F SPOIL.xls: Calculation of F value, Microbial Spoilage Rate, and Nutrient Retention, from Time...

F - SPOILCALCULATION OF F-VALUE, MICROBIAL SPOILAGE RATE, AND NUTRIENT RETENTION FROM TIME-TEMPERATURE DATA, BY ADDITION OF LETHALITIES

Protected; no password HAS University of Applies Sciences; NL

yellow cell = input values

interval = min. Constant interval between temperature measurements [in min.]Tref =z =

Micro- D-value of m.o. = min. D-value of micro-organism of concern [in min.]

organism Tref of m.o. =of z of m.o. =concern: Initial number a = total per can Initial total number of micro-organisms of concern in can

Input: Calculation results:Time Core Lethality F-value

temperature T factor(= min. at Tref. of

[min.] 0 [min.]0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!

For 4 worked examples of the use of this spreadsheet, and for its validation:

NB: If the product in the can is a SOLID, please calculate the spoilage rate twice:** first use the total initial number of micro-organisms in the can; the resulting number of spoiled cans b1 will be too high;** next use the initial number of micro-organisms in 1 gram; the resulting number of spoiled cans b2 will be too low.The correct number of spoiled cans, so the CORRECT SPOILAGE RATE b of the SOLID product then will be: b2 <

F: oC Reference temperature for F-value [in oC z-value for F-value [in oC]

oC Reference temperature for D of micro-organism [in oC] oC z-value for D-value of micro-organism [in oC]

[oC] oC)

0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!

0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!0 #DIV/0! #DIV/0!

F - SPOILWorked example 1:

pumpkin soup, which showed the following 15 time-temperature measurements, recorded at time intervals of 2 minutes:

yellow cell = input valuesAnswer Example 1: All cells in this example are protected, also the input cells!Fill up the yellow cells in the F section and the Micro-organism of concern section:

interval = 2 min. Constant interval between temperature measurements [in min.]Tref = 121.1z = 10

Micro- D-value of m.o. = 0.9 min. D-value of micro-organism of concern [in min.]

organism Tref of m.o. = 122.3of z of m.o. = 11concern: Initial number a = 1800 total per can Initial total number of micro-organisms of concern in can

When the food is a liquid, always type the TOTAL initial number of micro-organisms of concern in the whole can. In this example: 2 spores per ml, in a can of 900 ml; so 2 x 900 = 1800 spores per can.

In the yellow cells "Input: Core temperature T": type the food temperatures from the pumpkin soup time-temperature table:Input: Calculation results:

Time Core Lethality F-valuetemperature T factor

(= min. at Tref. of

F-value and Spoilage Rate b of Liquid Foods1a) Calculate the F0-value (= F at reference temperature Tref = 121.1 oC and z = 10 oC) of a heating process of liquid

1b) Also calculate the spoilage rate per can by a spore-forming spoilage micro-organism with D = 0.9 min at 122.3 z = 11 oC; initial number of spores of the micro-organism of concern = 2 per ml; volume of liquid in can = 900 ml.



[min.] 121.1 [min.]0 22.31 1.3213E-10 2.6426E-102 32 1.2303E-09 2.7248E-094 87.54 4.4055E-04 8.8111E-046 113.03 1.5596E-01 3.1279E-018 119.24 6.5163E-01 1.6160E+0010 120.41 8.5310E-01 3.3222E+0012 120.34 8.3946E-01 5.0012E+0014 120.27 8.2604E-01 6.6532E+0016 120.34 8.3946E-01 8.3322E+0018 120.42 8.5507E-01 1.0042E+0120 120.29 8.2985E-01 1.1702E+0122 102.93 1.5241E-02 1.1732E+0124 72.82 1.4859E-05 1.1733E+0126 55.87 2.9992E-07 1.1733E+0128 46.51 3.4754E-08 1.1733E+0130 7.7625E-13 1.1733E+01

The heating process in the table above results in a spoilage rate of about 7 cans per 100 000 000 cans, or 0.007 cans per 100 000 cans; 0.007 spoiled cans per 100 000 cans is less than 1 can per 100 000 cans.



