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 30C 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 the duct. Use a grid spacing with delta x = 2*(delta y), where delta y = 0.125L and L=150mm.
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 30C 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 the duct. Use a grid spacing with delta x = 2*(delta y), where delta y = 0.125L and L=150mm.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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 30C 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 the duct. Use a grid spacing with delta x = 2*(delta y), where delta y = 0.125L and L=150mm.
Transcribed Image Text:Air duct-
1.5L
-T₂ = 30°C
L
-T₁ = 80°C
-L-
-Concrete
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