Post on 03-Jun-2018
8/12/2019 Problem de Transfer.masa II
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EXTRA CCION LIQUIDO LIQUIDO
EXTRA CCION SOLIDO LIQUIDO
SECADO DE SOLIDOS
TR NSFERENCI DE M S II
.
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( − )
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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
= + =∝
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= ′ ( 1 + )
= ′ ( 1 + 0 )
= = 100
1 = 100
=0.4
=0
=indeterm.
= 0
′
. = 0
. =
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′
′
=0.4
=32.5
ver la ecu
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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
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= ( − )
= 100 (0.731716 − 0)
= 73.17157
= ( − )
=100 (32.5−0)
= 3250
=32.5
=0.731716
)
= 0
:
8/12/2019 Problem de Transfer.masa II
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) =0.5
= 0.5 3250 = 1625
= +′ + ′ = + 0
+ 0
= =0.4
= +′ + ′ = +
+ ′
= ′ = 1625
100 =16.25
Calculo del
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=0.223
′
′
′
=
=0.545
= + ′ = 0
Entonces M’=F’=100 Kg
=28.7
=0.659
C
=16.25
=0.4
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′ = ( − −
)
′ =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
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- EXTRACCION EN CONTRACORRIENTE)
100 / 40% (98% , 1.8%
0.2% . ) 98.5° a 70% 4% . ) . )
) )
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= + = 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%)
,
,
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∎ ′(.,)
∎′(0.4,0) =0.04 =0.70
′
′
=0.3355 =10.1667
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) ′ = . −
−
=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
′
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′
′
∆
=.
′
′
′
′
′
′
′
′
′
′
′
′
′
′ ′ = (8 )
′
= .
) .
%= ′ − ′ ′ 100
%= 100(0.4)−70.353(0.04)100(0.4) 100
%=92.965%
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( )
25° . 200 /
40% .
.
∶ a)
) 200 /
, . )
, 10% .
d) ,
30% .
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∗ 25° . 98%
2% .
∶
) f) 200 /
.
) % 20% .
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)
1
F=200 lb/h
=0.40
= 0
=0.366
=0.014
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= − −
=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
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=0.2
=0.4
=0.23
= . ( − −
=400. (0.0.1 = 133.33
= . (
− =400. (00.
=266.6
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%= . −
.
100
%= 200(0.4)−133.33(0.14)200(0.4) 100
%=76.667%
)
1
=0.40
F = 200 /
= 0
=
= . ( −
− )
=0.10
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= + = 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
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= . ( − )
−
= 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
= . ( −
− )
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=0.3
=0.25
= . (
− = 2 0 0 (0.20
=116.205 = + = 2 0 0 +
= 316.205
= . ( =316.2055. (0
0 = 142.90
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= . ( − )
−
=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
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=
=200 0
0. =19.0
=
=200
= 4
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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
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=0.21
=0.15
=0.241
= . ( − )
−
=400 (0.21−0.241)
0.15−0.241
= 136.2637 /
)
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)
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
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Dato =0
=0.30
=0.25
= . (
= 2 0 0 (0.0
=119.32 = + = 2 0 0 +
=319.3277
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%= . − .
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
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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
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= + = 100+40 =0.2143
= + = 140 /
∆
=0.378
5
( − )
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= . ( )
−
= . ( − )
−
%= . − .
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 = ( − )
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) =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 /
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( )150 5% , 4 ,
40 .
:
= 2.20
, ,
, ,
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1 2 3 4
′
′
′
′
′ ′ ′
′ ′ ′
*
∗
∶ E
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∶ = (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
′
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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′
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1
. ′ + . ′ = . ′ +.′
− = (−)
− − = (−)
−
= ′ −
′ −
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= −
;=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
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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
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′,
2
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2 = 142.5 kg
′ =0.03205
= 40
= 40
′ = 0
′
′
. ′ + . ′ = . ′ +.′
− = (−)
− − = (−)
−
= ′ −
′ −
= − = 142.540 =−3.56−3.5625= ′−0
′ −0.03205
′ =−3.5625′ +0.114178
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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
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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
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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
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′ =−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
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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
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′ =−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
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= (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
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= .
<= .
+ .
+ .
+ .
=′
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
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= (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
( −)
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, 100 20% , 100
/ 2/3 . Calcular la cantidad y composición del extracto y del refinado , así como el
porcentaje de aceite extraído
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+ 2
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= + = 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
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=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)
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⨀ M(0.1; 0.5)
:=0.4−
: = 1 −
: = 5
=0.06666 =0.33333
=0.16666 =0.833333
= ( − ) = ( − )
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= ( − ) = ( −
)
=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. −
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, , 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 −
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+
= 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
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=0.252 =0.348
=0.42 =0.58
= ( − ) = ( −
)
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( − ) −
% . = . − .
