A slab measuring L = 1.0 m, W = thickness = 30 cm, D = depth = 0.5 m is insulated on all sides except at the top where it is exposed to the surrounding air with T = 25°C and h = 20 W/m^2·K. Heat is uniformly generated inside the slab at the rate 3 kW/m^3 and transferred by heat conduction in the y-direction. The thermal conductivity of the slab is 0.5 W/m·K.
1- What is the top surface temperature, °C?
2- Explain why eq. 13.3-6 of the text expressed as
To = (qzL^2)/(2k) + Tw
Can be used with no modification to calculate the slab bottom temperature? Then use the equation to calculate that temperature.
TW
T = 25°C, h = 20 W/m^2·K
= 0.3 m
W = 1.0 m