100m

1. Q7 part 1 angle = arctan(30/100) = 16.7 deg 30m 100m Solar noon ω = 0 DegLatitude  = 51.46 DegDue South facing  = 0 Deg Lowest height at solar noon Winter Solstice = -23.45 Solve for h  h= 15 Deg Answer Yes, will block the sun for at least some of the time

2. Q7 part 2 “s” angle = arctan(120/100) = 50.2 deg 85m 120m 100m 156.2m “h” angle = arctan(85/156.2) = 28.7 deg Q7 part 3 Rearrange and solve for   = 1.375 degrees Latitude  = 51.46 Deg Rearrange and solve for DoY DoY = - 281 = 84 We know (or you can calculate) that Summer Solstice – DoY = 172Therefore we can say that Shadow will just clear 88 days before summer solstice and therefore will start to fall 88 days after summer solstice.ie from day 260. Shadow DoY 1 to 84 and then 260 to 365

3. Q7 part 4 Spring Equinox  = 0 deg. This is DoY 81 Solar noon ω = 0 DegLatitude  = 51.46 DegSolve for h = 38.5 deg Gsol is 1367 W/m2 Gh,oa calculated as 857 W/m2Gh is 800 W/m2kt is calculated as 0.934 From data in exam paper ψ is 0.165 Gbeam,h = 668 W/m2 Gdiffuse,h = 132 W/m2 α = incl of plane = 51.46 deg = angle of plane to sun = 0(I worked this out from first principles and diagrams but is true of equinox, solar noon, facing south and inclined at latitude deg) G beam, i = 1073 W/m2

4. Q7 part 5 From before Gsol is 1367 W/m2 Gh,oa is 857 W/m2Gh is 800 W/m2 We are toldGbeam,h = 350 W/m2Therefore, Gdiffuse,h = 550 W/m2 Tilting the plate to the more horizontal α (incl of plane) to 0 deg Will improve the Gdiffuse,I but reduce Gbeam,I as  will increase There is a mathematical solution for the optimum αbut this question is already too long!