Test: Radiative Heat Transfer Level - 3 - Mechanical Engineering MCQ

For emission from a black body,

= 13 W

∵ L = A_s/P = D/4 = 0.2/4 = 0.05 m

⟶ R_sL = gβ(T_s-T_ꝏ)L³

= 9.81 x 0.00245 x 230 x (0.05)³/27.4 x 39.7 x 10^-12

= 6.35 x 10⁵

= 10.5 W/m² K

∴ Q_conv = h A^s(T^s-T_ꝏ)

⇒ 75.7 W

∴ Electric power required,

⇒ P = 232 W

Consider outer surface of metal tank to be 1 & inner surface of enclosure to be surface 2.

By concept of thermal circuits

As the surfaces of 1 are planar, F₁₁ = 0

∴ F₂₁ = 1

Also ,

A₂ >> A₁

∴ (Q_1➝2)_net = σA₁ε₁(T₁⁴ - T₂⁴)

∴ Q ∝ ε₁

Q0.55/Q0.95 = 0.55/0.95 = 0.4211

Thus heat exchange rate decreases by 57.89 %, by painting the tank with aluminium paint.

Since the shield is placed midway between both the cylinders, therefore diameter of the shield

∴ d₃ = d₁+d₂/2 = 7.5 + 22.5/2

⇒ d₃ = 15.0 cm

A₁ = πd₁L

A² = πd₂L

A₃ = πd₃L

For the system to be steady,

(Q_1➝3)_net = (Q_3➝2)_net

By thermal circuits

R₁₃ = 11.166/7.5πl

Similarly,

R32 = 3/7.5πl

∴ T₃ = 717 K

Question_type 5

By definition of total emissivity

ε = E/E_b

Where E =

∴ E =

∴ E =

∴ E = 0.3

E = 0.3 E_b,0➝3 + 0.8E_b,3➝7 + 0.1E_b,7➝ꝏ

E = 0.3 f₃E_b + 0.8f_3➝7E_b + 0.1f_7➝ꝏE_ꝏ

E = [0.3f₃ + 0.8(f₇ -f₃₎ + 0.1(1- f₇)]E_b

∴ ε = [0.3f₃ + 0.8(f₇ - f₃ ) + 0.1(1-f₇)]

For T = 2400 μm-K

∴ f₃ = 0.140256

Similarly for λ = 7 μm

λT = 5600 μm-K

∴ f₇ = 0.701046

∴ ε = 0.3 x 0.140256 + 0.8

X (0.701046 - 0.140256) + 0.1 x (1 - 0.701046)

ε = 0.0420768 + 0.448632 + 0.0298954

ε = 0.5206042

≃ 0.52

Given l_a = 2 cm

L_b = 3 cm

L_c = 4 cm

By energyF conservation principle

F_aa + F_ab + F_ac = 1

F_ba + F_bb + F_bc = 1

F_ca + F_cb + F_cc = 1

As a,b,c are planar surfaces

F_aa, F_bb, F_cc, = 0

Therefore the equations are reduced to

F_ab + F_ac = 1 ….①

F_ba + F_bc + = 1 ...②

F_ca + F_cb = 1 ….③

By Reciprocity theorem

Substituting F_ba & F_bc in equation ②

Substituting F_cb & F_ca in equation ③

∴ 2F_ac - 6F_ab = 1

F_ac - 3F_ab = 1/2 ….Ⓢ

Solving equation ① & Ⓢ together

F_ab + F_ac =1

F_ac - 3F_ab = 1/2

____________

3F_ab + 3_ac = 3

3F_ac - 3F_ab = 1/2

+ + +

______________

6F_ac = 7/2

F_ac = 7/12

F_ac = 0.5833

Heat lost by outer surface of the lining to surroundings per unit area,

q = εσ(T₂⁴ - T_ထ⁴)

q = 0.85 x 5.67 10^-8[430⁴ - 300⁴]+1.5(430 -300)^0.33 x (430 - 300)

q = 1257.3116 + 971.932

q = 2229.24 W

n-shields

∴ Heat transfer from surface 1 to 2 through n-shields must be equal to ̇q

⇒ n+1 = 11.7

⇒ n = 10.7

⇒ n ≅ 11

Given F₁₃ = 0.06

Now F₁₁ +F₁₂ + F₁₃ = 1

⇒ F₁₂ = 0.94

⇒ F₂₁ = F₂₃

again A₁F₁₂ = A₂F₂₁

⇒ F₂₁ = 0.1175

From Symmetry F₂₁ = F₂₃

∴ F₂₂ = 1-(2 x 0.1175)

