Problem de Transfer.masa II

8/12/2019 Problem de Transfer.masa II

http://slidepdf.com/reader/full/problem-de-transfermasa-ii 1/105

EXTRA CCION LIQUIDO LIQUIDO

EXTRA CCION SOLIDO LIQUIDO

SECADO DE SOLIDOS

TR NSFERENCI DE M S II

.



( − )



solución

1

E1

Y1

R1

X1

S

ALIMENTACION :

= + = 40

6 0 + 4 0 =0.4

SOLVENTE PURO

= =

=indeterm

= + = 06 0 + 4 0 = 0

F=100 Kg

=0.40

= + =∝



= ′ ( 1 + )

= ′ ( 1 + 0 )

= = 100

1 = 100

=0.4

=0

=indeterm.

= 0

′

. = 0

. =



′

′

=0.4

=32.5

ver la ecu



y = -8.1683x6 + 16.851x5 - 10.596x4 + 2.0989x3 + 0.1654x2 + 0.3249x + 0.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

N x

X

D’

ND

=0.573134+0.324875+0.16533 +2.099318 −10.59696 +16.85184 −8.168

=0.4

=0.731716



= ( − )

= 100 (0.731716 − 0)

= 73.17157

= ( − )

=100 (32.5−0)

= 3250

=32.5

=0.731716

)

= 0

:



) =0.5

= 0.5 3250 = 1625

= +′ + ′ = + 0

+ 0

= =0.4

= +′ + ′ = +

+ ′

= ′ = 1625

100 =16.25

Calculo del



=0.223

′

′

′

=

=0.545

= + ′ = 0

Entonces M’=F’=100 Kg

=28.7

=0.659

C

=16.25

=0.4



′ = ( − −

)

′ =100 0.4−0.2230.545−0.223

′ = 54.96894

= ( 1 + )

=54.96894(1+28.7)

= 1632.5775

′ = ( − −

)

′ =100 0.545−0.40.545−0.223

′ = 45.031056

= ( 1 + )

=45.031056(1+0.65

= 74.714

= 1 +

= 1 +

C

E

=

0.223

1+0.659 =0.1344 = 0.545

1+28.7=0.01835



- EXTRACCION EN CONTRACORRIENTE)

100 / 40% (98% , 1.8%

0.2% . ) 98.5° a 70% 4% . ) . )

) )



= + = 40

6 0 + 4 0 =0.4

= + = 0

60+40 = 0

= ′ ( 1 + )

= = 100 /

E (

= + = 0.2

1.8+0.2 =0.1

=

=

.. =49

= . . = 1.8%

==98%

= cido oleico = 0.2% = 60% …

C= 40%

= 0.70 (70%)

=0.04(4%)

,

,



∎ ′(.,)

∎′(0.4,0) =0.04 =0.70

′

′

=0.3355 =10.1667



) ′ = . −

−

=100. 0.40−0.3355

0.3355−0.1

=27.3885

= ( 1 + )

=27.3885(1+49)

=1369.4267 /

E

) (GRAFICAMENTE)

1 2 3 =1 0 0

=0.4

=27.38

=0.1

= + ′ =100+27.38=1

′ = 100 (..

..)

′ = ( − −

)

′ =57.035

=0.7

)

′ = (

′ = 127.38 (′ =70.35 = ( 1 + )

=57.035(1+21.833)

= 1302.304 /

= ( 1 + =70.353(1+

=111.39

′



′

′

∆

=.

′

′

′

′

′

′

′

′

′

′

′

′

′

′ ′ = (8 )

′

= .

) .

%= ′ − ′ ′ 100

%= 100(0.4)−70.353(0.04)100(0.4) 100

%=92.965%



( )

25° . 200 /

40% .

.

∶ a)

) 200 /

, . )

, 10% .

d) ,

30% .



∗ 25° . 98%

2% .

∶

) f) 200 /

.

) % 20% .



