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Consider a person with a skin layer of L = 3 mm thickness and with thermal conductivity of 0.3 W/m K. Assume that the inner part of the skin is at 35 C whereas the outer part of the skin is exposed to the environment. The surface are ais about 1.8 m2 and the emissivity of the skin can be taken as 0.95. When the person is in still air at 297 K, what is the skin surface temperature and the rate of heat loss to the environment?

Respuesta :

The skin surface temperature is 307.2 K and the  rate of heat loss is 146 W.

How to find skin surface temperature?

The skin surface temperature may be obtained by performing an energy balance at the skin surface to give the formula;

k(T_i - T_s)/L = h(T_s - T_∞) + h_r(T⁴_s - T⁴_surr)

Now, T_surround = T_s and so making T_s the subject gives us;

T_s = [(k*T_i)/L + (h + h_r)T_∞]/((k/l)

Using the equation h_r = εσ(T_s + T_surr)(T²_s + T²_surr) we can find h_r with a guessed value of T_s = 305 K and T∞ = 297 K, to yield h_r = 5.9 W/m².K.

Then, substituting numerical values into the T_s equation, we have;

T_s = [(0.3 * 308)/(3 * 10⁻³) + (2 + 5.9)297]/[((0.3 * 308)/(3 * 10⁻³)) + (2 + 5.9)]

T_s = 307.2 K

Thus the skin temperature is 307.2 K.

The rate of heat loss can be found by evaluating the conduction through the skin/fat layer:

q_s = kA(T_i - T_s)/L

q_s = 0.3*1.8(308 - 307.2)/(3 * 10⁻³)

q_s = 146 W

Read more about Skin Surface Temperature at; https://brainly.com/question/14866809

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