A mountain climber wears down clothing 3.05 cmthick with total surface area 1.72
ID: 1757666 • Letter: A
Question
A mountain climber wears down clothing 3.05 cmthick with total surface area 1.72 m2. The temperatureat the surface of the clothing is -22.0°C and at the skin is34.3°C. Determine the rate of heat flow by conduction throughthe clothing assuming it is dry and that the thermal conductivity,k, is that of down.Determine the rate of heat flow by conduction throughthe clothing assuming the clothing is wet, so that k is that ofwater and the jacket has matted down to 0.590 cmthickness.
A mountain climber wears down clothing 3.05 cmthick with total surface area 1.72 m2. The temperatureat the surface of the clothing is -22.0°C and at the skin is34.3°C. Determine the rate of heat flow by conduction throughthe clothing assuming it is dry and that the thermal conductivity,k, is that of down.
Determine the rate of heat flow by conduction throughthe clothing assuming the clothing is wet, so that k is that ofwater and the jacket has matted down to 0.590 cmthickness.
Explanation / Answer
k(feathers) = 0.02 W/m/K ( I found this on the internet. Give methe exact parameter if you have something different) Flux = -k*dT/dx = -0.02 W/m/K *(-22 C - 34.3 C)/(3.05 cm - 0) *100cm/1 m = 36.92 W/m2 Heat flow = Flux *Area = 36.92 W/m2 *1.72 m2 = 63.5 Watts b) k(water) = 0.6 Flux = -k*dT/dx = -0.6 W/m/K *(-22 C - 34.3 C)/(0.590 cm - 0)*100cm/1m = 5725.42373 W/m2 Heat flow = Flux*Area = 5725.42373 W/m2 *1.72 m2 = 9848 W = 9.85kW
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.