)

1

F=200 lb/h

=0.40

= 0

=0.366

=0.014



= − −

=200

0.4−0.366

0.366−0

=18.579 /

= − −

=200 0.4−0.014

0.014−0

=5514.2857 /

)

1

= 200

F = 200 /

=0.40

= 0

= . +S.

+

= 200 0.4 + 200(0)200+200 =0.2



=0.2

=0.4

=0.23

= . ( − −

=400. (0.0.1 = 133.33

= . (

− =400. (00.

=266.6



%= . −

.

100

%= 200(0.4)−133.33(0.14)200(0.4) 100

%=76.667%

)

1

=0.40

F = 200 /

= 0

=

= . ( −

− )

=0.10



= + = 2 0 0 + 3 1 9 . 4 8

= . ( −

− )

= 2 0 0 (0.154−0.4

0−0.154 )

= 319.48 /

= . ( − )

−

=519.48 (0.154−0.17)

0.1−0.17

= 519.48 /

= 118.738 /

=0.154



= . ( − )

−

= 400.74 /

=519.48 (0.154−0.1)

0.17−0.1

%= . − .

100

%=200(0.4)−118.739(0.1)

200(0.4) 100

% = 85.16 %

1F = 200 /

=0.40

=0.30

=0.02

)

=0.30

= . ( −

− )



=0.3

=0.25

= . (

− = 2 0 0 (0.20

=116.205 = + = 2 0 0 +

= 316.205

= . ( =316.2055. (0

0 = 142.90



= . ( − )

−

=316.2055. (0.253−0.196)

0.3−0.196

= 173.3049 /

%= . − .

100

%=

200(0.4)−142.90(0.196)

200(0.4) 100

% = 64.985 %

1

)

F = 200 /

=0.40

=0.02



=

=200 0

0. =19.0

=

=200

= 4



f )

1

=0.02

F = 200 /

=0.40

= 200 /

= 200 0.4 + 200(0.02)200+200 =0.21

= . +S. +

= 200 0.4 + 200(0.02)200+200 =0.21

0 21



=0.21

=0.15

=0.241

= . ( − )

−

=400 (0.21−0.241)

0.15−0.241

= 136.2637 /

)



)

1F = 200 /

=0.40

=0.20

=0.02

=0.20

%= . − .

100

%=74.45 %

= . ( − )

−

=400 (0.21−0.15)

0.241−0.15

= 263.7361 /

%= 200(0.4)−136.2637(0.15)200(0.4) 100

D t 0



Dato =0

=0.30

=0.25

= . (

= 2 0 0 (0.0

=119.32 = + = 2 0 0 +

=319.3277



%= . − .

100

%=63.85 %

= . ( − )

−

=319.3277 (0.258−0.306)

0.20−0.306

=144.6 /

= . ( − )

−

=319.3277(0.258−0.20)

0.306−0.20

= 174.7264 /

%= 200(0.4)−144.6(0.20)200(0.4) 10

PROBLEMA



PROBLEMA 4

Se dispone de 100 lb/h de una solución acuosa de acetona , con un contenido de 30% peso de acetona ,solución q se someterá a una extracción en etapas múltiples y en contracorriente , con la finalidad dereducir la concentración de acetona a un 5% peso en el refinado final.

La extracción se llevara a cabo ala temperaturas de 25°C y para tal efecto se empleara metilisobutilcetona com

a) si se utiliza el solvente a razón de 40 lb/h, Determinar el numero de etapas requeridas , el flujo de los

productos y el porcentaje de acetona recuperada.

b) si se debe obtener un extracto con 35% peso de acetona ,determinar el numero de etapas y flujo desolvente necesario , así como flujo de los productos y el porcentaje de acetona recuperada

1F = 100 /

=0.30

= 40

= 0

2 3

= . +S. = 100 0.3 + 40(0) =0 2143



= + = 100+40 =0.2143

= + = 140 /

∆

=0.378

5

( − )



= . ( )

−

= . ( − )

−

%= . − .

