13–19 Determine the view factors F13 and F3 between the rectangular surfaces shown in Fig. P13–19.
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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 .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.Two large parallel plates with surface conditions approximating those of a blackbody are maintained at 816C and 260C, respectively. Determine the rate of heat transfer by radiation between the plates in W/m2 and the radiative heat transfer coefficient in W/m2K.
- Determine the total average hemispherical emissivity and the emissive power of a surface that has a spectral hemispherical emissivity of 0.8 at wavelengths less than 1.5m, 0.6 at wavelengths from 1.5to2.5m, and 0.4 at wavelengths longer than 2.5m. The surface temperature is 1111 K.11.68 Two infinitely large, black, plane surfaces are 0.3 m apart, and the space between them is filled by an isothermal gas mixture at 811 K and atmospheric pressure. The gas mixture consists of by volume. If one of the surfaces is maintained at 278 K and the other at 1390 K, calculate (a) the effective emissivity of the gas at its temperature, (b) the effective absorptivity of the gas to radiation from the 1390 K surface, (c) the effective absorptivity of the gas to radiation from the 278 K surface, and (d) the net rate of heat transfer to the gas per square meter of surface area.1.28 The sun has a radius of and approximates a blackbody with a surface temperature of about 5800 K. Calculate the total rate of radiation from the sun and the emitted radiation flux per square meter of surface area.
- Need detailed Solution ,Previous One not enough. 1) For two surfaces, labelled 1 and 2, the view factor F_12 is 0.4. Surface 2 has twice the area of surface 1, what is the view factor from surface 2 to surface 1?A spherical particle of diameter D = 10 mm is suspended by a thin wire within a cylindrical tube of inner diameter D2 length L = 0.5 m. = 30 mm anc A4 A2 D2 L. -A1 -A3 Determine the view factors F11, F12, F13, and F14 for S = 80 mm. The view factor between the sphere and the bottom end of the tube (a disk of radius r3) is F13 = 0.5 1 – (1 + R)-2 where R3 = r3/S. F11 = i F12 = i F13 = i F14 =A spherical particle of diameter D = 10 mm is suspended by a thin wire within a cylindrical tube of inner diameter D, = 30 mm and length L = 0.5 m. A4 A2 - D2 L A1 A3 Determine the view factors F1,F12, F13, and F14 for S = 30 mm. The view factor between the sphere and the bottom end of the tube (a disk of radius r3) is F13 = 0.5 1- (1 + R})-12 where R3 = r3/S. F11 i F12 = i F13 = F14 =
- Determine the view factors fusand fo between the rectangular surtaces shown in FigureCalculate the view factor FBA for the following shape. The radii for the hollow circle, circle A and circle B are 8m, 5m, and10m, respectively. Vertical distance is 15m. circle A radii : 8m , 5m circle B radius : 10m vertical distance : 15mFor two surfaces, labelled 1 and 2, the view factor F_12 is 0.4. Surface 2 has twice the area of surface 1, what is the view factor from surface 2 to surface 1?