how does the latitude of a location impact the insolation that it receives?
TRANSCRIPT
HOW DOES THE LATITUDE OF A LOCATION IMPACT THE
INSOLATION THAT IT RECEIVES?
LATITUDE AND ZENITH ANGLE AT NOON ON MARCHAND SEPTEMBER 21.
SUN
Equator
Center of Earth
N
S
LATITUDE AND ZENITH ANGLE AT NOON ON MARCHAND SEPTEMBER 21.
SUN
North PoleLatitude = 90°N
South PoleLatitude = 90°S
EquatorLat. = 90°
LATITUDE AND ZENITH ANGLE AT NOON ON MARCHAND SEPTEMBER 21.
SUN
North PoleLatitude = 90°N
South PoleLatitude = 90°S
EquatorLat. = 90°
Z = 90°
Z = 90°
LATITUDE AND ZENITH ANGLE AT NOON ON MARCHAND SEPTEMBER 21.
SUN
Southampton, UKLatitude = 52°N
Punta Arenas, ChileLatitude = 52°S
Equator
LATITUDE AND ZENITH ANGLE AT NOON ON MARCHAND SEPTEMBER 21.
SUN
Southampton, UKLatitude = 52°N
Punta Arenas, ChileLatitude = 52°S
Equator Lat = 52°
Z = 52°
LATITUDE AND ZENITH ANGLE AT NOON ON MARCHAND SEPTEMBER 21.
SUN
Equator Lat = 52°
Z = 52°
Geometry:- “Transversal Postulate.”
Angle of Latitude = Noontime Zenith Angle
LATITUDE AND ZENITH ANGLE AT NOON ON MARCHAND SEPTEMBER 21.
SUN
GainesvilleLatitude = 29.5°N
Durban, SA.Latitude = 29.5°S
Equator
Lat = 29.5°Z = 29.5°
LATITUDE AND ZENITH ANGLE AT NOON ON MARCHAND SEPTEMBER 21.
SUN
Quito, EcuadorLatitude = 0°
Equator Lat = 0° Z = 0°
CAN WE QUANTIFY THIS EFFECT OF ZENITH ANGLE AT NOON AND LATITUDE?
SUN
N
S
Imaginary solarcollector of1 m2
SUN
Ten parallel rays representing the solar constant (345Wm-2)
SUN
Z = 90°: 0/10 Rays intercepted = 0/10*345 = 0Wm-2
North and South Poles (90°)
SUN
Z = 52°: 6/10 Rays intercepted = 6/10*345 = 207 Wm-2
Southampton and Punta Arenas (52°)
SUN
Z = 29.5°: 9/10 Rays intercepted = 9/10*345 = 310.5 Wm-2
Gainesville and Durban (29.5°)
SUN
Z = 0°: 10/10 Rays intercepted = 10/10*345 = 345 Wm-2
Quito (0°)
SUN
Cos (90) = 00 * 345 = 0
Z = 90°: 0/10 Rays intercepted = 0/10*345 = 0Wm-2
North and South Poles (90°)
SUN
Z = 52°: 6/10 Rays intercepted = 6/10*345 = 207 Wm-2
Southampton and Punta Arenas (52°)
Cos (52) = 0.620.62 * 345 = 210
SUN
Z = 29.5°: 9/10 Rays intercepted = 9/10*345 = 310.5 Wm-2
Gainesville and Durban (29.5°)
Cos (29.5) = 0.870.87 * 345 = 310
SUN
Z = 0°: 10/10 Rays intercepted = 10/10*345 = 345 Wm-2
Quito (0°)
Cos (0) = 1.001.00 * 345 = 345
ANSWER
• Q = Cos(Z) * S
Where Q = Quantity of insolation (Wm-2)
Z = Zenith angle (noontime value controlled by latitude)
S = Solar constant (345 Wm-2)