[oC] oC)

Summary findings of Worked Example 1 (liquid food; calculate F-value and spoilage rate):1a) The F0-value (= F at reference temperature Tref = 121.1 oC and z = 10 oC) of the heating procesis equal to F 0 = F 10 121.1 = 1.1733E+01 min. ≈ 11.7 min. (see F-value column).Such a sterilized food is safe, because each can received more than the "Botulinum Cook" of F

1b) The spoilage rate per can = 7.3914E-08 ≈ 7.4 cans per 108 cans (see column Spoilage rate b).So at a total of 108 cans (= 100 000 000 cans), all together some 7 spores will survive. These 7 cans may spoil.For a spoilage micro-organism (so NOT for a pathogenic micro-organism), the generally accepted spoilage rate is 1 can per 105 cans = 1 can per 100 000 cans.

Conclusion of Worked Example 1: This heating process gives a generally accepted result:* the sterilized food is safe: F0 > 3min.; * the sterilized food has an acceptable low spoilage rate (< 1 can per 10

F-value and Nutrient Retention of Liquid Foods2a) Calculate the F0-value (= F at reference temperature Tref = 121.1 oC and z = 10 oC) of a heating process of liquid

yellow cell = input valuesAnswer Example 2: All cells in this example are protected, also the input cells!

interval = 2 min. Constant interval between temperature measurements [in min.]Tref = 121.1z = 10


organism Tref of m.o. = 120of z of m.o. = 29.7concern: Initial number a = 100 total per can Initial total number of micro-organisms of concern in can

In the yellow "Input: Core temperature T" cells: type the food temperatures from the pumpkin soup time-temperature table:



[min.] 121.1 [min.]0 22.31 1.3213E-10 2.6426E-102 32 1.2303E-09 2.7248E-094 87.54 4.4055E-04 8.8111E-046 113.03 1.5596E-01 3.1279E-018 119.24 6.5163E-01 1.6160E+0010 120.41 8.5310E-01 3.3222E+0012 120.34 8.3946E-01 5.0012E+0014 120.27 8.2604E-01 6.6532E+0016 120.34 8.3946E-01 8.3322E+0018 120.42 8.5507E-01 1.0042E+0120 120.29 8.2985E-01 1.1702E+0122 102.93 1.5241E-02 1.1732E+0124 72.82 1.4859E-05 1.1733E+0126 55.87 2.9992E-07 1.1733E+0128 46.51 3.4754E-08 1.1733E+0130 7.7625E-13 1.1733E+01

2b) Also calculate the nutrient retention of vitamin B1. For (destruction of) vitamin B1: D = 128.3 min at 120 z = 29.7 oC. Initial vitamin B1 content = 100%. [(Holdsworth) 1997; vitamin B1 in milk at food temperatures T = 35

2a) For calculating the F0: fill up the yellow cells in the F section as in Worked example 1. 2b) For calculating the nutrient retention: type input figures in section "Micro-organism of concern": D = D of Tref = reference temperature of D of nutrient; z = z of nutrient; initial number a = 100 (initial % of F:

oC Reference temperature for F-value [in oC z-value for F-value [in oC]


[oC] oC)

Conclusion of Worked Example 2 (liquid food; F-value and nutrient retention):2a) The F0-value (= F at reference temperature Tref = 121.1 oC and z = 10 oC) of the heating procesis equal to F 0 = F 10 121.1 = 1.1733E+01 min. ≈ 11.7 min. (see F-value column).Such a sterilized food is safe, because each can received more than the "Botulinum Cook" of F

2b) The nutrient retention, thus how much vitamin B1 is retained, = 7.4844E+01


sterilization process. He used experimental time-product temperature results, and made hiscalculations according to the "equal time interval"-method of Patashnik, applying lethality tables.

Answer Validation Example: All cells in this example are protected, also the input cells!

Tref = reference temperature of D of nutrient; z = z of nutrient; initial number = 100 (initial % of nutrient = 100).interval = 5 min. Constant interval between temperature measurements [in min.]Tref = 121.1z = 10

Micro- D-value of m.o. = min. D-value of micro-organism of concern [in min.]

organism Tref of m.o. =of z of m.o. =concern: Initial number a = total per can Initial total number of micro-organisms of concern in can

In the yellow "Input: Core temperature T" cells: type the food temperatures from the Validation time-temperature table:

Input: Calculation results:

Or: of the initial 100% of vitamin B1 in the food, some 25% was destructed during the heating process.

