Food Biochemistry and Food Processing (2 edition)

BLBS102-c38 BLBS102-Simpson March 21, 2012 14:17 Trim: 276mm X 219mm Printer Name: Yet to Come

738 Part 7: Food Processing

Table 38.7.Procedure in Calculation of Process Time Using Ball Method for Conductive

Heating Food

Process Time Calculation

1 jh 1.41

2 fh 60 min

3 Retort temperature (Tr) 245 ◦F

4z(◦F) value 18 ◦F

5 Lethality(Fo)5min

6 CUT 8min

7 Initial product temperature (Ti) 170 ◦F

8Ih=Tr−Ti 75 ◦F= 245 − 170

9jh∗Ih 105.75=1.41∗ 75

10 Log(jh∗Ih) 2.024

11 Fi= 10

( 250 −Tr

z

)

1.9= 10 (

250 − 18245 )

12 fh/U=fh/(Fo∗Fi)fh/U=60/(1.9∗5)=6.32

13 Logg From Ball Table 38. 6 by interpolation=0.821

14 Log(jh.Ih)−logg 1.203=2.024−0.821

15 B=fh[log(jh∗Ih)−log g)] 72.18 min= 60 ∗1.203

16 Pt=B−42%CUT 68.82 min=72.18−0.42∗ 8

conservative assumption from safety point of view but

from quality point of view it has an over processing ef-

fect. In practice, most of the heating is done with steam or

agitated water, while cooling is normally done in slowly

circulating water. Therefore, generallyfh<fcandfh=fc

gives a more conservative processing conditions.

3. The temperature difference between the retort temper-
   ature and cooling water temperature (Tr−Tcw)isthe
   driving force. This is expressed in terms ofIc, which is
   (Ic=Tic−Tcw). Ball assumed that the sum ofIc+g=
   Tr−Tcw= 180 ◦F (for steam) or 130◦F (for full

immersion water overpressure). Stumbo (1973) discov-

ered that, a 10◦F change inTr−Tcwcan result in 1% dif-

ference in calculated lethality (F 0 ) and can be corrected if

required.

4. CUT correction forjhvalue. Ball noticed that only about
   42% of the CUT can be considered as effective contribu-
   tion to the process time.


5. No further product heating after cooling starts. However,
   this is not always valid for the critical point of conduc-
   tion in heated canned foods. In such cases, considerable
   overprocessing can occur.

Table 38.8.Calculation of Process Lethality Using Ball Method From Hypothetical Given Data

Process Time Calculation

1 jh 2.05

2 fh 34.9 min

3 CUT 8min

4Pt 45 min

Process time B=Pt+42%CUT

5(B=Pt+42%CUT) B= 45 +0.42∗ 8 =48.36min

6 Retort Temperature (Tr) 260 ◦F

7 Initial temperature(Ti) 100 ◦F

8Ich=Tr−Ti 160 ◦F= 260 − 100

9jh∗Ich 328 ◦F=2.05∗ 160

10 log (jh∗Ih) 2.52

11 zvalue 18 ◦F

12 Fi= 10 (

250 −zTr)

0.278

13 B/fh 1.39=48.36/34.9

14 logg=log(jh∗Ih)−B/fh logg=1.13=2.52−1.39

15 fh/U From Ball Table 38.6 by interpolation=22.4

16 Fo=(fh/Ufh∗Fi) Fo=34.9/(22.4∗0.278)=5.60 min