Question

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.

Answer 1

The heat flux is 785,600 W/m^2, and the heat loss per hour is 1,110,912,000 J/h for the first metal. Using a material with a thermal conductivity of 1.8 W/m-K, the heat loss is 42,504,000 J/h. With increased thickness to 24 mm, the heat loss is 491,820,000 J/h.

Step-by-step explanation:

Firstly, you would calculate the heat flux through the metal sheet using Fourier's Law of Heat Conduction, which is expressed as:

q = -k * (T_2 - T_1) / d

In this case, the temperature difference (T_2 - T_1) is (350°C - 140°C), the thickness (d) of the metal sheet is 14 mm (which is 0.014 meters), and the thermal conductivity (k) is given as 52.4 W/m-K. Substituting these values in:

q = -52.4 * (350 - 140) / 0.014 = 785,600 W/m^2

For part (b), to calculate the heat loss per hour, we multiply the heat flux by the area of the sheet and convert seconds to hours:

Q/t = q * A = 785,600 * 0.42 m^2 * 3600 seconds = 1,110,912,000 J/h

For part (c), using a material with a thermal conductivity of 1.8 W/m-K, the process is similar, and the heat loss per hour would be:

Q/t = -1.8 * (350 - 140) / 0.014 * 0.42 * 3600 = 42,504,000 J/h

For part (d), with the increased thickness of 24 mm (0.024 meters), the heat loss per hour would be:

Q/t = -52.4 * (350 - 140) / 0.024 * 0.42 * 3600 = 491,820,000 J/h

Answer 2

Step-by-step 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