IMPROVEMENT WITH RADIATION SHIELD. A shield having an emissivity of 0.1 Example 8.13 is placed around the thermometer in Example 8.12 such that the area ratio A/A, is 0.3. The temperature indicated by the thermometer under these conditions is 5°C. The convection heat- transfer coefficient is assumed to be the same, and the shield may be assumed essentiallv to surround the thermometer. Calculate the true air temperature under these circumstances. And find estimate its uncertainty? , = 0.1, A h = 10 W/m2-C E= 0.9, 0.3, T, = 273 +5 278 K, T, = 263 K %3D %3D %3D A,

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Chapter1: Basic Modes Of Heat Transfer
Section: Chapter Questions
Problem 1.68P
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IMPROVEMENT WITH RADIATION SHIELD.
A shield having an emissivity of 0.1
Example 8.13
is placed around the thermometer in Example 8.12 such that the area ratio A/A, is 0.3. The
temperature indicated by the thermometer under these conditions is 5°C. The convection heat-
transfer coefficient is assumed to be the same, and the shield may be assumed essentiallv to
surround the thermometer. Calculate the true air temperature under these circumstances. And find estimate
its uncertainty?
h = 10 W/m - °C
E, = 0.1,
A
= 0.3,
A,
E = 0.9,
T, = 273 +5 = 278 K,
T, = 263 K
Transcribed Image Text:IMPROVEMENT WITH RADIATION SHIELD. A shield having an emissivity of 0.1 Example 8.13 is placed around the thermometer in Example 8.12 such that the area ratio A/A, is 0.3. The temperature indicated by the thermometer under these conditions is 5°C. The convection heat- transfer coefficient is assumed to be the same, and the shield may be assumed essentiallv to surround the thermometer. Calculate the true air temperature under these circumstances. And find estimate its uncertainty? h = 10 W/m - °C E, = 0.1, A = 0.3, A, E = 0.9, T, = 273 +5 = 278 K, T, = 263 K
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