Heat Loss with a Trial-and-Error Solution. The exhaust duct from a heater has an inside diameter of 114.3 mm with ceramic walls 6.4 mm thick. The average k 1.52 W/m K. Outside this wall, installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56 x 10 T°C (W/m K). The inside surface temperature of the ceramic is Ti 588.7 K, and the outside surface temperature of the insulation is T 311 K. Calculate the heat loss for 1.5 m of duct and the interface temperature T2 between the ceramic and the insu- lation. [Hint: The correct value of km for the insulation is that evaluated at the mean temperature of (T2+ T3)/2. Hence, for the first trial assume a mean temperature of, say, 448 K. Then, calculate the heat loss and T2. Using this new T2, calculate a new mean temperature and proceed as before.] an insulation of rock wool 102 mm thick is ds MIsenitysM WTOBTHMAA hOS SUTD clveT al Jolysl sbi hesnigne lsaibemola SOC 19 9T01a d di 199nigne loime. dhe GA 9120x

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
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Chapter4: Numerical Analysis Of Heat Conduction
Section: Chapter Questions
Problem 4.54P
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Heat Loss with a Trial-and-Error Solution. The exhaust duct from a heater has an
inside diameter of 114.3 mm with ceramic walls 6.4 mm thick. The average
k 1.52 W/m K. Outside this wall,
installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56 x 10 T°C
(W/m K). The inside surface temperature of the ceramic is Ti 588.7 K, and the
outside surface temperature of the insulation is T 311 K. Calculate the heat loss
for 1.5 m of duct and the interface temperature T2 between the ceramic and the insu-
lation. [Hint: The correct value of km for the insulation is that evaluated at the mean
temperature of (T2+ T3)/2. Hence, for the first trial assume a mean temperature of,
say, 448 K. Then, calculate the heat loss and T2. Using this new T2, calculate a new
mean temperature and proceed as before.]
an insulation of rock wool 102 mm thick is
ds
MIsenitysM
WTOBTHMAA
hOS
SUTD
clveT al
Jolysl
sbi
hesnigne lsaibemola
SOC
19
9T01a
d di
199nigne loime. dhe
GA
9120x
Transcribed Image Text:Heat Loss with a Trial-and-Error Solution. The exhaust duct from a heater has an inside diameter of 114.3 mm with ceramic walls 6.4 mm thick. The average k 1.52 W/m K. Outside this wall, installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56 x 10 T°C (W/m K). The inside surface temperature of the ceramic is Ti 588.7 K, and the outside surface temperature of the insulation is T 311 K. Calculate the heat loss for 1.5 m of duct and the interface temperature T2 between the ceramic and the insu- lation. [Hint: The correct value of km for the insulation is that evaluated at the mean temperature of (T2+ T3)/2. Hence, for the first trial assume a mean temperature of, say, 448 K. Then, calculate the heat loss and T2. Using this new T2, calculate a new mean temperature and proceed as before.] an insulation of rock wool 102 mm thick is ds MIsenitysM WTOBTHMAA hOS SUTD clveT al Jolysl sbi hesnigne lsaibemola SOC 19 9T01a d di 199nigne loime. dhe GA 9120x
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