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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 = 250^{o}C 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 (T_{f} = 408 K)
k = 0.0344 x W/mK
ν = 27.4 x 10^{6} m^{2}/s
α = 39.7 x 10^{6} m^{2}/s
Pr = 0.70
β = 0.00245 K^{6}
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})L^{3}
= 9.81 x 0.00245 x 230 x (0.05)^{3}/27.4 x 39.7 x 10^{12}
= 6.35 x 10^{5}
= 10.5 W/m^{2} K
∴ Q_{conv} = h A^{s}(T^{s}T_{ꝏ})
⇒ 75.7 W
∴ Electric power required,
⇒ P = 232 W
A metal storage tank with planar surfaces of external area contains water at and is located inside a large enclosure with walls at 65^{o}C . 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/m^{2}K^{4}
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, F_{11} = 0
∴ F_{21} = 1
Also ,
A_{2} >> A_{1}
∴ (Q_{1➝2})_{net} = σA_{1}ε_{1}(T_{1}^{4}  T_{2}^{4})
∴ Q ∝ ε_{1}
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.
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.
Since the shield is placed midway between both the cylinders, therefore diameter of the shield
∴ d_{3} = d_{1}+d_{2}/2 = 7.5 + 22.5/2
⇒ d_{3} = 15.0 cm
A_{1} = πd_{1}L
A^{2}_{ }= πd_{2}L
A_{3 }= πd_{3}L
For the system to be steady,
(Q_{1➝3})_{net} = (Q_{3➝2})_{net}
By thermal circuits
R_{13} = 11.166/7.5πl
Similarly,
R32 = 3/7.5πl
∴ T_{3} = 717 K
Question_type 5
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?
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 f_{3}E_{b} + 0.8f_{3➝7}E_{b} + 0.1f_{7➝ꝏ}E_{ꝏ}
E = [0.3f_{3} + 0.8(f_{7} f_{3) }+ 0.1(1 f_{7})]E_{b}
∴ ε = [0.3f_{3} + 0.8(f_{7}  f_{3 }) + 0.1(1f_{7})]
For T = 2400 μmK
∴ f_{3} = 0.140256
Similarly for λ = 7 μm
λT = 5600 μmK
∴ f_{7} = 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
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 __________.
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
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/m^{2} K
q = εσ(T_{2}^{4}  T_{ထ}^{4})
q = 0.85 x 5.67 10^{8}[430^{4}  300^{4}]+1.5(430 300)^{0.33} x (430  300)
q = 1257.3116 + 971.932
q = 2229.24 W
nshields
∴ Heat transfer from surface 1 to 2 through nshields must be equal to ̇q
⇒ 2229.24 =
⇒ n+1 = 11.7
⇒ n = 10.7
⇒ n ≅ 11
Question_type 5
An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.
Given F_{13} = 0.06
Now F_{11} +F_{12} + F_{13} = 1
⇒ F_{12} = 0.94
⇒ F_{21} = F_{23}
again A_{1}F_{12} = A_{2}F_{21}
⇒ F_{21} = 0.1175
From Symmetry F_{21} = F_{23}
∴ F_{22} = 1(2 x 0.1175)
⇒ F_{22} = 0.765
∴ Power required to drive furnace,
= A_{2}A_{21} σ(T_{2}^{4}  T_{1}^{4}) + A_{2}F_{23}σ(T_{2}^{4}  T_{3}^{4})
= A_{2}F_{21}σ[2T_{2}^{4 } T_{1}^{4}  T_{3}^{4}]
= 3.14(0.1)(0.2).0.1175 x 5.67
X 10^{8}[2 x (1800)^{4}  (2000)^{4}  300^{4}]
= 2.087 kW
Question_type 5
A furnace is shaped like a long equilateral triangular duct. The width of each side is 1 m. The base surface has an emissivity of 0.7 and is maintained at a uniform temperature of 600 K . The heated leftside surface closely approximates a blackbody at . The rightside surface is well insulated. Determine the rate in kW at which heat must be supplied to the heated side externally per unit length of the duct in order to maintain these operating conditions.
