Radiation Analysis Notes

1. λ =
c
ν
‘ ’
μ
–

2. ’
λ
λ
Ebλ (λ, T) =
C1 ∙ λ−5
(e
C2
λ∙T − 1)
W
m2 ∙ μm
C1 = 2 ∙ π ∙ h ∙ c0
2
= 3.74177 × 108
W ∙ μm4
/m2
C2 = h ∙
c0
k
= 1.439 × 104
μm ∙ K
λ
’
’

3. ’
(λT)max power = 2897.8 μm ∙ K
λ
⟶
d
dλ
[Ebλ(T)] = 0
⟶
d
dλ
[
C1 ∙ λ−5
(e
C2
λ∙T − 1)
] = 0
⟶ (e
C2
λ∙T − 1) ∙ C⃥1
∙ (−5λ−6) − C⃥1
λ⃥−5
∙
C2
T
∙ (
−1
λ2⃥
) e
C2
λ∙T = 0
⟶ −e
C2
λ∙T ∙ 5 + 5 +
C2
λT
e
C2
λ∙T = 0
⟶ −1 + e
−(
C2
λ∙T
)
+
1
5
∙ (
C2
λ ∙ T
) = 0
Let
C2
λT
= x,
⟶
x
5
+ e−x
− 1 = 0
x ≈ 4. .965
C2
λT
= 4.965 ⇒ λT =
C2
4.965
=
1.439 × 104
μm ∙ K
4.965
𝛌𝐓 = 𝟐𝟖𝟕𝟗. 𝟖 𝛍𝐦 ∙ 𝐊
λ
→ Eb = ∫ Ebλ
∞
0
(λ, T) dλ = ∫
C1 ∙ λ−5
(e
C2
λ∙T − 1)
∞
0
dλ
Let
C2
λT
= y → λ =
C2
yT
→ dλ = −
C2
T ∙ y2
dy
At λ = 0 → y = ∞ λ = ∞ → y = 0
→ ∫
C1 ∙ (
C2
yT)
−5
(ey − 1)
0
∞
∙ (−
C2
T ∙ y2
) dy ⟹
C1 ∙ T4
C2
4 ∫ y3 ∙ (ey − 1)−1
∞
0
dy
→
C1 ∙ T4
C2
4 ∫ y3 ∙ (e−y + e−2y + e−3y + ⋯ )
∞
0
dy
→
C1 ∙ T4
C2
4 ∫ y3 ∙ e−ny
∞
0
dy
→
C1 ∙ T4
C2
4 ×
3!
n4
=
C1 ∙ T4
C2
4 × 3! × (
1
14
+
1
24
+
1
34
+
1
44
+ ⋯ )
→
C1 ∙ T4
C2
4 × 6 ×
π
90
= (
C1
C2
4 ×
π
15
) × T4
= (
3.74177 × 108
(1.439 × 104
)
4 ×
π
15
) × T4
= (5.67 × 10−8) × T−4
= 𝛔 × 𝐓 𝟒
σ → Stephan Boltzmann Constant

4. – σ
λ λ ∞
λ λ
Eb,0−λ(T) = ∫ Eb,λ(λ, T)
λ
0
dλ
–λ
λ
f0−λ,T =
∫ Eb,λ(λ, T)
λ
0
dλ
∫ Eb,λ(λ, T)
∞
0
dλ
=
∫ Eb,λ(λ, T)
λ
0
dλ
σ ∙ T4
λ
λ λ
f0−λ,T =
∫ Eb,λ(λ, T)
λ
0
dλ
σ ∙ T4
= ∫
C1 ∙ λ−5
(e
C2
λ∙T − 1) ∙ (σ ∙ T4)
λ
0
dλ =
C1
σ
∫
(λT)−5
(e
C2
λT − 1)
λ
0
d(λT)
λ λ
λ λ λ λ
fλ1−λ2
(T) =
∫ Ebλ(λ, T)
λ2
0
dλ − ∫ Ebλ(λ, T)
λ1
0
dλ
σT4
= fλ2
(T) − fλ1
(T)
λ λ λ λ
μ
μ
λ1T = 0.4 × 5800 = 2320 μm − K, f0−λ = 0.125 (from table)
λ1T = 0.76 × 5800 = 4408 μm − K, f0−λ = 0.550 (from table)
fλ1−λ2
= 0.55 − 0.125 = 0.426 or 42.6%
ω

