Transient Heating of a Flat-Iron Baseplate: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2013, Q6

Question 6: Transient Heating of a Flat-Iron Baseplate (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.

Given. Baseplate area $A=0.03\ \text{m}^2$, thickness $0.5$ cm; $\rho=2770$, $c_p=875\ \text{J/kg}\cdot^\circ\text{C}$, $\alpha=7.3\times10^{-5}\ \text{m}^2/\text{s}$ (so $k=\alpha\rho c_p\approx168\ \text{W/m}\cdot\text{K}$); $T_\infty=22\ ^\circ\text{C}$, $h=12\ \text{W/m}^2\cdot^\circ\text{C}$; useful input $\dot E=0.85\times1000=850$ W.

Find. (a) heat-up time from 22 °C to 140 °C; (b) the off-interval while the plate cools from 140 °C to 125 °C.

Approach. Confirm the plate is a lumped mass (Biot number), then integrate the lumped energy balance — with generation for the heat-up (a) and pure convective cooling for the off-time (b).

Lumped-capacitance check. $L_c=V/A=\delta=0.005$ m, $\mathrm{Bi}=hL_c/k=12(0.005)/168=3.6\times10^{-4}\ll0.1$, so the plate is spatially isothermal and a lumped model applies. Its thermal mass is $mc=\rho V c_p=2770(1.5\times10^{-4})(875)=363.6\ \text{J/}^\circ\text{C}$, with time constant $\tau=mc/(hA)=363.6/0.36=1010$ s.
(a) Heat-up with generation. The balance $mc\,\dfrac{dT}{dt}=\dot E-hA(T-T_\infty)$ has steady value $T_\infty+\dot E/(hA)=22+2361=2383\ ^\circ\text{C}$, giving$$t_a=\tau\ln\frac{T_\text{ss}-T_i}{T_\text{ss}-T_1}=1010\ln\frac{2383-22}{2383-140}=\boxed{51.8\ \text{s}}.$$
(b) Off-time (convective cooling only). With no generation, $T(t)=T_\infty+(T_\text{hi}-T_\infty)e^{-t/\tau}$, so cooling from 140 °C to 125 °C takes$$t_b=\tau\ln\frac{140-22}{125-22}=1010\ln\frac{118}{103}=\boxed{137\ \text{s}}.$$

Check

The steady temperature (2383 °C) is far above the operating range, so heat-up is nearly linear and the 12 W/m²·°C convective loss is minor during the 52 s ramp — consistent with a real iron reaching soleplate temperature in under a minute. Convection is assumed from the full 0.03 m² face; if only part of the plate is exposed, $\tau$ and the off-time (b) scale inversely with the active area, while the heat-up time (a) barely moves — with $T_\text{ss}\gg140\ ^\circ\text{C}$ it reduces to $t_a\approx mc\,\Delta T/\dot E=363.6(118)/850=50.5$ s, which contains no $h$ or $A$ at all.

Quantity	Result
Biot number / time constant	3.6×10⁻⁴ / τ = 1010 s
(a) Time to reach 140 °C	≈ 51.8 s
(b) Heater off-interval (140→125 °C)	≈ 137 s

Concept & Theory

Topic. Lumped-capacitance transient with internal heat generation and thermostatic cycling.

Key relations. $\mathrm{Bi}=hL_c/k$; with generation $t=\tau\ln\!\frac{T_\text{ss}-T_i}{T_\text{ss}-T}$, $T_\text{ss}=T_\infty+\dot E/hA$; free cooling $T=T_\infty+(T_0-T_\infty)e^{-t/\tau}$; $\tau=\rho Vc_p/(hA)$.

Why this works. A small Biot number means conduction inside the plate is far faster than convection off it, so one uniform temperature describes the whole plate and the energy balance is an ordinary differential equation in time. The same $\tau$ governs both the powered heat-up and the unpowered cool-down; only the driving (steady) temperature changes.

Common pitfall. Using $\alpha$ where $k=\alpha\rho c_p$ is needed for Biot, or forgetting the 85 % factor so that $\dot E=1000$ W rather than 850 W.

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §5.1–5.3 (lumped capacitance, with and without generation).

Question 6 of 8: Transient Heating of a Flat-Iron Baseplate

Question 6: Transient Heating of a Flat-Iron Baseplate (equal value)