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xxMikexx [17]
3 years ago
10

Determine F12 and F21 for the following configurations: (a) A long semicircular duct with diameter of 0.1 meters: (b) A hemisphe

re-disk arrangement, with hemisphere diameter of 0.1 meters:
Engineering
1 answer:
uysha [10]3 years ago
6 0

Answer:

long duct: 1.0 and 0.424

Hemisphere 1.0 ; 0.125; 0.5

Explanation:

For a long duct:

By inspection, F_{12} = 1.0

Calculating by reciprocity, F_{21} = \frac{A_{1} }{A_{2}F_{12}  }  = \frac{2RL}{\frac{3}{4}*2\pi RL  }* 1.0\\                                                         = 0.424

Hemisphere:

By reciprocity gives = 0.125

using the summation rule: F_{21} + F_{22} + F_{23} = 1

However, because this is a hemisphere, the value will be= 0.5 * 1

                                                                                               = 0.5

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In order to investigate the relation between surface treatment and optical properties, stainless steels (AISI 304 and 430) which were submitted to different chemical and mechanical surface treatments, were used. To increase the spectral selectivity, these surfaces were treated in air and in vacuum at different temperatures and times. The optical properties of these films were investigated. Visual and infrared spectral absorptances were measured at room temperature. The thermal hemispherical emittance and absorptance were obtained by a calorimetric method at 200°C. It was noticed that these chemically and mechanically treated stainless steel surfaces have good spectral properties without further oxidations. This is very important for high temperature uses. The best values are found for samples 7 and 8 under vacuum and air. These two samples with mechanically ground surfaces retained their selectivity and specularity after several hours oxidation. One can conclude that the surface ground treatment confers good selectivity on the steel surfaces for use in concentrating solar collectors with a working temperature of 500°C.

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Explanation:

3 0
3 years ago
A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m
Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

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velocity of the car = 20 m/s

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F c = 2666. 67 N

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μ = 0.136

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3 years ago
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