Clase Repaso Psicrometria Y Secado
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CLASE REPASO PSICROMETRIA Y SECADO Problema psicrometría Solución: 1000
m3/hv 1
m1 30ºC 85% HR
m2
Condensador
10ºC m2
2
50% HR m1 30ºC
3
Calentador
I m3
m2 II m3
4
85% 50% 1
H1 H4
4
H2
2 3
10ºC
30ºC
Primero convertimos el volumen en masa v en (1) = 0.89 m3/Kg ρ en (1) = 1.1236 Kg/m3 1000 m3/h x 1.1236 Kg/m3 = 1123.6 Kg/hv = m1 Balance de masa I
Balance de água II
m1 = m2 + m3 (1)
m2H2 + m3H1 = m1H4
1123.6 Kg/m = m2 + m3 (1)
1123.6 x 0.0136 = m2 x 0.0078 + m3 x 0.0234 (2)
m2 = 1123.6 – m3 (1)
(1) en (2)
15.281 = (1123.6 – m3) x 0.0078 + m3 x 0.0234
6.52 = (0.0234 – 0.0078) m3 6.52 = m3 0.0156
m3 = 417.95 Kg/hv m2 = 705.65 Kg/hv En términos de volumen m2 = 705.65 Kg/hv x v = 705.65x0.891 = 628.73 m3/hr Problema 5 separata de secado # 1
t 0
∆t 0
X 9
Xeq 0.088
Xl Xl 8.912 -
∆ Xl -
LS/A 0.1395
2
5
10
7.72
0.088
7.632
8.272
1.28
0.1395 0.017856
3
15
10
6.45
0.088
6.362
6.997
1.27
0.1395
0.01772
4
25
10
5.18
0.088
5.092
5.727
1.27
0.1395
0.01772
5
35
10
3.899
0.088
3.811 4.4515 1.281 0.1395
0.01787
6
45
10
2.625
0.088
2.537
3.174
1.274 0.1395
0.01777
7
55
10
1.567
0.088
1.479
2.008
1.058 0.1395
0.01476
8
65
10
1.164
0.088
1.076 1.2775 0.403 0.1395
0.00562
9
75
10
0.865
0.088
0.777
0.299 0.1395
0.00417
10
85
10
0.636
0.088
0.548 0.6625 0.229 0.1395
0.00320
11
95
10
0.455
0.088
0.367 0.4575 0.181 0.1395
0.00253
12
105
10
0.309
0.088
0.221
0.294
0.146 0.1395
0.00204
13
115
10
0.189
0.088
0.101
0.161
0.12
0.1395
0.00160
14
125
10
0.088
0.088
0
-
-
0.1395
15
135
10
0.088
0.088
0
-
-
0.1395
16
145
10
0.088
0.088
0
-
-
0.1395
.9265
LS e = ρx % de materia seca x } por ambas caras A 2
RC -
C
Kg LS 10 0.003 = 930 x x = 0.1395 A 100 2 m2
En el dato #6 está la velocidad crítica Entonces tiempo constante = tc = 0.1395
LS ( X li − X lc ) A Rc
(8.912 − 2.537 ) = 49.41 min 0.018
Para el tiempo decreciente E xl Xlc 1.251
Ln E
t
0.224
45
0.7915
-0.234
55
0.50355
-0.686
65
0.4170
-0.875
75
0.1589
-1.840
85
0.1179
-2.138
95
0.0903
-2.405
105
0.0713
-2.641
115
0.0576
-2.854
125
0.0473
-3.051
135
Regresión de Ln E y t hallamos la pendiente m =
td =
Ln 8 / π 2 − Ln Ef m = +0.0379
[( 0.13 − 0.088 ) + 0.101 ] X lf 2 Ef = = X lc 2.537 → X lc
Ef = 0.02818
td =
Ln 0.81057 − Ln 0.02818 = m
td =
− 0.21 − ( −3.569 ) 3.359 = m m = 0.0379
tdecreciente = 88.6279 minutos Ttotal = 49.41 + 88.63 = 138.04 minutos 2
π Difusividad D = m 2x
2x =
e = 0.0015 2