5. ω
dω =
dS
r2
=
(r ∙ sinθ ∙ dϕ) ∙ (r ∙ dθ)
r2
ω
dω =
dAn
r2
=
dA ∙ cosθ
r2
ω
θ ϕ
θ ϕ
Ie(θ, ϕ) =
dQ̇ e
dAn ∙ dω
=
dQ̇ e
dA ∙ cosθ ∙ dω
=
dQ̇ e
dA cos θ ∙ sin θ dθ dϕ
dE =
dQ̇ e
dA
= Ie(θ, ϕ) × (cosθ ∙ sinθ dθ dϕ)
E = ∫ dE
hemisphere
= ∫ ∫ Ie(θ, ϕ) ∙ cos θ ∙ sin θ dθ dϕ
π
2
θ=0
2π
ϕ=0
W/m2
E = π ∙ Ie
λ

6. G = ∫ dG
Hemsiphere
= ∫ ∫ Ie(θ, ϕ) ∙ cosθ ∙ sinθ dθ dϕ
π
2
θ=0
2π
ϕ=0
W/m2
J = ∫ ∫ Ie+r(θ, ϕ) ∙ cos θ ∙ sinθ dθ dϕ
π
2
θ=0
2π
ϕ=0
W
m2
G = Gα + Gρ + Gτ
Gα
G
+
Gρ
G
+
Gτ
G
= 1
α + ρ + τ = 1
α → ρ → τ →
α = 1 ⟶ black body
ρ = 1 ⟶ white body
τ = 1 ⟶ transparent body
→ Total Hemi spherical absorptivity, α =
Gα
G
→ Total Hemi spherical reflectivity, ρ =
Gρ
G
→ Total Hemi spherical transmissivity, τ =
Gτ
G
→ Spectral Hemi spherical absorptivity, αλ =
Gαλ
G
→ Spectral Hemi spherical reflectivity, ρλ =
Gρλ
G
→ Spectral Hemi spherical transmissivity, τ =
Gτλ
G
→ Spectral directional absorptivity, αλ,θ =
Gαλ,θ
Gλ,θ
α → given spectral variation of irradiation
→ α =
Gα
G
=
∫ Gα,λdλ
∞
0
∫ Gλdλ
∞
0
, αλ =
Gα,λ
Gλ
Gα,λ = αλ ∙ Gλ
→ α =
∫ αλ ∙ Gλ dλ
∞
0
∫ Gλ dλ
∞
0

7. → ρ =
∫ ρλ ∙ Gλ dλ
∞
0
∫ Gλ dλ
∞
0
→ τ =
∫ τλ ∙ Gλ dλ
∞
0
∫ Gλ dλ
∞
0
ε ≤ ε ≤
ε
ελ,θ(λ, θ, ϕ, T) =
Iλ,e(λ, θ, ϕ, T)
Ib,λ(λ, T)
εθ(θ, ϕ, T) =
Ie(θ, ϕ, T)
Ib(T)

8. ελ,θ =
Eλ(λ, T)
Ebλ(λ, T)
ε(T) =
E(T)
Eb(T)
→ ε =
∫ ελ ∙ Ebλ(T)
∞
0
∙ dλ
∫ Ebλ(T)
∞
0
∙ dλ
→ ε =
∫ ε1 ∙ Ebλ(T)
λ1
0
∙ dλ
σ ∙ T4
+
∫ ε2 ∙ Ebλ(T)
λ2
λ1
∙ dλ
σ ∙ T4
+
∫ ε3 ∙ Ebλ(T)
∞
λ2
∙ dλ
σ ∙ T4
→ ε = ε1 ×
∫ Ebλ(T)
λ1
0
∙ dλ
σ ∙ T4
+ ε2 ×
∫ Ebλ(T)
λ2
λ1
∙ dλ
σ ∙ T4
+ ε3 ×
∫ Ebλ(T)
∞
λ2
∙ dλ
σ ∙ T4
→ ε = ε1 × f0−λ1
(T) + ε2 × fλ1−λ2
(T) + ε3 × fλ2−λ3
(T)
→ α =
∫ αλ ∙ Gλ ∙ dλ
∞
0
∫ Gλ ∙ dλ
∞
0
=
∫ αλ ∙ Ebλ(T) ∙ dλ
∞
0
∫ Ebλ(T) ∙ dλ
∞
0
→ Gλ ∝ Ebλ(T)
’
ε α
→ Gincident = Eb(T) = σT4
→ Gabsorbed = α ∙ σT4
→ Eemitted = ε ∙ σT4
Eλ = αλ ∙ Gλ →
Eλ
αλ
= Gλ
ελ ∙ Ebλ = αλ ∙ Gbλ
As ∙ ε ∙ σT4
= As ∙ α ∙ σT4

