Question

Just need help with D and E. Others have been solved

3. A south-facing collector at latitude 40° is tipped up at an angle equal to its latitude. Compute the following insolations for January 1st at solar noon: a. The direct beam insolation normal to the sun's rays b. Beam insolation on the collector c. Diffuse radiation on the collector d. Reflected radiation with ground reflectivity 0.2. e. Total insolation on the collector.

Answer 1

solution - As given lives in the problem: in Latitude = 40 Collector tipped up at angle of 48 (d) Reflected radiation le oeflected cadiation Striking the collector is given by DR = (fe) (Sirep +c) (e-casos) - . where, Ek = Reflected radiatica P - Goound reflectivity IB - Direct beam radiations ß - Altitude angle of Sun at Solar moon C = ske diffuse faclos Q = Tile angle of Colleclões Νοιώ, Direct beam radiation, Ib. can be calculated as EB= A 2 km - where. A = Apparent extraterrestrial fler = 11607 75 Sim (07-275)] . on is the day number of the year. I = Optical depth = 0.174 +0.035 sim (5 (97-1003] - © m = Aig-mass ratios and B = 900-L+8- 6. and 8 = 23.46 sim 3.60 (01-800] -0.

Now, sing equation (3, and putting n=g (ast day of January), A = 1160 + 75 som en (1-9750 A = 1160 +75 Sim =$160 + 75 Sim 1-270.246) HD = 1766775 (-0.062) = 1260-5.17 =1154.82 o Using equatioon © K = 0.174 +0.035 Sini) ? -0.474 +0.035 Sin 5-97,64] :0.174 +0.035 (0.948) } 50.174+ 0.00861 Preto prej ç 0.182 using eslations © ® and ☺ 8 = 28.45 Sim [ 260 C1-8139 = 23.45 Sim [-78.2] 523.45*0.3529 =8.26 B = 90-L+8 = 90° -40°+8,26° 58.26 m = sinB Sin 58.26 sig - 1.01

IB = A e lem on putting the value of A, u and in in equation ③ (0.192) (6.01) - 1154.82 e 51154.82 €©194 =1154,82 x 1.202 = 1388.11 wima. NOW, sley diffelse factors, C can be calcellared as, C=0.095+0.04 Stan [369 (1)-2001 ,00, C30.095+0.04 340 [-97.64] 70.095+0.04 (0.248) Beauty How, oor putting the value of variables in equation 0.10e have FR =P 10 (Stop+c) < 660SC) = 0.2 (1388.11) (sion se:26+0.05) (-cesto Jalan = 277.62 (1.095) (1-567) w/m2 = 253.37 W/m² Thus, the reflected radiation with goound reflectivity of 0.2 is 263.37 W/m².

nes Total Insolation on IBC is given as, isolation on collector, loc=IB Cose where, DB = Clear sky beam insolation Casofa cosp cos (450c) simo & Sm B Coso -6. Where, V is a Riometh angle of Sun measuoed relative to due south. this, 9-4,200 for collection facing south. putting the value of ß and o in equation ③ Cost = cos 58.2 coso sios tó + Sizz 58.28 Cos 4o on -0.764. Thus, the magnitude of total insolation on Collector is b c = 1388.11 x 0.764 wo/m2 = 1061.53 w/m²

Answer:

(d) Reflected radiation = 253.37 W/m2

(e) Total Insolation on collector = 1061.53 W/m2