Heat Loss with 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, an insulation of rock wool 102 mm thick is installed. The thermal conductivity of the rock wool is k 10-4T°C(W/m· K). The inside surface temperature of the ceramic is T, = 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 T, between the ceramic and the insulation. [Hint : The correct value of k, for the insulation is that evaluated at the mean temperature of (T, + T;)/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.] 0.046 + 1.56 x %3!
Heat Loss with 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, an insulation of rock wool 102 mm thick is installed. The thermal conductivity of the rock wool is k 10-4T°C(W/m· K). The inside surface temperature of the ceramic is T, = 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 T, between the ceramic and the insulation. [Hint : The correct value of k, for the insulation is that evaluated at the mean temperature of (T, + T;)/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.] 0.046 + 1.56 x %3!
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![Heat Loss with 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, an insulation of rock wool 102 mm thick is
installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56 ×
10- T°C(W/m·K). The inside surface temperature of the ceramic is
T, = 588.7 K, and the outside surface temperature of the insulation is T,
K. Calculate the heat loss for 1.5 m of duct and the interface tem perature T,
between the ceramic and the insulation. [Hint : The correct value of k, for the
insulation is that evaluated at the mean temperature of (T + T3)/2. Hence, for
the first trial assume a mean temperature of, say, 448 K. Then calculate the heat
loss and T,. Using this new T,, calculate a new mean temperature and proceed
as before.]
= 311](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35c74e5e-a302-4aa3-a502-0c0c0abba88b%2F8a17d69e-f547-4e90-a7c5-05b35c17a9f8%2Frnmhrel_processed.png&w=3840&q=75)
Transcribed Image Text:Heat Loss with 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, an insulation of rock wool 102 mm thick is
installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56 ×
10- T°C(W/m·K). The inside surface temperature of the ceramic is
T, = 588.7 K, and the outside surface temperature of the insulation is T,
K. Calculate the heat loss for 1.5 m of duct and the interface tem perature T,
between the ceramic and the insulation. [Hint : The correct value of k, for the
insulation is that evaluated at the mean temperature of (T + T3)/2. Hence, for
the first trial assume a mean temperature of, say, 448 K. Then calculate the heat
loss and T,. Using this new T,, calculate a new mean temperature and proceed
as before.]
= 311
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