9. → ε(T) = α(T)
→ ε =
∫ ελ ∙ Ebλ(T)
∞
0
∙ dλ
Eb
=
∫ αλ ∙ Ebλ(T) dλ
∞
0
Eb
= α (Gλ = Ebλ)
’ ’
→ Ebλ(T) =
C1 ∙ λ−5
e
C2
λT − 1
μ
1) λT ≫ C2 & 2) λT ≪ C2
①
→ e
C2
λT = 1 + (
C2
λT
) +
1
2!
(
C2
λT
)
2
+
1
3!
(
C2
λT
)
3
+ ⋯
Higher order terms of
C2
λT
are neglected as λT ≫ C2
→ e
C2
λT − 1 =
C2
λT
→ Ebλ(T) =
C1 ∙ λ−5
C2
λT
=
C1T
C2λ4
⟼ Rayleigh′
s Jean Law
②
→ e
C2
λT − 1 = e
C2
λT (C2 ≫ λT)
Ebλ(T) =
C1 ∙ λ−5
e
C2
λT
= C1λ−5
∙ e
−
C2
λT ⟶ Wein′
s Law
View factor =
dQ1
Q1
= F12
n1, n2 → normals to dA1 & dA2
θ1, θ2 → polar angles
r → distance
dω21 → solid angle subtended by dA2when viewed from dA1
I1 → radiation intensity emitted by surface dA1uniformly
dQ1 → radiation emitted by dA1that strikes dA2
Q1 → total radiation emitted by dA1

10. → dFdA1−dA2
=
dQ1
Q1
dQ1 = dA1 ∙ I1 cos θ1 ∙ dω1−2 = dA1 ∙ I1 cos θ1 ∙
dA2 ∙ cos θ2
r2
=
I1dA1dA2 cos θ1 cosθ2
r2
Also, Q1 = dA1 ∙ πI1
dFdA1−dA2
=
dQ1
Q1
=
I1dA1dA2 cos θ1 cos θ2
r2
×
1
dA1 ∙ πI1
=
cosθ1 cos θ2 ∙ dA2
πr2
→ dFdA1−A2
= ∫
cos θ1 cosθ2
πr2
A2
∙ dA2
→ dFA2−dA1
=
dA1
A2
∫
cos θ1 cos θ2
πr2
A2
∙ dA2
→ dFA2−A1
=
1
A2
∫ ∫
cos θ1 cos θ2
πr2
A1
∙ dA1dA2
A2
→ dFA1−A2
=
1
A1
∫ ∫
cos θ1 cos θ2
πr2
A2
∙ dA2dA1
A1
 …
𝐴𝑖 ∙ 𝐹𝐴𝑖−𝐴 𝑗
= 𝐴𝑗 ∙ 𝐹𝐴 𝑗−𝐴𝑖
∑ 𝐹𝐴 𝑗−𝐴 𝑘
𝑁
𝑘=1
= 1





