Lumped-Capacitance Transient Heating of an Iron Baseplate: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2013, Q6

Question 6: Lumped-Capacitance Transient Heating of an Iron Baseplate

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. Flat baseplate, thickness 0.5 cm, area 0.03 m², $\rho=2770$ kg/m³, $C_p=875$ J/kg·K (the source's printed "J/k C" is read as the standard J/kg·K), $\alpha=7.3\times10^{-5}$ m²/s, surface coefficient $h=12$ W/m²·K, 85% of the 1000 W heater output reaching the baseplate, starting from thermal equilibrium with $22\,{}^{\circ}\text{C}$ air.

Given data
Quantity	Symbol	Value
Baseplate thickness	$\delta$	0.5 cm
Surface area	$A$	0.03 m²
Density	$\rho$	2770 kg/m³
Specific heat	$C_p$	875 J/kg·K
Thermal diffusivity	$\alpha$	$7.3\times10^{-5}$ m²/s
Surface coefficient	$h$	12 W/m²·K
Heater power reaching plate	$\dot Q_{in}$	$0.85\times1000=850$ W
Ambient / initial temperature	$T_\infty$	$22\,{}^{\circ}\text{C}$

Find. (a) Time to heat from $22\,{}^{\circ}\text{C}$ to $140\,{}^{\circ}\text{C}$ with the heater on; (b) time for the plate to cool from $140\,{}^{\circ}\text{C}$ back down to $125\,{}^{\circ}\text{C}$ with the heater off.

Approach. The thin plate (Biot-number-favourable geometry) is treated with lumped-capacitance transient analysis. With the heater on, the energy balance $mC_p\,dT/dt=\dot Q_{in}-hA(T-T_\infty)$ integrates to an exponential approach toward a (very high, never-reached) steady-state temperature; with the heater off, it reduces to plain Newtonian cooling. Both parts use the same time constant $\tau=mC_p/(hA)$.

Mass and time constant. $$m = \rho A\delta = 2770\times0.03\times0.005 = 0.4155\text{ kg}$$ $$\tau = \frac{mC_p}{hA} = \frac{0.4155\times875}{12\times0.03} = \frac{363.56}{0.36} = 1009.9\text{ s}$$
Part (a) — time to heat to $140\,{}^{\circ}\text{C}$. With the heater on, the (fictitious, never-reached) steady-state temperature the exponential approaches is $$T_{ss} = T_\infty+\frac{\dot Q_{in}}{hA} = 22+\frac{850}{0.36} = 2383.1\,{}^{\circ}\text{C}$$ and the transient solution $T(t)-T_{ss}=(T_\infty-T_{ss})e^{-t/\tau}$ gives, solving for $t$ at $T=140\,{}^{\circ}\text{C}$: $$t = -\tau\ln\!\left(\frac{140-T_{ss}}{22-T_{ss}}\right) = -1009.9\ln\!\left(\frac{140-2383.1}{22-2383.1}\right) = -1009.9\ln(0.9500)$$ $$\boxed{t_{\text{heat}} = 51.8\text{ s}}$$
Part (b) — time cooling from $140\,{}^{\circ}\text{C}$ to $125\,{}^{\circ}\text{C}$. With the heater off, plain Newtonian cooling toward $T_\infty=22\,{}^{\circ}\text{C}$ applies, $T(t)-T_\infty=(140-T_\infty)e^{-t/\tau}$: $$t = -\tau\ln\!\left(\frac{125-22}{140-22}\right) = -1009.9\ln\!\left(\frac{103}{118}\right) = -1009.9\ln(0.8729)$$ $$\boxed{t_{\text{off}} = 137.3\text{ s}}$$

Lumped-capacitance transient response: heating phase (red) toward the fictitious $2383\,{}^{\circ}\text{C}$ asymptote, cut off at $140\,{}^{\circ}\text{C}$; cooling phase (blue) back to $125\,{}^{\circ}\text{C}$.

Question 6 — results
Quantity	Value
Time constant $\tau$	1009.9 s
(a) Time to reach $140\,{}^{\circ}\text{C}$	51.8 s
(b) Heater-off duration ($140\to125\,{}^{\circ}\text{C}$)	137.3 s

Topic. Lumped-capacitance transient conduction with internal heat generation (heating phase) and plain Newtonian cooling (heater-off phase).

Key relations. Energy balance $mC_p\,dT/dt=\dot Q_{in}-hA(T-T_\infty)$; heating solution $T(t)=T_{ss}+(T_\infty-T_{ss})e^{-t/\tau}$ with $T_{ss}=T_\infty+\dot Q_{in}/(hA)$; cooling solution $T(t)=T_\infty+(T_0-T_\infty)e^{-t/\tau}$; time constant $\tau=mC_p/(hA)$.

Why this works. Lumped capacitance is valid whenever the Biot number $hL_c/k\ll1$ (a thin, highly conductive metal plate with modest surface convection easily satisfies this), meaning internal conduction resistance is negligible next to surface convection resistance and the whole plate can be treated as a single uniform temperature at every instant. The heating-phase asymptote $T_{ss}=2383\,{}^{\circ}\text{C}$ is never physically reached (the thermostat cuts the heater off at $140\,{}^{\circ}\text{C}$) — it is simply the mathematical steady state the exponential is heading toward, and its size illustrates just how thermostat-limited (rather than convection-limited) this heating process really is.

Common pitfall. Using the heater's full 1000 W instead of the 85% that actually reaches the baseplate; and, for part (b), reusing the heating-phase steady-state temperature $T_{ss}$ instead of switching to the correct heater-off target $T_\infty=22\,{}^{\circ}\text{C}$.

Source. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 5 (lumped-capacitance transient conduction).

Question 6 of 8: Lumped-Capacitance Transient Heating of an Iron Baseplate

Question 6: Lumped-Capacitance Transient Heating of an Iron Baseplate