Determination of thermal conductivity. in determining the thermal conductivity of an insulating material, the temperatures were measured on both sides of a flat slab of 25 mm of the material and were 318.4 and 303.2 k. the heat flux was measured as 35.1 w/m2. calculate the thermal conductivity in btu/h-ft-ºf and in w/m-k q = 35.1 w/m 2 t 1 = 318.4 k t 2 = 303.2 k 25 mm

k=0.0577w/mk

k=0.03338BTU/h-ft-ºF

Explanation:

First we use the equation for heat transfer by conduction and solve for K

Q=KA(T2-T1)/x

Where

q=heat

A=area

T=temperature

K=conductivity

x=thickness

solving for k

k=Qx/A(T2-T1)

Q/A=Heat flux=q

k=qx/(T2-T1)

k=35.1*0.025/(318.4-303.2)

K=0.0577w/mk

remember that 1 btu / hftk equals 1,729 w / mk

K=0.0577/1.729=0.03338BTU/h-ft-ºF

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