A typical car's exterior consists of a thin layer of silica (SiO2) over an opaque painted metal panel.Silica is transparent in the visible wavelengths but offers high reflectance in the near- to mid-infrared wavelengths. The plot on the next page depicts the diffuse spectral reflectivity (pa) of thecar's surface:Spectral reflectivity, P₂0.80.60.4ལ0.200.11110Wavelength, λ(μm)100If the car's exterior temperature is T₁ = 77°C, determine both the total absorptivity (a) and the totalemissivity (a) of the silica-covered panel. Assume that the Sun's temperature is Tsun = 5800 K.

To determine the total absorptivity and total emissivity of the silica-covered panel, we can use the…

Answered: A typical car's exterior consists of a…

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter11: Heat Transfer By Radiation

Section: Chapter Questions

Problem 11.11P

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1.26 Repeat Problem 1.25 but assume that the surface of the storage vessel has an absorbance (equal to the emittance) of 0.1. Then determine the rate of evaporation of the liquid oxygen in kilograms per second and pounds per hour, assuming that convection can be neglected. The heat of vaporization of oxygen at –183°C is .
A tungsten filament is heated to 2700 K. At what wavelength is the maximum amount of radiation emitted? What fraction of the total energy is in the visible range (0.4to0.75m)? Assume that the filament radiates as a graybody.
11.31 A large slab of steel 0.1 m thick contains a 0.1 -m-di- ameter circular hole whose axis is normal to the surface. Considering the sides of the hole to be black, specify the rate of radiative heat loss from the hole. The plate is at 811 K, and the surroundings are at 300 K.
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An opaque surface with the prescribed spectral, hemispherical reflectivity distribution is subject to the spectral irradiation shown. Find: 1.0 600 0.5 0 05 10 15 λ (μm) G₂(W/m²•μm) 300 0 0 5 10 15 20 20 λ (μm) (1) Find the spectral absorptivity distribution (αλ). (2) Determine the total irradiation G on the surface (3) Determine the radiant flux that is absorbed by the surface. (4) The total absorptivity a.
Two point sources of coherent light with a wavelength of 700 nm are near each other. The light from these sources is interfering in space around them. If you determine there is destructive interference between the two sources at a point 70.35 μm away from the second source, how far away from this point might the first source be? (Note that the image shown is not drawn to scale.) Source 1 74.20 μm o 77.175 um - 80.85 um ○ 80.85 μm 80.85 μm ? Source 2 70.35 μm destructive interference here
Irradiation on a semi-transparent medium is at a rate of 640 W/m². If 160 W/m² of the irradiation is reflected from the medium and 130 W/m² is transmitted through the medium, 1) Determine the absorptivity of the medium. 2) Determine the reflectivity of the medium. 3) Determine the transmissivity of the medium.
The last portion asks you for "net radiant heat flux to the surface", meaning that positive net radiative heat flux means in and negative net radiative heat flux means out. This is opposite the typical sign convention - be aware of this
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! Required information Irradiation on a semi-transparent medium is at a rate of 640 W/m². If 160 W/m² of the irradiation is reflected from the medium and 130 W/m² is transmitted through the medium, Determine the absorptivity of the medium. The absorptivity of the medium is 0.75 X
Consider an ideal gas enclosed in a spectral tube. When a high voltage is placed across the tube, many atoms are excited, and all excited atoms emit electromagnetic radiation at characteristic frequencies. According to the Doppler effect, the frequencies observed in the laboratory depend on the velocity of the emitting atom. The nonrelativistic Doppler shift of radiation emitted in the x direction is f = f0(1 + vx / c). The resulting wavelengths observed in the spectroscope are spread to higher and lower values because of the (respectively) lower and higher frequencies, corresponding to negative and positive values of vx. We say that the spectral line has been Doppler broadened. This is what allows us to see the lines easily in the spectroscope, because the Heisenberg uncertainty principle does not cause signifi cant line broadening in atomic transitions. (a) What is the mean frequency of the radiation observed in the spectroscope? (b) To get an idea of how much the spectral line is…
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