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15 days ago

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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.

1 answer:

Daniel [329]15 days ago

6 0

Answer:

18.7 mm

Explanation:

The law of heat conduction as described by Fourier for a plate is:

q = -k/t * ΔT

In this equation

q: heat transfer per unit time and area

k: coefficient of thermal conductivity

t: plate's thickness

ΔT: difference in temperature

By rearranging the terms, we find:

t = -k/q * ΔT

t = -1.4/(-1200) * 16 = 0.0187 m = 18.7 mm (the negative value of q indicates heat is being released from the plate)

Thus, the concrete slab can have a maximum thickness of 18.7 mm.

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