A radiator in a domestic heating system operates at a surface temperature of . Assuming the radiator behaves as a black body, the rate at which it emits the radiant heat per unit area is
(Assume σ =5.67 x 10^{8} W/m^{2}k^{4})
As per Stefan Boltzmann law
E_{b} = Q/A = σ x T^{4}
E_{b} = 5.67 x 10^{8} x (55 + 273)^{4}
= 656.2 W/m^{2}
≃ 0.66 kW/m^{2}
Sun’s surface at emits maximum spectral power for wavelength of A grey surface at a will Demit maximum spectral power at a wavelength of
By Wien’s displacement law,
λm ∝ 1/T
∴ (λ_{m})_{2}/(λ_{m})_{1} = T_{1}/T_{2}
∴ (λ_{m})_{2} = (λ_{m})_{1}T_{1}/T_{2}
∴ (λ_{m})_{2} = 0.5 x 5800/(300 + 273)
(λm)_{2} = 5.06μm
≃ 5μm
Two radiating surfaces A^{1}= 6m^{2} and A^{2}=4m^{2} have shape factor Then shape factor will be
By Reciprocity theorem
A_{1}F_{12} = A_{2}F_{21}
= 0.15
What will be the view factor for the geometry as shown in the figure (sphere within a hollow cube)?
1 is a spherical surface
∴ F_{11} = 0
By summation rule,
F_{12} + F_{11} = 1
∴ F_{12} = 1
By Reciprocity theorem,
A_{1}F_{12} = A_{2}F_{21}
A radiation shield should
A radiation shield should reflect back as much radiation as possible. Hence radiation shield should have high reflective power.
For an opaque plane surface at steady state the irradiation, radiosity and emissive power are respectively 20, 12 and . What is the emissivity of the surface?
J = _{p}G + εE_{b}
For an opaque body
∵ α + ρ = 1
ρ = 1 α
∴ J = (1α)G + εE_{b}
By kirchoff’s law, α = ε
∴ J = (1α)G + εE_{b}
12 = (1 ε)20 + 10
ε = 0.9
The wavelength for which the blackbody emissive power is maximum for a temperature of 300 K is
By Wien’s displacement law,
∵ λmT = 2898 μmk
λm ≃ 9.7μm
Solar radiation is incident on a semitransparent body at a rate of . If of this incident radiation is reflected back and is transmitted across the body, the absorptivity of the body is
G = 500 W/m^{2} G_{r} = 150 W/m^{2}
G_{t} = 225 W/m^{2}
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 = 10.3  0.45 = 0.25
A grey body is defined such that
The minimum number of view factors that need to be known to solve a 10surface enclosure completely, is
For nsurface enclosure if ^{n}C_{2} 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
The spectral distribution of surface irradiation is as follows
What is the total irradiation in kW/m^{2} ?
Total irradiation is given as
i.e. it is equal to area under the curve Gλ Vs λ Hence,
x (2520) x 1000
∴ G = 20000 W/m^{2}
I.e. G = 20 kW/m^{2}
Consider two infinitely long blackbody concentric cylinders with a diameter ratio . The shape factor of the inner surface of outer cylinder with respect to itself will be
By Energy conservation
F_{12} + F_{22 }=1
∴ F_{22} = 1F_{21} ….①
By Reciprocity theorem
A_{1}F_{12} = A_{2}F_{21}
πD_{1}L x F_{12} = πD_{2}L x F_{21}
A large spherical enclosure has a small opening. The rate of emission of radiative flux through this opening is . The temperature at the inner surface of the sphere will be about (assume Stefan Boltzmann constant σ = 5.67 x 10^{8} W/m^{2}K^{4})
A small opening in a large spherical enclosure behaves as a black surface
∴E_{b} = σT^{4}
7350 = 5.67 x 10^{8} x T^{4}
T = 600K
I.e. T = 327^{o}C
Consider monochromatic emissive power (E_{λ}) Vs Wavelength(λ) of a black and grey surface both at same temperature. The ratio AB/AC is given as
AB & AC are monochromatic emissive power of grey and black surfaces for same wavelength. Hence, ratio is
which is monochromatic emissivity of grey surface.
If the temperature of a solid surface changes from then its emissive power will increases in the ratio of
By Stefan Boltzmann law
E ∝ T^{4}
Consider a hemispherical furnace. The view factor of it’s roof (hemisphere) with respect to itself is __________.
By reciprocity theorem
A_{1}F_{12} = A_{2}F_{12}
1 is a planar surface, i.e F^{11} = 0
∴ F^{12} = 1
Hence F^{21} = 0.5
F^{21} + F^{22} = 0.5
F^{22} + 10.5 = 0.5
Fraction of radiative energy leaving one surface that strikes the other surface is called
Ice is very close to a
The absorptivity of ice is 0.985, hence ice is very close to a black body
For a grey diffuse surface with emissivity ε and temperature T, the intensity of emitted radiation is given as __________. ( is StefanBoltzmann constant.)
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} = εσT^{4}/π
Solar radiation of falls on a grey opaque surface at steady state. The surface has a temperature of and emissivity of 0.8. Find radiosity from the surface?
J = E+G_{r}
E = εσT4
E = 0.8 x 5.67 x 108 x (50 + 273)4
E = 493.72 W/m2
Given G = 1200 W/m2
But as surface is at steady state
α = ε = 0.8
∴ ρ = 1 α = 0.2
∴ G_{r} = ρG
G_{r} = 0.2 x 1200 = 240 W/m^{2}
Therefore
J = E+G_{r}
J = 493.72 + 240
J = 733.72 W/m^{2}
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