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All questions of Radiation for Mechanical Engineering Exam

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Consider a surface at 0oC 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
  • a)
    0 W/m2
  • b)
    15 w/m2
  • c)
    314.94 W/m2
  • d)
    300 W/m2
Correct answer is option 'C'. Can you explain this answer?

Sarita Yadav answered
By definition radiosity is the rate at which a thermal radiation levels a surface per unit area
Ga = 300 W/m2
As surface is black, it completely absorbs the radiation incident on it,
Therefore
Gr = 0
By Stefan Boltzmann Law
Eb = σT4
= 5.67 x 10-8 x 2734
Eb = 314.94 W/m2
Radiosity is given as
J = Eb +Gr
J = 314.94 W/m2
Note that Eb>G because the difference Eb - Ga must be equal to rate of heat convection from the environment to surface.

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?
Correct answer is 'A'. Can you explain this answer?

Telecom Tuners answered
J = E+Gr
E = εσT4
E = 0.8 x 5.67 x 10-8 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
∴ Gr = ρG
Gr = 0.2 x 1200 = 240 W/m2
Therefore
J = E+Gr
J = 493.72 + 240
J = 733.72 W/m2

The minimum number of view factors that need to be known to solve a 10-surface enclosure completely, is
  • a)
    1
  • b)
    10
  • c)
    22
  • d)
    45
Correct answer is option 'D'. Can you explain this answer?

Sarita Yadav answered
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

Two infinitely long cylinders are placed coaxially. The inner cylinder has an outside diameter of 7.5 cm, and emissivity of 0.1 and is maintained at 1050 K. The outer cylinder has an inside diameter of 22.5 cm, an emissivity of 0.2 and is maintained at 315 K. A cylindrical shield (very thin so that its temperature is the same on both sides) is placed concentrically midway between the two cylinders and allowed to come to thermal equilibrium. It has an emissivity of 0.3 on both sides. Assuming that vacuum exists between the annular spaces. Find the steady state temperature attained by the cylindrical shield.
  • a)
    715.0
  • b)
    718.0
Correct answer is option 'A'. Can you explain this answer?

Sarita Yadav answered
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,
(Q1➝3)net = (Q3➝2)net
By thermal circuits
R13 = 11.166/7.5πl
Similarly,
R32 = 3/7.5πl
∴ T3 = 717 K
Question_type 5

An industrial furnace employs a hollow brick lining. The inside and outside surfaces of the hollow brick lining are maintained at 900 K and 430 K by placing radiation shields in between the hollow space. The heat loss occurs to the surroundings (300 K) by both radiation and natural convection. Calculate number of shields needed. The emissivity of walls and shields can be taken as 0.85. The convective heat transfer coefficient is governed by the expression
h = 1.5(ΔT)0.33 W/m2- K
  • a)
    11.0
  • b)
    11.0
Correct answer is option ''. Can you explain this answer?

Sarita Yadav answered
Heat lost by outer surface of the lining to surroundings per unit area,
q = εσ(T24 - Tထ4)
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
⇒ 2229.24 =
⇒ n+1 = 11.7
⇒ n = 10.7
⇒ n ≅ 11
Question_type 5

On an overcast day the directional distribution of the solar radiation incident on the earth’s surface may be approximated by an expression of the form where In = 80 W/m2is the total intensity of radiation directed normal to the surface and is the zenith angle. What is the solar irradiation at the earth’s surface?
  • a)
    250
  • b)
    252
Correct answer is between ' 250, 252'. Can you explain this answer?

Sarita Yadav answered
Consider a point on the earth’s surface receiving solar radiation from all hemispherical directions above the point.
By definition of intensity of radiation
dQ1 = I(dA1 cos ፀ)ωdA2➝dA1 where dA2 is a small area on hemispherical surface.
∴ ωdA2➝dA1 = (rsin ፀ dф)(rdፀ)/r2
= sin ፀ dፀ dф
∴ dQ1 - IdA1 sin θ dθ dф
dQ1 - Incos ө dA1 sin θ dθ dф
dQ1 - IndA1 sin 2θ/2 x dθ dф
G = Qi/dA1 =In x π
∴ G = 80 x π
G = 251.32 W/m2

What is the equivalent emissivity for radiant heat exchange between a small body ( in a very large enclosure (emissivity = 0.5)?
  • a)
    0.5
  • b)
    0.4
  • c)
    0.2
  • d)
    0.10
Correct answer is option 'B'. Can you explain this answer?

