23-Chem-A2 Unit Operations and Separation Processes · May 2014
Question 5 of 6: Wall Shear Stress from a Heat-Transfer Correlation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 04-Chem-A2 Mechanical and Thermal Operations, May 2014 — open-book, 3 hours. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); all problems are worth 25 marks. The rubric asks candidates to attempt two problems per section; all six are solved in full below as a study resource.
Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe-flow friction and loss coefficients, agitator power correlations (Fig. 9.13, Table 9.3), and filtration theory; de Nevers, Fluid Mechanics for Chemical Engineers (3rd ed.) and Brodkey & Hershey, Transport Phenomena — mechanical-energy balance, fitting equivalent lengths, and the appended Fanning chart; Coulson & Richardson, Chemical Engineering Vol. 2 — rotary-drum filtration; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) and Lienhard, A Heat Transfer Textbook — composite-wall resistance networks, the Colburn (Chilton–Colburn) analogy, and LMTD/ε–NTU cross-flow exchangers.
Property note. Water properties are as printed in each question; the appended Table A.2 loss coefficients ($k$) and Table A3 turbine constants ($K_T$) are used directly, and pipe-friction factors are taken from Colebrook (identical to the appended Fanning chart, Fig. A1) so the numbers are reproducible without reading a graph.
Question B2: Wall Shear Stress from a Heat-Transfer Correlation (25 marks)
Given. Air over a flat plate, local heat-transfer correlation supplied; use the Colburn (Chilton–Colburn) analogy to convert it to wall friction.
Quantity
Value
Free-stream velocity $U$
50 m/s
Position $x$
1 m
Air $\rho$
$1.1769\ \mathrm{kg/m^3}$
Air $\mu$
$1.8464\times10^{-5}\ \mathrm{kg/(m\,s)}$
Prandtl $Pr$
0.708
Correlation
$Nu_x=0.04\,Re_x^{0.9}Pr^{-1/3}$
Find. the local wall shear stress $\tau_w$ at $x=1\ \mathrm{m}$.
Figure B2 — Turbulent boundary layer over the plate. The Colburn analogy links the supplied Nusselt correlation directly to the wall shear at $x=1\ \mathrm{m}$.
Approach. The Chilton–Colburn analogy $St\,Pr^{2/3}=C_f/2$ ties heat transfer to momentum transfer; substitute the given $Nu_x$ into $St=Nu_x/(Re_x Pr)$ to get $C_f/2$, then $\tau_w=(C_f/2)\rho U^2$.
Reynolds number at $x=1\ \mathrm{m}$. $$Re_x=\frac{\rho U x}{\mu}=\frac{1.1769(50)(1)}{1.8464\times10^{-5}}=3.19\times10^{6}.$$ This is well into the turbulent regime, consistent with the $Re^{0.9}$ form of the correlation.
Colburn analogy. With $St=\dfrac{Nu_x}{Re_x Pr}$, the analogy gives $$\frac{C_f}{2}=St\,Pr^{2/3}=\frac{Nu_x}{Re_x Pr}Pr^{2/3}=\frac{0.04\,Re_x^{0.9}Pr^{-1/3}}{Re_x}Pr^{-1/3}=0.04\,Re_x^{-0.1}Pr^{-2/3}.$$ The Prandtl exponents combine to $-2/3$, leaving a clean function of $Re_x$ only.
Wall shear stress. By definition $C_f=\tau_w/(\tfrac12\rho U^2)$, so $$\tau_w=\frac{C_f}{2}\,\rho U^2=0.01126(1.1769)(50)^2=33.1\ \mathrm{Pa}.$$ $\tau_w\approx 33.1\ \mathrm{Pa}$ at $x=1\ \mathrm{m}$
Quantity
Result
Reynolds number $Re_x$
$3.19\times10^{6}$ (turbulent)
Friction group $C_f/2$
0.01126
Wall shear stress $\tau_w$
≈ 33.1 Pa
Check — viscosity units The paper labels $\mu$ with kinematic units ($\mathrm{m^2/s}$), but $1.8464\times10^{-5}$ is the standard dynamic viscosity of air at 300 K in $\mathrm{kg/(m\,s)}$ (kinematic $\nu$ would be $\approx1.57\times10^{-5}\ \mathrm{m^2/s}$). The solution uses it as dynamic viscosity, which is dimensionally required by $Re_x=\rho U x/\mu$.
