H = 0.186 x 10 4 atm/mol frac. P  A = Hx  A 224-4 A 22 kgCO18 kgmolHOP=0.186 x 10 atm x 0.9 x 10 0.06848 atmkgHO44 kgmol CO  x        5 A 1.01325 x 10P= 0.06848 atm x = 1 atm 3 6.939 x 10 Pa  10.3-1 Phase Rule for a Gas-Liquid System. For the systen SO 2 -air-water, then total pressure is set at 1 atm abs and the partial pressure of SO 2  in the vapor is set at 0.20 atm. Calculate the number of degrees of freedom, F. What variables are unspecified that can be arbitrarily set? GIVEN: SO 2    –  air  –  H 2 O system P  AT  = 1 atm P  A  of SO 2  = 0.2 atm REQUIRED: Degrees of freedom, F Variables that can be set SOLUTION: F = C  –  P + 2 F = 3  –  2 + 2 F = 3 Variables that can be set: 1. total pressure 2. temperature 3. mole fraction composition x  A of SO 2  10.3-2 Equilibrium Swtage Contact for Gas-Liquid System. A gas mixture at 2.026 x 10 5  Pa total pressure containing air and SO 2  is contacted in a single-stage equilibrium mixer with pure water at 293 K. The partial pressure of SO 2  in the srcinal gas is 1.52 x 10 4  Pa. The inlet gas contains 5.70 total kg mol and the inlet water 2.20 total kg mol. The exit gas and liquid leaving are in equilibrium. Calculate the amounts and compositions of the outlet phases. Use equilibrium data from Fig.10.2-1. GIVEN: Use equilibrium data in Fig. 10.2-1 P T  = 2.026 x 10 5  Pa = T = 293 K P  A  of SO 2  = 1.52 x 10 4  Pa = .15 atm Inlet gas = 5.70 kg mol Inlet H 2 O = 2.20 total kgmol REQUIRED X  A1 , y  A1 , L 1  V 1 SOLUTION: x  Ao  = 0 amount of entering acetone = y  AN+1 v  AN+1  = 0.01(30) = 0.30 = 29.7 kgmol/air h acetone leaving in Vi = 0.10(0.30) = 0.30 kgmol/h acetone leaving in Ln = 0.9 (0.30) = 0.27 kgmol/h    V 1  = 29.7 + 0.03 = 29.73 kgmolH 2 O + acetone/hr  A1 0.030y= =0.0010129.73  Ln = 108 + 0.27 = 108.27 kgmol H 2 O + acetone/hr     AN 0.27X = 0.002493108.27   Using equation: 1 1 LLo108 A= = =1.4358mvmv2.53 (29.73)    NN N+1 L108.27 A= = =1.4265mv2.53 (30)   1N  A=AA1.4358(1.4265) 1.4311       N+1o 11log y - mx1N = log A  A A        0.01 - 2.53(0)11log 10.00101 - 2.53(0)1.43111.4311N = log 1.4311         3.7662 stages  10.3-3 Absorption in a Countercurrent Stage Tower. Repeat example 10.3-2 using the same conditions but with the following change. Use a pure water flow to the tower of 108 kg mol H 2 O/h, that is, 20% above the 90 used in Example 10.3-2. Determine the number of stages required graphically. Repeat using the analytical Kremser equation. GIVEN: L O  = 108 kg mol H 2 O/hr Y  AN + 1 = 0.01 V N + 1 = 30.0 kg mol/h T = 300 K P T  = 101.3 kPa y  A  = 2.53 x  A REQUIRED: Theoretical stages graphically using analytical Kremser equation. SOLUTION: x  Ao  = 0 amount of enetering acetone = y  AN + 1  V N + 1 = 0.01(30) = 0.30 entering air = (1-y  AN + 1 ) V N + 1  = (1  –  0.01)(30) = 29.7 kg mol/air h acetone leaving in V 1  = 0.10(0.30) = 0.30 kg mol/h acetone leaving in L n  = 0.9(0.30) = 0.27 kg mol/h V 1  = 29.7 + 0.03 = 29.73 kg mol air + acetone /hr L n  = 108 + 0.27 = 108.27 kg mol H 2 O + acetone/hr 10.4-1 Interface Concentrations and Overall Mass-Transfer Coefficients. Use the same