Telecom Tuners answered
If one surface is completely enclosed by another surface,
If surface 2 is very large,
Also we have q = εeσ(T14 - T24) ...②
From ① & ② εe = 1/Rth
But Rth = 1 /ε1A1
∴ εe = ε1 = 0.4

Consider sun (radius R) to be a black body of temperature T. Find out the view factor of earth ( with respect to sun. The center to center distance between sun and earth is (l >>r,R).
Option~
(A)
(B)
(C)
(D)
Correct answer is 'A'. Can you explain this answer?

Sarita Yadav answered
By definition of view factor
Fse = Qes➝ne/Qes
Where Qes = 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 Fse =

The spectral, hemispherical absorptivity of an opaque surface and the spectral irradiation at the surface are as shown.
λ(μm)
λ(μm)
What is the total hemispherical absorptivity of the surface?
  • a)
    0.95
  • b)
    0.97
Correct answer is option ''. Can you explain this answer?

Telecom Tuners answered
Based on data given
Total irradiation on the surface is
i.e. the area under the curve Gλ 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
Ga = 1000+800+2000+100
Ga = 4800 W/m2
∴ α = Ga/G
α = 4800/5000
α = 0.96
Quesiton_type: 5

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)
    Q
  • b)
    R
  • c)
    S
  • d)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Sarita Yadav answered
A grey surface is one for which monochromatic emissivity ελ is independent of wavelength (λ ) But total emissivity (ε) of a surface is given as
ε = E/Eb
Where E & Eb represent total emissive power of the surface and a black surface at same temperature.
But as ελ ≠ f(λ)
∴ ε = ελEb/Eb
∴ ε = ελ
i.e. total emissivity is equal to monochromatic emissivity. Hence statement Q is also correct.
In the figure shown,
AB = Eλ &
AC = Ebλ
∴ AB\Ac = Eλ/Ebλ
i.e, AB\AC = ελ
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.

A small stationary surface of area A1 = 10-4m2 emits diffusely with a total intensity of . A second surface of equivalent area is located at a fixed distance from . The connecting line between the two surfaces remains perpendicular to while rotates at an angular frequency of
ፀ2 = dፀ2/dt = 2 rad/s about the axis as shown in the figure.
What is the amount of energy intercepted by the two sides of , during one complete revolution?
Option~
(A) 2 x 10-6j
(B) 4 x 10-6j
(C) 8.0 x 10-6j
(D) 16 x 10-6j
Correct answer is 'C'. Can you explain this answer?

Sarita Yadav answered
At time instant t the normal to surface of plate 2 which faces plate 1 makes an angle of ፀ2 with the direction of view ®
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 = Qc1➝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
4Ic1A1A2 / r2ፀ2
= 4 x 100 x 10-4 x 10-4/(0.5)2 x 2
= 8 x 10-6 j

Determine the shape factor between a small area and a parallel circular disc A2(diameter D) is located on the axis of the disc and the semi-vertex angle of the cone formed with the disc as base and as the vertex is α.
Option~
(A)
(B)
(C)
(D)
Correct answer is 'A'. Can you explain this answer?

Telecom Tuners answered
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.

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 non-participating 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/m2
  • a)
    6.76
  • b)
    6.80
  • c)
    6.81
  • d)
    6.91
Correct answer is option 'A'. Can you explain this answer?

Sarita Yadav answered
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

For a grey diffuse surface with emissivity ε and temperature T, the intensity of emitted radiation is given as __________. ( is StefanBoltzmann constant.)
 
  • a)
  • b)
  • c)
  • d)
    επσT4
Correct answer is option 'A'. Can you explain this answer?

Telecom Tuners answered
For a diffuse surface, intensity of emitted radiation (Ie) is same in all directions and is given as
Ie = E/π
Where E is total emissive power
Therefore
Ie = εσT4/π

The spectral hemispherical absorptivity of an opaque surface is as shown.
The surface is exposed to ir-radiation for which the spectral distribution is as shown.
What is the total hemispherical absorptivity of surface?
  • a)
    0.6
  • b)
    0.8
  • c)
    0.7
  • d)
    Data insufficient
Correct answer is option 'B'. Can you explain this answer?

