Question

A person with external body temperature 35°C is present in a room at temperature 25°C. Assuming the emissivity of the body of the person to be 0.5 and surface area of the body of the person as 2.0 m2, calculate the radiant power of the person.

Answer 1

The person radiates energy at a rate:
 
P = Q / ∆t =A[(T1)^4 - (T2)^4] , T1>T2

here, P is the radiant power of the person,
  = 0.5 emissivity of the body of the person,
  = 5.672 x10^-8 J/(s.m^2.K^4) is the Stefan – Boltzmann constant,
  = 2.0m^2 is the surface area of the body of the person
 and
T1 is the temperature of the person = 35+273 =308K
and T2 is the temperature of the surroundings = 25+273  =298K

Then, the radiant power of the person will be:

P = A[(T1)^4 - (T2)^4]
P = 0.5 x 5.672 x 10^-8 x 2 x[(308)^4 - (298)^4]
P = 5.672 x 10^-8 x 1113028080
P = 6313095269.76 x 10^-8
P = 63.13 W

i hope it will help you 
regards

Answer 2

Answer:

Skin temperature (T₁) = 38°C =311 K Environment temperature (T₂)-0°C = 273

Emissivity (e) = 0.6

Area (A) 1.5 m²

Power loss

K

o Ae (T₁₁ - T₂¹) = 5.67 x 108 x 1.5 0.6 (311-273¹)

= 193 93 Watt

[Supp. 2073]

Soln: