HW9 solution

# HW9 solution - 0.87 3 0.3 0 0 λ , μm τ λ 2400*3 = 7200...

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CHE/ME 109 Homework #9 Name____________________ 0.6 cm 6 cm glass bulb - 0.5 mm thick glass filament - 1 mm diameter Problem I. Halogen lamps have bulbs with a glass cylinder 0.7 cm in diameter x 6 cm long. A filament inside the glass cylinder is coiled tightly to create an inner cylinder that is 1 mm in diameter and the length of the bulb. The glass cylinder is filled with a low pressure inert gas. Ia. Assuming the temperature of the filament to be 2400K, what fraction of the radiation is emitted in the yellow wavelength range of light? Yellow wavelength range: 0.54 - 0.60 μm 2400*0.54= 1296 f λ = .004849 2400*0.60= 1440 f λ = .010176 f λ 2 (T) - f λ 1 (T) = . 010176-.004849 = 0.005327 answer is 0.53% Ib. What fraction of the total emitted power is transmitted through the glass if the transmissivity curve is as follows:

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Unformatted text preview: 0.87 3 0.3 0 0 λ , μm τ λ 2400*3 = 7200 f λ = .819217 2400*0.3 = 720 f λ = .000001 f λ 2 (T) - f λ 1 (T) = . 819217-.000001 = 0.819216 .819216*0.87 = 0.679 or 71.3% of the total power is transmitted through the glass. A 2 A 1 A 3 Problem III. Determine the view factors F13 and F23 between the rectangular surfaces shown in the figure above. Using figure 12-6 For F 13 , L 2 /W = 1/2 L 1 /W = 1/2 F 13 = 0.245 = F 31 F 3-&gt;(1+2) = F 31 + F 32 Find F 3-&gt;(1+2) L 2 /W = 1 L 1 /W = 1/2 F 3-&gt;(1+2) = 0.29 F 3-&gt;(1+2) = F 31 + F 32 .290 = 0.245 + F 32 Then F 32 = 0.045 = F 23 by reciprocity b b b a Problem IV. Determine the view factors from the very long grooves to the surroundings without using any view factor tables or charts. Neglect end effects. The grooves are as shown in the figure above....
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## This note was uploaded on 11/08/2010 for the course ME 109 taught by Professor Staff during the Spring '08 term at San Jose State University .

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HW9 solution - 0.87 3 0.3 0 0 λ , μm τ λ 2400*3 = 7200...

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