Sarita Yadav answered
Total absorptivity is defined as
α = Ga/G
Where Ga = ∫oooαλGλdλ
Ga = ∫06 0.6 x 0 dλ + ∫610 0.8 x 500 dλ + ∫1014 0.8 x 250 dλ + ∫1400 0.8 x 0 dλ
∴ Ga = 0 + 0.8 x 500 x (10-6) + 0.8 x 250 x (14-10) + 0
Ga = 2400 W/m2 ...①
and
G = ∫000 Gλdλ
I.e. G = Area under graph Gλ Vs λ
G = 500 x (10-6) + 250 x (14-10)
G = 2000 + 1000
G = 3000 W/m2 ...②
Therefore, from ① & ②
α = Ga/G = 2400/3000
α = 0.8

The dark surface of a ceramic stove top may be approximated as a blackbody. The “burners,” which are integral with the stove top, heated from below by electric resistance heaters.
Consider a burner of diameter D = 200mm operating at a uniform surface temperature of Ts = 250oC in ambient air at . Without a pot or pan on the burner, if the efficiency associated with energy transfer from the heaters to the burners is , what is the electric power requirement? (The relation between Nusselt number and Rayleigh number is given as NuL = (RaL)¼)
Air (Tf = 408 K)
k = 0.0344 x W/mK
ν = 27.4 x 10-6 m2/s
α = 39.7 x 10-6 m2/s
Pr = 0.70
β = 0.00245 K-6
  • a)
    202 W
  • b)
    222 W
  • c)
    212 W
  • d)
    232 W
Correct answer is option 'D'. Can you explain this answer?

Sarita Yadav answered
For emission from a black body,
= 13 W
∵ L = As/P = D/4 = 0.2/4 = 0.05 m
⟶ RsL = gβ(Ts-Tꝏ)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
∴ Qconv = h As(Ts-Tꝏ)
⇒ 75.7 W
∴ Electric power required,
⇒ P = 232 W

In the previous question if εshield = εsurface, then what will be the number of shields?
Option~
(A) 100
(B) 3
(C) 99
(D) 4
Correct answer is 'C'. Can you explain this answer?

Sarita Yadav answered
With introduction of n shields the rate of heat transfer reduces by 99%, therefore,
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.

Determine for the following configuration. Long inclined plates (point B is directly above the center of )
 
  • a)
    F12 = 0.5, F21 = 0.707
  • b)
    F12 = 0.707, F21 = 0.5
  • c)
    F12 = 0.5, F21 = 0.637
  • d)
    F12 = 0.637, F21 = 0.50
Correct answer is option 'A'. Can you explain this answer?

Sarita Yadav answered
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

The spectral distribution of surface irradiation is as follows
What is the total irradiation in kW/m2 ?
  • a)
    10
  • b)
    20
  • c)
    30
  • d)
    40
Correct answer is option 'B'. Can you explain this answer?

Telecom Tuners answered
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 surface of area emits radiation as a blackbody at temperature T1 = 600 K. A fraction of energy emitted by A1 strikes another small surface of area A2 = 5 cm2 oriented as shown in the figure
A1 = 3 cm2
T1 = 600 K
Find out view factor of surface with respect to A1 ?
Option~
(A) 0.6215 x 10-4
(B) 1.234 x 10-4
(C) 4.14 x 10-4
(D) 2.071 x 10-4
Correct answer is 'B'. Can you explain this answer?

Sarita Yadav answered
View factor (F12 ) is the fraction of energy leaving the surface that is received at surface But as is a black surface, energy leaving the surface is only the emitted radiation.
∴ F12 = qc1➝r2/Qc1
By Stefan-Blotzmann law
Qc1 = σA1T14
Also, using the definition of Intensity of emitted radiation,
Qc1➝r2 = Ic1A1 cos ፀ1ω2➝1
Where ω2➝1 is solid angle subtended by point surface 2 over point surface 1.
∴ ω2➝1 = A2cosፀ2/r2
For black surface
Ic1 = σT14/π
⇒ F12 = 1.243 x 10-4

The spectral emissivity of an opaque surface at 800 K is approximated as
Black body radiation function values for a black surface at 800 K are as follows
Find the total emissivity of the surface?
  • a)
    0.4
  • b)
    0.45
  • c)
    0.6
  • d)
    0.52
Correct answer is option 'D'. Can you explain this answer?