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
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U ñ 20% −
20% , , , , , , .
,
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, 30% .
, 200 , , 50% .
/ 1 0.30.
) )
−
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−
, 25%
l −
98.5°)
) 50%
¿ ?
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U (; )
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= . + . +
= 200(0.30)+(0)200+
= . + . +
= 200(0)+(1)2 0 0 +
= 200(0.30)200+ = 60
200+ = 2 0 0 +
= . −
−
= . −
−
0 5 =0.
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=0.
=0.
=0.5 =0.5
=
= 60.01
= . − −
= 2 0 0 . 0.
0
= . − −
= ( −
− )
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=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
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1
= ′ = 34.219
=0.25
= 0
=0.25
= =34.219
=0=indeterm.
= 0
. = 0
. =
′
′
=0.25
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′
=36.5
ver la ecu
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d) = 0.5
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= 0.5 1249 = 624.5
Calculo del
= +′ + ′ = + 0 + 0
= =0.25
= +′
+ ′ = +
+ ′
= ′ = 624.5
34.219 =18.25
=18.25
=0 25′
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=0.25
′ =0.115
=0.358
′ = 34
C
=0.614
C
=0.115
=0.25 ′ = ( − −
)
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′ = ( − − )
′ =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
( )
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.
0.28 / , 6 0.08 /
,
0.14 / ,
.
0.33 0.04 / .
0 28
:
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=0.28
=0.08
= 6
= +
= . 1
( − )
=
. − ∗
.(
∗ = 0
= +
= . 1
− +
. − ∗
.( − ∗ − ∗ )
=0.14
= . 1
− +
. − ∗
.( − ∗ − ∗ )
:
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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% ,
.
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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
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()=20%
100 25
= 75 25
100 20 80
= 20 80
()
()
= 3
=0.25
=1.8
∗ = 0 (humedad en equilibrio despreciable)
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= . 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
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= 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
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= .
= = 3
= =0.25
= 5
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∗ =0.2 =1.8 =0.25
= . 1
−
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= 7.55 3 − 1 . 8
=1.8
=
. − ∗
.( − ∗
− ∗)
=7.5 1 . 8 − 05 .( 1 . 8 − 0
0.25−0)
= 5.33
= +
= 7.13
=
<
<
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( )
, 4 4 ,
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, 4 4 , 2 ,
.
7 / 140° 85° . 50 , 2.5 / , 20 % .
=50
= 2080 =0.25 85° = 1045.8 /
= 2.5 /
=85° =140°
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=85°
=140°
=0.0135
= 0.0135 .
= ( − )
85°
=
= + +
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=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 / − − °
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=
=264.505
− 1045.8
=0.2529
=
. 1
( − )
= 5016 . 1
0.2529 (2.5−0.25)
= 27.8
=4.809(140−85)
= 264.505 / −
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∶ =0.005 /
= 6 0 % = 6 0 % = 6 0 %
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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°
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= =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 /
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+ = + =
. + . = . + . = .
. + . = . . + ( − ).
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. + . = .
. + . = ( +. + . = . .(−) = . (
= . ( − −
. + ( ). . + . − .
( − ) = (
= . ( − −
. + . = .
( M − ). + . = .
M. − . + . = .
. ( − ) = ( − )
= . ( − )
−
. + . = .
= . +S.
= . +S. +
′ + ′ + ′
= 0
= 0
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′ + ′ = + = ′
′. +′ . = . +
. = ′ .
′. +′ . = . + . = ′ .
=
′. +′ .
′ + ′
= ′. + ′
′
′= −
−
′′ = −
−
. = 0
. =
′ = + = ′
′. = . + . = ′ .
′. + = . +
=
= ′. +′ . ′ + ′
= ′ ( − )
. + = ′ .
′ = − = −
=′( − )
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− −
′ = −
− = −
−
=′( − )
= ′ ( 1 + )
S = ′ ( 1 + )
= ′( 1 + )
= ′ (1 + )
′ = − −
′ = −
−
P :
:
= 1 +
= 1 +
= (1 + ′)
= (1 + ′)
′1 + ′ =2.20 ′
1 + ′
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= (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 +
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<
=
< = + + +
1 +
−
+ = + =
+ +
= =
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. + . = . + . = .
. + . = . + . = .
= . + . +
= . + . +
= . − − = . − −
= . − −
= . − −
= . − −
= . − −
= ( 1 − ). + 1
= .
+ 1
= 1 −
= −
= ( − )
=
= + +
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= +
= . 1
( − )
= . − ∗
.( − ∗
− ∗ )
=
<
<
= .
= 11
+
+
= . = .
= . + 2 . + 2 . .
= . = .
a
= .
= ( − )
= = .
= = .
= ( ./ = (
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=
(
)
=(0.0252+0.0405)( +459.7)
= .
.
= . .
=0.0128(). /
= /