⇒ F₂₂ = 0.765

∴ Power required to drive furnace,

= A₂A₂₁ σ(T₂⁴ - T₁⁴) + A₂F₂₃σ(T₂⁴ - T₃⁴)

= A₂F₂₁σ[2T₂⁴ - T₁⁴ - T₃⁴]

= 3.14(0.1)(0.2).0.1175 x 5.67

X 10^-8[2 x (1800)⁴ - (2000)⁴ - 300⁴]

= 2.087 kW

Since the surface 3 is a reradiating surface, (R₃ = 0) the thermal circuit can be drawn as shown above. Since surface 1 is a black surface,

∵ Length of all sides is equal to 1 m and taking unit length of the duct

A₁ = A₂ = A₃ = 1 x 1 = 1 m²

Total thermal resistance between 2 and 2 is,

From symmetry of the triangle we have

F₁₂ = F₁₃ = F₂₃ = 0.5

∴ R₁ = 0 (for black surface)

R₁₂ = 1/A₁F₁₂ = 1/1 x 0.5 = 2

R₁₃ = 1/A₁F₁₃ = 1/1 x 0.5 = 2

R₂₃ = 1/A₂F₂₃ = 1/1 x 0.5 = 2

R₂ = 1-ε₂/A₂ε₂ = 1-0.7/1 x 0.7 = 0.43

= (0.5 + 0.25)^-1 + 0.43

= 1.7633

= 5.67 x 10^-8(1000⁴ - 600⁴)/1.7633

Q = 27.988 kW

= heat radiation from thermocouple to walls

∴ hA_T/C(T_A - T_C) = ε_T/C(T_C⁴ - T_w⁴). σ

(∵ Area of walls very large as compared to area of thermocouple)

60 x 70 = 0.5 x 5.67 x 10-8(1073⁴ - T_w⁴)

1073⁴ - T_w⁴ = 1.4815 x 10¹¹

T_w = 1041.67 K

T_w = 768.7^oC

R_th = (n+1)(2/ε -1) (for unit area)

Overall resistance of a surface is given by,

R_th = 1/ε_o (for unit area)

Where ε_o ￫ overall emissivity

∴ 1/ε_o = (n+1)(2/ε -1)

1/ε_o = 2n/ε - n + 2/ε - 1

1/ε_o = 2n - nε + 2-ε/ε

1/ε_o = n(2-ε) + 1( 2-ε )/ε

1/ε_o = (n+1)(2-ε)/ε

ε_o = ε/(n+1)(2-ε)

Q ∝ 1/R

where R is the resistance to radiation heat exchange between the plates. When there is no shield

R₁ = 1-ε₁/Aε₁ + 1/A + 1-ε₂/Aε₂

R₁ = 1-0.8/A x 0.8 + 1/A + 1-0.8//A x 0.8

R₁ = 1.5/A

When n shields are placed in between the plates

= 1.5 + 39 n/A

Based on given condition

Q₂/Q₁ = 0.01

∴ 1.5/1.5 + 39 n = 0.01

∴ n = 3.8

i.e.n ≃ 4

1.5/1.5+1.5n = 0.01

n = 1.5(1-0.01)/1.5 x 0.01

n = 99

This shows that if the emissivity of the surface is large then number of shields should be large as shield shall offer less resistance. Hence for shields to be effective their surface emissivity should be as low as possible.

F_se = Q_es➝ne/Q_es

Where Q_es = σT⁴ x 4πR²

Consider an imaginary sphere of radius l with its center at center of sun. As sun is a symmetric body it emits radiation uniformly in all direction which is received uniformly at surface of imaginary sphere. The energy received by earth is actually the energy to be received by AB portion of the surface of imaginary sphere, where AB is interaction of imaginary sphere and earth. l >> r As therefore AB is nearly a circular surface of area πR².

