17-Phys-B6 Applied Thermodynamics and Heat Transfer · May 2013
Question 6 of 8: Heat Loss from a Square Sheet-Metal Duct
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B6 Applied Thermodynamics and Heat Transfer,
National Examination May 2013 — an open-book examination (any textbook, reference or
non-communicating calculator permitted). Part A (Thermodynamics, Q1–4) and Part B (Heat
Transfer, Q5–8) together form the paper; a complete exam is five answers, drawn either
three-from-A/two-from-B or two-from-A/three-from-B. Every question is nonetheless answered in
full below so the paper remains a complete study resource. All eight questions carry equal
value in the grading.
Reference texts. Y. A. Çengel & M. A. Boles, Thermodynamics: An
Engineering Approach, 8th ed. (ideal-gas cycles, vapour power cycles with reheat, gas
compressors, vapour-compression refrigeration); F. P. Incropera & D. P. DeWitt,
Fundamentals of Heat and Mass Transfer, 7th ed. (critical radius of insulation,
internal/external forced convection correlations, natural convection from horizontal cylinders,
radiation exchange, LMTD/effectiveness–NTU heat-exchanger analysis).
Question 6: Heat Loss from a Square Sheet-Metal Duct (Part B, 20 marks)
Check: the 4 mm sheet-metal wall's own thermal conductivity is not
given, so its conduction resistance is treated as negligible next to the internal/external
convection resistances (standard for thin sheet metal). Duct-air properties are evaluated at the
converged bulk-mean temperature.
Given. Square duct, side $=0.5$ m, length $L=4$ m; air inside at
$\dot m=120{,}000$ kg/hr, $T_{in}=60^{\circ}\text{C}$; surrounding air at
$T_\infty=30^{\circ}\text{C}$, $h_o=5$ W/m²·K.
Find. Rate of heat transfer between the duct air and the surrounding air.
Approach. Model the duct as a single-stream "heat exchanger" against a
fixed-temperature ambient: get the internal convection coefficient from a Dittus–Boelter
correlation (duct air is being cooled), combine it in series with the given external
coefficient into an overall $UA$, then apply the exponential-decay outlet-temperature relation.
Geometry and internal flow. Hydraulic diameter of the square duct
$D_h=4A_c/P=$ side $=0.5$ m; surface area $A=P\times L=(4\times0.5)\times4=8$ m².
At the converged bulk mean temperature ($\approx60^{\circ}\text{C}$, since the duct barely cools
— see Step 3), air properties are $\rho=1.06$ kg/m³,
$\mu=2.01\times10^{-5}$ Pa·s, $k=0.0288$ W/m·K, $Pr=0.703$,
$c_p=1008$ J/kg·K. With $\dot m=33.33$ kg/s,
$$V=\frac{\dot m}{\rho A_c}=\frac{33.33}{1.06\times0.25}=125.8\ \text{m/s},\qquad
Re_{D_h}=\frac{\rho V D_h}{\mu}=3.32\times10^{6}\ (\text{highly turbulent}).$$
Overall $UA$ and outlet temperature. Wall conduction neglected (thin sheet
metal), inner and outer areas equal (thin wall):
$$\frac{1}{UA}=\frac{1}{h_iA}+\frac{1}{h_oA}=\frac{1}{196.3\times8}+\frac{1}{5\times8}
\ \Rightarrow\ UA=39.0\ \text{W/K}.$$
Treating the duct as a stream losing heat to a fixed-temperature surroundings,
$$T_{out}=T_\infty+(T_{in}-T_\infty)\,e^{-UA/(\dot mc_p)}
=30+30\,e^{-39.0/(33.33\times1008)}=\boxed{59.97^{\circ}\text{C}}$$
— the very high internal velocity (and hence very small $NTU=0.0012$) means this duct's
enormous flow rate barely cools at all over just 4 m.
Heat transfer rate.
$$\dot Q=\dot m\,c_p(T_{in}-T_{out})=33.33\times1.008\times(60-59.97)=\boxed{1.17\ \text{kW}}.$$