100

=140. (0.2143−0.378)

0.05−0.378

=140. (0.2143−0.05)

0.378−0.05

= 70.128 /

= 69.872 /

%=100(0.3)−69.872(0.05)

100(0.3) 1% = 88.355 %

) =0.35 =0.05 = ( − )



) =0.3 = 0

3 3.98 ≈ 4

= . ( − ) −

=75.7276 /

= . ( − ) −

=145.63 (0.206−0.35)

0.05−0.35 =145.63 (0.206−0.05)

0.35−0.05

=69.9024 /

=0.206

%= . − .

100

%= 100(0.3)−69.9(0.05)100(0.3) 100=

= . ( − )

= 1 0 0 . 0.206−0.3

0−0.206 = 45.63 /

= + = 145.63 /



( )150 5% , 4 ,

40 .

:

= 2.20

, ,

, ,



1 2 3 4

′

′

′

′

′ ′ ′

′ ′ ′

*

∗

∶ E



∶ = (1 + ′)

= (1 + ′)

= (1 + ′)

= (1 + ′)

′ =

1 −

′ = 0.051−0.05 =0.052632

= (1 + ′)

=

1 + ′ =

150

1+0.052632 = 142.5

= 40

′ = 0

= (1 + ′)

=

= 2.20

′1 + ′ =2.20 ′

1 + ′

= 2.20′1−1.20′

= 40

′



X’ Y’

0 0

0.01 0.02226

0.02 0.045082

0.03 0.06846

0.04 0.092437

0.05 0.117

0.06 0.142

0.07 0.168

0.08 0.1947

= 2.20′1−1.20′



1

. ′ + . ′ = . ′ +.′

− = (−)

− − = (−)

−

= ′ −

′ −



= −

;=1,2,3,4

= −

= 142.540 =−3.5625

−

= ′ −

′ −

−3.5625=

′−0

′ −0.052632

′ =−3.5625′ +0.1875015

= 2.20′1−1.20′

′ =−3.5

2.20′′ =−3.5625′ +0.1875015



1−1.20′ 3.5625 +0.1875015

′ =0.03205

.

′ = 2.20′1−1.20′ = 2.20(0.03205)

1−1.20(0.03205) =0.07332

1

′

= 40

′ =0.07332

′ =0.03205

= 142.5 kg

′ = 0

′, ′ =0.03205;0.07332

0=−3.5625′ +0.1875015

′ =0.052632

′ =−3.5625′ +0.1



′,

2



2 = 142.5 kg

′ =0.03205

= 40

= 40

′ = 0

′

′

. ′ + . ′ = . ′ +.′

− = (−)

− − = (−)

−

= ′ −

′ −

= − = 142.540 =−3.56−3.5625= ′−0

′ −0.03205

′ =−3.5625′ +0.114178



2.20′1−1.20′ =−3.5625′ +0.114178

′ =0.019633

′ =−3.5625′ +0.114178

′ =−3.5625(0.019633)+0.114178

′ =0.044235

.

′ =−3.5625′ +0.114178



X' Y'

0.03205 0

0.03 0.007303

0.025 0.0251155

0.019633 0.044235

0.015 0.0607405

′; ′ =0.019633;0.044235

3



3 = 142.5 kg

′ =0.019633

= 40

′ = 0

′

= 40

′

. ′ + . ′ = . ′ +.′

−

= ′ −

′ −

= −

= 142.540 =−3.5625

−3.5625= ′−0

′ −0.019633′ =−3.5625′ +0.06992.20′

1−1.20′ =−3.5625′ +0.069

′ =0.012069

.

′ 3 5625′ 0 06994



′ =−3.5625′ +0.06994

′ =−3.5625(0.012069)+0.06994

′ =0.026944

X' Y'

0.019633 0

0.017 0.0093775

0.015 0.0165025

0.012069 0.026944

0.011 0.0307525

′; ′ =0.012069;0.026944

40 4



4 = 142.5 kg

′ =0.012069

= 40

′ = 0

= 40

= 142.5 kg

′

′

. ′ + . ′ = . ′ +.′

−

= ′ −

′ −

−3.5625= ′−0

′ −0.012069

′ =−3.5625′ +0.04292.20′

1−1.20′=−3.5625′ +0.04

′ =0.007436

.

