Dairy and Food Unit Operations and Preservation: Blanching, Homogenization, Pasteurization, Sterilization, Microbial Death Kinetics, Cooling, Freezing and Refrigeration
2. Kinetics of microbial death: D, z and F values
At a constant lethal temperature, microorganisms die by first-order kinetics: the number of survivors falls by the same fraction in each equal time interval, log₁₀(N₀/N) = t/D. The decimal reduction time D is the time at that temperature to reduce the population tenfold (one log cycle): falling from 10⁶ to 10² in 8 min is four log cycles in 8 min, so D = 2 min. Because death is logarithmic, no finite process reaches zero survivors; a process is specified by the number of log reductions it achieves.
D falls as temperature rises. The z value is the temperature rise that reduces D tenfold: log₁₀(D₁/D₂) = (T₂ − T₁)/z. If D is 4 min at 115 °C and 0.5 min at 121 °C, log₁₀ 8 = 6/z and z = 6.64 °C. The lethal effect of a whole process is summarised by its F value, the equivalent time at a reference temperature: for sterilization F₀ refers to 121.1 °C with z = 10 °C, and a process at temperature T for time t contributes t × 10^((T − 121.1)/10) minutes. Holding at 115 °C for 20 min therefore gives F₀ = 20 × 10^(−0.61) = 4.91 min; for a varying temperature, F₀ = ∫10^((T − 121.1)/10) dt. The required F follows from the target reduction: F = D × (log N₀ − log N). The 12D concept for low-acid canned foods demands twelve log reductions of Clostridium botulinum spores; taking D₁₂₁ = 0.21 min for them, the minimum process is F₀ = 12 × 0.21 = 2.52 min.
3. Pasteurization and sterilization of milk and other liquid foods
Pasteurization destroys all pathogenic organisms and most spoilage organisms in milk, without sterilising it, so pasteurized milk must still be kept cold. The two standard milk processes are LTLT (low temperature, long time: 63 °C for 30 min, a batch or vat method) and HTST (high temperature, short time: 72 °C for 15 s, continuous). They were set to destroy the most heat-resistant non-spore-forming pathogen of milk, Coxiella burnetii. Adequate pasteurization is verified by the alkaline phosphatase test: the enzyme is slightly more heat-resistant than the target pathogens and is naturally present in raw milk, so its absence shows the milk reached the required time–temperature combination.
An HTST plant is a plate heat exchanger in sections: a balance tank feeds raw milk to the regeneration section, where it is warmed by the hot pasteurized milk flowing the other way (which is thereby cooled); then the heating section raises it to 72 °C with hot water; it passes through a holding tube sized for the holding time; a flow-diversion valve sends it back to the balance tank if its temperature is below the set point; and it returns through regeneration and a final cooling section with chilled water. The regeneration efficiency is the temperature rise gained in regeneration over the total rise: milk entering at 5 °C, leaving regeneration at 60 °C and pasteurized at 72 °C has (60 − 5)/(72 − 5) = 82 per cent regeneration. The holding tube must hold the fastest particle for the full time: in laminar flow the fastest travels at twice the mean velocity, so a tube carrying milk at a mean velocity of 0.5 m/s needs a length of 2 × 0.5 × 15 = 15 m for 15 s.
Sterilization destroys all viable organisms, including spores, to give a commercially sterile product — one in which no organism can grow under normal storage, though a vanishingly small number of dormant spores may survive. In-container sterilization heats the filled, sealed can or bottle in a retort (steam under pressure, about 121 °C), with the process time set by heat penetration to the can’s slowest-heating point. UHT (ultra-high temperature) processing heats the liquid alone to about 135–150 °C for a few seconds, by direct steam injection or infusion or in indirect exchangers, and then fills it aseptically into containers sterilised separately (for example with hydrogen peroxide), in a sterile zone; UHT milk keeps for months without refrigeration while suffering less browning and flavour change than retorted milk, because at high temperature microbial death accelerates much faster than the quality reactions (whose z values are larger).
4. Heat loads in pasteurization and the value of regeneration
The heat a pasteurizer must supply is an energy balance, Q = ṁc_pΔT. Heating 3000 kg/h of milk (c_p = 3.93 kJ/kg·K) from 5 °C all the way to 72 °C would take 3000 × 3.93 × 67/3600 = 219.4 kW. With 82 per cent regeneration the heating section supplies only the rise from 60 to 72 °C: 3000 × 3.93 × 12/3600 = 39.3 kW, and the cooling section’s load falls by the same proportion, since the incoming cold milk has already taken most of the heat out of the outgoing hot milk. This is why every continuous pasteurizer is built around its regeneration section.
