Exergy analysis of solar radiation processes - Solar Energy Sciences and Engineering Applications - page 27

used, e.g., in the case of surface radiating to the forward hemisphere is:

β = π/ 2 ϕ = 2 π

d 2 C

≡ cos β sin β dβ dϕ = π (2.2.57)

β ϕ β = 0 ϕ = 0

Exergy of arbitrary polarized radiation . The exergy b A , W/m 2 , of the arbitrary
polarized radiation originating from unknown surface A and arriving in point P of the
considered surface A per unit time and unit absorbing surface area, can be interpreted
in equation (2.2.3) as B

b A . Developing the whole equation (2.2.3) by including also
interpretation (2.2.49) - (2.2.52), after rearranging, it yields:

=

b A = ( i 0, ν ,min + i 0, ν ,max ) cos β sin β dβ dϕ dν

ϕ ν

β

σT 0
3 π

− ( L 0, ν ,min + L 0, ν ,max ) cos β sin β dβ dϕ dν + cos β sin β dβ dϕ

ϕ ν ϕ

β β

(2.2.58)

In order to utilize formula (2.2.58) one has to know the solid angle ω within which
the surface A is seen from point P on surface A, and to know (e.g. from measurements)
i 0, ν ,min and i 0, ν ,max as a function of frequency ν and direction defined by β and φ .
The total exergy B A → A of the considered arbitrary radiation arriving to the all
points of surface A is calculated as:

B A → A = b A d A (2.2.59)

A

Exergy of arbitrary non-polarized radiation . The formula for such radiation is
obtained after taking into account in formula (2.2.58), that for a non-polarized radia-
tion i 0, λ ,max = i 0, λ ,min, thus i 0, λ ,max + i 0, λ ,min = 2 · i 0, λ . Additionally also L 0, λ ,max = L 0, λ ,min

thus L 0, λ ,max + L 0, λ ,min = 2 · L 0, λ .

2 2 b A = i 0, ν cos β sin β dβ dϕ dν − L 0, ν cos β sin β dβ dϕ dν

ϕ ν ϕ ν

β β

σT 0
3 π

+ cos β sin β dβ dϕ (2.2.60)

ϕ

β

In order to utilize formula (2.2.60) one has to know the solid angle ω within
which the surface A is seen from point P on surface A, and to know i 0, ν as function of
frequency ν and direction defined by β and ϕ . Formulae (2.2.59) can be also useful.