# The azimuth of the sun

There is no special case that can easily be illustrated geometrically except, perhaps, the trivial one of the sun at noon, when the azimuth is zero by definition. The general case is capable of geometrical treatment but, since three dimensions are involved, it is preferred here to give the result:

Tan z = ^—T——— ,Sin h j. , (7.4)

Sin L cos h — cos L tan d

EXAMPLE 7.3

Calculate the azimuth of the sun in London on 21 June at 17.00 h British Summer Time, making use of equation (7.4).

As before, L is 51°, d is +23y° and h is 60°, so by equation (7.4)

Tan z = 0.866/(0.777 x 0.5 — 0.629 x 435)

= 7.53

Whence z is 82.5°W of S, or 262.5° from N.

If the hour angle, h, is given a negative sign for times before mid-day and a positive one

For times after, the sign of the azimuth is negative when it is east of south and positive

When it is west of south.

EXAMPLE 7.4

Calculate the solar altitude and azimuth on 21 January for Perth, Western Australia, at

15.0 h sun time, given that the latitude is 32°S.

The day number is 21 Hence, by equation (7.1),

D = 23.45 sin[360(284 + 21)/365]

= 23.45 sin[300.82]

= -20.14°

L = -32° and h = +45°. Hence, by equation (7.3)

Sin a = sin(-20.14) sin(-32) + cos(-20.14) cos(-32) cos(45)

= (-0.3443X~0.5299) + (0.9389)(0.8480)(0.7071)

= 0.7454

Whence

A = 48.2°

By equation (7.4)

Tan e = sin(45)/[sin(-32) cos(45) — cos(-32) tan(-20.14)]

= 0.7071/[(-0.5299)(0.7071) — (0.8480)(-0.3667)]

= 0.7071/(-0.0637)

= -11.10

Whence,

Z = -84.9°

Note that, since the sun is to the north of the observer in the southern hemisphere, the solar azimuth is 84.9° west of north.

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