Mild steel nails were driven through a solid wood wall consisting of two layers, each 15 mm thick, for reinforcement. If the total cross- sectional area of the nails is 0.75% of the wall area, determine the unit thermal conductance of the composite wall, total heat flow, heat flow through the nail alone and the percent of the total heat flow that passes through the nails when the temperature difference across the wall is 35°C. Neglect contact resistance between the wood layers. Wood (Pine) (kyw) = 0.15 W/(m K); Mild steel (1% C) (kg) = 43 W/(m K)

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter1: Basic Modes Of Heat Transfer
Section: Chapter Questions
Problem 1.37P: 1.37 Mild steel nails were driven through a solid wood wall consisting of two layers, each 2.5-cm...
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Mild steel nails were driven through a solid
wood wall consisting of two layers, each 15
mm thick, for reinforcement. If the total cross-
sectional area of the nails is 0.75% of the wall
area, determine the unit thermal conductance
of the composite wall, total heat flow, heat
flow through the nail alone and the percent of
the total heat flow that passes through the
nails when the temperature difference across
the wall is 35'C. Neglect contact resistance
between the wood layers. Wood (Pine) (ky) =
0.15 W/(m K); Mild steel (1% C) (kg) = 43 W/(m
K)
а.
ANSWER:
W/m²-K
b.
ANSWER:
Transcribed Image Text:Mild steel nails were driven through a solid wood wall consisting of two layers, each 15 mm thick, for reinforcement. If the total cross- sectional area of the nails is 0.75% of the wall area, determine the unit thermal conductance of the composite wall, total heat flow, heat flow through the nail alone and the percent of the total heat flow that passes through the nails when the temperature difference across the wall is 35'C. Neglect contact resistance between the wood layers. Wood (Pine) (ky) = 0.15 W/(m K); Mild steel (1% C) (kg) = 43 W/(m K) а. ANSWER: W/m²-K b. ANSWER:
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