5. Preservation by cooling and freezing; refrigeration and cold storage
Cooling slows microbial growth and enzymatic and respiratory activity; raw milk is chilled to about 4 °C soon after milking in bulk milk coolers, and fruits and vegetables are pre-cooled and held in cold storage at temperatures that avoid chilling injury. Freezing stops microbial growth altogether by removing liquid water as ice and lowering a_w. The freezing point of a food is depressed below 0 °C by its dissolved solids, and freezing proceeds over a range of temperatures. Slow freezing grows large ice crystals between cells that rupture tissue and cause drip loss on thawing; quick freezing — air-blast, plate (contact), fluidised-bed IQF (individually quick frozen) and cryogenic freezing in liquid nitrogen at −196 °C — forms many small crystals and preserves texture.
Refrigeration by the vapour-compression cycle circulates a refrigerant through four components: the evaporator, where it boils at low pressure and absorbs heat from the space (the refrigeration effect); the compressor, which raises its pressure and temperature; the condenser, where it rejects heat to air or water and liquefies; and the expansion valve, which drops its pressure back to the evaporator’s. The coefficient of performance is COP = refrigerating effect/work input: 150 kJ/kg extracted for 30 kJ/kg of compressor work is a COP of 5. Capacity is quoted in kW or in tons of refrigeration, 1 TR = 3.517 kW (211 kJ/min). Ammonia (R717) serves large cold stores and dairies for its efficiency (with toxicity managed by design); halocarbons and their low-global-warming replacements serve smaller plants. Vapour-absorption systems use heat instead of shaft work.
The refrigeration load of a cold store is the sum of the product load (sensible heat above freezing, latent heat of freezing, sensible heat below freezing, and for live produce the heat of respiration), the transmission load through walls, roof and floor, the infiltration (air-change) load when doors open, and internal loads from lights, fans, forklifts and people. Cooling 10 000 kg of apples (c = 3.6 kJ/kg·K) from 25 to 0 °C in 24 h is a product load of 10 000 × 3.6 × 25/86 400 = 10.4 kW before respiration. A bulk milk cooler bringing 2000 kg of milk (c_p = 3.93 kJ/kg·K) from 35 to 4 °C in 3 h removes 2000 × 3.93 × 31/10 800 = 22.6 kW, about 6.4 TR.
Key takeaways
- Blanching inactivates enzymes, checked by peroxidase; steam blanching leaches less than water blanching.
- log(N₀/N) = t/D; log(D₁/D₂) = ΔT/z; F₀ = t·10^((T − 121.1)/10) with z = 10 °C; F = D × log reductions (12D for botulinum).
- LTLT 63 °C/30 min, HTST 72 °C/15 s; the alkaline phosphatase test proves pasteurization.
- Regeneration efficiency = regeneration rise/total rise; the holding tube is sized on the fastest particle (2 × mean in laminar flow).
- Q = ṁc_pΔT for heat and refrigeration loads; COP = refrigerating effect/work; 1 TR = 3.517 kW.
Practice questions (15)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
At a constant temperature, a bacterial population falls from 10⁶ to 10² per mL in 8 min. What is the decimal reduction time D at that temperature, in min?
Numerical answer — type the value.
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Answer: 2
Four log cycles (10⁶ → 10²) in 8 min, so D = 8/4 = 2 min. Dividing by the ratio 10⁴ instead of the number of log cycles, or using natural logarithms, gives a meaningless value.The D value of a spore is 4 min at 115 °C and 0.5 min at 121 °C. What is its z value, in °C (to two decimal places)?
Numerical answer — type the value.
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Answer: 6.64
log₁₀(D₁/D₂) = (T₂ − T₁)/z, so z = 6/log₁₀(4/0.5) = 6/log₁₀ 8 = 6/0.9031 = 6.64 °C. Using the natural logarithm (ln 8 = 2.079) gives 2.89 °C, and dividing the temperature difference by the D ratio itself gives 0.75.Taking D₁₂₁ = 0.21 min for Clostridium botulinum spores, what is the minimum F₀ of a 12D process, in min?
Numerical answer — type the value.
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Answer: 2.52
F = D × number of log reductions = 0.21 × 12 = 2.52 min at 121 °C. Dividing by 12 (0.0175 min) inverts the relation; a larger number of log reductions always needs a longer process.A canned food is held at 115 °C for 20 min. Taking z = 10 °C and a reference temperature of 121.1 °C, what F₀ does the hold deliver, in min (to two decimal places)?
Numerical answer — type the value.
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Answer: 4.91
Lethal rate L = 10^((115 − 121.1)/10) = 10^(−0.61) = 0.2455, so F₀ = 20 × 0.2455 = 4.91 min. Taking the exponent the wrong way round, 10^(+0.61), gives 81.5 min, which would credit a lower temperature with more lethality than the reference.3000 kg/h of milk (c_p = 3.93 kJ/kg·K) is heated from 5 °C to 72 °C with no regeneration. What is the heat load, in kW (to one decimal place)?
Numerical answer — type the value.