Validation of Lethality Factors and of F-value

Kooiman (1980) calculated the F10121,1-value, received by a can of soup during the heating stage of a

3a) Use F-SPOIL.xls to calculate all lethality factors and the F0-value (= F at reference temperature Tref = 121.1 and z = 10 oC) of the heating process of Kooiman (see table above: 12 time-temperature measurements; equal time intervals of 5 min.; Tref = 121.1 oC; z = 10 oC ).3b) Compare and contrast the results of F-SPOIL.xls with Kooiman's lethality factors and Kooiman's F

3a) For calculating the F0: fill up the yellow cells in the F section: interval = 5 min.; Tref = 121.1




(= min. at Tref. of[min.] 121.1 [min.]

0 102.4 1.3490E-02 6.7448E-025 112.5 1.3804E-01 7.5764E-0110 114.4 2.1380E-01 1.8266E+0015 115.7 2.8840E-01 3.2686E+0020 116.9 3.8019E-01 5.1696E+0025 117.7 4.5709E-01 7.4550E+0030 118.3 5.2481E-01 1.0079E+0135 118.7 5.7544E-01 1.2956E+0140 119 6.1660E-01 1.6039E+0145 119.2 6.4565E-01 1.9268E+0150 119.3 6.6069E-01 2.2571E+0155 119.3 6.6069E-01 2.5874E+0160 72.82 1.4859E-05 2.5875E+0165 7.7625E-13 2.5875E+0170 7.7625E-13 2.5875E+0175 7.7625E-13 2.5875E+01

(see column F-value).

Table Validation);* The Lethality factors, calculated by F-SPOIL.xls for each product temperature (see column Lethality factor),

These Lethality factors are exactly equal to the Lethality factors, reported by Kooiman at the same

of F and of Lethalities by the method of "equal time intervals" of Patashnik, as reported by Kooiman.

Theory behind the Lethality factors:

Thus: how to calculate the Lethality factors to find the F-value at reference temperature Tref = 121.1 °C when z = 10 °C?

... at a particular z value (for example z = 10 °C).

[oC] oC)

3b) Findings regarding the validation of F-SPOIL:* The F0-value = F10121.1 of the heating process, calculated by F-SPOIL.xls, is 2.5875E+01 =

This F0-value value is exactly equal to the F0-value = F10121.1 = 25.875 min., calculated by Kooiman (see

are 1.3490E-02 = 0.01349 ≈ 0.013 min. at T = 102.4 oC; 1.3804E-1 = 0.138 min. at 112.5 2.1380E-01 = 0.21380 ≈ 0.214 min. at 114.4 oC; 2.8840E-01 = 0.28840 ≈ 0.288 min. at 115.7 oC

product temperatures (see Table Validation): 0.013 min. at T = 102.4 0.214 min. at T = 114.4 oC; 0.288 min. at T = 115.7 oC; etc.

CONCLUSION: The computer calculations by F-SPOIL.xls are in complete agreement with the manual calculations

How to calculate the Lethality factors for each product temperature, if you want to find the F

The “Lethality factor” is a “conversion factor”. It converts ........... 1 minute of product heating at product temperature Tp (for example at Tp = 102.4 °C),... to an equivalent number of minutes of product heating at a reference temperature Tref (f.e. Tref = 121.1

The conversion equation is: Lethality factor = 10(Tp - Tref)/z. This equation is used in column "Lethality factor".

Example calculation: The Lethality factor (for 1 minute product heating) at product temperature Tp = 102.4 if Tref = 121.1 °C, and if z = 10 °C, is equal to 10(Tp - Tref)/z = 10(102.4 - 121.1)/10 = 0.01349 min.

column Lethality factor).

F - SPOILWorked example 4A:

which showed the following 17 time-temperature measurements, recorded at time intervals of 3 minutes:

Answer Example 4A:

The resulting spoilage rate b1 of Answer 4A thus will be too large.

yellow cell = input valuesAll cells in this example are protected, also the input cells!