→ 𝑁2
− [
𝑁(𝑁 − 1)
2
+ 𝑁 + 𝑁] =
𝑁(𝑁 − 3)
2
𝐹11 + 𝐹12 = 1
𝐹21 + 𝐹⃥22 = 1 → 𝐹21 = 1
𝐴1 𝐹12 = 𝐴2 𝐹21 = 𝐴2
𝐹12 =
𝐴2
𝐴1

11. 𝐹11 = 1 − 𝐹12 = 1 −
𝐴2
𝐴1
𝐹12 =
Σ𝐶𝑟𝑜𝑠𝑠𝑒𝑑 𝑠𝑡𝑟𝑖𝑛𝑔𝑠 − Σ𝑢𝑛𝑐𝑟𝑜𝑠𝑠𝑒𝑑 𝑠𝑡𝑟𝑖𝑛𝑔𝑠
2 × 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 1
=
(𝐿5 + 𝐿6) − (𝐿3 + 𝐿4)
2 × 𝐿1
‘ ’
cos 𝜃1 = cos 𝜃2 = cos 𝜃3 =
𝐿
𝑟
𝑟 = √𝐿2 + 𝜌2
𝑑𝐴2 = 𝜌 ∙ 𝑑𝜌 ∙ 𝑑𝜙
𝐹𝑑𝐴1−𝐴2
= ∫
cos θ1 cosθ2
πr2
A2
∙ dA2 =
1
𝜋
∫ ∫
𝐿2
∙ 𝜌
(𝐿2 + 𝜌2)2
2𝜋
𝜙=0
𝑅
𝜌=0
𝑑𝜌 ∙ 𝑑𝜙 = 2𝐿2
∫
𝜌
(𝐿2 + 𝜌2)2
𝑑𝜌
𝑅
0
𝐹𝑑𝐴1−𝐴2
=
𝑅2
𝑅2 + 𝐿2
𝑄1−2 = (
𝑅𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 1
𝑡ℎ𝑎𝑡 𝑠𝑡𝑟𝑖𝑘𝑒𝑠 2 𝑑𝑖𝑟𝑒𝑐𝑡𝑙𝑦
) − (
𝑅𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 2
𝑡ℎ𝑎𝑡 𝑠𝑡𝑟𝑖𝑘𝑒𝑠 1 𝑑𝑖𝑟𝑒𝑐𝑡𝑙𝑦
)
= 𝐴1 𝐸 𝑏1
𝐹12 − 𝐴2 𝐹21 𝐸 𝑏2
𝑄12 = 𝐴1 𝐹12 (𝐸 𝑏1
− 𝐸 𝑏2
) = 𝐴1 𝐹12 𝜎(𝑇1
4
− 𝑇2
4
)
𝑄1 = ∑ 𝑄𝑖−𝑗
𝑁
𝑗=1
= 𝐴𝑖 ∙ 𝐹𝑖−𝑗 ∙ 𝜎(𝑇1
4
− 𝑇2
4)
Ji → radiosity of surface
Gi → irradiation of surface
qi → net radiation temperatire at surface
qi = Ji − Gi

12. Ji = emitted component + reflected component
Ji = ϵEbi
+ ρGi
Ji = ϵiEbi
+ (1 − αi)Gi
Ji = ϵEbi
+ (1 − ϵi)Gi
Ji − Gi = ϵi(Ebi
− Gi)
Gi =
Ji − ϵiEbi
1 − ϵi
Ji − Gi = qi = ϵi (Ebi
−
Ji − ϵiEbi
1 − ϵi
)
qi = ϵi (
Ebi
− ϵiEbi
− Ji + ϵiEbi
1 − ϵi
) = ϵi (
Ebi
− Ji
1 − ϵi
)
Net rate heat transfer rate Qi = Aiqi
Qi =
Ebi
− Ji
(
1 − ϵi
ϵ1Ai
)
(Ebi
− Ji) → potential difference
(
1 − ϵi
ϵ1Ai
) → surface resistance
Qi =
Ebi
− Ji
R
Q1−2 = A1F12σ(T1
4
− T2
4) = A1F12 ∙ (J1 − J2)
Q1−2 =
J1 − J2
(
1
A1F12
)
→ Q̇ 1 = Q̇ 12 = −Q̇ 2
Q1−2 =
Eb1 − Eb2
R1 + R2 + R3
=
σ(T1
4
− T2
4)
1 − ε1
A1ε1
+
1
A1F12
+
1 − ε2
A2ε2

13. Q0 =
A ∙ σ ∙ (T1
4
− T2
4
)
1
ε1
+
1
ε2
− 1
𝐴1 = 𝐴2 = 𝐴3 = 𝐴
R1 =
1 − ε1
A ∙ ε1
R2 =
1
A ∙ F13
R3 =
1 − ε31
A ∙ ε31
R4 =
1 − ε32
A ∙ ε32
R5 =
1
A ∙ F32
R4 =
1 − ε2
A ∙ ε2
Q =
A ∙ σ ∙ (T1
4
− T2
4
)
(
1
ε1
+
1
ε2
− 1) + (
1
ε31
+
1
ε32
− 1)
⟶ Radiation with shield
→ Qwithout shield =
A ∙ σ ∙ (T1
4
− T2
4
)
2
ε
− 1