equilibrium data and film coefficients k’ y   and k’ x  as in Example 10.4-1. However, use bulk concentrations of y  AG  = 0.25 and x  AL  = 0.05. Calculate the following: a. Interface concentrations y  Ai  and x  Ai and flux N  A . b. Overall mass transfer coefficients K’ y  and K y  and flux N  A . c. Overall mass transfer coefficient K’ x  and flux N  A . GIVEN: Equilibrium data y  Ao = 0.380 mol fraction x  AL  = 0.10 T = 298 K P = 1.013 x 10 5  Pa k y  = 1.465 x 10 -3  kg mol/s.m 3  mol fraction k x  = 1.967 x 10 -3 kg mol A/s.m 2  mol fraction    REQUIRED: K’ X , K X SOLUTION: Trial 1: -3 AiM-3 AiM k'x/(1-x)1.967 x 10 /1.0 - 1.342k'y/(1-y)1.465 x 10 /1.0    y = mx + b 0.38 = -0.10 (1.342) + b b = 0.5142 x  A1 = 0.247 y  Ai = 0.183       AiAG AiM  Ai AG (1-y) - (1-y)(1-0.183) - (1-0.380)(1-y)== = 0.7151-y1-0.183ln ln 1-y1-0.380                   ALAi AiM  AL Ai (1-x) - (1-x)(1-0.1) - (1-0.247)(1-x)== = 0.825 1-x1-0.1ln ln 1-x1-0.247            Trial 2: -3 AiM-3 AiM Ai  Ai k'x/(1-x)-1.967 x 10 /0.825 - 1.163k'y/(1-y)1.465 x 10 /0.715 x= 0.254 y= 0.194 0.38 = - 1.163 (0.10) + b b = 0   .4963       AiM (1-0.194) - (1-0.380)(1-y)= = 0.70891-0.194ln 1-0.380             AiM (1-0.1) - (1-0.254)(1-0.1) - (1-0.247)(1-x)== = 0.82061-0.11-0.1ln ln 1-.2541-0.247             -3 AiM-3 AiM Ai  Ai k'x/(1-x)-1.967 x 10 /0.8206 - 1.160k'y/(1-y)1.465 x 10 /0.7089 -1.163 -1.60 x= 0.254 y= 0.194      AG Ai* Ai-3-3 y- y0.38 - 0.194m'' = = 2.0217 - x0.346 - 0.254111= + K'xm''k'yk'x111 = + K'x2.0217 (1.465 x 10)1.967 x 10  A  x   2 K'x = 0.00118 kgmol/s m frac      **  A*M ALA A*M  AL A A*M K'x K'x = (1-x)(1-x)-(1-x)(1-x)= (1-x)ln(1-x)(1-0.10)-(1-0.346)(1-x)= 0.7705(1-0.10)ln(1-0.346)0.00118Kx = 0.001530.7705          *  AAAL A*M A  A K'xN (x-x)(1-x)0.00118N= (0.346-0.10)0.7705N=        2 0.000377 kgmol/sm    A*MA*MA*M AiAG AM Ai AG AM ALAi AM 111 K'x/(1-x)m''/(1-y)K'x/(1-x)(1-y) - (1-y) (1-y)i = (1-y)ln(1-y)(1-0.194) - (1-0.38) (1-y)i = (1-0.194)ln(1-038)(1-x) - (1-x) (1-x)i (ln   0.7089  AL Ai (1-0.1) - (1-0.254)1-x)(1-0.1)ln(1-x)(1-0.254)   0.8206   -3-3 A*M 111 K'x/(1-x)2.0217(1.465 x 10/0.70891.967 x 10/0.8206 = 656.5320% resistance in the gas film239.3425 = (100)656.5392    36.46%% resistance in the liquid film 417.1835 = (100)656.54  63.54%  10.4- 2 Use the same equilibrium data and film coefficients k’y and k’x as in Example 10.4 -1. However, use bulk concentrations of y  AG = 0.25 and x  AL  = 0.05. Calculate the following. (a) Interface concentrations y  Ai and x  Ai and flux N  A . (b) Overall mass- transfer coefficients K’y and Ky and flux N  A . (c) Overall mass- transfer coefficient K’x and flux N  A . GIVEN: y  AG = 0.25 x  AL   = 0.05 ky = 1.465 x10  – 3  kgmol A / sm 3  mol fraction kx = 1.967 x10  – 3 kgmol A / sm 2 mol fraction REQUIRED: a.) Interface concentrations y  Ai & x  Ai & N  A SOLUTION: Trial 1: -3 AiM-3 AiM AiAi k'x/(1-x)1.967 x 10 /1.0 - 1.342k'y/(1-y)1.465 x 10 /1.0 y = mx + b 0.25 = -1.342 (0.05) + b b = 0.3171 x = 0.1634 y = 0.1010  

Jul 23, 2017

#### Clean Steam in Pharma

Jul 23, 2017
Search
Similar documents

View more...
Tags

Related Search