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On a windy day, we can define the wind-chill temperature Twc as the air temperature that...

On a windy day, we can define the wind-chill temperature Twc as the air temperature that would give the same rate of heat loss if experienced on a calm day. Consider a day when the air temperature is 2°C and the wind speed is 12 m/s, giving a convective heat transfer coefficient of 45 W/m2⋅K (compared with 10 W/m2⋅K on a calm day). Assume a person has a 3 mm layer of skin (k = 0.37 W/m⋅K), a body surface area of 1.9 m2,clothing equivalent to 25 mm-thick cotton (k=0.06 W/m⋅K), and a body temperature of 37°C below the skin. Thermal radiation can be neglected.

a)Draw a thermal resistance network that represents the total heat transfer between the body’s internal temperature and the surroundings. Compute all thermal resistances (per unit area) for both the calm and windy days.You may ignore thermal radiation.

b)Compute the wind-chill temperature.

c)Compute the difference between the person’s internal temperature and skin surface temperature on the windy day.

d)Suppose a flask of coffee, initially at 37°C, is insulated with the same total thermal resistance as the person’s skin and clothing. Will the coffee freeze? Why or why not? (Assume the freezing point is the same as water, 0°C.)

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Answer #1

Draw a thermal resistance network. k,A h A k A Calculate the total thermal resistance per unit area for windy days 0.003 0.02Compute the wind chill temperature. Heat loss during windy days -Heat loss during calm days 37-T 37-2 0.447 0.525 37-2 (0.44737-7 T-T 0.447 1 37-7 T-7 0.447 45 37-7 0.447x 45 - 8.4914 °C 8.49 °OC Calculate the difference between the persons intermalT-T R, 37-2 0.447 7830 W/m Since, required amount of heat is greater than the heat loss from the body. Therefore, the coffee

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