Telecom Tuners answered
By definition of total emissivity
ε = E/Eb
Where E =
∴ E =
∴ E =
∴ E = 0.3
E = 0.3 Eb,0➝3 + 0.8Eb,3➝7 + 0.1Eb,7➝ꝏ
E = 0.3 f3Eb + 0.8f3➝7Eb + 0.1f7➝ꝏEꝏ
E = [0.3f3 + 0.8(f7 -f3) + 0.1(1- f7)]Eb
∴ 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

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.
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Telecom Tuners answered
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

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?
  • a)
    ε1ε2
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Telecom Tuners answered
The equivalent thermal resistance between two gray bodies is given by
Rth = 1-ε1/A1ε1+1/A1F12+1-ε2/A2ε2
Now for infinite parallel panes, F12 =1
A1 = A2 = A
For unit area, A = 1 m2
Rth = 1/ε1+1/ε2-1
Hence equivalent emissivity
Εe = 1/Rth

A small surface of area emits radiation as a blackbody. A fraction of this energy emitted is received by another small area Oriented as shown in figure.
A1 = 2 cm2
The rate at which radiation emitted by surface 1 strikes is measured to be 274 x 10-6 W. Determine the surface temperature of in K?
  • a)
    550
  • b)
    555
Correct answer is between ' 550, 555'. Can you explain this answer?

Sarita Yadav answered
As surface 1 is a black surface, the energy leaving the surface is only the energy emitted by the surface.
Also by definition of black surface (diffuse characteristics)
Ic1 = σT14/π
274 x 10-6 W = 5.67 x 10-1 W/m2- K4 x T14 x ( 2cm2 x 6 cm2) x cos 60o x cos 30o /(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

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)
  • a)
    1 + sin α
  • b)
    1 - sin(α/2)
  • c)
    1 - sin α
  • d)
    1+ sin(α/2)
Correct answer is option 'B'. Can you explain this answer?

Telecom Tuners answered
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)

A metal storage tank with planar surfaces of external area contains water at and is located inside a large enclosure with walls at 65oC . The tank is initially painted black on the outside and its emissivity is approximated to 0.95. How much percentage reduction in heat loss would occur if the outside surface of tank is coated with aluminium paint (instead of black) of emissivity 0.55? Take radiation constant
σb = 5.67 x 10-8 W/m2K4
  • a)
    58
  • b)
    42
  • c)
    29
  • d)
    21
Correct answer is option 'A'. Can you explain this answer?

Sarita Yadav answered
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
∴ (Q1➝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.

A cake is being baked in an hemispherical oven of radius 500 mm. The lower surface of the oven is maintained at 1000 K while the covering surface ( is maintained at 800 K. The diameter of the round cake is 0.25 m and its height can be neglected. Considering the lower surface of the oven and cake to be black bodies and neglecting the area covered by the cake, find the time required to bake the cake, if it requires 3.2123 MJ.
  • a)
    44.0
  • b)
    46.0
Correct answer is between ' 44.0, 46.0'. Can you explain this answer?

Sarita Yadav answered
’
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

Consider an infinitely long three sided triangular enclosure with sides 2 cm, 3 cm and 4 cm. The view factor from 2 cm side to 4 cm side is __________.
  • a)
    0.58
  • b)
    0.59
Correct answer is option ''. Can you explain this answer?

Sarita Yadav answered
Given la = 2 cm
Lb = 3 cm
Lc = 4 cm
By energyF conservation principle
Faa + Fab + Fac = 1
Fba + Fbb + Fbc = 1
Fca + Fcb + Fcc = 1
As a,b,c are planar surfaces
Faa, Fbb, Fcc, = 0
Therefore the equations are reduced to
Fab + Fac = 1 ….①
Fba + Fbc + = 1 ...②
Fca + Fcb = 1 ….③
By Reciprocity theorem
Substituting Fba & Fbc in equation ②
Substituting Fcb & Fca in equation ③
∴ 2Fac - 6Fab = 1
Fac - 3Fab = 1/2 ….Ⓢ
Solving equation ① & Ⓢ together
Fab + Fac =1
Fac - 3Fab = 1/2
____________
3Fab + 3ac = 3
3Fac - 3Fab = 1/2
+ + +
______________
6Fac = 7/2
Fac = 7/12
Fac 0.5833
Question_type 5

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