Test: Radiative Heat Transfer Level - 1 - Chemical Engineering MCQ

The radiant heat transfer is given as Q = σAT⁴

Q = 5.67 x 10^-8 x 1 × (328)⁴

= 656.2 W/m²

≃ 0.66 kW/m²

By Reciprocity theorem

A₁F₁₂ = A₂F₂₁

= 0.15

1 is a spherical surface

∴ F₁₁ = 0

By summation rule,

F₁₂ + F₁₁ = 1

∴ F₁₂ = 1

By Reciprocity theorem,

A₁F₁₂ = A₂F₂₁

A radiation shield should reflect back as much radiation as possible. Hence radiation shield should have high reflective power.

J = _pG + εE_b

For an opaque body

∵ α + ρ = 1

ρ = 1- α

∴ J = (1-α)G + εE_b

By kirchoff’s law, α = ε

∴ J = (1-α)G + εE_b

12 = (1- ε)20 + 10

ε = 0.9

By Wien’s displacement law,

∵ λmT = 2898 μm-k

λm ≃ 9.7μm

G = 500 W/m² G_r = 150 W/m²

G_t = 225 W/m²

By energy conservation principle,

G_a+G_e+G_t = G

Where G_a = absorbed radiation

G_r = reflected radiation

G_t = transmitted radiation

G = incident radiation

= transmissivity

α = 1-0.3 - 0.45 = 0.25

Concept:

Gray body: When the emissivity of the non-black surface is constant at all temperatures and throughout the entire range of wavelength, the surface is called a gray body.

Emissivity: It is the ratio of emissive power of the body to the emissive power of the black body at the same temperature i.e. 

Monochromatic emissivity

• The ratio of the monochromatic emissive power of a body to the monochromatic emissive power of a black body at the same wavelength and temperature i.e.

  

• The radiation spectrum for a gray body, though reduced in vertical scale, is continuous and identical to the corresponding curve for a perfectly black surface; there is no shift in the peak of the curves.

For n-surface enclosure if ⁿC₂ view factors are known directly, entire enclosure can be solved. Thus for 10 surfaces, number of view factors that need to be evaluated directly will be

Total irradiation is given as

i.e. it is equal to area under the curve Gλ Vs λ Hence,

x (25-20) x 1000

∴ G = 20000 W/m²

I.e. G = 20 kW/m²

A small opening in a large spherical enclosure behaves as a black surface

∴E_b = σT⁴

7350 = 5.67 x 10^-8 x T⁴

T = 600K

I.e. T = 327^oC

AB & AC are monochromatic emissive power of grey and black surfaces for same wavelength. Hence, ratio is

which is monochromatic emissivity of grey surface.

By Stefan Boltzmann law

E ∝ T⁴

The view factor (also called the shape factor or configuration factor) quantifies the fraction of radiative energy leaving one surface that directly strikes another surface. It is a geometric property dependent only on the shapes and relative orientations of the surfaces.

For a diffuse surface, intensity of emitted radiation (I_e) is same in all directions and is given as

I_e = E/π

Where E is total emissive power

Therefore

I_e = εσT⁴/π

Gases transfer heat by the collision of molecules.

As the temperature increases, the kinetic energy of molecules of gases also increases and eventually collision between molecules also increases, which increases the thermal conductivity of gases.

∴ As temperature increases the thermal conductivity of gases increases.

Important Point: 
For liquids and solids, generally as the temperature increases, the thermal conductivity decreases.

When thermal radiation falls onto an object,

• The radiation will be absorbed by the surface of the object, causing its temperature to change
• The radiation will be reflected from the surface of the body, causing no temperature change
• The radiation will pass completely through the object, causing no temperature change

Absorptivity (α) is a measure of how much of the radiation is absorbed by the body.

Reflectivity (ρ) is a measure of how much is radiation is reflected.

Transmissivity (τ) is a measure of how much radiation passes through the object.

Each of these parameters is a number that ranges from 0 to 1.

Absorptivity may be a function of wavelength and/or direction, and is related to the emissivity of the region by Kirchhoff's law. The absorptivity is identically equal to unity for black bodies and is independent of temperature and wavelength for gray bodies.