Electrically Heated Pipe — Power to Hold 30 °C: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2016, Q7

Question 7: Electrically Heated Pipe — Power to Hold 30 °C (equal value)

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.

Quantity	Value
Pipe diameter / surface temp	$D=0.30$ m, $T_s=30$ °C (303.15 K)
Ambient air temperature	$T_\infty=20$ °C (293.15 K)
Surroundings (radiation)	$T_\text{sur}=25$ °C (298.15 K)
Emissivity	$\varepsilon=0.8$
Air properties at $T_f=25$ °C	$\nu=1.558\times10^{-5}$ m²/s, $\alpha=2.202\times10^{-5}$ m²/s, $k=0.02625$ W/m·K, $Pr=0.707$

Figure 7 — The heated horizontal pipe loses heat by natural convection to the 20 °C air and by radiation to the 25 °C surroundings; the electrical input per metre must balance the sum of the two.

Approach. At steady state the electrical power per metre equals the natural-convection loss (Churchill–Chu horizontal-cylinder correlation) plus the radiation loss to the surroundings.

Rayleigh number. At film temperature $T_f=25$ °C, with $\beta=1/T_f$ and $\Delta T=10$ °C, $$Ra_D=\frac{g\beta\,\Delta T\,D^3}{\nu\alpha}=2.59\times10^{7}$$
Convection coefficient (Churchill–Chu). $$Nu_D=\left\{0.60+\frac{0.387\,Ra_D^{1/6}}{\big[1+(0.559/Pr)^{9/16}\big]^{8/27}}\right\}^{2}=37.5,\qquad h=\frac{Nu_D\,k}{D}=3.28\ \text{W/m}^2\text{}\cdot\text{K}$$
Convective loss per metre. $q'_\text{conv}=h(\pi D)(T_s-T_\infty)=3.28(\pi\cdot0.3)(10)=30.9$ W/m.
Radiative loss per metre. $$q'_\text{rad}=\varepsilon\sigma(\pi D)\big(T_s^4-T_\text{sur}^4\big)=0.8(5.67\times10^{-8})(\pi\cdot0.3)(303.15^4-298.15^4)=23.2\ \text{W/m}$$
Total electrical power per metre. $$q'=q'_\text{conv}+q'_\text{rad}=30.9+23.2$$ ==**$q'\approx54$ W/m must be supplied.**==

Check

Radiation contributes ≈ 43 % of the loss here even though $\Delta T$ is only 5–10 °C — at near-ambient temperatures the two mechanisms are comparable, so neglecting radiation would under-size the heating by nearly half. Air properties are at the 25 °C film temperature; the small 10 °C driving difference keeps $h$ modest (≈ 3.3 W/m²·K).

Quantity	Result
Rayleigh number $Ra_D$	≈ 2.6 × 10⁷
Convection coefficient $h$	≈ 3.28 W/m²·K
Convective loss	≈ 30.9 W/m
Radiative loss	≈ 23.2 W/m
Total power per metre	≈ 54 W/m

Concept & Theory

Topic. Combined free convection and radiation from a horizontal cylinder.

Key relations. $Ra_D=g\beta\Delta T D^3/\nu\alpha$; Churchill–Chu $Nu_D$; $q'_\text{conv}=h\pi D\,\Delta T$; $q'_\text{rad}=\varepsilon\sigma\pi D(T_s^4-T_\text{sur}^4)$ (temperatures in kelvin).

Why this works. With no forced flow the boundary layer is buoyancy-driven, so the Churchill–Chu correlation gives $h$ from the Rayleigh number alone; the surface simultaneously radiates to its surroundings. Because both losses act on the same outer surface and are independent, they simply add to the required electrical input.

Common pitfall. Using the air temperature (20 °C) as the radiation sink instead of the stated surroundings temperature (25 °C), or working the radiation term in °C rather than kelvin.

Source. Çengel & Ghajar, Heat and Mass Transfer §9 (natural convection — horizontal cylinder) with combined-mode surface energy balance.

Question 7 of 8: Electrically Heated Pipe — Power to Hold 30 °C

Question 7: Electrically Heated Pipe — Power to Hold 30 °C (equal value)