Fill up the yellow cells in the F section and the Micro-organism of concern section:interval = 3 min. Constant interval between temperature measurements [in min.]Tref = 121.1z = 10



In the yellow "Input: Core temperature T" cells: type the food temperatures from the time-temperature table:

This is exactly equal to the Lethality factor, calculated by F-SPOIL, at product temperature Tp = 102.4

F-value and Spoilage Rate b of Solid Foods: consider the total number of micro-organisms in the can

4a) Calculate the F0-value (= F at reference temperature Tref = 121.1 oC and z = 10 oC) of a heating process of a solid food

4b) Also calculate the spoilage rate per can by a spore-forming spoilage micro-organism with D = 0.5 min at 122.3 z = 11 oC; initial number of spores of the micro-organism of concern = 1 per ml; volume of solid in can = 450 ml.


The product to be heated is a SOLID. So the spoilage rate should be calculated twice!In Answer 4A we will use the total initial number of micro-organisms in the can.





[min.] 121.1 [min.]0 19.11 6.3241E-11 1.8972E-103 18.99 6.1518E-11 3.7428E-106 22.44 1.3614E-10 7.8271E-109 43.88 1.8967E-08 5.7684E-0812 74.58 2.2284E-05 6.6911E-0515 92.68 1.4388E-03 4.3833E-0318 104.46 2.1677E-02 6.9414E-0221 112.68 1.4388E-01 5.0105E-0124 115.84 2.9785E-01 1.3946E+0027 117.23 4.1020E-01 2.6252E+0030 115.51 2.7606E-01 3.4534E+0033 78.34 5.2966E-05 3.4536E+0036 65.35 2.6607E-06 3.4536E+0039 50.4 8.5114E-08 3.4536E+0042 44.44 2.1577E-08 3.4536E+0045 38.9 6.0256E-09 3.4536E+0048 34.72 2.3014E-09 3.4536E+0051 7.7625E-13 3.4536E+00

the generally accepted spoilage rate.

F - SPOILWorked example 4B:


[oC] oC)

Summary findings of Worked Example 4A (solid food; total number of m.o. in can; F-value and spoilage rate):4a) In its coldest point the solid food received a F0 = F10121.1 = 3.4536E+00 = This F0-value is higher that the "Botulinum Cook" of F0 = 3.0 min. So after heating,

4b) The spoilage rate b1 is 4.1634E-04 = about 4 surviving spoilage spore in a total of 10 000 cans.The generally accepted spoilage rate is 1 can per 100 000 cans (1 : 105 cans) by a F-SPOIL however calculated: per 100 000 cans over 40 cans may spoil. This spoilage rate is over 40x higher than

Conclusion 4A: Sterilization process 4A results in a safe food (F0 > 3.0 min., so no health risk), but with an unacceptable high spoilage rate of the spore-forming spoilage micro-organism (shelf life problem!).

F-value and Spoilage Rate b of Solid Foods: consider the number of micro-organisms per 1 gram of food

4c) Calculate the F0-value (= F at reference temperature Tref = 121.1 oC and z = 10 oC) of a heating process of a solid food

Answer Example 4B:The product to be heated is a SOLID. So the Spoilage rate should be calculated twice!

The resulting spoilage rate b2 of Answer 4B thus will be too small.

yellow cell = input valuesAll cells in this example are protected, also the input cells!

Fill up the yellow cells in the F section and the Micro-organism of concern section:interval = 3 min. Constant interval between temperature measurements [in min.]Tref = 121.1z = 10



In this example 4B of the solid food, the initial number of micro-organisms is taken per 1 gram (or per 1 ml).

In the yellow "Input: Core temperature T" cells: type the food temperatures from the time-temperature table:Input: Calculation results:


(= min. at Tref. of[min.] 121.1 [min.]

0 19.11 6.3241E-11 1.8972E-103 18.99 6.1518E-11 3.7428E-106 22.44 1.3614E-10 7.8271E-109 43.88 1.8967E-08 5.7684E-0812 74.58 2.2284E-05 6.6911E-0515 92.68 1.4388E-03 4.3833E-0318 104.46 2.1677E-02 6.9414E-0221 112.68 1.4388E-01 5.0105E-0124 115.84 2.9785E-01 1.3946E+00

4d) Also calculate the spoilage rate per can by a spore-forming spoilage micro-organism with D = 0.5 min at 122.3 z = 11 oC; initial number of spores of the micro-organism of concern = 1 per ml; volume of solid in can = 450 ml.


In this Answer 4B we will use the initial number of micro-organisms in 1 gram of the food in the can.



