Aseptic Holding-Tube Spoilage Probability: 22-Agric-B8 Food Process Engineering (Part 1), December 2013, Q1

Question 1: Aseptic Holding-Tube Spoilage Probability (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.

Holding-tube and thermal-resistance data
Quantity	Symbol	Value
Tube length	\(L\)	28.6 m
Pipe inside diameter	\(d\)	3.48 cm = 0.0348 m
Volumetric flow rate	\(Q\)	144.1 L/min
Density	\(\rho\)	1042 kg/m³
Viscosity	\(\mu\)	100 cP = 0.100 Pa·s
Process temperature	\(T\)	138°C
z-value	\(z\)	11°C
Reference decimal reduction time	\(D_{121}\)	1.2 min
Initial spore load	\(N_0\)	100 spores/can

Find. The probability of spoilage (surviving-spore fraction) delivered by this holding tube.

Approach. Confirm the flow regime, take the residence time of the fastest-moving fluid element (not the mean) as the process time since that particle sees the least lethal treatment, convert \(D_{121}\) to \(D_{138}\) via the z-value, and apply the survivor-ratio (Bigelow) equation.

Mean velocity and Reynolds number. \(A = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}(0.0348\ \text{m})^2 = 9.511\times10^{-4}\ \text{m}^2\). \(\bar v = Q/A = \dfrac{144.1\times10^{-3}/60\ \text{m}^3/\text{s}}{9.511\times10^{-4}\ \text{m}^2} = 2.525\ \text{m/s}\). \(Re = \dfrac{\rho \bar v d}{\mu} = \dfrac{1042 \times 2.525 \times 0.0348}{0.100} \approx 916\) — below 2100, so the flow is laminar.
Fastest-particle residence time. For laminar Newtonian flow the parabolic profile gives a centreline (maximum) velocity of \(v_{max} = 2\bar v = 5.050\ \text{m/s}\). Aseptic-process design uses the minimum residence time, since that is the time seen by the least-treated element: \(t_{min} = L/v_{max} = 28.6/5.050 = 5.663\ \text{s} = \boxed{0.0944\ \text{min}}\).
Decimal reduction time at process temperature. The z-value relation \(D_T = D_{121}\, 10^{(121-T)/z}\) gives \(D_{138} = 1.2 \times 10^{(121-138)/11} = 1.2 \times 10^{-1.545} = \boxed{0.0342\ \text{min}}\).
Survivor ratio and spoilage probability. The Bigelow lethality equation \(\log_{10}(N_0/N) = t/D_T\) gives a log-reduction of \(t_{min}/D_{138} = 0.0944/0.0342 = 2.762\), so \(N = N_0\, 10^{-2.762} = 100\times1.731\times10^{-3} = \boxed{0.173\ \text{spores/can}}\). Because \(N<1\), this value is read directly as the fraction of cans expected to contain a surviving spore.

Final results
Quantity	Value
Flow regime	Laminar, \(Re \approx 916\)
Fastest-particle residence time, \(t_{min}\)	5.66 s (0.0944 min)
\(D_{138}\)	0.0342 min
Probability of spoilage	\(\approx 0.173\) (17.3%)

Check: this holding tube / flow-rate combination does not achieve commercial sterility (spoilage risk far above the industry target of < \(10^{-9}\)); the numbers are taken exactly as given rather than adjusted, since the exam's evident intent is to test whether the candidate recognises an under-designed process.

Concept & Theory

Topic: Continuous (aseptic/HTST) thermal-process lethality and residence-time distribution.

Key relations: \(D_T = D_{ref}\,10^{(T_{ref}-T)/z}\) (thermal death-time z-relation); \(\log_{10}(N_0/N) = t/D_T\) (Bigelow survivor curve); \(v_{max}=2\bar v\) for laminar Newtonian pipe flow.

Why this works: in a continuous holding tube every fluid element must clear the tube before it is considered "processed," so the shortest possible residence time (the centreline streamline in laminar flow) sets the worst-case, and therefore design-controlling, lethality — using the mean residence time would overstate the lethality delivered and under-protect the product.

Common pitfall: using the average velocity (or, for turbulent flow, ignoring the flatter but still non-uniform velocity profile) to compute holding time, which silently credits the process with lethality the slowest-treated particle never received.

Source: Toledo, Fundamentals of Food Process Engineering, Ch. 8 (thermal process calculations, continuous systems).

Question 1 of 10: Aseptic Holding-Tube Spoilage Probability

Question 1: Aseptic Holding-Tube Spoilage Probability (15 marks)