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TAREA 1
PROCESOS DE ALIMENTOS I 2015-2 DEPARTAMENTO DE INGENIERIA QUIMICA Y AMBIENTAL
UTFSM, CASA CENTRAL, VALPARAISO
Alumno: Matías López. Profesor: Sergio Almonacid. Fecha: 26/10/2015
1. The following data were obtained from a thermal resistance experiment conducted on a spore suspension at 112°C:
Time[min] Number of survivors
0 1E+6 4 1,1E+5 8 1,2E+4 12 1,2E+3
Determine the D value of the microorganism. [4.1 min]
La pendiente de la gráfica LOG(N) en el tiempo es igual a:
𝑠𝑙𝑜𝑝𝑒 = −1𝐷
Entonces se gráfica la siguiente tabla:
Time[min] LOGN 0 6 4 5,041392685 8 4,079181246 12 3,079181246
La pendiente es de:
𝑠𝑙𝑜𝑝𝑒 = −0,2431
y = -‐0,2431x + 6,0086
0
1
2
3
4
5
6
7
0 2 4 6 8 10 12 14
LOG(N)
Tiempo[min]
LOG(N) v/s tiempo
𝐷 = −1
𝑠𝑙𝑜𝑝𝑒 = −1
−0,2431 = 4,11[𝑚𝑖𝑛]
2. The following data were obtained from a thermal resistance experiment conducted on a spore suspension at 112°C:
Time[min] Number of survivors
0 1E+6 15 2,9E+5 30 8,4E+4 45 2,4E+4 60 6,9E+3
Determine the D value of the microorganism. [27.8 min]
Se sigue el mismo procedimiento que el ejercicio anterior.
Time[min] LOGN 0 6 15 5,46 30 4,92 45 4,38 60 3,83
La pendiente es de:
y = -‐0,036x + 6,0031
0
1
2
3
4
5
6
7
0 10 20 30 40 50 60 70
LOG(N)
Tiempo[min]
LOG(N) v/s tiempo
𝑠𝑙𝑜𝑝𝑒 = −0,036
𝐷 = −1
𝑠𝑙𝑜𝑝𝑒 = −1
−0,036 = 27,77[𝑚𝑖𝑛]
3. Calculate the D value of an organism which shows 30 survivors from an initial
inoculum of 5·106 spores after 10 min at 121°C. [1.9 min]
Según el problema se tiene que:
Time[min] Nº Survivor
LOGN
0 5000000 6,69 10 30 1,47
Se determina la pendiente de la recta.
𝑠𝑙𝑜𝑝𝑒 =𝑌2− 𝑌1𝑋2− 𝑋1 =
6,69− 1,4710− 0 = −0,52
𝐷 = −1
𝑠𝑙𝑜𝑝𝑒 = −1
−0,52 = 1,91[𝑚𝑖𝑛]
4. The decimal reduction times D for a spore suspension were measured at several temperatures, as follows:
T[ºC] D[min] 104 27,5 107 14,5 110 7,5 113 4 116 2,2
Determine the thermal resistance constant z for the spores. [11.1°C]
Se grafican los linealizan los datos, es decir, aplicar logaritmo en base diez.
T[ºC] LOG(D) 104 1,43 107 1,16 110 0,87 113 0,60 116 0,34
Se tiene que:
𝑠𝑙𝑜𝑝𝑒 = −1𝑧
𝑧 = −1
𝑠𝑙𝑜𝑝𝑒 −1
−0,0918 = 10,9[º𝐶]
5. If the z value of a microorganism is 16.5°C and D121 = 0.35 min, what is D110? [1.62 min]
Igual que en el ejercicio anterior, se calcula a través de la ecuación de la recta (con los datos linealizados).
