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What is net radiation heat exchange per square meter per unit time for two very large plates at temperature values of 800 K and 500 K? Emissivities of hot and cold plates are 0.8 and 0.6 respectively. StefanBoltzmann constant is  5.67x10^{8}W/m^{2}K^{4}
For two very large parallel plates, the net radiation heat exchange (using thermal circuit) is given by,
= 10.26 kW/m^{2}
The spectral hemispherical absorptivity of an opaque surface is as shown.
The surface is exposed to irradiation for which the spectral distribution is as shown.
What is the total hemispherical absorptivity of surface?
Total absorptivity is defined as
α = G_{a}/G
Where G_{a} = ∫_{o}^{oo}α_{λ}G_{λ}d_{λ}
G_{a} = ∫_{0}^{6} 0.6 x 0 d_{λ} + ∫_{6}^{10} 0.8 x 500 d_{λ} + ∫_{10}^{14} 0.8 x 250 d_{λ} + ∫_{14}^{00} 0.8 x 0 d_{λ}
∴ G_{a} = 0 + 0.8 x 500 x (106) + 0.8 x 250 x (1410) + 0
G_{a} = 2400 W/m^{2} ...①
and
G = ∫_{0}^{00} G_{λ}d_{λ}
I.e. G = Area under graph G_{λ} Vs λ
G = 500 x (106) + 250 x (1410)
G = 2000 + 1000
G = 3000 W/m^{2 } ...②
Therefore, from ① & ②
α = G_{a}/G = 2400/3000
α = 0.8
Two grey surfaces that form an enclosure exchange heat with one another by thermal radiation. Surface 1 has a temperature of 400 K, an area of and a total emissivity of 0.4. Surface 2 has a temperature of 600 K, area of and a total emissivity of 0.6. If the view factor is 0.3, the rate of radiation heat transfer between the two surfaces is
Using thermal circuit, the net heat exchange between two surfaces is
Q = 223.36 W
Which of the following statements is false regarding spectral distribution of thermal radiation emitted by a grey surface P. Monochromatic emissivity is independent of wavelength Q. Total emissivity is equal to monochromatic emissivity.
R. (monochromatic emissive power) Vs λ( graph for the surface at a
given temperature is scaled down version of the graph for black body at same temperature.
S. The wavelength corresponding to maximum monochromatic emissive power is same as that of grey surface at same temperature.
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_{λ}E_{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, A^{1}B^{1}/A^{1}C^{1} = AB/AC = A^{2}B^{2}/A^{2}C^{2}
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.
Two infinitely large parallel plates 1 & 2, facing each other are separated by a very small gap. The plates are exchanging heat by thermal radiation and gap between them is filled with nonparticipating medium. Surface 1 is a black surface with temperature equal to and surface 2 is a grey surface with temperature equal to and emissivity equal to 0.8. Find out the radiosity for surface 2 in kW/m^{2}
Consider radiation heat exchange between the surface as shown.
Black Grey,ε_{2} = 0.8
T_{1} = 800 K T_{2} = 500 K
Therefore radiosity for surface 2 is given as
J_{2} = E_{2} + ρ_{2}E_{1}
J_{2} = E_{2} + (1α_{2})E_{1}
By Kirchoff’s law, α_{2} = ε_{2}
∴ J_{2} = (1ε_{2})E_{1}+ E_{2}
As ① is a black surface,
E1 = σT_{1}^{4}
and ② is a gray surface,
E_{2} = ε_{2}σT_{2}^{4}
∴ J_{2} = (1ε_{2})σT_{1}^{4} + ε_{2}σT_{2}^{4}
J_{2} = σ[(1ε)T_{1}^{4} + ε_{2}T_{2}^{4}]
J_{2} = 5.67 x 10^{8} + [(10.8) x 800^{4} + 0.6 x 500^{4}]
J_{2} = 6.77 kW/m^{2}
Two large parallel grey plates with a small gap between them, exchange radiation at the rate of when their emissivities are 0.5 each. By coating one plate, its emissivity is reduced to 0.25. Temperatures remain unchanged. The new rate of heat exchange shall become
By thermal circuits, net radiation heat exchange between two surfaces can be written as
Where
For two large parallel plates
A_{1} = A_{2} = A & A_{12} = 1
∴ Q_{2}/Q_{1} = R_{1}/R_{2}
R_{1} = 3/A
R_{2} = 5/A
∴ Q_{2} =Q_{1} x R_{1}/R_{2}
Q_{2} = 1000 x ⅗
∴ Q_{2} = 600 W/m^{2}
An enclosure consists of four surfaces 1, 2, 3 and 4. The view factors for radiation heat transfer (where the subscripts 1, 2, 3, 4 refer to the respective surfaces) are F11 = 04 and F13 = 0.25. The surface areas A_{1} and A_{4} are 4m^{2} respectively. The view factor is
F11 + F12 + F13 + !4 = 1
0.1 + 0.4 + 0.25 + F14 = 1
∴ F14 = 0.25
From reciprocity theorem, A_{1}F_{14} = A_{4}F_{41}
4 x 0.25 = 2 x F_{41}
∴ F_{41} = 0.5
The net radiation heat exchange rate between a heating element at a temperature of 800^{o}C and inner wall of a furnace maintained at . The heat exchange rate when the element temperature is increased to 1000^{o}C for the same furnace temperature is
By Thermal circuits,
Where R is thermal resistance.
