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23-Chem-B1 Transport Phenomena · Undated paper

Question 3 of 6: B1 — Initial free-convection heat loss from a hot square steel plate

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

Notes on this paper

Paper. National Examination (Engineers Canada) — 16-Chem-B1 Transport Phenomena, May 2019. Open book, 3 hours. Six 25-point problems in three sections — A Fluid Mechanics (A1, A2), B Heat Transfer (B1, B2), C Mass Transfer (C1, C2); one problem from each section must be attempted, plus a fourth from any section. All six problems are solved and fully worked below.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change, the power-law slit-flow momentum balance and the film/annular diffusion balances (A2, B2, 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 external-flow convection/mass-transfer correlations (A1, B1, C1); F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — Churchill–Chu free convection, horizontal-plate correlations, slug-flow internal convection and air properties (B1, B2); C. J. Geankoplis, Transport Processes and Separation Process Principles (Prentice Hall) — boundary-layer mass transfer and the film model (C1); R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.

Question 3: B1 — Initial free-convection heat loss from a hot square steel plate (25 marks: 15 + 10)

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 (square), thickness$L\times L,\ t$$2.5\ \text{m}\times2.5\ \text{m},\ 2.5\ \text{mm}$
Initial surface / air temperature$T_s,\ T_\infty$$430\ \text{K},\ 295\ \text{K}$ ($\Delta T=135$ K)
Film temperature$T_f=(T_s+T_\infty)/2$$362.5\ \text{K}$
Air @ $T_f$ (Table 1, 360–370 K): $\rho,\ \mu$—$0.9739\ \text{kg/m}^3,\ 21.36\times10^{-6}\ \text{Pa}\cdot\text{s}$
$\nu=\mu/\rho,\ k,\ \alpha,\ Pr=\nu/\alpha$—$21.93\times10^{-6}\ \text{m}^2/\text{s},\ 0.03056\ \text{W/m}\cdot\text{K},\ 31.07\times10^{-6}\ \text{m}^2/\text{s},\ 0.706$
Expansion coefficient$\beta=1/T_f$$2.759\times10^{-3}\ \text{K}^{-1}$

Find. The initial convective loss $q=\bar h A\,(T_s-T_\infty)$ at $T_s=430$ K, first horizontal (upper and lower faces treated separately) then vertical. The plate hangs freely, so both large faces are exposed, each of area $6.25\ \text{m}^2$; the edges ($4\times2.5\times0.0025=0.025\ \text{m}^2$, 0.2 % of the face area) are neglected. The thin steel plate is taken as uniform at 430 K at the instant of removal.

(a) horizontal (Tₛ = 430 K) top: rising plume (good) bottom: trapped, stratified (poor) (b) vertical buoyant film height L = 2.5 m Lc = A/P = 0.625 m
Fig. B1: A hot horizontal plate convects vigorously from its upper face (rising plume, $\overline{Nu}=0.15Ra^{1/3}$) but weakly from its stably-stratified lower face ($\overline{Nu}=0.27Ra^{1/4}$). Hung vertically, buoyancy drives a single boundary layer up each face (Churchill–Chu). Both faces are active in each case.

Approach. Evaluate air properties at $T_f=362.5$ K from the appended table and use $\beta=1/T_f$. For the horizontal plate use the characteristic length $L_c=A_s/P=0.625$ m with separate upper-face and lower-face correlations; for the vertical plate use the height $L=2.5$ m and Churchill–Chu. Then $q=\bar hA\,\Delta T$ with $\Delta T=135$ K.

  1. Property check. $g\beta/(\nu\alpha)=9.81(2.759\times10^{-3})/[(21.93\times10^{-6})(31.07\times10^{-6})]=3.97\times10^{7}\ \text{m}^{-3}\text{K}^{-1}$, matching the table’s own $g\beta/(\nu\alpha)$ column interpolated at 362.5 K ($3.98\times10^7$).
  2. (a) Horizontal Rayleigh number. With $L_c=A_s/P=(2.5)^2/(4\times2.5)=0.625$ m, $$Ra_{L_c}=\frac{g\beta\,\Delta T\,L_c^3}{\nu\alpha}=3.97\times10^{7}(135)(0.625)^3=1.31\times10^{9}.$$
  3. (a) Upper and lower faces. The hot upper face ($10^7\le Ra\le10^{11}$) uses $\overline{Nu}=0.15\,Ra^{1/3}=164.1$; the hot lower face (stably stratified, $10^5\le Ra\le10^{10}$) uses $\overline{Nu}=0.27\,Ra^{1/4}=51.4$. With $k/L_c=0.03056/0.625=0.0489\ \text{W/m}^2\text{K}$, $$h_{\text{top}}=8.02,\qquad h_{\text{bot}}=2.51\ \ \text{W/m}^2\text{K}.$$
  4. (a) Horizontal heat loss. Each face has area $6.25\ \text{m}^2$: $q_{\text{top}}=8.02(6.25)(135)=6.77$ kW and $q_{\text{bot}}=2.51(6.25)(135)=2.12$ kW, so $$\boxed{\;q_a=(h_{\text{top}}+h_{\text{bot}})\,(6.25)\,(135)\approx8.89\times10^{3}\ \text{W}\approx8.9\ \text{kW}.\;}$$
  5. (b) Vertical Rayleigh number. The characteristic length is the height $L=2.5$ m: $$Ra_L=\frac{g\beta\,\Delta T\,L^3}{\nu\alpha}=3.97\times10^{7}(135)(2.5)^3=8.38\times10^{10},$$ well above $10^9$, so most of each face carries a turbulent boundary layer.
  6. (b) Churchill–Chu coefficient. Valid for all $Ra_L$: $$\overline{Nu}_L=\left\{0.825+\frac{0.387\,Ra_L^{1/6}}{\bigl[1+(0.492/Pr)^{9/16}\bigr]^{8/27}}\right\}^2=496,$$ so $h_b=\overline{Nu}_L\,k/L=496(0.03056)/2.5=6.07\ \text{W/m}^2\text{K}$ on each face.
  7. (b) Vertical heat loss. Total area $=2(6.25)=12.5\ \text{m}^2$: $$\boxed{\;q_b=h_b\,(12.5)\,(135)=6.07(12.5)(135)\approx1.02\times10^{4}\ \text{W}\approx10.2\ \text{kW}.\;}$$ The vertical plate loses about 15 % more than the horizontal one, because the horizontal average is dragged down by its poorly-convecting underside.
Orientation$Ra$$\bar h$ (W/m²K)Initial loss
(a) Horizontal (top + bottom)$1.31\times10^9$$8.02$ / $2.51$$\approx8.9\ \text{kW}$
(b) Vertical$8.38\times10^{10}$$6.07$$\approx10.2\ \text{kW}$
Check — both faces, convection only

The plate hangs freely, so both $2.5\times2.5$ m faces convect (total $12.5\ \text{m}^2$); halve the results if only one face is intended. No emissivity is given, so radiation is not included — for an oxidised steel plate ($\varepsilon\approx0.7$) radiation to 295 K surroundings would add roughly 13 kW from the two faces, comparable to or larger than convection, so the numbers above are the free-convection loss that the question’s data support. The lower-face correlation $0.27Ra^{1/4}$ is the classic McAdams form.