Calculate the rate of heat loss per cm² through a stabilized zirconia (ZrO₂) lining of a high-temperature laboratory furnace operated at 1,400 degrees Celsius. The external face of the lining is at 100 degrees Celsius, and its thickness is 1 cm. Assume the data for stabilized zirconia in Table 7.4 are linear with temperature and can be If needed, extrapolated to 1,400 degrees Celsius. use the information on the following page.
Calculate the rate of heat loss per cm² through a stabilized zirconia (ZrO₂) lining of a high-temperature laboratory furnace operated at 1,400 degrees Celsius. The external face of the lining is at 100 degrees Celsius, and its thickness is 1 cm. Assume the data for stabilized zirconia in Table 7.4 are linear with temperature and can be If needed, extrapolated to 1,400 degrees Celsius. use the information on the following page.
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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![Calculate the rate of heat loss per cm² through a stabilized zirconia (ZrO2)
lining of a high-temperature laboratory furnace operated at 1,400 degrees Celsius. The
external face of the lining is at 100 degrees Celsius, and its thickness is 1 cm. Assume the
data for stabilized zirconia in Table 7.4 are linear with temperature and can be
If needed,
extrapolated to 1,400 degrees Celsius.
use the information on the following page.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14ad990f-dd7d-47e0-a13f-1f780fa26f58%2F4a330aaf-947b-469e-b29c-1e928325198e%2F8ajia1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Calculate the rate of heat loss per cm² through a stabilized zirconia (ZrO2)
lining of a high-temperature laboratory furnace operated at 1,400 degrees Celsius. The
external face of the lining is at 100 degrees Celsius, and its thickness is 1 cm. Assume the
data for stabilized zirconia in Table 7.4 are linear with temperature and can be
If needed,
extrapolated to 1,400 degrees Celsius.
use the information on the following page.
![TABLE 7.4
Values of Thermal Conductivity for a Variety of Materials
Material
Metalsa
Aluminum
Copper
Gold
Iron
Nickel
Silver
Titanium
Tungsten
Ceramics and glassesa, b
Mullite (3Al2O3-2SiO2)
Porcelain
Fireclay refractory
Al2O3
Spinel (MgO-Al2O3)
MgO
ZrO₂ (stabilized)
Tic
Silica glass
Soda-lime-silica glass
Temperature = 27°C (300 K)
237
398
315
80
91
427
22
178
k[J/(s.m.K)]
100°C
5.9
1.7
1.1
30.0
15.0
38.0
2.0
25.0
2.0
1.7
527°C (800 K)
220
371
292
43
67
389
20
128
1,000°C
3.8
1.9
1.5
6.3
5.9
7.1
2.3
5.9
2.5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14ad990f-dd7d-47e0-a13f-1f780fa26f58%2F4a330aaf-947b-469e-b29c-1e928325198e%2Foa6mrr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:TABLE 7.4
Values of Thermal Conductivity for a Variety of Materials
Material
Metalsa
Aluminum
Copper
Gold
Iron
Nickel
Silver
Titanium
Tungsten
Ceramics and glassesa, b
Mullite (3Al2O3-2SiO2)
Porcelain
Fireclay refractory
Al2O3
Spinel (MgO-Al2O3)
MgO
ZrO₂ (stabilized)
Tic
Silica glass
Soda-lime-silica glass
Temperature = 27°C (300 K)
237
398
315
80
91
427
22
178
k[J/(s.m.K)]
100°C
5.9
1.7
1.1
30.0
15.0
38.0
2.0
25.0
2.0
1.7
527°C (800 K)
220
371
292
43
67
389
20
128
1,000°C
3.8
1.9
1.5
6.3
5.9
7.1
2.3
5.9
2.5
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