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23-Chem-B1 Transport Phenomena · December 2019

Question 3 of 6: B1 — Semicircular-tube air heater (radiation + convection)

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 3: B1 — Semicircular-tube air heater (radiation + convection) (25 marks: 20 + 5)

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
Tube radius$r$$0.02\ \text{m}$
Plane-surface (hot) temperature$T_1$$1000\ \text{K}$
Air mean temperature$T_m$$400\ \text{K}$
Air mass flow rate$\dot m$$0.01\ \text{kg/s}$
Emissivity (both surfaces)$\varepsilon_1,\varepsilon_2$$0.8$
Air properties at 400 K$\mu,k,Pr$$2.301\times10^{-5},\ 0.0338,\ 0.690$

Find. (a) the equilibrium temperature $T_2$ of the insulated curved surface; (b) the heat supply per unit length $q'$ to the plane surface.

plane surface 1: T₁ = 1000 K, ε = 0.8 curved surface 2 (insulated), T₂ = ? air: ṁ = 0.01 kg/s Tᾢ = 400 K r = 2 cm radiation 1→2 convection to air (both surfaces)
Fig. B1: Semicircular duct cross-section. The hot flat surface radiates to the reradiating curved surface (view factor $F_{12}=1$) and convects to the air. The insulated curved surface is in balance: the radiation it absorbs is re-emitted as convection to the air, which fixes its temperature $T_2$.

Approach. First get the internal-flow convection coefficient from the hydraulic diameter and a turbulent correlation. Then treat the two surfaces as a two-zone gray radiation enclosure ($F_{12}=1$). The insulated surface’s energy balance — radiation gained equals convection lost — is one equation for $T_2$; the plane surface’s balance then gives the required heat supply.

  1. Hydraulic diameter of the semicircle. Cross-section area $A_c=\tfrac12\pi r^2=6.283\times10^{-4}\ \text{m}^2$ and wetted perimeter $P=\pi r+2r=0.1028\ \text{m}$ (arc plus diameter), so $$D_h=\frac{4A_c}{P}=\frac{4(6.283\times10^{-4})}{0.1028}=0.02444\ \text{m}.$$
  2. Reynolds number and convection coefficient. $$Re_{D}=\frac{\dot m\,D_h}{A_c\,\mu}=\frac{0.01(0.02444)}{(6.283\times10^{-4})(2.301\times10^{-5})}=1.69\times10^{4}\ (\text{turbulent}).$$ Dittus–Boelter (heating, $n=0.4$): $Nu_D=0.023\,Re^{0.8}Pr^{0.4}=0.023(1.69\times10^4)^{0.8}(0.690)^{0.4}=47.8$, giving $$h=\frac{Nu_D\,k}{D_h}=\frac{47.8(0.0338)}{0.02444}=66.1\ \text{W/m}^2\text{K}.$$
  3. Radiation resistance of the two-zone enclosure. Per unit length the areas are $A_1'=2r=0.04$ and $A_2'=\pi r=0.0628\ \text{m}^2/\text{m}$; the flat surface sees only the curved one, so $F_{12}=1$. The series gray-body resistance is $$\mathcal R=\frac{1-\varepsilon_1}{\varepsilon_1A_1'}+\frac{1}{A_1'F_{12}}+\frac{1-\varepsilon_2}{\varepsilon_2A_2'}=6.25+25.0+3.98=35.23\ \text{m}^{-1}.$$
  4. (a) Energy balance on the insulated surface. Being well-insulated, surface 2 loses by convection exactly what it gains by radiation: $$\frac{\sigma(T_1^4-T_2^4)}{\mathcal R}=h\,A_2'\,(T_2-T_m).$$ Substituting $\sigma=5.67\times10^{-8}$, $\sigma/\mathcal R=1.61\times10^{-9}$ and $hA_2'=4.15$, $$1.61\times10^{-9}(1000^4-T_2^4)=4.15\,(T_2-400).$$ Solving numerically, $\boxed{T_2\approx696\ \text{K}.}$
  5. Net radiation to surface 2. $$q'_{rad}=\frac{\sigma(T_1^4-T_2^4)}{\mathcal R}=1.61\times10^{-9}\bigl(10^{12}-696^4\bigr)=1231\ \text{W/m},$$ which indeed equals $hA_2'(T_2-T_m)=4.15(296)=1231\ \text{W/m}$ — the balance closes.
  6. (b) Heat supplied to the plane surface. The heater feeds both the radiation to surface 2 and the direct convection from surface 1 to the air: $$q'=q'_{rad}+h\,A_1'\,(T_1-T_m)=1231+66.1(0.04)(600)=1231+1588\ \Longrightarrow\ \boxed{q'\approx2.82\ \text{kW/m}.}$$
QuantityResult
Hydraulic diameter$D_h=0.0244\ \text{m}$
Reynolds number / $Nu$ / $h$$1.69\times10^{4}$ / $47.8$ / $66.1\ \text{W/m}^2\text{K}$
Radiation resistance$\mathcal R=35.2\ \text{m}^{-1}$
(a) Insulated-surface temperature$T_2\approx696\ \text{K}$
(b) Heat supplied per unit length$q'\approx2.82\ \text{kW/m}$
Check — single convection coefficient, gray-diffuse surfaces

One internal-flow $h$ (based on $D_h$ and the bulk $T_m$) is applied to both surfaces, the standard simplification for a duct cross-section. The surfaces are taken gray, diffuse and isothermal per unit length, with $F_{12}=1$ (the flat surface can only radiate to the curved wall). Air properties are evaluated at $T_m=400$ K; $Re\approx1.7\times10^4$ safely exceeds the $Re>10^4$ validity floor of Dittus–Boelter.