where C is the number of times the solar radiation is concentrated. If a focusing mirror concentrates C times and there are x multiple passes, then C = C'x. Equation (1) differs from Ref. 1 in that, instead of Br2, the expression Br2 - Br* is written to include the effect of depleting the Br2. The effect of depletion of Br2 is small for C<100 and pressures <1 Torr. The rate equation for the CO2 (001) = Nm is The first parentheses include the effect of depleting the CO2, which causes large differences to the results when the CO2 concentration is small. Deexcitation by Br2 is included; the latter was mentioned in Ref. 1 but not included in the final expression. Its inclusion with k6 = 8 x 10 15 cm3 s-1 reduces the results for the inversion population by about an order of magnitude. Using the steady-state solution of Br* from Eq. 1: II.2. Temperature of the Medium The lower lasing level is assumed to be the CO2 (100) state, at E100 = 0.17 eV above ground level. If the temperature of the gases is T, then by Boltzmann statistics Moo = Mooexp(-£1OO/AD = (CO2 - MxJexpf-1972/7) , (4) with k = Boltzmann's constant, T in K. The depletion of CO2 if Moi becomes appreciable at very high radiance is included. Clearly T must be kept as low as possible. As a first consideration the container wall will increase in temperature. If the average wall transparency is t, then a fraction (1 - r) of the incident radiation will be absorbed. However, the largest heat contribution to the wall comes from the gases. The collector of area Ac receives 1.4 kW m~2. If the overall laser efficiency is p, and pA is the absorption efficiency or fraction of the solar radiation absorbed, then 1.4AcpA (1 -p) kW have to be dissipated by radiation into space. Assume the heat is perfectly conducted to a radiator of area Ar and emissivity e, which is at approximately the same temperature (Tf) as the wall. Then by Stefan's law: The quantity t)a (l-r))/e = pA — 0.2 if p is small. The area Ar could be made equal to Ac by using the back of the collector to radiate, in which case Tw would be about 300 K; if Ac/Ar = 40, then, Tw = 700 K. However, the heat has to be conducted efficiently to the radiator, and so the problem of weight arises. The temperature of the gases can be roughly estimated by considering the energy
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