Exergy analysis of solar radiation processes - Solar Energy Sciences and Engineering Applications - page 68

Second group losses. As results from formula (2.4.48) the external exergy loss
δB 1 − 0 is equal to the escaping exergy of three unabsorbed emissions at tempera-
tures T 1 = T S , T 2 and T 3 reflected to the environment. By multiplying these emissions

respectively by the exergy/energy ratio the three losses can be expressed as follows:

B 1 − 1 = Q 1 − 1 ψ S (2.4.54)

B 2 − 1 = Q 2 − 1 ψ 2 (2.4.55)

B 3 − 1 = Q 3 − 1 ψ 3 (2.4.56)

where ψ S , ψ 2 and ψ 3 are calculated from formula (2.2.45) for T S , T 2 and T 3 , respec-
tively, and Q 1 − 1 , Q 2 − 1 and Q 3 − 1 are the sums of the unabsorbed portion of the
respective emissions of surfaces 1, 2 and 3. Thus, heat Q 1 − 1 represents energy portions
from many reflections of emission E 1 of temperature T S , at the concave surface 2, and
arriving in surface 1:

Q 1 − 1 = E 1 ϕ 1 − 2 ρ 2 ϕ 21 + E 1 ϕ 12 ρ 2 ϕ 2 − 2 ρ 2 ϕ 2 − 1 + E 1 ϕ 12 ρ 2 ϕ 2 − 2 ρ 2 ϕ 2 − 2 ρ 2 ϕ 2 − 1 +···

(2.4.57)

The portions in equation (2.4.57) can be expressed as the sum of the terms of the
infinite geometric progression with the common ratio φ 2 − 2 · ρ 2 , thus

1

Q 1 − 1 = E 1 ϕ 1 − 2 ρ 2 ϕ 2 − 1 (2.4.58)

1 − ϕ 2 − 2 ρ 2

φ 2 − 1 of emission E 2 of surface 2 which
directly arrives at surface 1 and the portions in results of many reflections of emission
E 2 at surface 2, arriving at surface 1:

Heat Q 2 − 1 ( T 2 ) represents the portion E 2 ·

Q 2 − 1 = E 2 ϕ 2 − 1 + E 2 ϕ 2 − 2 ρ 2 ϕ 2 − 1 + E 2 ϕ 2 − 2 ρ 2 ϕ 2 − 2 ρ 2 ϕ 2 − 1 + ... (2.4.59)

and

1

Q 2 − 1 = E 2 ϕ 2 − 1 (2.4.60)

1 − ϕ 2 − 2 ρ 2

φ 3 − 1 of emission E 3 of surface 3 which
arrives at surface 1 as the direct radiation, and the portions as a result of many
reflections of emission E 3 at surface 2:

Heat Q 3 − 1 ( T 3 ) represents the portion E 3 ·

Q 3 − 1 = E 3 ϕ 3 − 1 + E 3 ϕ 3 − 1 ρ 2 ϕ 2 − 1 + E 3 ϕ 3 − 1 ρ 2 ϕ 2 − 2 ρ 2 ϕ 2 − 1

+ E 3 ϕ 3 − 1 ρ 2 ϕ 2 − 2 ρ 2 ϕ 2 − 2 ρ 2 ϕ 2 − 1 + ... (2.4.61)

and

E 3 ϕ 3 − 1

1

Q 3 − 1 = + ϕ 3 − 2 ρ 2 ϕ 2 − 1 (2.4.62)

−

1 ϕ 2 − 2 ρ 2