[oC] oC)

27 117.23 4.1020E-01 2.6252E+0030 115.51 2.7606E-01 3.4534E+0033 78.34 5.2966E-05 3.4536E+0036 65.35 2.6607E-06 3.4536E+0039 50.4 8.5114E-08 3.4536E+0042 44.44 2.1577E-08 3.4536E+0045 38.9 6.0256E-09 3.4536E+0048 34.72 2.3014E-09 3.4536E+0051 7.7625E-13 3.4536E+00

F-value and spoilage rate):

and an acceptable low spoilage rate of the spore forming spoiling micro-organism (no shelf life problems).

General conclusion about this sterilization process for this solid food:

there was more heating than required for the "Botulism Cook".

So the heating process may be insufficient for a proper food shelf life.

Summary findings of Worked Example 4B (solid food; number of m.o. in 1 gram of the canned food;

4c) In its coldest point, the solid food received an F0 = F10121.1 = 3.4536E+00 = This F0-value is higher that the "Botulinum Cook" of F0 = 3.0 min. So after heating,

4d) The spoilage rate b2 = 9.2520E-07, or more than 9 surviving spoilage spores in a total of 10 000 000 cans;which is almost equal to 10 spores per 10 000 000 cans, or 1 spore per 1 000 000 cans (1 spore per 10The generally accepted spoilage rate is 1 can per 100 000 cans (1 : 105 cans), by a So this sterilization process sufficiently destructs of the spore-forming spoilage micro-organisms of concern

Conclusion Example 4B: Sterilization process 4B results in a safe food (F0 > 3.0 min., so no health risk),

1) The food is safe: In its coldest spot the food received a F0 > 3.0 min. so

2) The spoilage rate b by the sporeforming spoilage micro-organism will be in the range of4 • 10-4 < b < 1 • 10-6. The generally accepted spoilage rate by spoilage micro-organisms, however, is 1

Version 2.5 (September 25, 2013)CALCULATION OF F-VALUE, MICROBIAL SPOILAGE RATE, AND NUTRIENT RETENTION FROM TIME-TEMPERATURE DATA, BY ADDITION OF LETHALITIES

[email protected]

blue cell = calculation results; equations

Constant interval between temperature measurements [in min.]

D-value of micro-organism of concern [in min.]

Initial total number of micro-organisms of concern in can

Calculation results:(F for Spoilage rate bmicro- after heating

organism) [= number of surviving m.o. per can]

#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502

validation: see from row 205 onward.

, please calculate the spoilage rate twice: initial number of micro-organisms in the can; the resulting number of spoiled cans b1 will be too high;

; the resulting number of spoiled cans b2 will be too low. of the SOLID product then will be: b2 < b < b1.

Reference temperature for F-value [in oC]z-value for F-value [in oC]

Reference temperature for D of micro-organism [in oC]

z-value for D-value of micro-organism [in oC]

#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502

#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502


blue cell = calculation results; equationsAll cells in this example are protected, also the input cells!



Initial total number of micro-organisms of concern in canWhen the food is a liquid, always type the TOTAL initial number of micro-organisms of concern in the whole can.

In the yellow cells "Input: Core temperature T": type the food temperatures from the pumpkin soup time-temperature table:Calculation results:

(F for Spoilage rate bmicro- after heating

organism) [= number of

Spoilage Rate b of Liquid FoodsC and z = 10 oC) of a heating process of liquid

by a spore-forming spoilage micro-organism with D = 0.9 min at 122.3 oC and C; initial number of spores of the micro-organism of concern = 2 per ml; volume of liquid in can = 900 ml.




surviving m.o. per can]1.62566103E-09 1.8000E+031.3983402E-08 1.8000E+03

0.001383675925 1.7936E+030.288661910303 8.6008E+021.342680274045 5.7997E+012.689197190406 1.8504E+004.016127693019 6.2074E-025.323756685547 2.1877E-036.650687188159 7.3387E-058.00002566551 2.3246E-06

9.313140546184 8.0786E-089.347823885501 7.3927E-089.347887396388 7.3915E-089.347889224232 7.3914E-089.347889481882 7.3914E-089.347889481897 7.3914E-08

The heating process in the table above results in a spoilage rate of about 7 cans per 100 000 000 cans, or 0.007 cans per 100 000 cans; 0.007 spoiled cans per 100 000 cans is less than 1 can per 100 000 cans.


