Answer:
The total rate of heat transfer is 138.8894 W
Explanation:
Step 1: Given data
⇒ Temperature in the room Tr= 23° C
⇒ Exposed surface area As= 1.7 m²
⇒ Person's skin temperature Ts= 32 ° C
⇒ heat transfer coefficient h : 3.5 W/m²⋅K
⇒ Emissivity ε = 0.9
⇒ ( Stefan Boltzmann constant σ = 5.67×10^−8 W⋅m^−2⋅K^−4
Step 2: Calculate Qrad
Heat transfer Q = Qrad + Qconv (Thermal properties of people are assumed to be constant. The heat transfer will take place due to both convective and radiation heat loss)
⇒ Qrad = Heat transfer from the person to the surroundings due to radiation
Q = ε*σ*As * (Ts ^4 - Tr ^4 ) + h* As*ΔT
⇒ Qrad =ε*σ*As * (Ts ^4 - Tr ^4)
Qrad = 0.9 * (5.7 * 10 ^ -8) * 1.7 * ((32 + 273.15)^4 -(18 +273.15)^4)
Qrad = 85.3394 W
<u>Step 3</u>: Calculate Qconv
Qconv = The heat transfer from the person to the surroundings due to convection
Qconv = h* As*ΔT
⇒ Q conv = 3.5 W/m²⋅K * 1.7 m² * (32 - 23)
Qconv = 3.5 * 1.7 * 9 = 53.55 W
<u>Step 4: </u>Calculate Q
Q = Qrad + Qconv = 85.3394 + 53.55 = 138.8894 W
The total rate of heat transfer is 138.8894 W