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Answer: 219.4
Q = ṁc_pΔT = (3000/3600) × 3.93 × 67 = 0.8333 × 263.3 = 219.4 kW. Leaving the flow in kg/h gives 789 930 "kW", three thousand six hundred times too large.In an HTST pasteurizer, milk enters at 5 °C, leaves the regeneration section at 60 °C and is pasteurized at 72 °C. What is the regeneration efficiency, in per cent (to one decimal place)?
Numerical answer — type the value.
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Answer: 82.1
Regeneration efficiency = (60 − 5)/(72 − 5) = 55/67 = 82.1 per cent. Dividing by 72 °C instead of by the total rise (60/72 = 83.3 per cent) measures temperatures from zero rather than from the inlet.In the same pasteurizer (3000 kg/h of milk, c_p = 3.93 kJ/kg·K), regeneration brings the milk to 60 °C. What heat must the heating section then supply to reach 72 °C, in kW (to one decimal place)?
Numerical answer — type the value.
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Answer: 39.3
Q = (3000/3600) × 3.93 × (72 − 60) = 0.8333 × 47.16 = 39.3 kW, against 219.4 kW with no regeneration. Using the full 5-to-72 °C rise ignores what regeneration has already done.Milk flows in laminar flow through an HTST holding tube at a mean velocity of 0.5 m/s. The required holding time is 15 s. What minimum tube length ensures that every particle is held for at least 15 s, in m?
Numerical answer — type the value.
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Answer: 15
In laminar flow the centre-line velocity is twice the mean, 1.0 m/s, and that fastest particle must stay 15 s: L = 1.0 × 15 = 15 m. Sizing on the mean velocity (7.5 m) holds the fastest milk for only 7.5 s, under-pasteurizing it.In a vapour-compression refrigerator, the refrigerating effect is 150 kJ/kg of refrigerant and the compressor work is 30 kJ/kg. What is the coefficient of performance?
Numerical answer — type the value.
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Answer: 5
COP = refrigerating effect/work = 150/30 = 5. Dividing the heat rejected in the condenser (180 kJ/kg) by the work gives 6, the COP of the same machine used as a heat pump.10 000 kg of apples (specific heat 3.6 kJ/kg·K) are to be cooled from 25 °C to 0 °C in 24 h. Ignoring respiration and other loads, what is the product cooling load, in kW (to one decimal place)?
Numerical answer — type the value.
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Answer: 10.4
Heat to remove = 10 000 × 3.6 × 25 = 900 000 kJ; over 86 400 s that is 10.4 kW. Dividing by 24 alone gives a rate in kJ/h (37 500), not kW; the respiration heat of live fruit would be added on top.A bulk milk cooler must bring 2000 kg of milk (c_p = 3.93 kJ/kg·K) from 35 °C to 4 °C in 3 h. What average refrigeration capacity does this need, in kW (to one decimal place)?
Numerical answer — type the value.
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Answer: 22.6
Q = 2000 × 3.93 × 31 = 243 660 kJ over 3 × 3600 = 10 800 s, i.e. 22.6 kW — about 6.4 tons of refrigeration at 3.517 kW per ton. Dividing by 1 h instead of 3 h gives 67.7 kW, three times too high, and using the initial temperature alone (35 K) instead of the 31 K drop overstates it slightly.The adequacy of milk pasteurization is routinely checked by the:
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Answer: A — alkaline phosphatase test
Alkaline phosphatase, naturally present in raw milk, is destroyed by a correct time–temperature combination and slightly outlasts the target pathogens, so its absence confirms pasteurization. Peroxidase is the blanching indicator, and methylene-blue and clot-on-boiling tests judge the quality of raw milk.The standard HTST pasteurization of milk is:
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Answer: A — 72 °C for 15 s
HTST is 72 °C for 15 s, continuous. 63 °C for 30 min is the batch LTLT method, around 135 °C for seconds is UHT, and 121 °C for many minutes is a retort sterilization, not a pasteurization.Which of the following statements about blanching are correct?
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Answer: A — Its main purpose before freezing is to inactivate enzymes; B — Peroxidase is commonly used as the indicator of adequate blanching; D — It expels air from the tissue
Blanching stops the enzymes that spoil frozen and dried vegetables, is judged by peroxidase inactivation, and drives out intercellular air. Immersion in hot water leaches more solubles than steam, so the third statement has the comparison reversed.Which of the following statements about UHT processing and sterilization are correct?
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Answer: A — In UHT processing the product is sterilized before it is filled into containers; C — In aseptic packaging the containers are sterilized separately from the product; D — UHT milk suffers less browning than in-container sterilized milk
UHT sterilizes the liquid in a flow system and fills it aseptically into separately sterilized packs; the very high temperature for seconds kills spores far faster than it drives browning, so quality is better than after a long retort process. Commercial sterility means no organism can grow in normal storage — not the absence of every spore — so the second statement is false.