17-Phys-B6 Applied Thermodynamics and Heat Transfer · December 2017
Question 6 of 8: Combined Internal/External Convection — Tube Length for a Target Outlet Temperature
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 December 2017 — a three-hour open-book examination;
candidates are expected to bring both a thermodynamics text and a heat-transfer text to make use
of the property tables. A complete examination is five questions — either three from Part A
(Thermodynamics, Q1–Q4) and two from Part B (Heat Transfer, Q5–Q8), or two from Part A
and three from Part B — every question carrying equal value; all eight are solved below as a
complete study set. Candidates are invited to state any assumptions where a question is open to
interpretation; this licence is used explicitly in Question 4 (the exam's own printed text is
ambiguous about whether the piston displacement is 1.00 m³, read here as the intended value)
and Question 6 (the external air stream is treated as an effectively infinite, constant-temperature
reservoir since no air mass flow rate or duct is specified).
Reference texts. Y. A. Çengel and M. A. Boles, Thermodynamics: An
Engineering Approach, 8th ed. (two-phase closed systems, flash chambers, steam turbines,
vapour-compression refrigeration, reciprocating compressors); F. P. Incropera and D. P. DeWitt,
Fundamentals of Heat and Mass Transfer, 7th ed. (composite cylindrical conduction, internal
and external forced convection correlations, natural convection from a vertical plate,
heat-exchanger LMTD analysis). Saturation and superheat property values below were computed (Bell et al., IAPWS-95 / REFPROP-quality equations of state for water, ammonia and R134a)
and cross-checked against the printed appendix tables on pages 5–8 of the source exam, which
they matched to 3–4 significant figures throughout.
Question 6: Combined Internal/External Convection — Tube Length for a Target Outlet
Temperature
Given. Hot water flows turbulently inside a bare (thin-walled) horizontal tube,
losing heat to a fast external crossflow of cooler air, which is large enough to be treated as an
effectively constant-temperature reservoir.
Given data
Quantity
Symbol
Value
Tube diameter
$D$
2 cm
Water inlet temperature
$T_{hi}$
$77\,{}^{\circ}\text{C}$
Target water outlet temperature
$T_{ho}$
$55\,{}^{\circ}\text{C}$
Internal water velocity
$V$
6 m/s
Ambient air temperature
$T_\infty$
$27\,{}^{\circ}\text{C}$
External air crossflow velocity
$V_\infty$
30 m/s
Find. The tube length $L$ needed for the water to leave at $55\,{}^{\circ}\text{C}$.
Horizontal tube: hot water flows inside (cooled along its length); air crosses
the outside of the tube perpendicular to its axis.
Approach. Evaluate the internal water-side coefficient $h_i$ (Dittus–Boelter,
turbulent, cooling exponent) at the water's mean bulk temperature, and the external air-side
coefficient $h_o$ (Churchill–Bernstein correlation for crossflow over an isothermal cylinder)
at the air's free-stream temperature. Combine as a thin-wall overall coefficient
$U=(1/h_i+1/h_o)^{-1}$, then treat the water as a single stream cooling exponentially toward the
effectively constant air temperature to solve for $L$.
Internal (water-side) coefficient. At the mean bulk temperature
$T_m=(77+55)/2=66\,{}^{\circ}\text{C}$: $\rho=980.0\text{ kg/m}^3$, $\mu=4.267\times10^{-4}\text{ Pa}\cdot\text{s}$,
$k=0.6564\text{ W/m}\cdot\text{K}$, $Pr=2.722$. Then
$$Re_D=\frac{\rho VD}{\mu}=\frac{980.0\times6\times0.02}{4.267\times10^{-4}}=275{,}573\quad(\text{turbulent})$$
Dittus–Boelter (fluid being cooled, $n=0.3$):
$$Nu_D=0.023Re_D^{0.8}Pr^{0.3}=698.9 \quad\Rightarrow\quad
h_i=\frac{Nu_D\,k}{D}=\frac{698.9\times0.6564}{0.02}$$
$$\boxed{h_i=22{,}939\text{ W/m}^2\cdot\text{K}}$$
Water mass flow rate: $\dot m=\rho V(\pi D^2/4)=980.0\times6\times(\pi\times0.02^2/4)=1.847\text{ kg/s}$.
External (air-side) coefficient. At $T_\infty=27\,{}^{\circ}\text{C}$:
$\nu=1.576\times10^{-5}\text{ m}^2/\text{s}$, $k=0.02640\text{ W/m}\cdot\text{K}$, $Pr=0.7070$.
$$Re_D=\frac{V_\infty D}{\nu}=\frac{30\times0.02}{1.576\times10^{-5}}=38{,}062$$
Churchill–Bernstein (valid for all $Re\cdot Pr\gt0.2$):
$$Nu_D=0.3+\frac{0.62\,Re_D^{1/2}Pr^{1/3}}{\left[1+(0.4/Pr)^{2/3}\right]^{1/4}}
\left[1+\left(\frac{Re_D}{282{,}000}\right)^{5/8}\right]^{4/5}=116.0$$
$$\boxed{h_o=\frac{Nu_D\,k}{D}=\frac{116.0\times0.02640}{0.02}=153.1\text{ W/m}^2\cdot\text{K}}$$
Overall coefficient and required tube length. With a thin tube wall (no wall
data given, so its resistance is neglected):
$$U=\left(\frac1{h_i}+\frac1{h_o}\right)^{-1}=152.1\text{ W/m}^2\cdot\text{K}$$
The air behaves as an effectively infinite reservoir, so the water cools exponentially along the
tube:
$$\frac{T_{ho}-T_\infty}{T_{hi}-T_\infty}=\exp\!\left(-\frac{U\pi DL}{\dot mc_p}\right)
\quad\Rightarrow\quad
L=-\ln\!\left(\frac{T_{ho}-T_\infty}{T_{hi}-T_\infty}\right)\frac{\dot mc_p}{U\pi D}$$
$$L=-\ln\!\left(\frac{55-27}{77-27}\right)\times\frac{1.847\times4187.8}{152.1\times\pi\times0.02}
=0.5798\times\frac{7735}{9.556}$$
$$\boxed{L=469.5\text{ m}}$$
Check: this length is impractically long for a single straight tube in a real
installation — it is a direct consequence of the modest air-side coefficient (roughly
1/150th of the water-side coefficient) controlling the overall $U$; in practice a finned tube or a
tube bank would be used to shrink the required length. The result follows honestly from the stated
idealised single-bare-cylinder-in-crossflow model, per the exam's own invitation to state
assumptions.