T[ºC] D LOG(D) 121 0,35 -‐0,455 110 -‐ -‐
Se tiene que,
𝑠𝑙𝑜𝑝𝑒 =𝐿𝑂𝐺(𝐷2)− 𝐿𝑂𝐺(𝐷1)
𝑇2− 𝑇1
𝑠𝑙𝑜𝑝𝑒 = −1𝑧
−1𝑧 =
𝐿𝑂𝐺(𝐷2)− 𝐿𝑂𝐺(𝐷1)𝑇2− 𝑇1
−1𝑧 ∗ (𝑇2− 𝑇1) = 𝐿𝑂𝐺(
𝐷2𝐷1)
𝐷2 = 𝐷1 ∗ 10!!!∗(!!!!!)
y = -‐0,0918x + 10,979
0 0,2 0,4 0,6 0,8 1
1,2 1,4 1,6
102 104 106 108 110 112 114 116 118
LOG(D)
Temperatura [ºC]
Log(D) v/s Temperatura
𝐷2 = 0,35 ∗ 10!!
!",!∗(!!"!!"!) 𝐷2 = 1,62[𝑚𝑖𝑛]
6. Estimate the spoilage probability of a 50 min process at 113°C when D113 = 4 min
and the initial microbial population is 104 per container. [3.16·10-‐8]
Se tiene que,
𝑠𝑙𝑜𝑝𝑒 = −1𝐷 =
𝐿𝑜𝑔(𝑁/𝑁𝑜)𝑡 − 𝑡𝑜
−1𝐷 ∗ (𝑡 − 𝑡𝑜) = 𝐿𝑜𝑔(𝑁/𝑁𝑜)
𝑁 = 𝑁𝑜 ∗ 10!!!∗(!!!")
𝑁 = 10! ∗ 10!!!∗(!") = 3,16 ∗ 10!!
7. An F0 value of 7 min provides an acceptable economic spoilage for a given product. Determine the process time at 115°C. [27.9 min]
Se pueden relacionar el F y F de referencia de la siguiente forma:
𝐹𝐹!"#
=𝐷𝐷!"#
𝐷𝐷!"#
= 10!!"#!!
!
Se puede reescribir como:
𝐹𝐹!"#
= 10!!"#!!
!
𝐹 = 𝐹!"# ∗ 10!!"#!!
! = 7 ∗ 10!"!!!!"
!" 𝐹 = 27,86[𝑚𝑖𝑛]
8. The F value at 121.1°C for a 99.999% inactivation of a strain of Clostridium Botulinum is 1.2 min. What is the D value of this organism at 121.1°C? [0.24 min].
El número de reducciones,
𝑥 = 𝐿𝑜𝑔𝑁𝑜𝑁𝑓
y
𝑥 ∗ 𝐷 = 𝐹
según el enunciado el número final de sobrevivientes es igual a:
𝑁𝑓 =0,001100 ∗ 𝑁𝑜
reemplaza:
𝑥 = 𝐿𝑜𝑔𝑁𝑜
0,001100 ∗ 𝑁𝑜
= 5
5 ∗ 𝐷 = 1,2
𝐷 = 0,24[𝑚𝑖𝑛] 9. Considering that 12 is the minimum sterilizing value for the inactivation of Clostridium Botulinum spores, calculate F0 based on inactivation of Clostridium Botulinum. [2.88 min]
Se debe trabajar con el D calculado en el ejercicio anterior para obtener una esterilización adecuada (mínimo requerido):
𝐹! = 12 ∗ 𝐷
𝐹! = 12 ∗ 0,24 = 2,88 [𝑚𝑖𝑛] 10. A process is based on an F0 = 2.88 min. If a can contained 10 spores of an organism having D0 = 1.5 min, calculate the probability of spoilage from the latter organism. [12%]
𝑥 = 𝐿𝑜𝑔𝑁𝑜𝑁𝑓
𝑥 ∗ 𝐷 = 𝐹
𝐿𝑜𝑔𝑁𝑜𝑁𝑓 =
𝐹𝐷
𝑁𝑜𝑁𝑓 = 10
!!
𝑁𝑓 =𝑁𝑜
10!!=
10
10!,!!!,!
