22-Agric-B8 Food Process Engineering (Part 1) · May 2017
Question 6 of 10: F-value for a target spoilage rate, and the Ball formula process time
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-Agric-B8, Food Process Engineering (Part 1) — National Exams, May 2017. 3 hours duration, closed book (one aid sheet permitted). Ten questions are printed, grouped into four sections (I–IV); a complete exam paper requires six. All ten are solved below as a complete study set.
Reference texts: Toledo, R.T., Fundamentals of Food Process Engineering (this exam's own cited source for its Stumbo g-table and steam-table appendix); Geankoplis, C.J., Transport Processes and Separation Process Principles (evaporator design and steam economy); Incropera, F.P. & DeWitt, D.P., Fundamentals of Heat and Mass Transfer (unsteady-state Heisler-chart conduction, composite cylindrical walls); Singh, R.P. & Heldman, D.R., Introduction to Food Engineering, and Cleland, A.C., Food Refrigeration Processes (freezing-time prediction, modified Plank equation).
Question 6: F-value for a target spoilage rate, and the Ball formula process time (15 marks)
Given (a). Initial inoculum N₀=10 spores/can, D₀(at 250°F reference)=1.2 min, target spoilage 1 can in 100,000, z=18°F.
Find (a). The reference F-value (at 250°F) for the target inactivation, and the equivalent F at 280°F.
Approach (a). The survivor equation F=D·log₁₀(N₀/N᷇) gives the reference process value; converting F between temperatures uses the same z-value TDT relation as D.
F needed for the target spoilage rate. A spoilage rate of 1 can in 100 000 means the expected surviving-spore count per can must fall to N᷇=1/100 000=10⁻⁵.
$$F = D_0\log_{10}\!\left(\frac{N_0}{N_f}\right)=1.2\log_{10}\!\left(\frac{10}{10^{-5}}\right)=1.2\log_{10}(10^6)=1.2(6)=\boxed{7.2\ \text{min (at 250}^\circ\text{F)}}$$
Convert to F at 280°F. The same z-based scaling that relates D at different temperatures applies to F (both are lethality-equivalent times): $$F_{280}=F_{250}\times 10^{(250-280)/z}=7.2\times10^{-30/18}=7.2\times0.02154=\boxed{0.155\ \text{min}}$$ A much shorter time suffices at the higher temperature — consistent with the steep lethality-vs-temperature relationship z=18°F implies.
Final results (a)
Quantity
Value
F at 250°F
7.20 min
F at 280°F
0.155 min
Given (b). fℎ=55.6 min, jℎ=jᵏ=1.2, retort temperature 240°F, z=18°F, target F₀=8 min, initial product temperature 120°F, g-vs-fℎ/U table on page 8 (z=18 column).
Find (b). The process (come-up-corrected heating) time B at 240°F retort temperature.
Approach (b). Ball's formula method: convert the target F₀ (defined at the 250°F reference) into an equivalent time U at the actual (lower) retort temperature, use fℎ/U with the page-8 table to read the "unaccomplished-lethality" factor g, then solve the heating-curve equation for the process time B.
Convert F₀ to the retort's own temperature basis, U. $$U=F_0\times10^{(250-T_{retort})/z}=8\times10^{(250-240)/18}=8\times10^{0.5556}=8\times3.594=\boxed{28.75\ \text{min}}$$
Look up g from the page-8 table. $$\frac{f_h}{U}=\frac{55.6}{28.75}=1.934$$ Interpolating the z=18 column between fℎ/U=1.0 (g=0.523, Δg/Δj=0.192) and fℎ/U=2.0 (g=1.93, Δg/Δj=0.68) at fℎ/U=1.934 gives g(j=1)=1.837, Δg/Δj=0.648. Table 9.12's own footnote formula adjusts from j=1 to the actual lag factor jᵏ=1.2:
$$g_{j=1.2}=g_{j=1}+(j-1)\left(\frac{\Delta g}{\Delta j}\right)=1.837+(0.2)(0.648)=\boxed{1.967\ ^\circ\text{F}}$$
Solve Ball's heating-curve equation for the process time B. $$B=f_h\log_{10}\!\left[\frac{j_c(T_{retort}-T_i)}{g}\right]=55.6\log_{10}\!\left[\frac{1.2(240-120)}{1.967}\right]=55.6\log_{10}(73.2)$$
$$=55.6(1.8645)=\boxed{103.7\ \text{min}}$$
