The pasteurisation ladder: D, z and F

Thermal processing · D, z, F on an HTST line

The lab establishes the baseline: held at 63 °C, the target organism falls 4 logs in 12 minutes. A second run at 72 °C measures a D-value of 0.27 minutes. Out on the HTST line, the holding tube gives 15 seconds — but today's chart shows it running at 71.7 °C instead of the specified 72. Build the ladder: the D-value at 63 °C, the z-value the two temperatures imply, and the F-value that says what today's 15 seconds at 71.7 °C is actually worth in 72 °C time.

TT_reft_pDzTF

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • t = 12 minHolding time in the 63 °C run
  • LR = 4 logsReduction achieved in that run
  • D₂ = 0.27 minMeasured D-value at 72 °C
  • t_p = 15 sHolding tube residence time
  • T = 71.7 °CToday's process temperature
  • T_ref = 72 °CSpecified reference temperature
Determine
  1. (a)the D-value at 63 °C
  2. (b)the z-value across the two temperatures
  3. (c)today's process time as equivalent minutes at 72 °C
Step 1 of 3(a) · solve for D-value

A D-value is the price of one log: 12 minutes bought 4 logs, so each factor of ten costs 3 minutes at 63 °C. It is the survival curve compressed to a single number — and it silently assumes the curve IS a straight line in log-count, which real organisms mostly honour and spores test.

DN/N0tLR
Rearranged for D
D=tLRD = \frac{t}{\mathrm{LR}}
Your values, in your units
D=(12 min)(4 logs)D = \frac{\left( 12\ \text{min} \right)}{\left( 4\ \text{logs} \right)}
Answer
D=180 sD = 180\ \text{s}

Carried onward at full precision, not this rounded figure.

Open the D-Value (Decimal Reduction Time) solver →

Step 2 of 3(b) · solve for z-value

Nine degrees dropped the D-value from 3 minutes to 0.27 — a factor of 11.1, just over one log — so z comes out at 8.6 C°: the temperature climb that makes killing ten times faster. Everything about trading time against temperature runs through this one constant.

D1D2zT1T2DT
Rearranged for z
z=T2T1log10(D1/D2)z = \frac{T_2 - T_1}{\log_{10}(D_1 / D_2)}
180 scarried from step 1
Your values, in your units
z=(72 C)(63 C)log10((180 s)/(0.27 min))z = \frac{\left( 72\ ^{\circ}\text{C} \right) - \left( 63\ ^{\circ}\text{C} \right)}{\log_{10}(\left( 180\ \text{s} \right) / \left( 0.27\ \text{min} \right))}
Converted to base units
z=(72 C)(63 C)log10((3 min)/(0.27 min))z = \frac{\left( 72\ ^{\circ}\text{C} \right) - \left( 63\ ^{\circ}\text{C} \right)}{\log_{10}(\left( 3\ \text{min} \right) / \left( 0.27\ \text{min} \right))}
Answer
z=8.6062 Kz = 8.6062\ \text{K}

Carried onward at full precision, not this rounded figure.

Open the z-Value (Thermal Resistance Constant) solver →

Step 3 of 3(c) · solve for F-value (equivalent time)

The F-value converts today's process into reference currency: 15 seconds at 71.7 °C is worth only 13.8 seconds at 72 °C. A drift of three-tenths of a degree quietly took 8 % of the lethality — the process LOOKS on-spec on time and is short on kill.

TrefTFtt
Rearranged for F
F=t×10(TTref)/zF = t \times 10^{\,(T - T_{\mathrm{ref}})/z}
8.6062 Kcarried from step 2
Your values, in your units
F=(15 s)×10((71.7 C)(72 C))/(8.6062 K)F = \left( 15\ \text{s} \right) \times 10^{\,(\left( 71.7\ ^{\circ}\text{C} \right) - \left( 72\ ^{\circ}\text{C} \right))/\left( 8.6062\ \text{K} \right)}
Converted to base units
F=(0.25 min)×10((71.7 C)(72 C))/(8.6062 C)F = \left( 0.25\ \text{min} \right) \times 10^{\,(\left( 71.7\ ^{\circ}\text{C} \right) - \left( 72\ ^{\circ}\text{C} \right))/\left( 8.6062\ \text{C}^{\circ} \right)}
Answer
F=13.843 sF = 13.843\ \text{s}

Carried onward at full precision, not this rounded figure.

Open the F-Value (Equivalent Time at Reference Temperature) solver →

Answer

Therefore D₆₃ = 3 minutes, z = 8.6 C°, and today's 15 seconds at 71.7 °C delivers an F of only 13.8 equivalent seconds — the line is running 8 % short of its specified lethality on a temperature error small enough to miss on a gauge.

Why this order

The three constants answer three different questions and only make sense in this order. D asks: at one temperature, how fast do they die? z asks: how does that speed change with temperature? F asks: given both, what is this actual, imperfect process worth in the reference temperature's currency? The ladder matters because regulation is written at the top of it — a pasteurisation standard is an F requirement — while everything a plant can measure lives at the bottom, in times and temperatures.

The ending is the lesson: lethality is exponential in temperature with z in the exponent, so small temperature errors are never small. Here −0.3 C° cost 8 %; a full degree low would cost 23 %. That asymmetry — time errors are linear, temperature errors are exponential — is why holding-tube thermometers are the most audited instruments in the building, and why the compliant-looking chart with a slightly lazy sensor is the most dangerous document in it.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.