Test: Radiative Heat Transfer Level - 2 - Mechanical Engineering MCQ

For two very large parallel plates, the net radiation heat exchange (using thermal circuit) is given by,

= 10.26 kW/m²

Total absorptivity is defined as

α = G_a/G

Where G_a = ∫_o^ooα_λG_λd_λ

G_a = ∫₀⁶ 0.6 x 0 d_λ + ∫₆¹⁰ 0.8 x 500 d_λ + ∫₁₀¹⁴ 0.8 x 250 d_λ + ∫₁₄⁰⁰ 0.8 x 0 d_λ

∴ G_a = 0 + 0.8 x 500 x (10-6) + 0.8 x 250 x (14-10) + 0

G_a = 2400 W/m² ...①

and

G = ∫₀⁰⁰ G_λd_λ

I.e. G = Area under graph G_λ Vs λ

G = 500 x (10-6) + 250 x (14-10)

G = 2000 + 1000

G = 3000 W/m² ...②

Therefore, from ① & ②

α = G_a/G = 2400/3000

α = 0.8

Using thermal circuit, the net heat exchange between two surfaces is

Q = 223.36 W

A grey surface is one for which monochromatic emissivity ε_λ is independent of wavelength (λ ) But total emissivity (ε) of a surface is given as

ε = E/E_b

Where E & E_b represent total emissive power of the surface and a black surface at same temperature.

But as ε_λ ≠ f(λ)

∴ ε = ε_λE_b/E_b

∴ ε = ε_λ

i.e. total emissivity is equal to monochromatic emissivity. Hence statement Q is also correct.

In the figure shown,

AB = E_λ &

AC = E_bλ

∴ AB\Ac = E_λ/E_bλ

i.e, AB\AC = ε_λ

i.e. ratio of line segment AB and AC is monochromatic emissivity for surface corresponding to wavelength (λ) But as (ελ≠f(λ))

this ratio is same for all λ

i.e, A¹B¹/A¹C¹ = AB/AC = A²B²/A²C²

Thus graph of grey surface is simply scaled down version of graph of black surface. This also suggests that corresponding to maxima point for both surfaces is same.

Hence statement R & S are also correct.

Thus all statements are correct & option (D) is the correct answer.

Consider radiation heat exchange between the surface as shown.

Black Grey,ε₂ = 0.8

T₁ = 800 K T₂ = 500 K

Therefore radiosity for surface 2 is given as

J₂ = E₂ + ρ₂E₁

J₂ = E₂ + (1-α₂)E₁

By Kirchoff’s law, α₂ = ε₂

∴ J₂ = (1-ε₂)E₁+ E₂

As ① is a black surface,

E1 = σT₁⁴

and ② is a gray surface,

E₂ = ε₂σT₂⁴

∴ J₂ = (1-ε₂)σT₁⁴ + ε₂σT₂⁴

J₂ = σ[(1-ε)T₁⁴ + ε₂T₂⁴]

J₂ = 5.67 x 10^-8 + [(1-0.8) x 800⁴ + 0.6 x 500⁴]

J₂ = 6.77 kW/m²

By thermal circuits, net radiation heat exchange between two surfaces can be written as

Where

For two large parallel plates

A₁ = A₂ = A & A₁₂ = 1

∴ Q₂/Q₁ = R₁/R₂

R₁ = 3/A

R₂ = 5/A

∴ Q₂ =Q₁ x R₁/R₂

Q₂ = 1000 x ⅗

∴ Q₂ = 600 W/m²

F11 + F12 + F13 + !4 = 1

0.1 + 0.4 + 0.25 + F14 = 1

∴ F14 = 0.25

From reciprocity theorem, A₁F₁₄ = A₄F₄₁

4 x 0.25 = 2 x F₄₁

∴ F₄₁ = 0.5

By Thermal circuits,

Where R is thermal resistance.

As T₂ increases from 800^oC to 1000^oC, the net heat exchange increases n times

Where

n = 2.068

Thus the rate of heat exchange for

T1 = 1000⁰C

Q’ = Q x n

= 8 x 2.068

= 16.544 kW/m²

At steady state electric power supply to the bulb is equal to net radiation heat exchange between bulb and enclosure

here ∈₁ = ∈₂ = 1 and

F₁₂ = 1; filament is completely enclosed

Q = A₁F₁₂σ(T₁⁴ - T₂⁴)

Q = (πDL) x 1 x 5.67 x 10^-8 (T₁⁴ - 343⁴)

= 75 = 3.14 x 0.0001 x 0.05 x 1 x 5.67 x 10^-8 (T₁⁴= 343⁴)

T₁ = 3029.4 K

By definition radiosity is the rate at which a thermal radiation levels a surface per unit area

G_a = 300 W/m2

As surface is black, it completely absorbs the radiation incident on it,

Therefore

G_r = 0

By Stefan Boltzmann Law

E_b = σT⁴

= 5.67 x 10^-8 x 273⁴

E_b = 314.94 W/m²

Radiosity is given as

J = E_b +G_r

J = 314.94 W/m²

Note that Eb>G because the difference E_b - G_a must be equal to rate of heat convection from the environment to surface.

