Embed Size (px)
Citation preview
0 10 20 30 40 50 60 70 80 90 1000.0
0.1
0.2
0.3
Fig. 2: VELOCIDAD LETAL EN FUNCIÓN DEL TIEMPO PARA HALLAR EL F0 (EL ÁREA BAJO LA CURVA)
t (min)
L
Tiempo Tº TºRetorta pmf
t T(min) (ºC) (ºC)
1 0 20.0 16.42 5 82.5 28.13 6 94.4 30.44 8 104.6 55.75 10 114.1 60.36 12 115.9 65.97 14 116.1 71.28 16 116.1 76.89 18 115.5 82.3
10 20 114.9 86.811 22 114.6 89.512 24 114.7 94.213 26 114.6 96.814 28 114.7 99.315 30 115.2 102.016 32 115.4 104.017 34 115.5 105.718 36 115.5 107.319 38 115.6 108.720 40 115.5 109.721 42 115.3 110.622 44 115.3 111.423 46 115.3 112.024 48 115.2 112.525 50 115.2 113.026 52 115.2 113.427 54 115.2 113.728 56 115.2 113.929 58 115.3 114.230 60 115.4 114.431 62 115.3 114.632 64 115.4 114.733 66 115.3 114.834 70 115.3 114.935 74 115.3 115.136 76 115.2 115.137 78 32.7 89.838 80 29.0 82.639 82 29.3 83.440 84 26.5 81.541 86 24.3 77.442 90 21.7 66.6
PROBLEMA: Se esteriliza una salsa de tomate en el autoclave.
TR
MÉTODO GENERAL
Tiempo Tº F Lpmf Δt Fx((L1+L2)/2) Acumulado
t T (min) (min)
(min) (ºC)1 0 16.4 0.0 0.02 5 28.1 5 0.0 0.0 0.03 6 30.4 1 0.0 0.0 0.04 8 55.7 2 0.0 0.0 0.05 10 60.3 2 0.0 0.0 0.06 12 65.9 2 0.0 0.0 0.07 14 71.2 2 0.0 0.0 0.08 16 76.8 2 0.0 0.0 0.09 18 82.3 2 0.0 0.0 0.0
10 20 86.8 2 0.0 0.0 0.011 22 89.5 2 0.0 0.0 0.0
Para el C. botulinum: Z = 10 ºC y temperatura de 121.1 ºC
F0 F0
10 (1/Z)(T-121.1)
0 10 20 30 40 50 60 70 80 90 1000
20
40
60
80
100
120
140
Fig. 1 : MONITOREO DE LA TEMPERATURAS DE LA RETORTA Y DEL PRODUCTO, EN FUNCIÓN DEL TIEMPO
TR T
T (min)
TR
y T
12 24 94.2 2 0.0 0.0 0.013 26 96.8 2 0.0 0.0 0.014 28 99.3 2 0.0 0.0 0.015 30 102.0 2 0.0 0.0 0.016 32 104.0 2 0.0 0.0 0.117 34 105.7 2 0.0 0.0 0.118 36 107.3 2 0.0 0.1 0.219 38 108.7 2 0.1 0.1 0.320 40 109.7 2 0.1 0.1 0.421 42 110.6 2 0.1 0.2 0.622 44 111.4 2 0.1 0.2 0.823 46 112.0 2 0.1 0.2 1.024 48 112.5 2 0.1 0.3 1.325 50 113.0 2 0.2 0.3 1.626 52 113.4 2 0.2 0.3 1.927 54 113.7 2 0.2 0.4 2.228 56 113.9 2 0.2 0.4 2.629 58 114.2 2 0.2 0.4 3.030 60 114.4 2 0.2 0.4 3.431 62 114.6 2 0.2 0.4 3.932 64 114.7 2 0.2 0.5 4.333 66 114.8 2 0.2 0.5 4.834 70 114.9 4 0.2 0.9 5.735 74 115.1 4 0.3 1.0 6.736 76 115.1 2 0.3 0.5 7.237 78 89.8 2 0.0 0.3 7.538 80 82.6 2 0.0 0.0 7.539 82 83.4 2 0.0 0.0 7.540 84 81.5 2 0.0 0.0 7.541 86 77.4 2 0.0 0.0 7.542 90 66.6 4 0.0 0.0 7.5
7.5 F0 =
F0
MÉTODO DE BALL
Tiempo Tº TºRetorta pmf
t T(min) (ºC) (ºC)
1 0 20.0 16.42 5 82.5 28.13 6 94.4 30.44 8 104.6 55.75 10 114.1 60.36 12 115.9 65.97 14 116.1 71.28 16 116.1 76.89 18 115.5 82.3
10 20 114.9 86.811 22 114.6 89.512 24 114.7 94.213 26 114.6 96.814 28 114.7 99.315 30 115.2 102.016 32 115.4 104.017 34 115.5 105.7
