Question

Consider the base plate of an 800-W household iron with a thickness of L = 0.6 cm, base area of A =160 cm2, and thermal conductivity of k = 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron.

Answer 1

Given-

Power, P = 800W

Thickness, L = 0.6cm

Area, A = 160cm²

Thermal conductivity, k = 60W/mK

The heat conduction would be

$\frac{d^2T}{dx^2} + \frac{d^2T}{dy^2} + \frac{d^2T}{dz^2} + \frac{e(gen)}{k} = \frac{1}{\alpha } \frac{dT}{dt}$

Except $\frac{d^2T}{dx^2}$ all the values are 0.

Therefore,

$\frac{d^2T}{dx^2} = 0$

Thus, the boundary conditions here would be

1. $Q_x_=_0 = -kA \frac{dT (0)}{dx} = Q_o$

2. $T(L) = T_L$