Radiation parallel plates

For two large parallel plates of equal areas, and separated by a small distance, it may be assumed that all of the radiation leaving plate 1 falls on plate 2, and similarly all of the... 

Our experimental techniques have been described extensively in earlier papers (2, 13). The gamma ray irradiations were carried out in a 50,000-curie source located at the bottom of a pool. The photoionization experiments were carried out by krypton and argon resonance lamps of high purity. The krypton resonance lamp was provided with a CaF2 window which transmits only the 1236 A. (10 e.v.) line while the radiation from the argon resonance lamp passed through a thin ( 0.3 mm.) LiF window. In the latter case, the resonance lines at 1067 and 1048 A. are transmitted. The intensity of 1048-A. line was about 75% of that of the 1067-A. line. The number of ions produced in both the radiolysis and photoionization experiments was determined by measuring the saturation current across two electrodes. In the radiolysis, the outer wall of a cylindrical stainless steel reaction vessel served as a cathode while a centrally located rod was used as anode. The photoionization apparatus was provided with two parallel plate nickel electrodes which were located at equal distances from the window of the resonance lamp. 

Using two pulsed tunable dye lasers, Na atoms in a beam are excited to an optically accessible ns or ml state as they pass between two parallel plates. Subsequent to laser excitation the atoms are exposed to millimeter wave radiation from a backward wave oscillator for 2-5 [is, after which a high voltage ramp is applied to the lower plate to ionize selectively the initial and final states of the microwave transition. For example, if state A is optically excited and the microwaves induce the transition to the higher lying state B, atoms in B will ionize earlier in the field ramp, as shown in Fig. 16.5. The A-B resonance is observed by monitoring the field ionization signal from state B at fB of Fig. 16.5 as the microwave frequency is swept. 

Two large parallel plates with grey surfaces are situated 75 mm apart one has an emissivity of 0.8 and is at a temperature of 350 K and the other has an emissivity of 0.4 and is at a temperature of 300 K. Calculate the net rate of heat exchange by radiation per square metre taking the Stefan-Boltzmann constant as 5.67 x 10-8 W/m2 K4. Any formula (other than Stefan s law) which you use must be proved. 

For two large parallel plates with grey surfaces, the heat transfer by radiation between them is given by putting Ai = A2 in equation 150 to give ... 

Several parameters are used to characterize the interaction of microwave radiation and matter the complex permittivity (e ), the dielectric constant O ). and the loss tangent (tan S). The dielectric constant, i-. can be thought of in a straightforward manner, as shown in Figure 5.15. Two parallel plates have a given capacitance, Co, when there is no material between them a vacuum. When the vacuum is replaced by a nonconducting medium, a dielectric, the new capacitance, C, is greater than Cq. The dielectric constant, e, is the ratio of these two capacitances ... 

Problem Two extremely large parallel plates at 800°C and 500°C exchange heat via radiation. Determine the heat transfer per unit area. Assume that s = 1 for the plates. 

Find an expression for the view factor between two parallel plates, Fi-2, of width L and a distance d apart, with an intervening plate placed at a distance d/2 from each of them as shown. There are 2-D openings spaced uniformly in the intervening plate that allows radiation from the one plate to the other, such that the openings and obstructions are of width L/9 each. Intuitively, explain what would happen to Fi 2 when the holes become very small, and show that the expression obtained provides the mathematical basis for your intuition. 

Note that the numerator expresses a potential and the denominator a resistance. If a series of n radiation shields with emittance cs is interspersed between the two infinite parallel plates, the equivalent resistance can be shown to be... 

When tl c end effects are negligible, radiation heat transfer between two large parallel plates at temperatures T and 7j is expressed as (see Chapter 13 for detail.s)... 

Minimize radiation heat transfer through the air space. This can be done by reducing the einissivity of glass surfaces by coating them with low-eiTiissivity (or "low-e" for short) material. Recall that the effective emissivity of two parallel plates of emissivities e, and 2 is given by... 

At high. surface temperatures (typically above 300°C), heat transfer acros.s the vapor film by radiation becomes significant and needs to be considered (fig. 10-12). Treating the vapor film as a transparent medium sandwiched between two large parallel plates and approximating the liquid as a blackbody, radiation heat transfer can be determined from... 

SOLUTION Two large parallel plates are maintained at uniform temperatures. The net rate of radiation heat transfer between the plates Is to be determined. Assumptions Both surfaces are opaque, diffuse, and gray. 

Radiation heat transfer between two large parallel plates of emissivi-lies e, and e maintained at uniform temperatures T and is given by Eq. 13-38 ... 

The radiation shield placed between two parallel plates and the radiation network associated with it. 

A thin aluminum sheet with an emissivity of 0.1 on both sides is placed between two very large parallel plates that are maintained at uniform temperatures Ti = 800 K and T2 = 500 K and have emissivities ci = 0.2 and et 0.7, respectively, as shown in Fig. 13-32. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the.result to that without the shield. 

Discussion Note that the rate of radiation heat transfer reduces to about one-fourth of v/hat it was as a result of placing a radiation shield between.the two parallel plates. 

Radiation heat transfer between two surfaces can be reduced greatly by inserting belsveen the two siirface.s thin, high-reflectivity (low emissivity) sheets of material called radiation shields. Radiation heat transfer between two large parallel plates separated by N radiation shields is... 

Two very large parallel plates are maintained at uniform temperatures of T, 1000 K and Tj = 800 K and have emissiviiies of e, = Cj = 0.5, respectively. It is desired to reduce the net rate of radiation heat transfer betweert the two plates to one-fifih by placing thin aluminum sheets with an emissivity of 0.1 on both sides between the plates. Defermine ihe number of sheets that need to be inserted. 

A radiation shield that has Ihe same emissivity 63 on both sides is placed between two large parallel plates, which are maintained at uniform temperatures of T, = 650 K and l = 400 K and have emissivities of e, = 0,6 and Sj = 0.9, respectively. Determine the emissivity of the radiation shield if... 

Two thin radiation shields with emissivilies of Cj = 0.10 and = 0.15 on both sides are placed between two very large parallel plates, which arc maintained at uniform temperatures 7 = 600 K and Tj = 300 K and have cmissivities e, = 0.6 and 62 — 0.7, respectively. Deteiinine the net rates of radiation heat transfer between the two plates with and without (he shields per unit surface area of the plates, and the temperatures of the radiation shields in steady operation. 

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