′ =−3.5625′ +0.042996



′ =−3.5625(0.007436)+0.042996

′ =0.0165053

X'

Y'

0.012069 0

0.01 0.007371

0.009 0.01093355

0.007436 0.0165053

0.006 0.021621

′; ′ =0.007436;0.016505



= (1 + ′)

= 1 + = 40 1 + 0.07332 = 42.9328

= 1 + = 40 1 + 0.044235 = 41.7694

= 1 + = 40 1 + 0.026944 = 41.0778

= 1 + = 40 1 + 0.0165053 = 40.6602 kg

=

<= + + +

= + + + = 166.4402



= .

<= .

+ .

+ .

+ .

=′

1 + ′

= ′1 + ′ = 0.07332

1+0.07332 =0.06831

= ′1 + ′

= 0.0442351+0.044235 =0.04236

= ′1 + ′= 0.0269441+0.026944 =0.026237

= ′1 + ′

= 0.01650531+0.0165053 =0.0162373

= .

+ .

+ .

+ .

=0.038457



= (1 + ′)

= 142.5 1 + =142.5(1+0.007436)

= 143.56875 )

%.= . − . .

100

%.= 150(0.05) −143.56875(0.007436)150(0.05) 100

%.=85.75 %

. − . =6.4312

( −)



, 100 20% , 100

/ 2/3 . Calcular la cantidad y composición del extracto y del refinado , así como el

porcentaje de aceite extraído



+ 2



= + = 2

3

= . + 1

= 1 − . + 1

1

, = . +

E

0 2/3

0.5 2/3

1 2/3

0 0.4

0.2 0.2

0.4 0

A

= 100



=0.20

= 0

=0.80

= 0

= 1

= .

+ .

+

= 100.(0.20)+100.(0)200 =0.1

= . + . +

= 100.(0)+100.(1)200 =0.5

; =(0 1;0 5)



⨀ M(0.1; 0.5)

:=0.4−

: = 1 −

: = 5

=0.06666 =0.33333

=0.16666 =0.833333

= ( − ) = ( − )



= ( − ) = ( −

)

=200( 0.16666−0.10.16666−0.06666)

= 133.3332

=200( 0.1−0.06660.16666−0.06666)

= 66.6668

% . = . − .

100

% . = 100(0.20)−133.3332(0.06666)1000(0.20) 100

% . = 144.44355%

7. −



, , 100 40% , 100

97% 3%

.Calcular la cantidad y composición del extracto como del refinado , así c

porcentaje de aceite extraído

: =2/3 ()

= 100 =0.40

= 0

=0.60

=0.03

=0.97

= 100

= . + . = = 1 −



+

= 100(0.4)+100(0.03)100+100

=0.35

= . + . +

= 100(0) +100(0.97)

100+100

=0.485

E

0 3/2

0.5 3/2

1 3/2

0

0.3

0.6

= . + 1 1

= + = 3

2



=0.252 =0.348

=0.42 =0.58

= ( − ) = ( −

)



( − ) −

% . = . − .

100

=200( 0.42−0.350.42−0.252) =200(0.35−0.252

0.42−0.252)

% . =100(0.4) − 83.333 (0.252)

100(0.4) 100

= 83.333 = 116.667

% . = 47.5 %

8. (. − . − )

U



U ñ 20% −

20% , , , , , , .

,



, 30% .

, 200 , , 50% .

/ 1 0.30.

) )

−



−

, 25%

l −

98.5°)

) 50%

¿ ?



U (; )



= . + . +

= 200(0.30)+(0)200+

= . + . +

= 200(0)+(1)2 0 0 +

= 200(0.30)200+ = 60

200+ = 2 0 0 +

= . −

−

= . −

−

0 5 =0.



=0.

=0.

=0.5 =0.5

=

= 60.01

= . − −

= 2 0 0 . 0.

