African solar systems part 5: solar system tracking
Introduction
As we saw in part 4, heat can make up a significant fraction of the total energy budget and the total energy budget per day is far less than the incident solar energy. Hence, in principle, a CSP system might make sense for a small village. Such CSP systems are called micro-CSP's to contrast with the large CSP farms found in the western United States (among other places) that produce mega-Watts of power.
A CSP uses a mirror to focus sunlight onto a receiver, which absorbs the incoming radiation and heats up. For a given curved mirror, there's a slight complication - the sun moves across the sky as the earth rotates. Therefore the location where the light focuses constantly changes!
The constant motion of the focus point makes it impossible for certain mirror shapes (but not all) to provide concentrated sunlight without an automated tracking system installed on the mirror. A automated tracker is a combination of hardware and software that keeps the axis of the mirror aligned with the sun as it moves across the sky. Automated tracking systems decrease reliability, as they are an extra, complicated part and a CSP that requires automated tracking will simply stop operating if the tracking fails. For rural regions that both rely on the CSP for critical power and are far from technical expertise, such a point of failure is problematic. Therefore CSP systems can also be designed as a parabolic trough, where the sun tracks along the axis of the trough during the day. The trough still must be adjusted seasonally, but this is much less wear and tear.
The question
Consider a spherical concave mirror (not a trough) with a radius of curvature of 2 m. At noon on the equator the mirror is oriented so that the axis of the mirror is pointed directly at the sun. Let the focal point of the mirror be denoted by . After 15 minutes, the sun's rays focus at a new point . How far in meters is from ? You may assume the sun's rays are always perfectly parallel to each other and that the period of rotation of the earth is 24 hours.
The answer is 0.0654.
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1 solution
f = 2 R . . . . . . ( 1 ) 3 6 0 θ = 2 4 h r 1 5 m i n ( 6 0 m i n 1 h r ) θ = 2 4 h r ( 3 6 0 ) ( 1 / 4 ) = 3 . 7 5 . . . ( 2 ) t a n ( θ ) = l f ⟹ l = 2 R ( t a n ( 3 . 7 5 ) ≈ 0 . 0 6 5