14. → Qwith shield =
A ∙ σ ∙ (T1
4
− T2
4
)
2 (
2
ε − 1)
‘ ’
→ QN =
A ∙ σ ∙ (T1
4
− T2
4
)
(N + 1) (
2
ε − 1)
→
QN
Q0
=
1
N + 1
→ QN = Q0 (
1
N + 1
)
Qo =
σ(T1
4
− T2
4)
1 − ε1
A1ε1
+
1
A1F12
+
1 − ε2
A2ε2
=
σ ∙ A1 ∙ (T1
4
− T2
4)
1
ε1
− 1 +
1
F12
+
A1
A2
(
1
ε2
− 1)
=
σ ∙ A1 ∙ (T1
4
− T2
4)
1
ε1
+
A1
A2
(
1
ε2
− 1)
Qshield =
σ(T1
4
− T2
4)
1 − ε1
A1ε1
+
1
A1F13
+
1 − ε31
A3ε31
+
1 − ε32
A3ε32
+
1
𝐴3 𝐹31
+
1 − ε2
A2ε2
Qshield =
σ(T1
4
− T2
4)
1
𝜀1
+
𝐴1
𝐴2
(
1
𝜀2
− 1) +
𝐴1
𝐴3
(
1
𝜀31
+
1
𝜀32
− 1)

15. Eb1 − J1
R1⏟
Q1
+
J2 − J1
R12⏟
−Q12
+
J3 − J1
R13⏟
−Q13
= 0
J1 − J2
R12
+
Eb2 − J2
R23
+
J3 − J2
R23
= 0
J1 − J3
R13
+
J2 − J3
R23
+
Eb3 − J3
R3
= 0
Q̇ i = ∑ Q̇ i→j
N
j=1
= ∑ AiFi→j(Ji − Jj)
N
j=1
= ∑
(Ji − Jj)
(1/AiFi→j)
N
j=1
’

16. R̅ = R1 + Req + R2
1
Req
=
1
R12
+
1
R13 + R23
R̅ = (
1 − ε1
A1ε1
) + (
1
1
A1F12
+
1
A1F13 + A2F23
) + (
1 − ε2
A2ε2
)
Q12 =
Eb1
− Eb2
R̅

17. λ
λ λ λ
dIλ(s)
ds
= (
emission per
unit volume
) − (
absorption per
unit volume
)
(
Absorption per
unit volume
) = κλ ∙ Iλ(s)
𝜿 𝝀
’
𝜿 𝝀
(
Emission per
unit volume
) = κλ Ibλ(T)
λ
𝑑𝐼𝜆(𝑠)
𝑑𝑠
+ 𝜅 𝜆 𝐼𝜆(𝑠) = 𝜅 𝜆 𝐼 𝑏𝜆(𝑇)
𝐼𝜆(𝑠) = 𝐼𝜆(0) 𝑎𝑡 𝑠 = 0

