Retort Come-Up Estimate — Pudding Can Heat Penetration: 22-Agric-B8 Food Process Engineering (Part 1), May 2014, Q1

Question 1: Retort Come-Up Estimate — Pudding Can Heat Penetration (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.

Can and process data
Quantity	Symbol	Value
Can diameter	\(D\)	7 cm → \(r_0=3.5\) cm
Can length	\(L\)	8.5 cm → half-length \(L_c=4.25\) cm
Density	\(\rho\)	1020 kg/m³
Thermal conductivity	\(k\)	0.32 W/(m·K)
Specific heat	\(C_p\)	3.6 kJ/(kg·K)
Surface coefficient	\(h\)	8000 W/(m²·K)
Initial temperature	\(T_i\)	28°C
Retort (steam) temperature	\(T_\infty\)	130°C
Claimed process time	\(t\)	30 min
Target centre temperature	\(T_c\)	93°C

Find. Whether 30 minutes in the retort is really enough for the geometric centre of the can to reach 93°C.

Fig. A — finite-cylinder can: radius \(r_0\), half-length \(L_c\), uniform steam boundary condition on all surfaces.

Approach. The can is a finite cylinder heated from a very large, essentially infinite surface coefficient, so its dimensionless centre response is the product of an infinite-cylinder solution (radial) and an infinite-slab solution (axial); because the two Fourier numbers are below 0.2 the single-term Heisler approximation is not reliable here, so the centre temperature is found from a converged multi-term series (the same physics the supplied Gurney-Lurie charts, Fig. 1, plot graphically) and then solved for the time that actually reaches 93°C.

Biot numbers — confirm near-zero surface resistance. \(Bi_r = hr_0/k = 8000(0.035)/0.32 = 875\) and \(Bi_L = hL_c/k = 8000(0.0425)/0.32 = 1063\), i.e. \(k/(hr_0)\) and \(k/(hL_c)\) are both \(\approx 0.001\) — the "0" curve on Fig. 1, consistent with the stated assumption that the steel can wall and its resistance are negligible.
Fourier numbers at 30 min. \(\alpha = k/(\rho C_p) = 0.32/(1020\times3600) = 8.71\times10^{-8}\ \text{m}^2/\text{s}\). At \(t=1800\ \text{s}\): \(Fo_r = \alpha t/r_0^2 = 0.128\) and \(Fo_L=\alpha t/L_c^2 = 0.0868\). Both are below 0.2, so a converged Heisler series (14 terms; the Bessel-zero eigenvalues for the cylinder and the odd-multiple-of-\(\pi/2\) eigenvalues for the slab, both at the \(Bi\to\infty\) limit) is used rather than the single-term approximation.
Dimensionless centre temperatures at 30 min. \(Y_{cyl}=\sum_n \dfrac{2}{\lambda_n J_1(\lambda_n)}e^{-\lambda_n^2 Fo_r} = 0.7425\), \(Y_{slab}=\sum_n \dfrac{2(-1)^{n+1}}{\lambda_n}e^{-\lambda_n^2 Fo_L} = 0.9672\). The finite-cylinder rule gives \(Y = Y_{cyl}\,Y_{slab} = \boxed{0.7181}\).
Predicted centre temperature at 30 min. \(T_c = T_\infty - Y(T_\infty-T_i) = 130 - 0.7181(130-28) = \boxed{56.8^\circ\text{C}}\) — well short of the claimed 93°C.
Time actually needed to reach 93°C. Solving the same series for the time at which \(T_c=93^\circ\text{C}\) (root-found numerically) gives \(t_{93} = 3222\ \text{s} = \boxed{53.7\ \text{min}}\).

Final results
Quantity	Value
\(Y_{cyl}\), \(Y_{slab}\), \(Y_{finite\ cyl}\)	0.7425, 0.9672, 0.7181
Centre temperature after 30 min	56.8°C
Time actually required to reach 93°C	53.7 min
Engineer's estimate	not accurate — understates the required hold time by ≈79% (about 24 min short)

Check: at \(Fo<0.2\) a single Heisler term over-predicts \(Y\) (it can even exceed 1); the 14-term converged series used above is the numerically exact analogue of reading Fig. 1's Gurney-Lurie chart at these low-\(Fo\) abscissas.

Concept & Theory

Topic: Transient conduction in a finite cylinder via the product solution.

Key relations: \(\theta_{finite\,cyl} = \theta_{\infty\,cyl}\cdot\theta_{\infty\,slab}\) (Newman's rule); \(\theta = \sum_n C_n e^{-\lambda_n^2 Fo}\) with \(\lambda_n\), \(C_n\) the shape's eigenvalues/coefficients at the appropriate Biot number.

Why this works: in a homogeneous solid with a single-valued diffusivity, the linear heat-conduction PDE separates into independent radial and axial problems whose product satisfies the original 2-D boundary-value problem exactly — this is why a finite cylinder can be built from two 1-D chart solutions instead of a full 2-D solve.

Common pitfall: using the single-term Heisler approximation at low Fourier numbers (\(Fo<0.2\)), where the series has not yet converged and the first term alone can overshoot the true \(Y\) — exactly the regime this can sits in at the claimed 30-minute process time.

Source: Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, transient-conduction chapter (finite cylinder as a product of infinite cylinder and infinite slab); Toledo, Fundamentals of Food Process Engineering, Ch. on unsteady-state heat transfer (this exam's own Gurney-Lurie chart, Fig. 1).

Question 1 of 10: Retort Come-Up Estimate — Pudding Can Heat Penetration

Question 1: Retort Come-Up Estimate — Pudding Can Heat Penetration (15 marks)