Question

Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of 20C

Answer 1

Answer: Rate of heat transfer q = 517.4W

Step-by-step explanation:

Thermal conductivity is a material property that describes its ability to conduct heat. Thermal conductivity is the quantity of heat transmitted through a unit thickness of a material, in a direction normal to a surface of unit area,due to a unit temperature gradient under steady state conditions. It can be shown mathematically using the equation below;

Q/t = kA(∆T)/d

q = kA(∆T)/d

q = Q/t = rate of heat transfer

Q = total amount of heat transfer

t = time

k = thermal conductivity of material

A = Area

d= distance

For the case above, the material used is still air.

k for still air at 20°C and 1bara = 0.02587W/mK

A = 1m^2

d = 1mm = 0.001m

∆T = 20°C = 20K

q = 0.02587×1×20/0.001

q = 517.4W

Therefore, the rate of heat conduction through the still air is 517.4W

Answer 2

The rate of heat conduction through the layer of still air is 517.4 W

Step-by-step explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

`Q =(KA(\delta T))/(L)`

`Q =(25.87*1*20)/(1)`

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W