B1 — Turbulent air heating in a pipe (constant heat flux): 23-Chem-B1 Transport Phenomena, May 2016, Q3

Question 3: B1 — Turbulent air heating in a pipe (constant heat flux) (25 marks)

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.

Quantity	Symbol	Value
Pressure / inlet temperature	$P,\ T$	101.325 kPa, 150 °C (423.15 K)
Pipe diameter	$D$	5.08 cm $=0.0508$ m
Air velocity	$v$	8 m/s
Wall–air temperature difference	$\Delta T_{w}$	20 °C
Thermal conductivity of air	$k$	$3.52\times10^{-2}\ \text{W/m}\!\cdot\!\text{K}$
Viscosity of air	$\mu$	$2.38\times10^{-5}\ \text{kg/m}\!\cdot\!\text{s}$
Specific heat of air	$c_p$	$1017\ \text{J/kg}\!\cdot\!\text{K}$
Molecular weight	$M$	29 g/mol

Find. (a) the heat transfer per unit length $q'$; (b) the bulk temperature rise over $L=2$ m.

Fig. B1: Fully-developed turbulent internal flow with a uniform wall–fluid temperature offset of 20 °C. The convective coefficient follows from Dittus–Boelter; the bulk temperature climbs linearly along the pipe under constant wall flux.

Approach. Get the air density from ideal gas, form $Re$ and $Pr$, apply the Dittus–Boelter correlation (heating exponent 0.4) for $h$, then $q'=h\,\pi D\,\Delta T_w$; a bulk energy balance gives the temperature rise over 2 m.

Air density (ideal gas). $\rho=\dfrac{PM}{RT}=\dfrac{101325(0.029)}{8.314(423.15)}=0.836\ \text{kg/m}^3.$
Reynolds and Prandtl numbers. $Re=\dfrac{\rho vD}{\mu}=\dfrac{0.836(8)(0.0508)}{2.38\times10^{-5}}\approx1.43\times10^{4}$ (turbulent); $Pr=\dfrac{c_p\mu}{k}=\dfrac{1017(2.38\times10^{-5})}{3.52\times10^{-2}}\approx0.688.$
Dittus–Boelter, heating. $Nu=0.023\,Re^{0.8}Pr^{0.4}=0.023(1.43\times10^{4})^{0.8}(0.688)^{0.4}\approx41.6.$ Hence $h=\dfrac{Nu\,k}{D}=\dfrac{41.6(3.52\times10^{-2})}{0.0508}\approx28.8\ \text{W/m}^2\text{K}.$
(a) Heat transfer per unit length. With the wall a fixed $\Delta T_w=20$ °C above the local air, $$q'=h(\pi D)\Delta T_w=28.8(\pi\cdot0.0508)(20)\;\Longrightarrow\;\boxed{q'\approx92\ \text{W/m}}.$$
Mass flow rate. $\dot m=\rho v\dfrac{\pi D^2}{4}=0.836(8)\dfrac{\pi(0.0508)^2}{4}\approx1.355\times10^{-2}\ \text{kg/s}.$
(b) Bulk temperature rise over 2 m. Energy balance $\dot m c_p\,\Delta T_b=q'L$: $$\Delta T_b=\frac{q'L}{\dot m c_p}=\frac{92.2(2.0)}{1.354\times10^{-2}(1017)}\;\Longrightarrow\;\boxed{\Delta T_b\approx13.4\ \text{°C}}.$$

Quantity	Result
Air density	$\rho\approx0.836\ \text{kg/m}^3$
$Re$ / $Pr$	$1.43\times10^{4}$ (turbulent) / $0.688$
$Nu$ / $h$	$41.6$ / $28.8\ \text{W/m}^2\text{K}$
(a) Heat per unit length	$q'\approx92\ \text{W/m}$
(b) Bulk rise over 2 m	$\Delta T_b\approx13.4\ \text{°C}$

Concept & Theory

Topic. Forced-convection internal flow with a uniform wall heat flux; Dittus–Boelter correlation.

Key relations. $Re=\rho vD/\mu$, $Pr=c_p\mu/k$, $Nu=0.023Re^{0.8}Pr^{n}$ ($n=0.4$ heating), $h=Nu\,k/D$; per-length duty $q'=h\pi D\,\Delta T_w$; bulk balance $\dot m c_p\Delta T_b=q'L$.

Why this works. Under constant wall flux the wall–fluid difference stays fixed once the flow is thermally developed, so a single $h$ describes the whole pipe. The bulk temperature then rises linearly with length, and the exponent $n=0.4$ (rather than 0.3 for cooling) is Dittus–Boelter’s empirical allowance for the temperature-dependent fluid properties in the wall layer, which shift one way when the wall is hotter than the bulk and the other way when it is colder.

Common pitfall. Using the cooling exponent $n=0.3$, or evaluating $q'$ per unit area instead of per unit length — the question asks for $q'=h(\pi D)\Delta T$ in W/m.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 8 (internal flow, constant heat flux, Dittus–Boelter).

Question 3 of 6: B1 — Turbulent air heating in a pipe (constant heat flux)

Question 3: B1 — Turbulent air heating in a pipe (constant heat flux) (25 marks)