Question

A heating cable is embedded in a concrete slab for snow melting. The heating cable is heated electrically with joule heating to provide the concrete slab with a uniform heat of 1200 W/ m2. The concrete has a thermal conductivity of 1.4 W/m⋅K. To minimize thermal stress in the concrete, the temperature difference between the heater surface (T1) and the slab surface (T2) should not exceed 16°C (2015 ASHRAE Handbook—HVAC Applications, Chapter. 51). Formulate the temperature profile in the concrete slab, and determine the thickness of the concrete slab (L) so that T1 − T2 ≤ 16°C.

Answer 1

Answer:

18.7 mm

Explanation:

Fourier's law for heat conduction on a plate is:

q = -k/t * ΔT

Where

q: heat conducted per unit of time and surface

k: thermal conductivity

t: thickness of the plate

ΔT: temperature difference

Rearranging:

t = -k/q * ΔT

t = -1.4/(-1200) * 16 = 0.0187 m = 18.7 mm (q is negative because it is heat leaving the plate)

The maximum thickness of the concrete slab is 18.7 mm.