18. →
𝑑𝐼𝜆(𝑠)
𝑑𝑠
+ 𝜅 𝜆 𝐼𝜆(𝑠) = 𝜅 𝜆 𝐼 𝑏𝜆(𝑇)
→
𝑑𝐼𝜆(𝑠)
(𝐼 𝑏𝜆(𝑇) − 𝐼𝜆(𝑠))
= 𝜅 𝜆 𝑑𝑠
→ ∫
𝑑𝐼𝜆(𝑠)
( 𝐼 𝑏𝜆( 𝑇) − 𝐼 𝜆( 𝑠))
𝐼 𝜆(𝑠)
𝐼 𝜆(0)
= ∫ 𝜅 𝜆 𝑑𝑠
𝑠
0
⟹ ln( 𝐼 𝑏𝜆( 𝑇) − 𝐼 𝜆( 𝑠)) |
𝐼𝜆(𝑠)
𝐼𝜆(0)
= −𝜅 𝜆 ∙ 𝑠
→ ln
𝐼 𝑏𝜆( 𝑇) − 𝐼 𝜆( 𝑠)
𝐼 𝑏𝜆( 𝑇) − 𝐼 𝜆(0)
= −𝜅 𝜆 ∙ 𝑠 ⟹
𝐼 𝑏𝜆( 𝑇) − 𝐼 𝜆( 𝑠)
𝐼 𝑏𝜆( 𝑇) − 𝐼 𝜆(0)
= 𝑒−𝜅 𝜆∙𝑠
→ (𝐼 𝑏𝜆(𝑇) − 𝐼𝜆(0)) ∙ 𝑒−𝜅 𝜆∙𝑠 = 𝐼 𝑏𝜆(𝑇) − 𝐼𝜆(𝑠)
𝑰 𝝀(𝒔) = 𝑰 𝝀(𝟎) ∙ 𝒆−𝜿 𝝀∙𝒔
+ (𝟏 − 𝒆−𝜿 𝝀∙𝒔)𝑰 𝒃𝝀(𝑻)
𝐼𝜆(𝐿) = 𝐼𝜆(0) ∙ 𝑒−𝜅 𝜆∙𝐿
+ (1 − 𝑒−𝜅 𝜆∙𝐿)𝐼𝑏𝜆(𝑇)
λ
𝐼𝜆(𝐿) = 𝐼 𝜆(0) ∙ 𝑒−𝜅 𝜆∙𝐿
τλ
𝜏 𝜆 =
𝐼𝜆(𝐿)
𝐼𝜆(0)
= 𝑒−𝜅 𝜆∙𝐿
𝜏 𝜆 + 𝛼 𝜆 = 1
𝛼 𝜆
𝛼 𝜆 = 1 − 𝑒−𝜅 𝜆∙𝐿
’ 𝛼 𝜆 ελ.
𝜀 𝜆 = 𝛼 𝜆 = 1 − 𝑒−𝜅 𝜆∙𝐿
𝐼𝜆(𝐿) = (1 − 𝑒−𝜅 𝜆∙𝐿) 𝐼 𝑏𝜆( 𝑇)

19. ’
εw = Cw ∙ εw,1 atm
Cw = 1 for P = 1 atm and thus
Pw + P
2
≅ 0.5

20. εg = εc + εw − Δε
εg = Cc ∙ εc,1 atm + Cw ∙ εw,1 atm − Δε
Δε
Δε
Le = 3.6 ×
V
A
V → Volume of gas body , A → surface area of enclosure
αg = αc + αw − Δα
Δα = Δε
CO2: αc = Cc × (
Tg
Ts
)
0.65
× εc (Ts,
PcLTs
Tg
)
H2O: αw = Cw × (
Tg
Ts
)
0.65
× εw (Ts,
PwLTs
Tg
)
α = ε Ts = Tg
𝑄 𝑒 = 𝐴 𝑠 ∙ 𝜀 𝑔 ∙ 𝜎𝑇𝑔
4
𝐴 𝑠 𝜎𝑇4
𝑄̇ 𝑎 = 𝐴 𝑠 ∙ 𝛼 𝑔 ∙ 𝜎𝑇4
𝑄̇ 𝑛𝑒𝑡 = 𝑄 𝑒 − 𝑄̇ 𝑎 = (𝐴 𝑠 ∙ 𝜀 𝑔 ∙ 𝜎𝑇𝑔
4
) − (𝐴 𝑠 ∙ 𝛼 𝑔 ∙ 𝜎𝑇4
)
𝑄̇ 𝑛𝑒𝑡 = 𝐴 𝑠 ∙ 𝜎 × (𝜀 𝑔 𝑇𝑔
4
− 𝛼 𝑔 𝑇4
) → 𝐵𝑙𝑎𝑐𝑘 𝐸𝑛𝑐𝑙𝑜𝑠𝑢𝑟𝑒

21. εs
𝑄̇ 𝑛𝑒𝑡,𝑔𝑟𝑎𝑦 =
𝜀 𝑠 + 1
2
𝑄̇ 𝑛𝑒𝑡,𝑏𝑙𝑎𝑐𝑘 =
𝜀 𝑠 + 1
2
× 𝐴 𝑠 ∙ 𝜎 × (𝜀 𝑔 𝑇𝑔
4
− 𝛼 𝑔 𝑇4
)