As T_{2} increases from 800^{o}C to 1000^{o}C, the net heat exchange increases n times
Where
n = 2.068
Thus the rate of heat exchange for
T1 = 1000^{0}C
Q’ = Q x n
= 8 x 2.068
= 16.544 kW/m^{2}
A filament of 75 W light bulb may be considered as a black body radiating into a black enclosure at . The filament diameter is 0.1 mm and length is 5 cm. Considering only radiation, the filament temperature is
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
F_{12} = 1; filament is completely enclosed
Q = A_{1}F_{12}σ(T_{1}^{4}  T_{2}^{4})
Q = (πDL) x 1 x 5.67 x 10^{8} (T_{1}^{4}  343^{4})
= 75 = 3.14 x 0.0001 x 0.05 x 1 x 5.67 x 10^{8} (T_{1}^{4}= 343^{4})
T_{1} = 3029.4 K
Consider a surface at 0^{o}C that may be assumed to be a blackbody in an environment at . If of radiation is incident on the surface, the radiosity of this black surface is
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} = σT^{4}
= 5.67 x 10^{8} x 273^{4}
E_{b} = 314.94 W/m^{2}
Radiosity is given as
J = E_{b} +G_{r}
J = 314.94 W/m^{2}
Note that Eb>G because the difference E_{b}  G_{a} must be equal to rate of heat convection from the environment to surface.
A solar flux of intensity directly strikes a space vehicle surface which has an absorptivity of 0.4 and emissivity of 0.6. The equilibrium temperature of this surface in space at 0 K is
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 T_{1}^{4}
⇒ 0.4 x 1400 = 0.6 x 5.67 x 10^{8} x T_{1}^{4}
⇒ T_{1} = 358.2 K
The shape factor of a cylindrical cavity (open from top) with respect to itself is __________. L & D are length and diameter of the cavity respectively.
