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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


−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


−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].


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]


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]


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

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