22-Agric-B8 Food Process Engineering (Part 1) · December 2013
Question 9 of 10: Can Heating in a Retort — Centre Temperature by Heisler Charts
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 December 2013 — a three-hour open-book exam (any non-communicating
calculator permitted). Ten questions are set in four sections (I–IV), each with a
"choose N of M" instruction; candidates who follow the choice rule answer 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 — 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, vapour recompression); R.P. Singh and
D.R. Heldman, Introduction to Food Engineering, 5th ed. (freezing-time estimation,
modified Plank equation, 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).
Question 9: Can Heating in a Retort — Centre Temperature by Heisler Charts (15 marks)
Find. The can's geometric-centre temperature after 30 minutes.
A finite cylinder is the intersection of an infinite cylinder (radius \(r\))
and an infinite slab (half-height \(L\)) — Newman's rule multiplies their two
dimensionless centre-temperature ratios.
Approach. Check the Biot number (it is very large here, i.e. surface
resistance is negligible, corresponding to the chart's "0" curve); compute the Fourier numbers
for the radial and axial directions; read the finite-cylinder centre response as the
product of the infinite-cylinder and infinite-slab one-term Heisler solutions
(Newman's rule).
Thermal diffusivity, Fourier and Biot numbers.
\(\alpha = k/(\rho C_p) = 1.6/(961\times2.8) = 5.946\times10^{-4}\ \text{m}^2\text{/h}\).
Radius \(r=0.04\) m, half-height \(L=0.05\) m:
\(Fo_r=\alpha t/r^2 = 0.1858\), \(Fo_L=\alpha t/L^2=0.1189\).
\(Bi_r = hr/k = 12270(0.04)/1.6 = 307\), so \(k/(hr)=0.0033\approx0\) — effectively the
chart's zero-surface-resistance curve (retort surface temperature reached almost instantly).
Infinite-cylinder and infinite-slab centre ratios (one-term, \(Bi\to\infty\)).
Cylinder: \(\lambda_1=2.4048\), \(A_1=1.6021\):
\(Y_{cyl}=A_1 e^{-\lambda_1^2 Fo_r}=1.6021\,e^{-5.783(0.1858)}=\boxed{0.5470}\).
Slab: \(\lambda_1=\pi/2\), \(A_1=4/\pi\):
\(Y_{slab}=A_1 e^{-\lambda_1^2 Fo_L}=1.2732\,e^{-2.467(0.1189)}=\boxed{0.9495}\).
Newman's rule and centre temperature.
\(Y_{can}=Y_{cyl}Y_{slab}=0.5470\times0.9495=\boxed{0.5194}\).
\(Y=(T_{ret}-T)/(T_{ret}-T_i)\), so
\(T = T_{ret}-Y(T_{ret}-T_i) = 110-0.5194(110-60) = \boxed{84.0^\circ\text{C}}\).