TR
0 10 20 30 40 50 60 70 80 90 1000.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
Fig. 3: F0 ACUMULADO
t (min)
F0
acu
mu
lad
o
18 36 115.5 107.319 38 115.6 108.720 40 115.5 109.721 42 115.3 110.622 44 115.3 111.423 46 115.3 112.024 48 115.2 112.525 50 115.2 113.026 52 115.2 113.427 54 115.2 113.728 56 115.2 113.929 58 115.3 114.230 60 115.4 114.431 62 115.3 114.632 64 115.4 114.733 66 115.3 114.834 70 115.3 114.935 74 115.3 115.136 76 115.2 115.137 78 32.7 89.838 80 29.0 82.639 82 29.3 83.440 84 26.5 81.541 86 24.3 77.442 90 21.7 66.6
SOLUCIÓN
ES CONSTANTE) T1 Y LA Tº pmf
Tiempo TR Tpmf Diferenciat T T1 - T T1 = 115.3
(min) (ºC) (ºC) (ºC) 16.40 20.0 16.4 98.95 82.5 28.1 87.26 94.4 30.4 84.98 104.6 55.7 59.6
10 114.1 60.3 55.012 115.9 65.9 49.414 116.1 71.2 44.116 116.1 76.8 38.518 115.5 82.3 33.020 114.9 86.8 28.522 114.6 89.5 25.824 114.7 94.2 21.126 114.6 96.8 18.528 114.7 99.3 16.0
PASO 1.- SE HALLA LA DIFERENCIA ENTRE LA Tº DE LA RETORTA (CUANDO ESTA
T0 =
30 115.2 102.0 13.332 115.4 104.0 11.334 115.5 105.7 9.636 115.5 107.3 8.038 115.6 108.7 6.640 115.5 109.7 5.642 115.3 110.6 4.744 115.3 111.4 3.946 115.3 112.0 3.348 115.2 112.5 2.850 115.2 113.0 2.352 115.2 113.4 1.954 115.2 113.7 1.656 115.2 113.9 1.458 115.3 114.2 1.160 115.4 114.4 0.962 115.3 114.6 0.764 115.4 114.7 0.666 115.3 114.8 0.570 115.3 114.9 0.474 115.3 115.1 0.276 115.2 115.1 0.278 32.7 89.8 25.580 29.0 82.6 32.782 29.3 83.4 31.984 26.5 81.5 33.886 24.3 77.4 37.990 21.7 66.6 48.7
PASO 2.- GRAFICANDO EL PERFIL DE TEMPERATURAS
0 10 20 30 40 50 60 70 80 90 1000
20
40
60
80
100
120
140
36; 115.16; 115.9
FIG. 4: PERFIL DE TEMPERATURAS, TR Y T
T1 T
t (min)
T (
ºC)
FIG. 5: TÉRMINOS USADOS PARA BALL
En base a los terminos definidos para el método de Ball
CUT = Coming Up Time o Tiempo de levante de la temperatura de la retorta.
CUT = Tc 12
76
th - 0,58*CUT = 76 - 0.58x12 = 69.0
GRAFICARLOS EN PAPEL SEMILOGARÍTMICO
Tomamos los puntos solo de calentamiento
Tiempo Diferenciat T1 - T
(min) (ºC)0 98.95 87.26 84.98 59.6
10 55.012 49.414 44.116 38.518 33.020 28.522 25.824 21.1
th =
tB =
PASO 3.- TOMANDO EL TIEMPO Y LA DIFERENCI DE Tº CON LA FINALIDAD DE
0 10 20 30 40 50 60 70 80 90 1000
20
40
60
80
100
120
140
36; 115.16; 115.9
FIG. 4: PERFIL DE TEMPERATURAS, TR Y T
T1 T
t (min)
T (
ºC)
26 18.528 16.030 13.332 11.334 9.636 8.038 6.640 5.642 4.744 3.946 3.348 2.850 2.352 1.954 1.656 1.458 1.160 0.962 0.764 0.666 0.570 0.474 0.276 0.2
10 20 30 40 50 60 70 80
0.1
1.0
10.0
100.0
6
FIG. 6: DIFERENCIA DE TEMPERATURAS EN FUNCIÓN DEL TIEMPO
t (min)
(T1-
T),
ºC
CON LA FINALIDAD DE HALLAR LA CURVA AJUSTADA.
