NivaarExam PrepOfficial exam papers ↗

22-Agric-B8 Food Process Engineering (Part 1) · May 2014

Question 7 of 10: Aseptic Holding-Tube Length and Spoilage Probability

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 04-Agric-B8 Food Process Engineering (Part 1), National Exams May 2014 — a three-hour open-book exam (any non-communicating calculator permitted). Ten questions are set in four sections (I–IV), each with a "do one/any N of M" instruction; a candidate following the choice rules answers six questions for a 100-mark paper. All ten are worked here so the set is a complete study resource.

Reference texts. R.T. Toledo, Fundamentals of Food Process Engineering, 3rd ed. (thermal-process lethality, D and z values, Ball/Stumbo process calculation, aseptic holding-tube residence time, evaporator design — this is the exam's own appendix source); C.J. Geankoplis, Transport Processes and Separation Process Principles, 4th ed. (evaporator heat and mass balances, multiple-effect steam economy); R.P. Singh and D.R. Heldman, Introduction to Food Engineering, 5th ed. (freezing-time estimation, modified Plank and Cleland-Earle equations, unsteady-state heat transfer in canned foods); A.C. Cleland, Food Refrigeration Processes: Analysis, Design and Simulation (Plank/Cleland-Earle freezing-time correlations); F.P. Incropera and D.P. DeWitt, Fundamentals of Heat and Mass Transfer (transient conduction, Heisler charts, composite-wall resistance).

Check: this paper's four roman-numeral section headers ("I. Heat transfer", "II. Food freezing and freeze concentration", "III. Thermal processing", "IV. Several assumptions (retort come-up correction factor, reference temperature for spore D-values, evaporator steam temperature reused for Question 9) are flagged inline where the source leaves a value implicit.

Question 7: Aseptic Holding-Tube Length and 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 data
QuantitySymbolValue
Tube inside diameter\(d\)3.48 cm = 0.0348 m
Volumetric flow rate\(Q\)113.6 L/min
Density\(\rho\)1042 kg/m³
Product 1 viscosity\(\mu_1\)10 cP
Product 2 viscosity\(\mu_2\)100 cP
Process temperature\(T\)280°F
Reference D-value at 250°F\(D_0\)1.2 min
z-value, tube sizing\(z_1\)50°F
z-value, spoilage check\(z_2\)20°F
Initial inoculum\(N_0\)100 spores/can

Find. The required holding-tube length for a 5D reduction of the first product, then the probability of spoilage of a second, more viscous product run through the SAME tube.

Product in vᵇᴸ (parabolic/turbulent core) fastest streamline, vₓ›ₓ (sets minimum hold time) Product out L = holding-tube length (solved for), D = 3.48 cm
Fig. C — holding tube: the fastest (centreline) streamline sets the minimum, worst-case residence time that must clear the required lethality.

Approach. \(D_0=1.2\) min is the reference decimal reduction time at the industry-standard 250°F, so it is z-shifted to the actual 280°F process temperature before the 5D target time is set. The mean velocity and Reynolds number fix the flow regime and hence the velocity ratio used to find the FASTEST streamline's residence time (the least-treated fluid element governs tube sizing); that length is then reused, unchanged, for the second (more viscous, now laminar) product to find its own residence time and resulting spoilage probability.

  1. Flow geometry and Reynolds number (product 1, 10 cP). \(A=\dfrac{\pi}{4}d^2=9.511\times10^{-4}\ \text{m}^2\); \(\bar v = Q/A = \dfrac{113.6\times10^{-3}/60}{9.511\times10^{-4}}=1.991\ \text{m/s}\); \(Re=\rho\bar v d/\mu = 1042(1.991)(0.0348)/0.010=\boxed{7218}\) — turbulent (\(Re>4000\)), so \(v_{max}\approx1.2\bar v=2.389\ \text{m/s}\).
  2. Target D-value and 5D hold time. \(D_{280}=D_0\,10^{(250-280)/50}=1.2\times10^{-0.6}=\boxed{0.3014\ \text{min}}\); \(5D=5(0.3014)=1.507\ \text{min}=\boxed{90.4\ \text{s}}\).
  3. Required tube length. The fastest streamline must still spend at least the 5D time in the tube: \(L=t_{5D}\,v_{max}=90.4\times2.389=\boxed{216\ \text{m}}\).
  4. Flow regime, product 2 (100 cP, same tube/flow rate). \(Re=1042(1.991)(0.0348)/0.100=\boxed{722}\) — laminar (\(Re<2100\)), so this time \(v_{max}=2\bar v=3.981\ \text{m/s}\) (parabolic profile).
  5. Residence time and log reduction, product 2. \(t_{min}=L/v_{max}=216.0/3.981=54.26\ \text{s}=0.9043\ \text{min}\). At \(z_2=20^\circ\text{F}\), \(D_{280}=1.2\times10^{-30/20}=\boxed{0.03795\ \text{min}}\), so the log-reduction is \(t_{min}/D_{280}=0.9043/0.03795=\boxed{23.83}\).
  6. Spoilage probability. \(N=N_0\,10^{-23.83}=100\times10^{-23.83}\approx\boxed{1.5\times10^{-22}}\) spores/can — effectively zero; the system is comfortably sterile for the second product too.
Final results
QuantityValue
Flow regime, product 1 (10 cP)Turbulent, \(Re=7218\)
Required tube length \(L\)216 m
Flow regime, product 2 (100 cP)Laminar, \(Re=722\)
Product-2 residence time54.3 s (0.904 min)
Product-2 spoilage probability\(\approx1.5\times10^{-22}\) (negligible)
Check: the source states "D0 = 1.2 min" without repeating its reference temperature; it is taken as the standard 250°F reference, which is what makes the z-shift to 280°F meaningful and gives a physically reasonable ~90 s target hold time. If D0 were instead already the D-value AT 280°F, the 5D time (and tube length) would scale up by roughly a factor of four.