29901988 sample problem 5

2
5/26/15 1 Using molecular sieves, water vapor was removed from nitrogen gas in a packed bed at 28.3°C. The column height was 0.268 m, with the bulk density of the solid being equal to 712.8 kg/m 3 . The initial water concentration in the solid was 0.01 kg water/kg solid and the mass velocity of the nitrogen gas was 4052 kg/m 2 h. The initial water concentration in the gas was c 0 = 926 kg water/kg nitrogen. Sample Problem #5: t (h) 0 9 9.2 9.6 10 10.4 c (kgH 2 O/kgN 2 ) <0.6 0.6 2.6 21 91 235 t (h) 10.8 11.25 11.5 12 12.5 12.8 c (kgH 2 O/kgN 2 ) 418 630 717 855 906 926 The breakthrough data are as follows: A value of c/c0 = 0.02 is desired at the break point. Do as follows: a) Determine the break point time, the fraction of total capacity used up to the break point, and length of the unused bed. b) For a proposed column length = 0.40 m, calculate the break point time and the fraction of the total capacity used. Solution: Plot c/c 0 vs t t c c/c0 0 0 0.000 9 0.6 0.001 9.2 2.6 0.003 9.6 21 0.023 10 91 0.098 10.4 235 0.254 10.8 418 0.451 11.25 630 0.680 11.5 717 0.774 12 855 0.923 12.5 906 0.978 12.8 926 1.000 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 2 4 6 8 10 12 14 t c/c0 t=9.6 c/c 0 =0.02 A 1 A 2 From data and plot, at c/c 0 = 0.02, breakpoint time (t b ) 9.6 From plot, the area (A 1 ) above the curve 9.6 × 1 = 9.6

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  • 5/26/15

    1

    Using molecular sieves, water vapor was removed from nitrogen gas in a packed bed at 28.3C. The column height was 0.268 m, with the bulk density of the solid being equal to 712.8 kg/m3. The initial water concentration in the solid was 0.01 kg water/kg solid and the mass velocity of the nitrogen gas was 4052 kg/m2h. The initial water concentration in the gas was c0 = 926 kg water/kg nitrogen.

    Sample Problem #5:

    t (h)

    0

    9

    9.2

    9.6

    10

    10.4 c (kgH2O/kgN2)

  • 5/26/15

    2

    A2 is the sum of the areas of trapezoid above data points

    A2= (12.8-9.6)(1)-0.5[(0.098+0.023)(109.6)+(0.254+0.098)

    (10.4-10)+(0.451+0.254)(10.8-10.4)+(0.68+0.451)

    (11.25-10.8)+(0.774+0.68)(11.5-11.25)+(0.923+0.774)

    (12-11.5)+(0.978+0.923)(12-11.5)+(1+0.978)(12.8-12.5)]

    Therefore, A2 = 1.33

    Fraction of bed used up to the breakpoint time =

    1

    1 2

    9.6 0.889.6 1.33

    AA A

    = =+ +

    Length of unused bed =

    2

    1 2

    1.33 0.268 0.0339.6 1.33

    A H mA A

    = =+ +

    For a proposed length of 0.4m

    By ratio and proportion:

    b b

    old new

    t tH H

    =

    9.6 0.40.268

    therefore, tb,new = =14.33h

    new A1 = 14.33, A2 is still 1.33

    Fraction of total capacity used =

    1

    1 2

    14.33 0.914.33 1.33

    newAnewA A

    = =+ +