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# Physical Meteorology ESCI 340

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This 11 page Class Notes was uploaded by Cordie Miller on Thursday October 15, 2015. The Class Notes belongs to ESCI 340 at Millersville University of Pennsylvania taught by Alex DeCaria in Fall. Since its upload, it has received 43 views. For similar materials see /class/223514/esci-340-millersville-university-of-pennsylvania in Earth Sciences at Millersville University of Pennsylvania.

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Date Created: 10/15/15

ESCI 340 Physical Meteorology Radiation Lesson 7 Solar Radiation References An Introduction to Atmospheric Radiation Liou Radiation and Cloud Processes in the Atmosphere Liou Atmospheric Science An Introductory Survey Wallace and Hobbs Reading Petty Sections 741 through 743 DIRECT AND DIFFUSE RADIATION 0 We categorize solar radiation as either 0 Direct This radiation comes directly from the Sun without scattering or absorption 0 Diffuse This is solar radiation that has been scattered at least once Sun 0 THE SOLAR CONSTANT O The solar constant So is defined as the total solar flux at the top of the Earth s atmosphere on a surface perpendicular to the Sun s rays at a distance of the Earth s mean distance to the sun 1 AU The solar constant is really not constant it varies with the power output from the Sun itself which isn t constant 0 The solar constant is around 1368 Wmz One method of measuring the solar constant is to make use of Beer s law for a planeparallel atmosphere 0 Beer s Law for direct solar radiation passing through the atmosphere and reaching the ground is 10 31 Mira where m is the monochromatic solar intensity at the ground st is the monochromatic solar intensity at the top of the atmosphere and I is the optical depth of the entire atmosphere Taking the natural log of the above equation gives 10g0 10gIS1 7zu 0 If measurements are taken on a clear day at several different sun angles and log 30 is plotted against u the plot should be a straight line with slope of 7k and yintercept of log 151 this assumes that the optical depth remains fairly constant 0 This method not only determines the monochromatic solar intensity but can determine the amount of absorbers along the path as well This method is one way to determine the overhead column amount of things like ozone 0 If measurements are taken at many wavelengths and the results summed the total solar intensity at the top of the atmosphere and hence the solar constant with corrections made for Far UV and Far IR radiation which doesn t make it to the ground 0 Once the solar intensity is known the solar ux S at the top of the atmosphere can be found 0 Since the ux at the top of the atmosphere varies inversely with the square of the distance from the sun d the solar constant S0 is related to S via d 2 where dm is the mean distance of Earth from the Sun 1 AU 0 The distance to the Sun can be found from i i dm 1 ec0san where sis the eccentricity of the Earth s orbit 0167 n is the number of days from aphelion and dis the orbital rate of the Earth 09863 day INSOLATION O Insolation is defined as the solar irradiance upon a horizontal area not necessarily perpendicular to the Sun s rays 0 Insolation is a contraction for INcoming SOLar radiATION 2 The solar insolation F S at the top of the atmosphere is the total solar ux perpendicular to the solar rays multiplied by the cosine of the zenith angle d 2 FSSCOSQS0 7 c050 The insolation integrated with time over the hours of daylight will give the energy per unit area per day received at a point on the Earth The relationship between the insolation and the total energy per unit area Q is FS dQdt so that Q j FS dt The total energy per unit area per day received at the top of the atmosphere is then found by tI d 2 t QM Sdt S cos 0tdt 1 o d J where t is the time of sunrise and ts is the time of sunset From the expression we have for zenith angle we get d 2 quot QM S0 J sinism 6 cos lcos 6coshdt The diagram below shows a latitude circle The shaded region represents the nighttime portion of the latitude circle The hour angle at sunset is hs H while the hour angle at sunrise is h 27r H H is the halfday angle 0 Sun h r hr2H 272H sunset smug O The angular velocity of the Earth is Q illa2 so that we can substitute a M0 to get 2 9 g A lmxlm5coico5cohdh 0 Upon integration we get Q 2h 7h xmlxmticoico5 Q d mh mm and from the relations h 7h 2H 7 27 2H and th rxmh th 71n2 rHmH me 2th we get Q EVAT HunhmJc0ic05mH Q d o u formula anquot me Mr cosH rtanltanti VERTICAL PROFILE OF ABSORPTION and heats the atmosphere 0 At the levels where absorption is occurring illdz gt 0 0 At levels where there is little absorption 411 a39z will be very small 0 The diagram below shows possible profiles of and dIidz and their relationship to absorption 111112 absurptmn 111112 absurptmn maximum absorption maximum absorption 111112 absurptmn 111112 absurptmn I tilldz some levelz is given by Mz 11ZJIAM whereh z is the intensity at level z andIAw is the monochromatic intensity at the top of the atmosphere From this we see that E K amp dz dz 39 