On the June solstice the Earth's declination is

PHYSICS · Multi-part · Fermi University Physics · Earth, Climate & Energy Physics · Energy resources and conversion · Energy resources and conversion

Energy from Physics, Ch The solar resource resolved: geometry, spectrum and the limits of concentration

Problem

\delta = 23.44^\circ

. At latitudes

\phi \ge 90^\circ - \delta

the Sun does not set at all on that day: the zenith angle stays below

90^\circ

for the full

24\,\mathrm{h}

, so the sunset hour angle is

\omega_s = \pi

. ((a)) Starting from the daily extraterrestrial insolation formula

H_0 = \dfrac{S_0T_{\rm day}}{\pi}\left(\dfrac{\bar d}{d}\right)^2\left[\omega_s\sin\phi\sin\delta + \cos\phi\cos\delta\sin\omega_s\right]

(with

\omega_s

in radians,

T_{\rm day} = 24\,\mathrm{h}

and the June orbital factor

(\bar d/d)^2 = 0.967

), write down the simplified form of

H_0(\phi)

that holds throughout the region of continuous daylight, and use it to identify the latitude in the northern hemisphere at which

H_0

is greatest on 21 June. ((b)) Estimate the value of

H_0

at that latitude, using

S_0 = 1361\,\mathrm{W\,m^{-2}}

, and compare it with the

52^\circ\mathrm{N}

value of

11.5\,\mathrm{kWh\,m^{-2}}

quoted for the same day on the same basis.