exp ( κh/ cos θ) whereas as that of the diffuse source is never zero (expect as h ).

Transcription

1 Homework 3: Due Feb Solution done in class The transmissivity along any dection is exp ( κh/ cos θ) where h is the slab thickness and θ is the angle between that dection and the normal to the slab. This is the transmissivity of a mono-dectional beam. The transmissivity of a diffuse source is exp ( κh/ cos θ ), where cos θ 1/1.7. The transmissivity of a normal incident beam will be greater than that of the diffuse source because the path length through the slab is least. But for some angle of incidence of the beam, the path length will be greater than the average path length for the source (about 53 degrees), and thus transmissivity of the beam will be less than that of the diffuse source. At glancing incidence the transmissivity is zero, whereas as that of the diffuse source is never zero (expect as h ). If reflection is included, assume that the reflectivity of the slab increases with increasing angle of incidence to 1 at glancing incidence. Again, the transmissivity of the normally incident beam is even greater than that of the diffuse source, but at some angle of incidence will be less and will decrease to zero at a faster rate because the reflectivity is increasing to As noted in the question, the trivial solution is with the sun below the horizon. Expect for scattering by a molecules, no ultraviolet radiation from the sun reaches Earth. So take the sun to be above the horizon but low in the sky. Fst consider what happens without a scattering layer. Because of the long slant path through the ozone, attenuation of UV because 1

2 of absorption is large. Now add a scattering layer above the ozone. This layer does scatter some incident UV radiation away from Earth. But it aslo scatters some UV radiation toward Earth, and the average path length of this radiation through the ozone may be less than the slant-path length. Scattering takes away but also gives in the form of shorter path lengths through absorbing ozone. Because attenuation by absorption is an exponential function of path length, the decreased attenuation because of shorter path lengths in dections toward Earth can be greater than scattering in dections away from Earth. Where we would most likely expect this to occur is at high latitudes at times of the year when the sun is low in the sky. When the sun is high in the sky, a scattering layer above the ozone can only decrease the UV radiation reaching the Earth The essential ingredient here is to take the equation Q = Mc dt dt where M is the coffee mass, c is the heat capacity, and Q the oven power. Thus the time requed to change the temperature by T is τ Mc T Q Taking the mass of an 8 ounce cup of coffee as about 0.22 kg, for T = 65 C and Q = 1500 W, the result is t 40 s. This is in accord with what is observed: it doesn t take long, about a minute, to heat a cup of coffee in a microwave oven The normal emissivity of a cloud of thickness h is ε = 1 exp ( κh) = 1 exp ( NC abs h) 2

3 where N is the number of droplets per unit volume (assumed uniform) and C abs is the absorption cross section of a droplet. Multiply and divide the argument of the exponential by v, the volume of a cloud droplet: ε = 1 exp( NvC abs h/v) The quantity Nv is the volume of cloud water per unit cloud volume. If A ios the crosssectional area of the cloud, the total volume of cloud water is NvhA. Divide this by A to obtain the liquid water path h W, the thickness of a slab of liquid water that would result from compressing the cloud into a liquid layer. Thus the emissivity of the cloud is ε = 1 exp ( h W C abs /v) From Fig in the text it follows that the volumetric absorption cross-section of a cloud droplet is approximately equal to the bulk absorption coefficient of water, κ b over a wide range of wavelengths. The emissivity of a slab of liquid water is approximately ε = 1 exp ( κ b h W ) if reflection is neglected. Thus the emissivity of a cloud is approximately that of a liquid layer with thickenss equal to the cloud liquid water path Ignore the terrestrial radiation balance of droplets at the top of the fog., which surely must be negative. This will overestimate the total (net) radiation absorbed by fog droplets. The rate of absorption of radiation by an illuminated droplet is an integral over the spetrum F (λ) C abs (λ) dλ 3

4 where F (λ) is the spectral radiance. This quantity integrated over time t is the total energy absorbed by the droplet. Set this equal to the amount of energy requed to completely evaporate the droplet: t F (λ) C abs (λ) dλ = ρvl v where l v is the latent heat, ρ its density, and v the droplet volume. This gives t = l v ρ F (λ) (C abs (λ) /v) dλ Given that we are interested in morning fog, the sun will be low in the sky. So let s assume a total radiance of 1000 W/m2, which is high That is F dλ = 1000 W m 2 Take half of this to be in the infrared. F (λ) (C abs (λ) /v) dλ = F (λ) (C abs (λ) /v) dλ + F (λ) (C abs (λ) /v) dλ vis From inspection of Fig and the spectrum Fig. 1.7 it seems evident that the major contribution to the total is over the infrared. Thus F (λ) (C abs (λ) /v) dλ F (λ) (C abs (λ) /v) dλ Using the mean value theorem, we can write F (C abs (λ) /v) dλ = C abs (λ) /v F (λ) dλ = C abs (λ) /v 500W/m2 4

5 where <> denotes a mean. The the time for evaporation is τ 1400 C abs /v hr The biggest unceratinty here is the value to choose for C abs /v due to its rapid change over the IR. This quantity increases more or less exponentially with wavelenght, but the spectral radiance decreases more or less exponentially. For τ to be, say, 0.5 hr C abs /v must be /m or 30 cm 1. This corresponds to a wavelength of about 2 µm, which is implausible given that the radiance is low(relative to the peak) at this wavelength (Fig. 1.7). So the conclusion is that morning fog cannnot dissipate morning fog in times of order one half hour In the prior problem a time was estimated for how long it would take to evaporate a droplet through absorption. Here, the goal is to determine the radiative equilibrium temperature of the droplet. Set up a radiative energy balance. Let T d be the temperature of the droplet T e the radiative temperature of its envonment, and F the incident spectral radiance. The energy balance is F C abs dλ + 4πa 2 σt 4 e = 4πa 2 σt 4 d Assume an emissivity of 1 for a droplet, which is not strictly correct but good enough. Divide both sidees by the droplet cross sectional area to obtain 1 4 F Q abs dλ + σt 4 e = σt 4 d The integral of F is 1000 W/m2, and about half is in the IR. We have Q abs = 2d/3C abs /v where d is the diameter. Take it to be 10 3 cm (10 microns) and so Q abs = (C abs /v). Form Fig the volumeteric absorption cross-section over the spectrum varies considerably, rising rapidly with increasing wavelength. Thus the contribution to the inte- 5

6 gral comes mostly from the IR. This F Q abs dλ F Q abs dλ where F dλ 500 W m 2 Now use the mean value theorem to obtain C abs /v + σt 4 e = σt 4 d where the units of C abs /v are in /cm. The envonmental absorption term for a temperature of 240 K is around 220 W/m2. So for the absorption term to be comparable would reque C abs /v to be around 10 3 /cm. This is not plausible given that from Fig the value of C abs /v in the region around λ = 2 µm is about 10 1 /cm, and this is at a wavelength where the infrared radiance has dropped by almost a factor of 10 from its value at λ = 1 µm. 6