B) What is the person\'s net radiation loss to the walls? The emissivity of skin

Question

B) What is the person's net radiation loss to the walls? The emissivity of skin is 0.97. The surface area of an adult human is about 1.8 m^2 . Suppose a person with a skin temperature of 34 degrees C is standing nude in a room where the air is 25 degrees C but the walls are 17 degrees C. There is a "dead-air" layer next to your skin that acts as insulation. If the dead-air layer is 5.0 mm thick, what is the person's rate of heat loss by conduction? What is the person's net radiation loss to the walls? The emissivity of skin is 0.97.

Explanation / Answer

For radiation, P = sigma*A*e*(T^4-To^4)----------------------1

sigma is the constant 5.67x10^-8, A is the surface area of the object, e is the emissivity,

T is the temperature of the skin (convert it to Kelvin),

and To is the temperature of the background (walls, convert to Kelvin).

For conduction, P=k*A*(Th-Tc)/L--------------------------2

Here, L is the thickness of the dead air layer,

Th is the hot temp (in Kelvin, probably), Tc is the cold temp, k is the coefficient of thermal conduction and A is the surface area again.

the value of coefficient of conductivity k is needed to solve this.

Then you add the power from conduction and radiation, and compare that number to 160W, and if they add up to less than 160 they feel cold, more they feel hot, close they feel comfortable, etc.

b. rate of loss of heat H = dQ/dt = A e sigms (T2^4 - T1^4)

H = 1.8* 0.97 * 5.67*10^-8 (307^4 -290^4)

H = 179.18 Watts

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