Question

A wood fire has a temperature of 600◦C. A person with a total surface area of 1.6 m2 and at a temperature of 34◦C stands next to the fire. If the emissivity is 0.95, determine the rate of radiative heat transfer, in kW. Assume that only one-half of the total person surface area is exposed to the fire.

Answer 1

We need to find the rate of radiative heat transfer, Q

Q = ε x σ x A x (T^{​​​​​​4}​​​​​ - t^{​​​​​​4})

Where,

ε = Emissivity = 0.95

σ = Stefan Boltzmann constant = 5.67 x 10^-8 W/m^{​​​​​​2}.K⁴

A = Exposed surface area = 1.6/2 = 0.8 m^{​​​​​​2}

T = Temperature of wood fire = 600°C = 600 + 273 = 873 K

t = Temperature of person = 34°C = 34 + 273 = 307 K

Substitute all the values

Therefore,

Q = 0.95 x 5.67 x 10^-8 x 0.8 x (873⁴ - 307⁴)

Q = 24647 W

1000 W = 1 kW

So, Q = 24.65 kW

Hope it helps.