Radiative Cooling of a Brass Rod: 23-Chem-A2 Unit Operations and Separation Processes, May 2013, Q6

Question B3: Radiative Cooling of a Brass Rod (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. Lumped body, radiation only. $\rho=8526\ \mathrm{kg/m^3}$, $C_p=382\ \mathrm{J/(kg\,K)}$, $\sigma=5.67\times10^{-8}\ \mathrm{W/(m^2K^4)}$; rod $D=1$ m, $L=2$ m; radiating fraction 0.95 (trestle shades 5%); oxidized-brass emissivity $\varepsilon\approx0.60$ (Fig. B2 at $\sim\!660\,$°F); $T_i=400\,$°C$=673.15$ K, $T_f=300\,$°C$=573.15$ K.

Find. (a) the governing ODE; (b) the cooling time from 400 to 300 °C.

Figure B3 — The rod is treated as a lumped body radiating from 95% of its surface. The $T^4$ loss gives a separable ODE whose integral yields the cooling time.

Approach. Energy conservation on the lumped body gives the ODE; separate variables and integrate the $T^{-4}$ term between the two temperatures, using the cylinder’s volume-to-(radiating)area ratio.

(a) Governing equation. The stored-enthalpy loss rate equals the radiant loss to the (cold) surroundings: $$\rho V C_p\frac{dT}{dt}=-A\varepsilon\sigma\big(T^4-T_{\text{surr}}^4\big).$$ With $T_{\text{surr}}^4\ll T^4$ (hot body, cool room), this reduces to the required form $$\boxed{(\rho V C_p)\frac{dT}{dt}+A\varepsilon\sigma T^4=0.}$$
Separate and integrate. $$\int_{T_i}^{T_f}\frac{dT}{T^4}=-\frac{A\varepsilon\sigma}{\rho V C_p}\int_0^t dt\;\Rightarrow\;t=\frac{\rho V C_p}{3\,\varepsilon\sigma A}\left(\frac{1}{T_f^{3}}-\frac{1}{T_i^{3}}\right).$$
Geometry. $V=\tfrac{\pi}{4}D^2L=1.571\ \mathrm{m^3}$; total surface $A_{\text{tot}}=\pi DL+2\cdot\tfrac{\pi}{4}D^2=7.85\ \mathrm{m^2}$; radiating area $A=0.95A_{\text{tot}}=7.46\ \mathrm{m^2}$, so $V/A=0.2105$ m.
Evaluate. With $\varepsilon=0.60$, $\sigma=5.67\times10^{-8}$: $$t=\frac{8526(382)}{3(0.60)(5.67\times10^{-8})}(0.2105)\!\left(\frac{1}{573.15^{3}}-\frac{1}{673.15^{3}}\right)=\boxed{1.37\times10^{4}\ \mathrm{s}\approx3.8\ \mathrm{h}.}$$

Quantity	Result
Governing ODE	$(\rho VC_p)\dot T+A\varepsilon\sigma T^4=0$
Volume / radiating area	1.571 m$^3$ / 7.46 m$^2$
Emissivity (oxidized brass)	$\approx$ 0.60
Cooling time (400→300 °C)	≈ $1.37\times10^{4}$ s (3.8 h)

Check — lumped assumption & chart reads The 1-m diameter is taken as printed. Lumped capacitance requires a small radiative Biot number; for high-conductivity brass ($k\approx110\ \mathrm{W/mK}$) $Bi=\varepsilon\sigma T^3 (D/2)/k\approx0.06\ll0.1$, so it is defensible. The time scales as $1/\varepsilon$; reading $\varepsilon=0.55$–0.65 from Fig. B2 shifts $t$ by $\pm8\%$.

Concept & Theory

Topic. Transient lumped-capacitance cooling dominated by radiation.

Key relations. Stefan–Boltzmann $q=\varepsilon\sigma A(T^4-T_{\text{surr}}^4)$; lumped energy balance $\rho VC_p\dot T=-q$; integral $t=\dfrac{\rho VC_p}{3\varepsilon\sigma A}\!\left(T_f^{-3}-T_i^{-3}\right)$.

Why this works. When internal conduction is fast relative to surface loss (low Biot number), the body has one uniform temperature and the balance is a single ODE. The $T^4$ law makes it separable; the cubic-reciprocal integral is the radiation analogue of the exponential decay seen with convection.

Common pitfall. Working in °C instead of kelvin in the $T^4$ term (a fatal error), forgetting the 5% shaded area, or assuming a convection-style exponential decay — radiation gives the $T^{-3}$ integral, not $e^{-t/\tau}$.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, §5.1–5.3 (lumped capacitance, radiation); Özişik, Radiative Transfer (emissivity data, Fig. B2).

Question 6 of 6: Radiative Cooling of a Brass Rod

Section A — Mechanical Operations

Question B3: Radiative Cooling of a Brass Rod (25 marks)