d t Mz gt 0 level z o Thelarger dbldz the greater the absorption 0 If dtl dz is zero then there is no absorption Since we know from Beer s Law that tl expTd1Iu then the measure of absorption is given as dt 1 d 7d A 7 7 iex 7 dz dz dl a and we also know from Lesson 4 that d1 7 k dz 0 1 so the measure of absorption is 0 k1 ex T dz P M If we want to nd at what optical depth the absorption will be maximum we need to take the derivative of the above equation with optical depth and set it equal to zero If we do this we get 2 0 dz u39o Equation 1 tells us that the maximum absorption will occur at the altitude where the vertical gradient of density apaz is equal to k1 u 02 This altitude will depend on the density distribution of the atmosphere 1 LEVEL OF MAXIMUM ABSORPTION IN AN EXPONENTIAL ATMOSPHERE To a reasonable approximation the vertical density profile of Earth s atmosphere is given by NJ po eXpzHa where H is called the scale height of the atmosphere and is given by H Rdrg where T is a representative average temperature In an exponential atmosphere the optical depth is see exercises 70 ltpo Putting 2 into 1 results in 2 3 ltan l1 a which from 3 is just TM 4 Equation 4 gives us the very important result that in an exponential atmosphere the maximum absorption will occur at that level where the optical depth is equal to the cosine of the zenith angle The path optical thickness from the top of the atmosphere to some level 1 is related to the optical depth via 751 Tanl1 a 1 1 so we can also state that in an e the level of absorption will occur at that altitude where the path optical thickness from the top of the atmosghere is equal to one A word of caution In the above analysis we ve assumed that the molecules causing the extinction are well mixed This is not always the case particularly for the variable gasses SOLAR HEATING RATES Solar radiation at any level in the atmosphere consists of a diffuse component and a direct parallel beam component At any level in the atmosphere there is radiation arriving from both the upward and downward directions Denoting the downward uxes with a minus sign and upward uxes with a plus sign we have F 7 F direct FF diffuse F dim 0 Note Remember that ux and irradiance are the same thing We are using the ux notation here because physically it is more intuitive but we could just as well called this the upward and downward irradiances 0 Also keep in mind that we are only dealing with solar fluxes solar radiation in this section We are not accounting for any radiation emitted by the Earth or atmosphere o The net ux at any level is qu F z7quotz am not the other way is that the downward ux is likely to he larger than the upward ux and wewould like the net ux to be a positive number It could still work out 0 FM is positive if downward and negative if upward o hut Mimiquot a easy to work with as shown below we F1 51 o The heating rate is proportional to the power ahsorhed via E E 7 F it 1 it 1 capacity cl dQdT this becomes A m 7 no aha F 0 We also can write the heat capacity in terms of specific heat area and Jz as c erp pAJz 2 so that dT 1 F712 an2t z z E pop 61 0 In the limit as 61 gt 0 then net net Fz z Fz an 6zgt0 6Z dz this is nothing more than the definition of a derivative from the rst Fundamental Theorem of Calculus Therefore we have the result that the heating rate is directly proportional to the derivative of the net ux with height dT L d quot 3 d pop dz EXERCISES 1 Calculate the difference in solar radiation at the top of the atmosphere between perihelion and aphelion 2 a nd the amount of energy received per square meter in one day at the top of the atmosphere at the latitudes and times in the table below Assume that d dm Date Latitude Energy per area per day Equinox 0 60N 90N Summer Solstice 90N 0 90S b At the Summer Solstice the sun angle is higher at the equator than at the North Pole and yet the maximum in energy is at the North Pole Why 5quot a Show that by differentiating jig p7 exp rM with respect to z and setting it equal to zero that it results in ffij pz 1 b Using the de nition of optical depth from Lesson 4 show that in an exponential atmosphere the optical depth is 7 k ApH 3 c Put the density pro le for an exponential atmosphere equation 2 into equation 1 above to show that it results in kApH u 10 d Show that in an exponential atmosphere that the altitude of maximum absorption is given by Zmax H1nk1HpoIu 4 UV radiation of wavelength 025um penetrates to an altitude of about 45 km above the surface of the Earth a Estimate the massextinction coefficient using the equation for zmax above Use a temperature of 288 K when calculating the scale height and assume a sealevel density of air of 123 kgm3 Also assume a zenith angle of zero b Radiation at 025um is absorbed primarily by ozone in the stratosphere Do you think your estimate of extinction coefficient in part a is very accurate If not explain why not 5 Using the data in the table below find the solar heating rates in Cday at 5 15 25 and 35 km You can find density by assuming a scale height of 81 km and a surface density of 123 kgm3 Use c 1006 Jkg39lK l

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