5D Process Time and Spoilage Under a Mis-Stated z-Value: 22-Agric-B8 Food Process Engineering (Part 1), December 2013, Q3

Question 3: 5D Process Time and Spoilage Under a Mis-Stated z-Value (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Process design vs. actual organism resistance
Quantity	Value
\(D_{121}\)	3 min (fixed property, both parts)
Design z-value	10°C (18°F)
Actual z-value	20°F = 11.11°C
Process temperature	138°C (280°F)
Target reduction (design)	5D
Initial inoculum, \(N_0\)	100 spores/can

Find. (a) The heating time for a designed 5D process; (b) the actual probability of spoilage once that fixed time is delivered to an organism whose true z is 20°F rather than 18°F.

Approach. Compute \(D_{138}\) under the design z, size the 5D time from it, then hold that same physical time and recompute the actual log-reduction with the organism's true (larger) z-value, which changes \(D_{138}\).

5D process time (design z = 10°C). \(D_{138} = D_{121}\,10^{(121-138)/10} = 3\times10^{-1.7} = \boxed{0.0599\ \text{min}}\). \(t_{5D} = 5D_{138} = 5\times0.0599 = \boxed{0.2993\ \text{min}}\) (\(\approx\)18.0 s).
Actual decimal reduction time at 138°C with the true z. \(z=20^\circ \text{F}=11.11^\circ\text{C}\): \(D_{138,actual} = 3\times10^{(121-138)/11.11} = 3\times10^{-1.530} = \boxed{0.0885\ \text{min}}\). A larger z means the organism's resistance falls off more slowly with temperature, so at 138°C it is actually harder to kill than the design assumed (\(D_{138,actual} > D_{138,design}\)).
Actual spoilage probability. Using the fixed process time \(t_{5D}=0.2993\) min against the higher \(D_{138,actual}\): log-reduction \(= 0.2993/0.0885 = 3.380\), so \(N = 100\times10^{-3.380} = \boxed{0.0417\ \text{spores/can}}\).

Final results
Quantity	Value
Designed 5D heating time	0.299 min (18.0 s)
\(D_{138}\), actual z	0.0885 min
Actual probability of spoilage	\(\approx 0.0417\) (4.17%)

Check: even though the true z (20°F) is less conservative in the sense that the exam might expect "wrong z = worse outcome," the arithmetic result here is that the fixed 5D time still delivers noticeably fewer than 5 decimal reductions (\(3.38D\) actual vs. \(5D\) intended) — the process is under-lethal relative to design, consistent with the general lesson of this section that mis-estimating z has first-order consequences for delivered sterility.

Concept & Theory

Topic: Sensitivity of thermal-process lethality to the z-value.

Key relations: \(D_T=D_{ref}\,10^{(T_{ref}-T)/z}\); \(N=N_0\,10^{-t/D_T}\).

Why this works: the 5D target is only achieved if the process time is sized against the correct \(D_T\); since \(D_T\) itself is exponential in \(1/z\), a modest error in the assumed z-value (18 vs. 20°F here) produces a non-trivial shift in \(D_{138}\) and therefore in the delivered log-reduction, even though the process time and \(D_{121}\) are unchanged.

Common pitfall: assuming a larger z automatically means "safer" or "less safe" without recomputing \(D_T\) at the actual process temperature — the direction of the effect depends on whether \(T \gt T_{ref}\) or \(T \lt T_{ref}\).

Source: Toledo, Fundamentals of Food Process Engineering, Ch. 8.

Question 3 of 10: 5D Process Time and Spoilage Under a Mis-Stated z-Value

Question 3: 5D Process Time and Spoilage Under a Mis-Stated z-Value (15 marks)