Total energy emitted by sum = σ(4πR²)T⁴

∴ Energy received per unit area of imaginary sphere

= σ(4πR²)T⁴/4πl²

= σR²T⁴/l²

∴ Energy received by earth (i.e. surface AB)

= σR²T⁴/l² x πr²

Hence F_se =

Radiosity from surface 2

= reflected energy from ② + emitted energy from ②

⇒ J₂ = ρ₂σA₁T₁⁴ + ε₂ A₂ σT₂⁴

Ε₂ = 0.8 ⇒ α₂ = 0.8

∴ α₂+ρ₂+τ₂ = 1

⇒ ρ₂ = 0.2

⇒ J₂ = 38,071.644 W

Energy incident on the cake, G₃ = F₂₃J₂

A₃F₃₂ = A₂F₂₃

∴ G₃ = 0.03125 x 38071.644

G₃ = 1189.74 W

∴ Time required to bake the cake,

t = 3.2123 x 10⁶/1189.74

⇒ t = 2700 s

⇒ t ≅ 45 min

Considering an elementary ring dA₂ of width dr at a radius r,

dA₂ = 2 πr dr

Now, r = L tanΦ₁

Dr = L sec²Ф₁dФ₁

dA₂ = 2πL² tan Φ₁ sec² Φ₁dΦ₁

L/P = cos Φ₁ or P = L/cosΦ₁

= sin² α = D²/4/D²/4 + L² = D²/D² + 4L²

where D is the diameter of the disc.

∴ F₁₂ = q_c1➝r2/Q_c1

By Stefan-Blotzmann law

Q_c1 = σA₁T₁⁴

Also, using the definition of Intensity of emitted radiation,

Q_c1➝r2 = I_c1A₁ cos ፀ₁ω_2➝1

Where ω_2➝1 is solid angle subtended by point surface 2 over point surface 1.

∴ ω_2➝1 = A₂cosፀ₂/r²

For black surface

I_c1 = σT₁⁴/π

⇒ F₁₂ = 1.243 x 10^-4

Also by definition of black surface (diffuse characteristics)

I_c1 = σT₁⁴/π

274 x 10^-6 W = 5.67 x 10^-1 W/m²- K⁴ x T₁⁴ x ( 2cm² x 6 cm²) x cos 60^o x cos 30^o /(100 cm)²

274 x 10^-6 = 5.67 x 10^-8 x 12 x 10^-4/ 10⁴ x 1/2 x ✓3/4 x T₁⁴

274 x 10^-6 x 10^-8 x 4 / 5.67 x 12 x ✓3 = T₁⁴

∴ T₁⁴ = 9.30 x 10¹⁰

T₁ = 552.23 K

Thus for the given time instant the rate at which energy emitted by 1 strikes at 2 is

Thus for the given time instant the rate at which energy emitted by 1 strikes at 2 is

dQ = Q_c1➝r2 dt

Where,

dt = dፀ₂/θ²

This face of the surface 2 is in front of surface

1 for θ₂ = π/2 to π/2

Thus total heat received is

Similarly second face of surface 2 will also come in front of surface 1 during remaining half of revolution Thus total energy received by surface 2 during one complete revolution is

4I_c1A₁A₂ / r²ፀ₂

= 4 x 100 x 10^-4 x 10^-4/(0.5)² x 2

= 8 x 10^-6 j

By definition of intensity of radiation

dQ₁ = I(dA₁ cos ፀ)ω_dA2➝dA1 where dA₂ is a small area on hemispherical surface.

∴ ω_dA2➝dA1 = (rsin ፀ dф)(rdፀ)/r²

= sin ፀ dፀ dф

∴ dQ₁ - IdA₁ sin θ dθ dф

dQ₁ - I_ncos ө dA₁ sin θ dθ dф

dQ₁ - I_ndA₁ sin 2θ/2 x dθ dф

G = Q_i/dA₁ =I_n x π

∴ G = 80 x π

G = 251.32 W/m²

Based on data given

Total irradiation on the surface is

i.e. the area under the curve G_λ vs λ.

G = 0+1/2 x 500 x (6-2) + 500 x (12-6) + 1/2 x 500 x (16-12) + 0

G = 0 + 1000 + 3000 + 1000 + 0

G = 5000 W/m²

Similarly radiation absorbed (in W/m² ) is given as

G_a = 1000+800+2000+100

G_a = 4800 W/m²

∴ α = G_a/G

α = 4800/5000

α = 0.96

Quesiton_type: 5