HTST Batch Sterilization — Spore Kill vs. Heat-Labile Vitamin Destruction: 23-Chem-B4 Biochemical Engineering, May 2016, Q2

Question 2: HTST Batch Sterilization — Spore Kill vs. Heat-Labile Vitamin Destruction (20 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.

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As printed in the paper: the bacterial SPORES carry the LOWER activation energy (E_od=10 kcal/g-mol) and the heat-labile VITAMIN carries the HIGHER one (E_od=65 kcal/g-mol). This is the reverse of the usual HTST design premise (spore E_od ≫ nutrient E_od, which is what lets a brief high-temperature pulse kill spores while sparing the nutrient); the numbers are used exactly as given, and the physical consequence is worked out in Step 4 below.

Given.

Quantity	Symbol	Value
Initial spore count	N₀	10⁹/L
Spore activation energy / pre-exponential	E_od,spore, α_spore	10 kcal/g-mol, 1×10⁴ min^-1
Vitamin activation energy / pre-exponential	E_od,vit, α_vit	65 kcal/g-mol, 1×10³⁶ min^-1
Initial vitamin concentration	C_0,vit	30 mg/L
Acceptable failure probability	p	0.0001 (1 spore / 10,000 L)
Sterilization temperature	T	120°C (393.15 K)

Find. The holding time required to meet the spore-kill design criterion, and the fraction (and mass) of vitamin destroyed over that same hold.

Approach. Evaluate both first-order death-rate constants from the Arrhenius equation at 120°C; convert the stated failure probability into the standard sterilization design (∇, "del") factor — a volume-independent log-ratio of spore counts; use it with the SPORE'S kd to get the required isothermal holding time; then apply the VITAMIN'S kd over that same time to find how much of the vitamin survives.

Evaluate the spore and vitamin death-rate constants at 120°C. With R=1.987 cal/(mol·K) and T=393.15 K (Ea values converted kcal→cal): $$k_{d,spore}=(1\times10^4)\exp\!\left(\frac{-10{,}000}{1.987\times393.15}\right)=\boxed{0.0276\ \text{min}^{-1}}$$ $$k_{d,vit}=(1\times10^{36})\exp\!\left(\frac{-65{,}000}{1.987\times393.15}\right)=\boxed{0.731\ \text{min}^{-1}}$$ Expressed as decimal reduction times $D=\ln10/k_d$: D_spore≈83.5 min but D_vit≈3.15 min — at this temperature the vitamin decays roughly 26× FASTER than the spores, the opposite of the usual HTST intent.
Design (∇) factor from the failure-probability criterion. A 0.0001 failure probability means an expected 1 surviving spore per 10,000 L; expressed per litre (so volume cancels), N_f=1/10,000=10^-4/L: $$\nabla=\ln\frac{N_0}{N_f}=\ln\frac{10^9}{10^{-4}}=\ln(10^{13})=\boxed{29.93}$$
Required isothermal holding time from the spore kinetics. First-order death gives $\nabla=k_{d,spore}\,t$: $$t=\frac{\nabla}{k_{d,spore}}=\frac{29.93}{0.0276}=\boxed{1085\ \text{min}}\ (=18.1\ \text{hr})$$
Vitamin destroyed over that same 1085-minute hold. Applying the vitamin's OWN kd over the spore-dictated hold time: $$k_{d,vit}\,t=(0.731)(1085)=793\qquad \frac{C}{C_0}=e^{-793}\approx0$$ This exponent corresponds to roughly 344 decimal reductions (793/ln10) — astronomically beyond "essentially zero." For all practical purposes, the entire 30 mg/L of vitamin is destroyed (fraction destroyed &boxed;≈1.00, i.e. ∼100%) well before the spore-kill criterion is even approached, since the vitamin's own decimal reduction time (3.15 min) is more than 300× shorter than the required hold.

Fig. 3 — semi-log spore survivor curve: the design point (∇=29.9) is reached at t≈1085 min, the isothermal hold this problem requires.

Quantity	Value
k_d,spore at 120°C	0.0276 min^-1 (D_spore≈83.5 min)
k_d,vit at 120°C	0.731 min^-1 (D_vit≈3.15 min)
Design factor, ∇	29.93
Required holding time, t	1085 min (≈18.1 hr)
Vitamin destroyed	≈100% (≈30 mg/L; residual undetectable, e^-793)

Topic: Batch thermal-death kinetics (Arrhenius k_d, decimal reduction time D, design/∇ factor) applied simultaneously to microbial sterilization and nutrient/vitamin degradation.

Key relations: $k_d=\alpha\exp(-E_{od}/RT)$; first-order death $N=N_0e^{-k_dt}$; design factor $\nabla=\ln(N_0/N_f)=k_dt$; decimal reduction time $D=\ln10/k_d$.

Why this works: HTST processing normally relies on the spore-inactivation reaction having a MUCH higher activation energy than the nutrient-degradation reaction, so a brief excursion to high temperature kills spores disproportionately faster (in relative, not absolute, terms) than it degrades the nutrient — letting a short high-T hold sterilize while sparing quality. With the Arrhenius parameters exactly as given here, that ordering is reversed: the vitamin degrades faster in absolute terms at 120°C, so whatever time the (here, low-E_od, comparatively slow) spore-kill criterion demands, the vitamin cannot survive it.

Common pitfall: Assuming HTST automatically spares a heat-labile component regardless of the specific kinetic parameters — the "short time" benefit only materializes when the labile component's own k_d is much smaller than the spore's over the required hold; and mixing R units (Ea must be converted to the SAME energy unit as R, here cal/mol, before dividing).

Source: Shuler & Kargi, Bioprocess Engineering, 2nd ed., Ch. 9 (batch sterilization kinetics, del factor, decimal reduction time D); Bailey & Ollis, Biochemical Engineering Fundamentals, 2nd ed., Ch. 9 (thermal death kinetics).

Question 2 of 5: HTST Batch Sterilization — Spore Kill vs. Heat-Labile Vitamin Destruction

Question 2: HTST Batch Sterilization — Spore Kill vs. Heat-Labile Vitamin Destruction (20 marks)