(liquid food; calculate F-value and spoilage rate):-value (= F at reference temperature Tref = 121.1 oC and z = 10 oC) of the heating proces

, because each can received more than the "Botulinum Cook" of F0 = F10121.1 = 3.0 min.

cans (see column Spoilage rate b). some 7 spores will survive. These 7 cans may spoil.

micro-organism (so NOT for a pathogenic micro-organism), the generally accepted

: This heating process gives a generally accepted result:

* the sterilized food has an acceptable low spoilage rate (< 1 can per 105 cans), so the food is shelf stable.

Nutrient Retention of Liquid FoodsC and z = 10 oC) of a heating process of liquid





In the yellow "Input: Core temperature T" cells: type the food temperatures from the pumpkin soup time-temperature table:



0.001027553414 9.9998E+010.003205599338 9.9994E+010.164678892561 9.9705E+011.329743101915 9.7642E+013.215304897287 9.4393E+015.279899128246 9.0959E+017.333319220174 8.7668E+019.375625650516 8.4513E+0111.42904574244 8.1455E+0113.49524123508 7.8490E+0115.54071684526 7.5661E+0116.0731713687 7.4941E+01

16.12475094937 7.4872E+0116.13861118892 7.4853E+0116.14531959311 7.4844E+0116.14550182566 7.4844E+01

For (destruction of) vitamin B1: D = 128.3 min at 120 oC and 1997; vitamin B1 in milk at food temperatures T = 35 oC - 150 oC.]

: fill up the yellow cells in the F section as in Worked example 1. : type input figures in section "Micro-organism of concern": D = D of nutrient;

; initial number a = 100 (initial % of nutrient = 100).




(liquid food; F-value and nutrient retention):-value (= F at reference temperature Tref = 121.1 oC and z = 10 oC) of the heating proces

, because each can received more than the "Botulinum Cook" of F0 = F10121.1 = 3.0 min.

, thus how much vitamin B1 is retained, = 7.4844E+01 ≈ 75% (column "spoilage rate b").

sterilization process. He used experimental time-product temperature results, and made hiscalculations according to the "equal time interval"-method of Patashnik, applying lethality tables.

All cells in this example are protected, also the input cells!

Tref = reference temperature of D of nutrient; z = z of nutrient; initial number = 100 (initial % of nutrient = 100).Constant interval between temperature measurements [in min.]



In the yellow "Input: Core temperature T" cells: type the food temperatures from the Validation time-temperature table:

Calculation results:

some 25% was destructed during the heating process.

of Lethality Factors and of F-value

-value, received by a can of soup during the heating stage of a

(= F at reference temperature Tref = 121.1 oC C) of the heating process of Kooiman (see table above: 12 time-temperature measurements;

Compare and contrast the results of F-SPOIL.xls with Kooiman's lethality factors and Kooiman's F0.

: fill up the yellow cells in the F section: interval = 5 min.; Tref = 121.1 oC; z = 10 oC.






#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502#DIV/0! Err:502

* The Lethality factors, calculated by F-SPOIL.xls for each product temperature (see column Lethality factor),

These Lethality factors are exactly equal to the Lethality factors, reported by Kooiman at the same

of F and of Lethalities by the method of "equal time intervals" of Patashnik, as reported by Kooiman.

Thus: how to calculate the Lethality factors to find the F-value at reference temperature Tref = 121.1 °C when z = 10 °C?

of the heating process, calculated by F-SPOIL.xls, is 2.5875E+01 = 25.875 min. = F 0

10121.1 = 25.875 min., calculated by Kooiman (see

; 1.3804E-1 = 0.138 min. at 112.5 oC; ; 2.8840E-01 = 0.28840 ≈ 0.288 min. at 115.7 oC; etc.

0.013 min. at T = 102.4 oC; 0.138 min. at T = 112.5 oC;

The computer calculations by F-SPOIL.xls are in complete agreement with the manual calculations

How to calculate the Lethality factors for each product temperature, if you want to find the F10121.1 of a process?

of product heating at product temperature Tp (for example at Tp = 102.4 °C),... to an equivalent number of minutes of product heating at a reference temperature Tref (f.e. Tref = 121.1 °C),

This equation is used in column "Lethality factor".

