Question

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/m^2. Calculate the thermal conductivity in btu/h*ft*Fahrenheit and in W/m*Kelvins

Answer 1

Part a

Thickness of slab x = 25 mm x 1m/1000mm = 0.025 m

Temperature T1 = 318.4 K

Temperature T2 = 303.2 K

Heat flux Q/A = 35.1 W/m2

From the Fouriers law of heat conduction

Q/A = - k(T2 - T1) /x

35.1 W/m2 = - k (303.2 - 318.4)K / 0.025 m

35.1*0.025 = 15.2k

Thermal conductivity k = 0.0577 W/m-K

Part b

Thickness of slab x = 25 mm x 1m/1000mm

= 0.025 m * 3.28ft/m

= 0.082 ft

T(F) = 1.8 * (T(K) - 273) + 32

Temperature T1 = 1.8 * (318.4 - 273.15) + 32 = 113.45 F

Temperature T2 = 1.8 * (303.2 - 273.15) + 32 = 86.09 F

Heat flux Q/A = 35.1 W/m2

= 35.1 J/s-m2 * (1BTU/1055.06J) * (3600s/h) * (1m/3.28ft)^2

= 11.132 BTU/h-ft2

From the Fouriers law of heat conduction

Q/A = - k(T2 - T1) /x

11.132 BTU/h-ft2 = - k (86.09 - 113.45)F/ 0.082 ft

11.132*0.082 = 27.36 k

Thermal conductivity k = 0.0334 BTU/h-ft-F

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