A common arrangement for heating a large surface area is to move warm air through rectangular ducts below the surface The ducts are square and located midway between the top and bottom surfaces that are exposed to room air and insulated, respectively For the condition when the floor and duct temperatures are 30 and 80C, respectively, and the thermal conductivity of concrete is 1 4 W m K, calculate the heat rate from each duct, per unit length of duct Use a grid spacing with x 2y, where y 0 125L and L 150mm Ty 30C Air duct Concrete 7 80C 3 1 5L

Question: A common arrangement for heating a large surface area is to move warm air through rectangular ducts below the surface. The ducts are square and

A common arrangement for heating a large surface area is to move warm air through rectangular ducts below the surface. The ducts are square and located midway between the top and bottom surfaces that are exposed to room air and insulated, respectively.

-Ty = 30°C Air duct- Concrete -7 = 80°C !3! 1.5L

For the condition when the floor and duct temperatures are 30 and 80Â°C, respectively, and the thermal conductivity of concrete is 1.4 W/m ∙ K, calculate the heat rate from each duct, per unit length of duct. Use a grid spacing with ∆x = 2∆y, where ∆y = 0.125L and L = 150mm.

-Ty = 30C Air duct- Concrete -7 = 80C !3! 1.5L

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KNOWN Rectangular air ducts having surfaces at 80C in a concrete slab with an insulated bottom and upper surface maintained at 30C FIND Heat rate from each duct per unit length of duct q SCHEMATIC Air ... View full answer

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