: The Lethality factor (for 1 minute product heating) at product temperature Tp = 102.4 °C,






In the yellow "Input: Core temperature T" cells: type the food temperatures from the time-temperature table:

This is exactly equal to the Lethality factor, calculated by F-SPOIL, at product temperature Tp = 102.4 oC (see

Spoilage Rate b of Solid Foods: number of micro-organisms in the can

C and z = 10 oC) of a heating process of a solid food

by a spore-forming spoilage micro-organism with D = 0.5 min at 122.3 oC and C; initial number of spores of the micro-organism of concern = 1 per ml; volume of solid in can = 450 ml.








1.24799304E-09 4.5000E+022.46502806E-09 4.5000E+024.97078785E-09 4.5000E+022.27829874E-07 4.5000E+020.000137929573 4.4971E+020.00622479895 4.3728E+02

0.077889133563 3.1437E+020.478364493302 4.9715E+011.254340918879 1.3949E+002.292376317329 1.1707E-023.016559653815 4.1696E-043.016862176272 4.1638E-043.016882121937 4.1634E-043.016882994421 4.1634E-043.016883244997 4.1634E-043.016883323576 4.1634E-043.016883356333 4.1634E-043.016883356356 4.1634E-04


(solid food; total number of m.o. in can; F-value and spoilage rate): = 3.4536E+00 = 3.45 min. = F 10 121.1 (see column F-value).

= 3.0 min. So after heating, the food product is SAFE.

4 surviving spoilage spore in a total of 10 000 cans.The generally accepted spoilage rate is 1 can per 100 000 cans (1 : 105 cans) by a spoilage micro-organism.

may spoil. This spoilage rate is over 40x higher than

> 3.0 min., so no health risk), of the spore-forming spoilage micro-organism (shelf life problem!).

Spoilage Rate b of Solid Foods: consider the number of micro-organisms per 1 gram of food in the can

C and z = 10 oC) of a heating process of a solid food




Initial total number of micro-organisms of concern in canIn this example 4B of the solid food, the initial number of micro-organisms is taken per 1 gram (or per 1 ml).

In the yellow "Input: Core temperature T" cells: type the food temperatures from the time-temperature table:Calculation results:



1.24799304E-09 1.0000E+002.46502806E-09 1.0000E+004.97078785E-09 1.0000E+002.27829874E-07 1.0000E+000.000137929573 9.9937E-010.00622479895 9.7174E-01

0.077889133563 6.9859E-010.478364493302 1.1048E-011.254340918879 3.0997E-03

by a spore-forming spoilage micro-organism with D = 0.5 min at 122.3 oC and C; initial number of spores of the micro-organism of concern = 1 per ml; volume of solid in can = 450 ml.



initial number of micro-organisms in 1 gram of the food in the can.




2.292376317329 2.6016E-053.016559653815 9.2658E-073.016862176272 9.2529E-073.016882121937 9.2520E-073.016882994421 9.2520E-073.016883244997 9.2520E-073.016883323576 9.2520E-073.016883356333 9.2520E-073.016883356356 9.2520E-07

F-value and spoilage rate):

and an acceptable low spoilage rate of the spore forming spoiling micro-organism (no shelf life problems).

General conclusion about this sterilization process for this solid food:

So the heating process may be insufficient for a proper food shelf life.

(solid food; number of m.o. in 1 gram of the canned food;

= 3.4536E+00 = 3.45 min. = F 10 121.1 (see column F-value). = 3.0 min. So after heating, the product is SAFE.

than 9 surviving spoilage spores in a total of 10 000 000 cans;which is almost equal to 10 spores per 10 000 000 cans, or 1 spore per 1 000 000 cans (1 spore per 106 cans).The generally accepted spoilage rate is 1 can per 100 000 cans (1 : 105 cans), by a spoilage micro-organism.

sufficiently destructs of the spore-forming spoilage micro-organisms of concern.

Sterilization process 4B results in a safe food (F0 > 3.0 min., so no health risk),

The food is safe: In its coldest spot the food received a F0 > 3.0 min. so

The spoilage rate b by the sporeforming spoilage micro-organism will be in the range of

micro-organisms, however, is 1•105.

F SPOIL.xls: Calculation of F value, Microbial Spoilage Rate, and Nutrient Retention, from Time...

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