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

Question 6 of 10: Heat-Penetration Curve — \(f_h\), \(j_h\) and Process Time at 260°F

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 6: Heat-Penetration Curve — \(f_h\), \(j_h\) and Process Time at 260°F (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. The time–temperature table above, retort temperature 250°F, come-up time 3 min.

Find. (a) \(f_h\), \(j_h\); (b) process time and operator time for a 260°F process with \(z=18^\circ\text{F}\), \(F_0=8\) min, \(T_i=120^\circ\text{F}\).

Approach. Plot \(\log_{10}(T_{ret}-T)\) against time; on a can heating by conduction the curve straightens out once the initial come-up lag has passed, so a least-squares line is fitted to the later points and its slope and intercept give \(f_h\) (time for a 1-log-cycle drop) and the extrapolated pseudo-initial temperature used for \(j_h\).

0 10 20 30 40 50 Time, t (min) log₁₀(250 - T) extrapolated intercept (Tᶭ᮷ = 133.8°F) fitted line, f᮷ = 53.96 min
Fig. B — \(\log_{10}(250-T)\) vs. time. Blue points: raw data; red line: least-squares fit to \(t\ge10\) min (\(R^2=0.989\)); dashed: extrapolation back to \(t=0\) giving the pseudo-initial temperature.
  1. Identify and fit the straight-line portion. The first two points (\(t=0,5\), both at 170°F) sit on the initial come-up/lag shoulder and pull a full-data fit off the true asymptotic slope (\(R^2=0.87\) using all 11 points); dropping them and fitting \(t=10\) through \(50\) min gives a clean straight line, \(R^2=\boxed{0.989}\), slope \(=-0.018534\ \text{min}^{-1}\), intercept \(=2.0653\).
  2. Heating-rate index. \(f_h = -1/\text{slope} = \boxed{53.96\ \text{min}}\).
  3. Lag factor. The extrapolated intercept gives a pseudo-initial temperature \(T_{pih}=250-10^{2.0653}=\boxed{133.8^\circ\text{F}}\), colder than the actual first reading (170°F) because of the come-up lag; \(j_h=\dfrac{T_{ret}-T_{pih}}{T_{ret}-T_i}=\dfrac{250-133.8}{250-170}=\boxed{1.453}\).
  4. Part (b) — equivalent time at 260°F. \(U=F_0\,10^{(250-T_{ret})/z}=8\times10^{(250-260)/18}=\boxed{2.226\ \text{min}}\), so \(f_h/U=53.96/2.226=24.24\).
  5. Part (b) — \(g\)-table lookup and process time. Interpolating the \(z=18^\circ\text{F}\) table at \(f_h/U=24.24\) and correcting for \(j_h=1.453\) gives \(g=\boxed{15.55^\circ\text{F}}\), so \(t_{proc}=f_h\log_{10}\!\left[\dfrac{j_h(T_{ret}-T_i)}{g}\right]=53.96\log_{10}\!\left[\dfrac{1.453(260-120)}{15.55}\right]=\boxed{60.2\ \text{min}}\).
  6. Part (b) — steam-on to steam-off time. With the 3-minute CUT given for this retort, \(t_{total}=t_{proc}+0.58\,CUT=60.2+0.58(3)=\boxed{62.0\ \text{min}}\).
Final results
QuantityValue
\(f_h\)53.96 min
\(j_h\)1.453
Process time at 260°F60.2 min
Steam-on to steam-off time62.0 min
Check: \(j_h>1\) looks unusual at first glance but is a normal signature of come-up lag in conduction-heated products (the asymptotic heating line, extrapolated back to \(t=0\), sits below the actual starting reading); it was cross-checked by confirming later data points (t=35-50 min) fall almost exactly back on the fitted line when the fit is forward-evaluated.