0

= . − −

= ( −

− )



=0.30

= 0

= ( − − )

M = + = 2 0 0 + 6 0 = 2 6 0

=260(0.23077−0.19

0.5−0.19)

= 34.219

=260(0.5−0.230770.5−0.19 )

=225.8058

% . = . − .

100

% . = 200(0.3)−225.8058(0.19)200(0.3) 100

% .=28.49 %

−

′ 34 219



1

= ′ = 34.219

=0.25

= 0

=0.25

= =34.219

=0=indeterm.

= 0

. = 0

. =

′

′

=0.25



′

=36.5

ver la ecu



d) = 0.5



= 0.5 1249 = 624.5

Calculo del

= +′ + ′ = + 0 + 0

= =0.25

= +′

+ ′ = +

+ ′

= ′ = 624.5

34.219 =18.25

=18.25

=0 25′



=0.25

′ =0.115

=0.358

′ = 34

C

=0.614

C

=0.115

=0.25 ′ = ( − −

)



′ = ( − − )

′ =34.219 0.25−0.1150.358−0.115

0.25

=0.358

’=’+’=34.219

′ = 19.01

E

= ( 1 + )

=19.01(1+34)

= 665.37

′ =34.219 0.358−00.358−0.1

= ( 1 + ) E

′ = 15.209

=15.029(1+0.614) = 859. 125

%= ′ − ′ ′ 100

%= 34.219(0.25)−15.029(0.1134.219(0.25)

%=19.95 %

P

( )



.

0.28 / , 6 0.08 /

,

0.14 / ,

.

0.33 0.04 / .

0 28

:



=0.28

=0.08

= 6

= +

= . 1

( − )

=

. − ∗

.(

∗ = 0

= +

= . 1

− +

. − ∗

.( − ∗ − ∗ )

=0.14

= . 1

− +

. − ∗

.( − ∗ − ∗ )

:



6 = . 1

0.28 − 0.14 +

. 0.14−0

.(0.14−00.08−0)

. =27.479285

=0.33

=0.04

∗ = 0

= ?

= . 1

− +

. − ∗

.( − ∗ − ∗ )

= 27.479285 0.33 − 0.14 + 27.479285. 0.14 − 0 . (0.14−00.04−0)

=10.0406

=0.14

( ) 1.5 31.25 , 75% 20% ,

.



1200 /

25 501.25 , ,

186° 20% , é

W SH ,Kg N ,Kg/m2*h

7.500 5.00

5.625 5.00

3.750 5.00

3.500 4.50

3.375 4.00

3.200 3.70

3.000 3.00

2.900 2.78

2.750 2.35

2.625 2.00

2.250 1.00

1.8 / , á .

, ú

: ()=75%

100 75 CONVIERTIEND



()=20%

100 25

= 75 25

100 20 80

= 20 80

()

()

= 3

=0.25

=1.8

∗ = 0 (humedad en equilibrio despreciable)



= . 1

− +

. − ∗

.( − ∗ − ∗ )

= +

= .

P 1.5 3 0.0125

= 1.5 3 0.0125 = 0.05625 =1200 /

=67.5

= 2 1.5 3 = 9 = 67.5

9 = 7.5 /

0.25 0.5 0.0125



= 0.25 0.5 0.0125 = 1.5625 10

= .

= 1200 / 1.562510

= 1.875 ss

= −

W SH ,Kg N ,Kg/m2*h

7.500 5.00

5.625 5.00

3.750 5.00

3.500 4.50

3.375 4.00

3.200 3.70

3.000 3.00

2.900 2.78

2.750 2.35

2.625 2.00

2.250 1.00

X=(W SH -W SS

3

2

1

0.8667

0.8

0.7067

0.6

0.5467

0.4667

0.4

0.2



= .

= = 3

= =0.25

= 5



∗ =0.2 =1.8 =0.25

= . 1

−



= 7.55 3 − 1 . 8

=1.8

=

. − ∗

.( − ∗

− ∗)

=7.5 1 . 8 − 05 .( 1 . 8 − 0

0.25−0)

= 5.33

= +

= 7.13

=

<

<



( )

, 4 4 ,



, 4 4 , 2 ,

.

