Calculate the heat flux (in w/m^2) through a sheet of a metal 14 mm thick if the temperatures at the two faces are 350 and 140°c. assume steady-state heat flow and that the thermal conductivity of this metal is 52.4 w/m-k. (b) what is the heat loss per hour (in j/h) if the area of the sheet is 0.42 m2? (c) what will be the heat loss per hour (in j/h) if a material with a thermal conductivity of 1.8 w/m-k is used? (d) calculate the heat loss per hour (in j/h) if the first metal is used and the thickness is increased to 24 mm.

Explanation:

The rate of conductive heat transfer in watts is:

q = (k/s) A ΔT

where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.

a)

Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:

q/A = (52.4 / 0.014) (210)

q/A = 786,000 W/m²

b)

Given that A = 0.42 m², we can find q:

q = (0.42 m²) (786,000 W/m²)

q = 330,120 W

A watt is a Joule per second.  Convert to Joules per hour:

q = 330,120 J/s * 3600 s/hr

q = 1.19×10⁹ J/hr

c)

If we change k to 1.8 W/m/K:

q = (k/s) A ΔT

q = (1.8 / 0.014) (0.42) (210)

q = 11,340 J/s

q = 4.08×10⁷ J/hr

d)

If k is 52.4 W/m/K and s is 0.024 m:

q = (k/s) A ΔT

q = (52.4 / 0.024) (0.42) (210)

q = 192,570 J/s

q = 6.93×10⁸ J/hr

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the results is a decrease in air temperature with an increase of altitude.

explanation: