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23-Chem-B1 Transport Phenomena · December 2019

Question 4 of 6: B2 — Initial cooling rate of a heated aluminium plate

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 16-Chem-B1 Transport Phenomena, December 2019 — open-book, 3 hours, six problems (25 marks each) in three sections: A Fluid Mechanics (A1, A2), B Heat Transfer (B1, B2), C Mass Transfer (C1, C2). The rubric asks the candidate to attempt one problem from each section plus a fourth; all six are worked here as a complete study resource. Pages 9–12 of the paper are an appendix of the equations of change (continuity, Navier–Stokes, energy, species) and are used as the reference tables rather than re-derived.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the film/annular differential balances (A1, C1, C2); J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, the Moody chart and the internal-/external-flow convection correlations (A2, B1, B2); F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — radiation enclosures, Dittus–Boelter, Churchill–Chu natural convection and air properties (B1, B2); C. J. Geankoplis, Transport Processes and Separation Process Principles (Prentice Hall) — the film model for diffusion with reaction (C1); R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties of air and water.

Question 4: B2 — Initial cooling rate of a heated aluminium plate (25 marks: 12 + 13)

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.

QuantitySymbolValue
Plate face dimensions$L\times L$$1.2\ \text{m}\times1.2\ \text{m}$
Surface / air temperature$T_s,\ T_\infty$$370\ \text{K},\ 290\ \text{K}$
Air stream velocity (part a)$U$$2\ \text{m/s}$
Film temperature$T_f$$330\ \text{K}$
Air props @ 330 K ($\nu,k,Pr,\alpha$)—$18.91\times10^{-6},\ 0.02852,\ 0.703,\ 26.9\times10^{-6}$

Find. The initial convective heat-loss rate $q=hA\,(T_s-T_\infty)$ for (a) forced convection over a horizontal plate and (b) natural convection from a vertical plate. Both large faces are exposed, so $A=2(1.2)^2=2.88\ \text{m}^2$.

(a) horizontal, forced U = 2 m/s flow over both faces, L = 1.2 m (b) vertical, natural buoyant plume Tₛ = 370 K
Fig. B2: (a) The plate lies in a 2 m/s air stream — a forced-convection flat-plate boundary layer grows on each face. (b) Hung vertically in still air, buoyancy drives an upward natural-convection boundary layer on each face. In both cases both 1.2 m×1.2 m faces lose heat.

Approach. Evaluate air properties at the film temperature $T_f=(T_s+T_\infty)/2=330$ K. For (a) form the plate Reynolds number and use the laminar flat-plate average Nusselt correlation; for (b) form the Rayleigh number and use the Churchill–Chu vertical-plate correlation. Then $q=hA\,\Delta T$ with $A$ counting both faces.

  1. (a) Plate Reynolds number. $$Re_L=\frac{UL}{\nu}=\frac{2(1.2)}{18.91\times10^{-6}}=1.27\times10^{5}<5\times10^{5},$$ so the boundary layer is laminar over the whole plate.
  2. (a) Forced-convection coefficient. Average laminar flat plate: $$\overline{Nu}_L=0.664\,Re_L^{1/2}Pr^{1/3}=0.664(356)(0.889)=210,\qquad h_a=\frac{\overline{Nu}_L\,k}{L}=\frac{210(0.02852)}{1.2}=5.0\ \text{W/m}^2\text{K}.$$
  3. (a) Heat-loss rate. With both faces exposed, $A=2(1.2)^2=2.88\ \text{m}^2$ and $\Delta T=80\ \text{K}$: $$\boxed{q_a=h_aA\,\Delta T=5.0(2.88)(80)\approx1.15\ \text{kW}.}$$
  4. (b) Rayleigh number, vertical plate. With $\beta=1/T_f=3.03\times10^{-3}\ \text{K}^{-1}$ and characteristic length $L=1.2$ m (height), $$Ra_L=\frac{g\beta\,\Delta T\,L^3}{\nu\alpha}=\frac{9.81(3.03\times10^{-3})(80)(1.2)^3}{(18.91\times10^{-6})(26.9\times10^{-6})}=8.1\times10^{9}.$$ This exceeds $10^9$, so the boundary layer becomes turbulent near the top.
  5. (b) Churchill–Chu coefficient. Valid over all $Ra_L$: $$\overline{Nu}_L=\left\{0.825+\frac{0.387\,Ra_L^{1/6}}{\bigl[1+(0.492/Pr)^{9/16}\bigr]^{8/27}}\right\}^2=\bigl(0.825+14.5\bigr)^2=236,$$ giving $h_b=\dfrac{\overline{Nu}_L\,k}{L}=\dfrac{236(0.02852)}{1.2}=5.6\ \text{W/m}^2\text{K}.$
  6. (b) Heat-loss rate. $$\boxed{q_b=h_bA\,\Delta T=5.6(2.88)(80)\approx1.29\ \text{kW}.}$$ The vertical natural-convection loss slightly exceeds the 2 m/s forced case because the tall plate develops a strong buoyant plume.
Case$Re$ or $Ra$$h$ (W/m²K)Heat loss
(a) Horizontal, forced (2 m/s)$Re_L=1.27\times10^5$$5.0$$\approx1.15\ \text{kW}$
(b) Vertical, natural (still air)$Ra_L=8.1\times10^9$$5.6$$\approx1.29\ \text{kW}$
Check — both faces cooled; radiation omitted

The plate is thin (1 cm) and freely exposed, so both 1.2 m×1.2 m faces convect — the heat-transfer area is $2\times1.44=2.88\ \text{m}^2$ (halve the results if only one face is intended). No emissivity is supplied, so radiation is not included; for a bare aluminium plate ($\varepsilon\!\approx\!0.1$) radiation would add only a small correction, but an oxidised/painted surface could roughly double the loss. The aluminium’s high conductivity keeps the plate nearly isothermal, justifying a uniform $T_s$ for this initial-rate estimate.