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vodka [1.7K]
4 years ago
12

A liquid stream containing solute A and carrier liquid C enters a packed column. The solute is to be stripped using pure gas B.

The liquid stream enters the column at a rate of 150 kmole/hr and this stream contains 7 mole % A. The stripping gas B enters at a rate of 500 kmole/hr. The design calls for 1 mole % A in the exiting liquid. For the phase equilibrium relationship use yA= 0.4XA. At the conditions of interest we have Kza = 75kmole/(hr *m3)
Determine the height of packing required for this separation if the column cross sectional area is 1m².
Engineering
1 answer:
Gennadij [26K]4 years ago
7 0

Answer:

the height of packing required for this separation is 15.85 m

Explanation:

Given that :

liquid stream (A+C) enters a packed column which the solute is stripped by using a pure gas B

The solute flowrate (L_s) = 150 kmole/hr

Mole fraction of the solute (x_2) = 7% = 0.07

(A+C)liquid = x_1 = 0.01

Pure gas (G_s) = 500 kmole/hr

y_1 = 0

The delivery force in the gas phase & liquid phase is expressed as:

(K_x)(x_2-x_1) = k_y(y_2-y_1)

Given that (K_x)_a = 75 kmole/hr m³ and equilibrium relationship y_A = 0.4x_A

Then :

K_{xa}(x_2-x_1) = (k_y)_a(y_2-y_1)

75(0.07-0.01)=(k_y)_a(y_2-0)

where ;

y_2 = 0.4x_2

= 0.4(0.07)

= 0.028

75(0.06)=(k_y)_a(0.028)

(k_y)_a = \frac{4.5}{0.028}

(ky)_a = 160.71 \ kmol/hr.m^3

NOW; the overall mass transfer coefficient on the liquid phase is :-

\frac{1}{(K_{ox})_a} = \frac{1}{(K_x)_a}+ \frac{1}{m(K_y)_a}

where m= slope = 0.4

\frac{1}{(K_{ox})_a} = \frac{1}{75}+ \frac{1}{0.4(160.71)}

{(K_{ox})_a} = 34.61 \  kmol/hr.m^3

Finally; the height of the tower (z) is = (HTU)_{oL}(NTU)_{oL}

(HTU)_{oL} = \frac{\frac{L_s}{s} }{(K_{ox})_a}

where :

s = 1m²

L_s = 150 kmol/hr

(HTU)_{oL} = \frac{\frac{150}{1} }{34.61} = 4.33 m

(NTU)_{oL} = e^x [\frac{(\frac{x_2-\frac{y_1}{m} }{x_1-\frac{y_1}{m} })(1-A)+A }{1-A} ]

where A  = \frac{L_s}{G_s *m} = \frac{150}{500(0.4)}

= 0.75

Then:

(NTU)_{oL} = e^x [\frac{(\frac{0.07-0 }{0.01-0 })(1-0.75)+0.75 }{1-0.75} ]

(NTU)_{oL} =3.66 \ m

the height of the tower (z) is = (HTU)_{oL}(NTU)_{oL}

= (4.33)(3.66)

=15.85 m

Thus, the height of packing required for this separation is 15.85 m

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Revenue from T2 = 40,000 x $250 = $10,000,000

Total Revenue from T1 and T2 = $36,400,000

b) Production Level (in units) for T1 and T2

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Total Units sold             160,000           40,000

Add Closing Inventory   25,000             9,000

Units Available for sale 185,000           49,000

less opening inventory  20,000             8,000

Production Level          165,000 units 41,000 units

c) Total Direct Material Purchases (in dollars):

Cost of direct materials used    T1                T2

A:       (165,000 x 4 x $12)   $7,920,000   $2,460,000 (41,000 x 5 x $12)

B:       (165,000 x 2 x $5)       1,650,000          615,000 (41,000 x 3 x $5)

C:                                                           0          123,000 (41,000 x 1 x$3)

Total cost                            $9,570,000     $3,198,000 Total = $12,768,000

Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2

Cost of direct materials used for production $12,768,000

Cost of closing direct materials:

                 A  (36,000 x $12)  $432,000

                 B (32,000 x $5)        160,000

                 C (7,000 x $3)            21,000             $613,000

Cost of direct materials available for prodn   $13,381,000

Less cost of beginning direct materials:

                 A  (32,000 x $12)        $384,000

                 B  (29,000 x $5)            145,000

                 C  (6,000 x $3)                18,000        $547,000

Cost of direct materials purchases               $12,834,000

d) The Total Direct Manufacturing Labor Cost (in dollars):

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Direct labor cost per unit   $24                          $48

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less closing inventory of materials     =      $613,000

Cost of materials used for production    12,768,000

add Labor cost                                           5,928,000

add Overhead cost                                    9,060,000

Total production cost                            $27,756,000

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Total Production cost             =    $27,756,000

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Total cost of goods sold            $26,960,000

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Gross profit                             $9,440,000

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Total Expected Operating Income = $9,040,000

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T1 = 20,000 x ($58 + 24 + 40) = $2,440,000

T2 = 8,000 ($78 + 48 + 60) = $1,488,000

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T2 = 9,000 ($78 + 48 + 60) = $1,674,000

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