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September 17, 2014

September 17, 2014

Posted by **KAIT** on Thursday, August 18, 2011 at 4:12am.

i.Calculate the rate of conduction of heat loss in watts if the skin temperature is 28 C at a distance of 5 cm.

- PHYSICS -
**Jai**, Thursday, August 18, 2011 at 5:59amrecall that the rate of energy being transferred is given by:

q = -kA(dT/dx)

where

q = energy per unit time (in Watts)

k = thermal conductivity (W/m-K)

T = temperature (in Kelvin)

x = distance (in meters)

assuming steady-state heat transfer (meaning, at a particular location, the variable does not vary with time), the formula becomes:

q = -kA(T2 - T1)/(x2 - x1)

where subscripts 2 and 1 refer to final and initial values, respectively.

for air, k = 0.025 W/(m-K)

substituting,

q = -0.025*3*(22-28)/(0.05)

q = 0.0225 W

..this is the amount of heat that is being transferred from the skin to air at a distance of 0.05 meters.

*note that for change in temp, you may not convert Celsius to Kelvin since the conversion factor (which is, + 273) will just cancel each other, and answers will be the same:

in Kelvin: (22+273) - (28+273) = 22-28

hope this helps~ :)

- PHYSICS -
**drwls**, Thursday, August 18, 2011 at 6:56amI have no idea what the 5 cm "distance" represents. You cannot assume that the heat is conducted through a 5 cm layer of air, with a linear temperature gradient in between. That is not what happens. This is a natural convection problem. There is a dimensionless equation involving a parameter called the "Grashof Number" that will allow you to estimate the natural convection heat loss. It should be find a reference to it online.

- PHYSICS -
**drwls**, Thursday, August 18, 2011 at 7:11amFor a discussion of how to perform a natural convection problem correctly, see

http://en.wikipedia.org/wiki/Natural_convection#Natural_Convection_from_a_Vertical_Plate

What you want to calculate, to estimate the heat transfer, is a Newtonian "film coefficient" h or a "Nusselt number" Nu. To do this will require calculating the Grashof number first.

You may get about the same answer as you would get assuming heat is concucted through a 5 cm layer of still air, but this would be a lucky coincidence.

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