For equilibrium,

Rate of heat absorbed = Rate of heat emitted

As the space vehicle is radiating to space at 0 K

∴ αAG_r = ∊ A σ T₁⁴

⇒ 0.4 x 1400 = 0.6 x 5.67 x 10^-8 x T₁⁴

⇒ T₁ = 358.2 K

The cavity is the shape

2 is a planar surface

∴ F₂₂ = 0 & F₂₁ = 1

By Reciprocity theorem

A₁A₁₂ = A₂F₂₁

Also, F₁₁ = 1- F₁₂

F₁₁ = 1 - D²/D²+4DL

1-D/D+4L

∴ F₁₁ = 4L/D+4L

Let us imagine the rectangular gap as a surface ③. !s the energy going out of triangular gaps can be neglected, but for rectangular gap, it is quite significant

Length of side 3 = 2 ℓ sin(α/2)

∴ A₃ = (2 ℓ sin α/2 x b) = 2 ℓb sin (α/2)

A₁ = ℓ x b = ℓb

A₂ = ℓ x b = ℓlb

As 3 is a planar surface

F₃₃ = 0

Also F₃₁ + F₃₂ + F₃₃ =1

But because of symmetric arrangement of ①

& ② with respect to ③

F₃₁ = F₃₂

∴ F₃₁ = F₃₂ = 0.5

By Reciprocity theorem

A₁A₁₃ = A₃A₃₁

F₁₃ = ain(α/2)

F₁₁ + F₁₂ + F₁₃ = 1

0 + F₁₂ + sin (α/2) = 1

F₁₂ = 1-sin(α/2)

By thermal circuits,

Q ∝ 1/R

Where R is resistance to radiation heat exchange between the surfaces.

When there is no sheet between the plates

T1 = 750 K T2 = 500 K

ε1 = 0.85 ε20.7

R₁ = 1-ε₁/Aε₁+1/A+1-ε₂/Aε₂

R₁ = 1-0.85/0.854A+1/A+1-0.7/0.7A

R₁ = 1.605/A

When a sheet is placed between the surfaces

Where ε_s is the emissivity of single shield

surface

∴ R₂ = 2.605/A+2(1-ε_s)/Aε_s

∴ Q₂/Q₁ = R₁/R₂

It is given that by introducing the sheet in between the plates reduces the heat exchange by 90%

∴ Q₂/Q₁ = 0.1

∴ R₁/R₂ = 0.1

∴ 1.605 = 0.2605 + 0.2 (1-ε_s)/ε_s

6.7225 ε_s = 1-ε_s

ε_s = 0.1295

ε_s ≃ 0.13

The equivalent thermal resistance between two gray bodies is given by

R_th = 1-ε₁/A₁ε₁+1/A₁F₁₂+1-ε₂/A₂ε₂

Now for infinite parallel panes, F₁₂ =1

A₁ = A₂ = A

For unit area, A = 1 m²

R_th = 1/ε₁+1/ε₂-1

Hence equivalent emissivity

Ε_e = 1/R_th

The base of the pyramid (side 1) and the other four surfaces form an enclosure From symmetry, we have.

F₁₂ = F₁₃ = F₁₄ = F₁₅

Again we have,

F₁₁+F₁₂+F₁₃+F₁₄+F₁₅ = 1

Now F₁₁ = 0

∴ F₁₂ = F₁₃ = F₁₄ = F₁₅ = 0.25

Closing the remaining side by a hypothetical surface 3, we have

F₁₁ + F₁₂ + F₁₃ = 1

F₁₁ = 0; By symmetry we have,

F₁₂ = F¹³

∴ F₁₂ = F₁₃ = 0.5

A₁F₁₂ = A₂F₂₁(by reciprocity relation)

(considering depth of plates ⊥ to plane of paper to be l)

∴ F₂₁ = 1.4142 x 0.5

F₂₁ = 0.707

Q_1➝2 = Q_3➝2 + Q_4➝2

Q_1➝2 /Q₁ = Q_3➝2 /Q₁+ Q_4➝2/Q₁

If one surface is completely enclosed by another surface,

If surface 2 is very large,

Also we have q = ε_eσ(T₁⁴ - T₂⁴) ...②

From ① & ② ε_e = 1/R_th

But R_th = 1 /ε₁A₁

∴ ε_e = ε₁ = 0.4

∴ f_λ1➝λ2 = f_λ2 - f_λ1

Where f_λ is black body radiation function for wavelength λ.

∴ f_0.4-0.76 = f_0.76 - f_0.4

= 0.550015 - 0.124509

= 0.4255

∴ i.e, ≃ 42.58%