A partir de t = 28 min la curva es lineal.
Tiempo Diferenciat T1 - T
(min) (ºC)28 16.030 13.332 11.334 9.636 8.038 6.640 5.642 4.744 3.946 3.348 2.850 2.352 1.954 1.656 1.458 1.160 0.962 0.764 0.666 0.570 0.474 0.276 0.2
PASO 4.- TOMANDO SOLO LOS DATOS QUE SE ASEMEJAN A UNA LINEA RECTA,
10 20 30 40 50 60 70 80
0.1
1.0
10.0
100.0
6
FIG. 6: DIFERENCIA DE TEMPERATURAS EN FUNCIÓN DEL TIEMPO
t (min)
(T1-
T),
ºC
20 30 40 50 60 70 80
0.1
1.0
10.0
100.0
f(x) = 221.308723145274 exp( − 0.0920225842419571 x )R² = 0.99785025767747
FIG. 7: CURVA AJUSTADA
t (min)
(T1
-T),
ºC
a b valor absoluto de b
221.31 -0.092 0.092
SIMULTÁNEAMENTE
Tiempo Diferencia Diferencia ajustada
t T1 - T T1 - T(min) (ºC) (ºC)
0 98.9 221.35 87.2 139.76 84.9 127.48 59.6 106.0
10 55.0 88.212 49.4 73.414 44.1 61.016 38.5 50.818 33.0 42.220 28.5 35.122 25.8 29.224 21.1 24.326 18.5 20.228 16.0 16.830 13.3 14.032 11.3 11.734 9.6 9.736 8.0 8.138 6.6 6.740 5.6 5.642 4.7 4.644 3.9 3.946 3.3 3.248 2.8 2.750 2.3 2.252 1.9 1.954 1.6 1.5
y = a e b x
PASO 5.- GRAFICANDO LOS DATOS EXPERIMENTALES Y LOS AJUSTADOS
20 30 40 50 60 70 80
0.1
1.0
10.0
100.0
f(x) = 221.308723145274 exp( − 0.0920225842419571 x )R² = 0.99785025767747
FIG. 7: CURVA AJUSTADA
t (min)
(T1
-T),
ºC
56 1.4 1.358 1.1 1.160 0.9 0.962 0.7 0.764 0.6 0.666 0.5 0.570 0.4 0.474 0.2 0.276 0.2 0.2
Sacando Ln y dividiendo entre 2.303 para transformarlo a Log decimal
2.303
Log (T1-T) = Log 221.31 - 0.092/2.303 t
Log (T1-T) = Log 221.31 - 0.0399 t
Ecuación de Ball:Log (T1-T) = Log JI - (1/fh) t
1 69.04
2 25.0
3 115.3
(T1 -T) = 221.31 e -0.092 t
PASO 6.- ELABORAR LA SIGUIENTE TABLA:
t, min (th - 0,58*CUT = 76 - 0.58x12)
fh, min
T1
0 10 20 30 40 50 60 70 80
0
1
10
100
1000
FIG. 8: CURVA AJUSTADA Y DATOS EXPERIMENTALES
Column D
t (min)
(T1
-T).
ºC
(JI)
4 16.4
5 I = (T1 - T0) 98.9
6 JI 221.31
7 0.3
8 J = JI/I 2.2
9 Log JI 2.35
10 t/fh 2.76
11 -0.413
12 0.386
13 0.886
14 7.4
TABLAS DE BALL
GRÁFICO PARA LOS CÁLCULOS DEL TRATAMIENTO TÉRMICO POR EL MÉTODO DE BALL
m + g = Tr - CW =100ºC; donde Tr = Tº de la retorta y CW = Tº del agua de enfriamiento No existe, mucha diferencia aunque se aplique a m + g = 70 a 110 ºC
t0 , ºF
L = 10 (T1 - 121.1)/Z =10 (T1 - 121.1)/10
Log g = Log JI - t/fh
g = 10 Log g
USE TABLA O GRÁFICO PARA OBTENER "R (fh/U)" CUANDO "Log g" ES CONOCIDO
R = fh/U
F0 = (fh x L) / R , min
![UNIT 1: Overview of General Contract Principles - Software... · Overview of General Contract Principles. ... Carlill v Carbolic Smoke Ball Co [1893] 1 QB 256 The Carbolic Smoke Ball](https://img.pdfslide.net/doc/110x75/5b772c777f8b9a47518c75a4/unit-1-overview-of-general-contract-software-overview-of-general-contract.jpg)