= 0,12
11. A canned food prior to processing contains 1000 spores per can. The spores have a D value of 1.5 min at 121.1°C. If the process is carried out to an equivalent process time of 10 min at 121.1°C, what would be the probability of spoilage? [2/10000]
Del ejercicio anterior se sabe que:
𝑁𝑓 =𝑁𝑜
10!!=1000
10!"!,!
= 0,0002 =2
10000
12. If the most probable spore load in a product is 100 spores per can and the D0 = 1.5 min, calculate the time of heating at 121.1°C necessary to achieve a probability of
spoilage from this organism of one in 105 cans. Under these conditions, the D0 of Clostridium Botulinum Type B is 0.2 min. Is the calculated process time F0 equivalent to or more than what is required for a 12D reduction of Clostridium Botulinum?
[F0 = 10.5 min]
𝑥 = 𝐿𝑜𝑔𝑁𝑜𝑁𝑓 = 𝐿𝑜𝑔
110!100 = 7
𝐹 = 7 ∗ 1,5 = 10,5 [𝑚𝑖𝑛]
El valor calculado de F es mucho mayor que el F0 (2,54[min]).
13. What level of inoculation of PA 3679 (D0 = 1.2 min) is required such that a probability of spoilage of one in 100 is attributed to PA 3679 would be equivalent to 12 D inactivation of Clostridium Botulinum? Assume the same temperature process and the same z values for both organisms. The D0 value of Cl. Botulinum is 0.22 min. [N0=1.6]
Se sabe que:
𝐹 = 12 ∗ 𝐷 = 12 ∗ 0,22 = 2,64[𝑚𝑖𝑛]
y Do es igual a 1,2 [min].
𝐿𝑜𝑔𝑁𝑜𝑁𝑓 =
𝐹𝐷
𝑁𝑜 = 𝑁𝑓 ∗ 10!! = 0,01 ∗ 10
!,!"!,! = 1,6
14. For an initial inoculum of 10 spores/g of product (D121°C = 1.2 min), a spoilage
rate of one can in 105 is desired. Calculate an F value for the process that would give the desired level of inactivation. Calculate F138°C for a z value of 10°C. [F0=7.2 min F138,10=0.14 min]
𝐿𝑜𝑔𝑁𝑜𝑁𝑓 =
𝐹𝐷
𝐹! = 𝐷 ∗ 𝐿𝑜𝑔𝑁𝑜𝑁𝑓 = 1,2 ∗ 𝐿𝑜𝑔
10110!
= 7,2[𝑚𝑖𝑛]
𝐹𝐹!"#
= 10!!"#!!
!
𝐹 = 𝐹!"# ∗ 10!!"#!!
! = 7,2 ∗ 10!"!!!"#
!" 15. For an initial inoculum of 100 spores/can of product (D121°C = 1.5 min, z = 8°C), a
spoilage rate of one can in 104 is desired. Calculate F115°C value for the process that would give the desired level of inactivation. [50.6 min]
𝐷(115)𝐷(121) = 10
!!"#!!!
𝐷(115) = 𝐷(121) ∗ 10!!"#!!
! = 1,5 ∗ 10!"!!!!"
! 𝐷(115) = 8,4[𝑚𝑖𝑛]
𝐿𝑜𝑔𝑁𝑜𝑁𝑓 =
𝐹𝐷
𝐹 = 8,4 ∗ 𝐿𝑜𝑔100110!
= 50,64[𝑚𝑖𝑛]
16. Realice el mismo cálculo hecho en problema 15, pero utilice k y Ea. Son los resultados iguales. Explique.
Por Arrhenius:
𝐸𝑎 =𝑇! ∗ (𝑇! − 𝑧) ∗ 2.303 ∗ 𝑅
𝑧 =394 ∗ (121− 8) ∗ 2.303 ∗ 𝑅
281 = 364,85 ∗ 𝑅
𝑘 = 𝑘𝑟 ∗ 𝑒!!"! ∗ !!!
!!"
𝐷 =𝐿𝑛(10)𝑘