7 / 140° 85° . 50 , 2.5 / , 20 % .

=50

= 2080 =0.25 85° = 1045.8 /

= 2.5 /

=85° =140°



=85°

=140°

=0.0135

= 0.0135 .

= ( − )

85°

=

= + +



=140°

=85°

=0.0128().

= ( )

=(0.0252+0.0405(0.0135))(140+459.7)

=15.44

.

= (0.0252 + 0.0405 )( +459.7)

= . = 0.06564(

)7

= .. (

)

=0.06564 (

=1654.158 −

= . (/2. )

=0.0128(1654.158).

= 4.809 / − − °



=

=264.505

− 1045.8

=0.2529

=

. 1

( − )

= 5016 . 1

0.2529 (2.5−0.25)

= 27.8

=4.809(140−85)

= 264.505 / −



∶ =0.005 /

= 6 0 % = 6 0 % = 6 0 %



calentador I II III77°

= =

83°

= =

95.2° 102.7°

107°

77° 125°

= =0.005 1 2

3 = =0.0145 4

5

7

6 = =0 . .0 2 4

= =0 . .0 2 8 8

9 =0..0314

= 6 0 %

= = 0.005 /

= =0.024 /

= =0.0145 /

=77° =83° =95.2°



= =0.028 /

= 0. . 0314 /

=

=15

= (0.0252 + 0.0405 )( +459.7)

=102.7 =107°

=(0.0252+0.0405 (0.0134))(107+459.7)

∶=160

= = 10.6655 /

= ( − )

= 10.6655 (0.0314 − 0.05)

= 0.2815 /



+ = + =

. + . = . + . = .

. + . = . . + ( − ).



. + . = .

. + . = ( +. + . = . .(−) = . (

= . ( − −

. + ( ). . + . − .

( − ) = (

= . ( − −

. + . = .

( M − ). + . = .

M. − . + . = .

. ( − ) = ( − )

= . ( − )

−

. + . = .

= . +S.

= . +S. +

′ + ′ + ′

= 0

= 0



′ + ′ = + = ′

′. +′ . = . +

. = ′ .

′. +′ . = . + . = ′ .

=

′. +′ .

′ + ′

= ′. + ′

′

′= −

−

′′ = −

−

. = 0

. =

′ = + = ′

′. = . + . = ′ .

′. + = . +

=

= ′. +′ . ′ + ′

= ′ ( − )

. + = ′ .

′ = − = −

=′( − )



− −

′ = −

− = −

−

=′( − )

= ′ ( 1 + )

S = ′ ( 1 + )

= ′( 1 + )

= ′ (1 + )

′ = − −

′ = −

−

P :

:

= 1 +

= 1 +

= (1 + ′)

= (1 + ′)

′1 + ′ =2.20 ′

1 + ′



= (1 + ′)

= (1 + ′)

′ = 1 −

= 1 + ′

′ = 0

=

= 2.20′1−1.20′

. ′ + . ′ = . ′ +.′

−

= ′ −

′ −

= − ;=1,2,3,4

′ =−3.5625′ +0.1875015

2.20′1−1.20′ =−3.5625′ +0.1875015

= .

= .

+ . + .

+ . = ′

1 +



<

=

< = + + +

1 +

−

+ = + =

+ +

= =



. + . = . + . = .

. + . = . + . = .

= . + . +

= . + . +

= . − − = . − −

= . − −

= . − −

= . − −

= . − −

= ( 1 − ). + 1

= .

+ 1

= 1 −

= −

= ( − )

=

= + +



= +

= . 1

( − )

= . − ∗

.( − ∗

− ∗ )

=

<

<

= .

= 11

+

+

= . = .

= . + 2 . + 2 . .

= . = .

a

= .

= ( − )

= = .

= = .

= ( ./ = (



=

(

)

=(0.0252+0.0405)( +459.7)

= .

.

= . .

=0.0128(). /

= /

Problem de Transfer.masa II

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Transcript of Problem de Transfer.masa II