F-value by the graphical (General) method: 22-Agric-B8 Food Process Engineering (Part 1), May 2017, Q7

Question 7: F-value by the graphical (General) method (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. A hand-drawn centre-temperature-vs-time curve for a can processed in a retort: it starts at 180°F, rises to a peak of about 244°F held from roughly t=23 to 27 min, then falls back to 180°F by t=45 min. z=18°F.

[Figure not reproduced: Digitized centre-temperature curve, read at ~3 minute intervals off the source graph. See the official exam paper.]

Find. The process F-value (referenced to 250°F) by the graphical (Bigelow General) method.

Approach. The general method integrates the instantaneous lethal rate L(t)=10^((T(t)−250)/z) over the whole time–temperature history; since no analytical T(t) is given, the curve is digitized at closely spaced time points and the area under L(t) is evaluated numerically (trapezoidal rule).

Check: the temperature-time pairs used below are read directly off the supplied graph at roughly 3-minute spacing (the exam's own intended "graphical method" input) — a reader working from the same hand-drawn curve would obtain closely similar, but not bit-identical, digitized values.

Digitize the curve. Reading centre temperature T (°F) at time t (min):

t (min) 0 3 6 9 12 15 18 21 23 25 27 29 31 33 36 39 42 44 45

T (°F) 180 193 205 216 225 233 239 243 244 244 244 241 233 222 206 195 184 180.5 180
Compute the lethal rate at every point. $$L(t)=10^{\frac{T(t)-250}{18}}$$ e.g. at the peak (T=244°F): L=10^((244−250)/18)=10^(−0.333)=0.464; at T=225°F: L=10^(−1.389)=0.0408; at T≤205°F, L drops below 0.006 and contributes negligibly.
Integrate L(t) over the whole curve (trapezoidal rule). $$F=\int_0^{45} L(t)\,dt \approx \sum \frac{L_i+L_{i+1}}{2}(t_{i+1}-t_i)$$ Summing all seventeen intervals (weighted most heavily by the ~10-minute window either side of the plateau, where L is largest) gives: $$F \approx \boxed{6.0\ \text{min}}$$

t (min)	0	3	6	9	12	15	18	21	23	25	27	29	31	33	36	39	42	44	45
T (°F)	180	193	205	216	225	233	239	243	244	244	244	241	233	222	206	195	184	180.5	180

Final results
Quantity	Value
Peak lethal rate, Lᵣᶟᵏᵳ	0.464 (at 244°F)
Process F-value (graphical method)	≈6.0 min

Topic: The graphical (Bigelow General) method for computing process lethality from a measured heat-penetration curve.

Key relations: instantaneous lethal rate L(t)=10^((T(t)−Tᵣᵤᵤ)/z); F=∫L(t)dt, evaluated numerically off the digitized curve.

Why this works: unlike the Ball formula (which assumes an idealized exponential heating/cooling curve so it can be solved in closed form), the general method makes no shape assumption at all — it works directly from the measured curve, which is why it is the reference method against which formula methods like Ball's are validated.

Common pitfall: integrating temperature instead of lethal rate, or forgetting that L(t) falls off extremely fast below the reference temperature (a 10°F drop below 250°F at z=18°F already more than halves the lethal rate) — the tails of the curve near 180°F contribute almost nothing to F even though they cover many minutes of elapsed time.

Source: Toledo, Fundamentals of Food Process Engineering, the general (graphical) method for thermal process evaluation.

Question 7 of 10: F-value by the graphical (General) method

Question 7: F-value by the graphical (General) method (15 marks)