2
π D = 0.0379 0.0015
D = 8.64012 x 10-9
m2 min
m3/hv 1
m1 30ºC 85% HR
m2
Condensador
10ºC m2
2
50% HR m1 30ºC
3
Calentador
I m3
m2 II m3
4
85% 50% 1
H1 H4
4
H2
2 3
10ºC
30ºC
Primero convertimos el volumen en masa v en (1) = 0.89 m3/Kg ρ en (1) = 1.1236 Kg/m3 1000 m3/h x 1.1236 Kg/m3 = 1123.6 Kg/hv = m1 Balance de masa I
Balance de água II
m1 = m2 + m3 (1)
m2H2 + m3H1 = m1H4
1123.6 Kg/m = m2 + m3 (1)
1123.6 x 0.0136 = m2 x 0.0078 + m3 x 0.0234 (2)
m2 = 1123.6 – m3 (1)
(1) en (2)
15.281 = (1123.6 – m3) x 0.0078 + m3 x 0.0234
6.52 = (0.0234 – 0.0078) m3 6.52 = m3 0.0156
m3 = 417.95 Kg/hv m2 = 705.65 Kg/hv En términos de volumen m2 = 705.65 Kg/hv x v = 705.65x0.891 = 628.73 m3/hr Problema 5 separata de secado # 1
t 0
∆t 0
X 9
Xeq 0.088
Xl Xl 8.912 -
∆ Xl -
LS/A 0.1395
2
5
10
7.72
0.088
7.632
8.272
1.28
0.1395 0.017856
3
15
10
6.45
0.088
6.362
6.997
1.27
0.1395
0.01772
4
25
10
5.18
0.088
5.092
5.727
1.27
0.1395
0.01772
5
35
10
3.899
0.088
3.811 4.4515 1.281 0.1395
0.01787
6
45
10
2.625
0.088
2.537
3.174
1.274 0.1395
0.01777
7
55
10
1.567
0.088
1.479
2.008
1.058 0.1395
0.01476
8
65
10
1.164
0.088
1.076 1.2775 0.403 0.1395
0.00562
9
75
10
0.865
0.088
0.777
0.299 0.1395
0.00417
10
85
10
0.636
0.088
0.548 0.6625 0.229 0.1395
0.00320
11
95
10
0.455
0.088
0.367 0.4575 0.181 0.1395
0.00253
12
105
10
0.309
0.088
0.221
0.294
0.146 0.1395
0.00204
13
115
10
0.189
0.088
0.101
0.161
0.12
0.1395
0.00160
14
125
10
0.088
0.088
0
-
-
0.1395
15
135
10
0.088
0.088
0
-
-
0.1395
16
145
10
0.088
0.088
0
-
-
0.1395
.9265
LS e = ρx % de materia seca x } por ambas caras A 2
RC -
C
Kg LS 10 0.003 = 930 x x = 0.1395 A 100 2 m2
En el dato #6 está la velocidad crítica Entonces tiempo constante = tc = 0.1395
LS ( X li − X lc ) A Rc
(8.912 − 2.537 ) = 49.41 min 0.018
Para el tiempo decreciente E xl Xlc 1.251
Ln E
t
0.224
45
0.7915
-0.234
55
0.50355
-0.686
65
0.4170
-0.875
75
0.1589
-1.840
85
0.1179
-2.138
95
0.0903
-2.405
105
0.0713
-2.641
115
0.0576
-2.854
125
0.0473
-3.051
135
Regresión de Ln E y t hallamos la pendiente m =
td =
Ln 8 / π 2 − Ln Ef m = +0.0379
[( 0.13 − 0.088 ) + 0.101 ] X lf 2 Ef = = X lc 2.537 → X lc
Ef = 0.02818
td =
Ln 0.81057 − Ln 0.02818 = m
td =
− 0.21 − ( −3.569 ) 3.359 = m m = 0.0379
tdecreciente = 88.6279 minutos Ttotal = 49.41 + 88.63 = 138.04 minutos 2
π Difusividad D = m 2x
2x =
e = 0.0015 2
2
π D = 0.0379 0.0015
D = 8.64012 x 10-9
m2 min
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