The cavity is the shape
2 is a planar surface
∴ F_{22} = 0 & F_{21} = 1
By Reciprocity theorem
A_{1}A_{12} = A_{2}F_{21}
Also, F_{11} = 1 F_{12}
F_{11} = 1  D^{2}/D^{2}+4DL
1D/D+4L
∴ F_{11} = 4L/D+4L
What is the view factor for inclined parallel plates of equal width and a common edge? The width of plates to the plane of paper is b. (b >>L)
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)
∴ A_{3} = (2 ℓ sin α/2 x b) = 2 ℓb sin (α/2)
A_{1} = ℓ x b = ℓb
A_{2} = ℓ x b = ℓlb
As 3 is a planar surface
F_{33} = 0
Also F_{31} + F_{32} + F_{33} =1
But because of symmetric arrangement of ①
& ② with respect to ③
F_{31} = F_{32}
∴ F_{31} = F_{32} = 0.5
By Reciprocity theorem
A_{1}A_{13} = A_{3}A_{31}
F_{13} = ain(α/2)
F_{11} + F_{12} + F_{13} = 1
0 + F_{12} + sin (α/2) = 1
F_{12} = 1sin(α/2)
Two very large parallel plates separated by vacuum are maintained at uniform temperatures T_{1}= 750K & T_{2}=500K and have emissivities ε_{1}= 0.85 and ε_{2} = 0.7 respectively. If a thin aluminium sheet with same emissivity on both sides is placed in between the plates, the rate of heat transfer between the plates reduces by 90%. The emissivity of aluminium sheet is given as
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
R_{1} = 1ε_{1}/Aε_{1}+1/A+1ε_{2}/Aε_{2}
R_{1} = 10.85/0.854A+1/A+10.7/0.7A
R_{1 }= 1.605/A
When a sheet is placed between the surfaces
Where ε_{s }is the emissivity of single shield
surface
∴ R_{2} = 2.605/A+2(1ε_{s})/Aε_{s}
∴ Q_{2}/Q_{1} = R_{1}/R_{2}
It is given that by introducing the sheet in between the plates reduces the heat exchange by 90%
∴ Q_{2}/Q_{1} = 0.1
∴ R_{1}/R_{2} = 0.1
∴ 1.605 = 0.2605 + 0.2 (1ε_{s})/ε_{s}
6.7225 ε_{s} = 1ε_{s}
ε_{s} = 0.1295
ε_{s }≃ 0.13
For the radiation between two infinite parallel planes of emissivity respectively, which one of the following is the expression for emissivity factor or equivalent emissivity?
The equivalent thermal resistance between two gray bodies is given by
R_{th} = 1ε_{1}/A_{1}ε_{1}+1/A_{1}F_{12}+1ε_{2}/A_{2}ε_{2}
Now for infinite parallel panes, F_{12} =1
A_{1} = A_{2} = A
For unit area, A = 1 m^{2}
R_{th} = 1/ε_{1}+1/ε_{2}1
Hence equivalent emissivity
Ε_{e} = 1/R_{th}
Determine the view factors from the base of the pyramid shown in figure to each of its four side surfaces. The base of the pyramid is a square and its side surfaces are isosceles triangles.
The base of the pyramid (side 1) and the other four surfaces form an enclosure From symmetry, we have.
F_{12} = F_{13} = F_{14} = F_{15}
Again we have,
F_{11}+F_{12}+F_{13}+F_{14}+F_{15} = 1
Now F_{11} = 0
∴ F_{12} = F_{13} = F_{14} = F_{15} = 0.25
Determine for the following configuration. Long inclined plates (point B is directly above the center of )
Closing the remaining side by a hypothetical surface 3, we have
F_{11} + F_{12} + F_{13} = 1
F_{11} = 0; By symmetry we have,
F_{12} = F^{13}
∴ F_{12} = F_{13} = 0.5
A_{1}F_{12} = A_{2}F_{21}(by reciprocity relation)
(considering depth of plates ⊥ to plane of paper to be l)
∴ F_{21} = 1.4142 x 0.5
F_{21} = 0.707
Consider surfaces ① & ② as shown in the figure.
Surface ① is divided into two parts ③ & ④. Which of the following options is correct expression for F_{12}
Q_{1➝2} = Q_{3➝2} + Q_{4➝2}
Q_{1➝2} /Q_{1} = Q_{3➝2} /Q_{1}+ Q_{4➝2}/Q_{1}
What is the equivalent emissivity for radiant heat exchange between a small body ( in a very large enclosure (emissivity = 0.5)?
If one surface is completely enclosed by another surface,
If surface 2 is very large,
Also we have q = ε_{e}σ(T_{1}^{4}  T_{2}^{4}) ...②
From ① & ② ε_{e} = 1/R_{th}
But R_{th} = 1 /ε_{1}A_{1}
∴ ε_{e} = ε_{1} = 0.4
Determine the percentage of radiation emitted by the sun within the visible spectrum ( Wavelengths from 0.4 μm (violet) to 0.76 μm(red), considering the sun to be black body and its effective surface temperature to be 5800 K. Given black body radiation function for 2320μmk is 0.124509 and for 4408μmk is 0.550015 .
∴ f_{λ1➝λ2} = f_{λ2}  f_{λ1}
Where f_{λ} is black body radiation function for wavelength λ.
∴ f_{0.40.76} = f_{0.76}  f_{0.4}
= 0.550015  0.124509
= 0.4255
∴ i.e, ≃ 42.58%
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