Since the surface 3 is a reradiating surface, (R_{3 }= 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
A_{1} = A_{2} = A_{3} = 1 x 1 = 1 m^{2}
Total thermal resistance between 2 and 2 is,
From symmetry of the triangle we have
F_{12} = F_{13} = F_{23} = 0.5
∴ R_{1} = 0 (for black surface)
R_{12} = 1/A_{1}F_{12} = 1/1 x 0.5 = 2
R_{13 }= 1/A_{1}F_{13} = 1/1 x 0.5 = 2
R_{23} = 1/A_{2}F_{23} = 1/1 x 0.5 = 2
R_{2} = 1ε_{2}/A_{2}ε_{2} = 10.7/1 x 0.7 = 0.43
= (0.5 + 0.25)^{1} + 0.43
= 1.7633
= 5.67 x 10^{8}(1000^{4}  600^{4})/1.7633
Q = 27.988 kW
Question_type 5
Air is flowing inside a cylindrical tube in which a thermocouple is kept. Convective heat transfer coefficient between air and thermocouple is 60 W/m^{2} C, ε_{thermocouple }= 0.5, ε_{wall} = 0.6. If T_{a} = 970^{o }C and^{ }T_{c} = 800^{o}C then, T_{w} is ________. Air is a nonparticipating medium in terms of radiation
(T_{c})
Option~
(A) 1041.7_{o}C
(B) 1000_{o}C
(C) 768.7_{o}C
(D) 509_{o}C
= heat radiation from thermocouple to walls
∴ hA_{T/C}(T_{A}  T_{C}) = ε_{T/C}(T_{C}^{4}  T_{w}^{4}). σ
(∵ Area of walls very large as compared to area of thermocouple)
60 x 70 = 0.5 x 5.67 x 108(1073^{4}  T_{w}^{4})
1073^{4}  T_{w}^{4} = 1.4815 x 10^{11}
T_{w} = 1041.67 K
T_{w} = 768.7^{o}C
If is the emissivity of surfaces and shields and n is the number of shields introduced between the two surfaces, then overall emissivity is given by
Option~
(A) n/ε
(B) 1/(2nε)
(C) nε
(D) ε/[(n+1)(2ε)]
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 + 2ε/ε
1/ε_{o} = n(2ε) + 1( 2ε )/ε
1/ε_{o} = (n+1)(2ε)/ε
ε_{o }= ε/(n+1)(2ε)
The net radiation from the surfaces of two parallel plates maintained at is to be reduced by at least 99%. Number of shields which should be placed between surfaces to achieve this reduction is (ε_{shield} = 0.05), (ε_{surface} = 0.8)
Option~
(A) 3
(B) 4
(C) 5
(D) 6
Q ∝ 1/R
where R is the resistance to radiation heat exchange between the plates. When there is no shield
R_{1} = 1ε_{1}/Aε_{1 }+ 1/A + 1ε_{2}/Aε_{2}
R_{1} = 10.8/A x 0.8 + 1/A + 10.8//A x 0.8
R_{1} = 1.5/A
When n shields are placed in between the plates
= 1.5 + 39 n/A
Based on given condition
Q_{2}/Q_{1} = 0.01
∴ 1.5/1.5 + 39 n = 0.01
∴ n = 3.8
i.e.n ≃ 4
In the previous question if ε_{shield} = ε_{surface}, then what will be the number of shields?
Option~
(A) 100
(B) 3
(C) 99
(D) 4
1.5/1.5+1.5n = 0.01
n = 1.5(10.01)/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.
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)
F_{se} = Q_{es➝ne}/Q_{es}
Where Q_{es} = σT^{4} x 4πR^{2}
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 πR^{2}.
Total energy emitted by sum = σ(4πR^{2})T^{4}
∴ Energy received per unit area of imaginary sphere
= σ(4πR^{2})T^{4}/4πl^{2}
= σR^{2}T^{4}/l^{2}
∴ Energy received by earth (i.e. surface AB)
= σR^{2}T^{4}/l^{2} x πr^{2}
Hence F_{se}_{ } =
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.
Radiosity from surface 2
= reflected energy from ② + emitted energy from ②
⇒ J_{2} = ρ_{2}σA_{1}T_{1}^{4} + ε_{2} A_{2} σT_{2}^{4}
Ε_{2} = 0.8 ⇒ α_{2} = 0.8
∴ α_{2}+ρ_{2}+τ_{2} = 1
⇒ ρ_{2} = 0.2
⇒ J_{2} = 38,071.644 W
Energy incident on the cake, G_{3} = F_{23}J_{2}
A_{3}F_{32} = A_{2}F_{23}
∴ G_{3} = 0.03125 x 38071.644
G_{3} = 1189.74 W
∴ Time required to bake the cake,
t = 3.2123 x 10^{6}/1189.74
⇒ t = 2700 s
⇒ t ≅ 45 min
Determine the shape factor between a small area and a parallel circular disc A_{2}(diameter D) is located on the axis of the disc and the semivertex angle of the cone formed with the disc as base and as the vertex is α.
Option~
(A)
(B)
(C)
(D)
Considering an elementary ring dA_{2} of width dr at a radius r,
dA_{2} = 2 πr dr
Now, r = L tanΦ_{1}
Dr = L sec^{2}Ф_{1}dФ_{1}
dA_{2} = 2πL^{2} tan Φ_{1} sec^{2} Φ_{1}dΦ_{1}
L/P = cos Φ_{1} or P = L/cosΦ_{1}
= sin^{2} α = D^{2}/4/D^{2}/4 + L^{2} = D^{2}/D^{2} + 4L^{2}
where D is the diameter of the disc.
A small surface of area emits radiation as a blackbody at temperature T_{1} = 600 K. A fraction of energy emitted by A_{1} strikes another small surface of area A_{2} = 5 cm^{2} oriented as shown in the figure
A^{1} = 3 cm^{2}
T^{1} = 600 K
Find out view factor of surface with respect to A^{1} ?
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}
∴ F_{12} = q_{c1➝r2}/Q_{c1}
By StefanBlotzmann law
Q_{c1} = σA_{1}T_{1}^{4}
Also, using the definition of Intensity of emitted radiation,
Q_{c1➝r2} = I_{c1}A_{1 }cos ፀ_{1}ω_{2➝1}
Where ω_{2➝1} is solid angle subtended by point surface 2 over point surface 1.
∴ ω_{2➝1} = A_{2}cosፀ_{2}/r^{2}
For black surface
I_{c1} = σT_{1}^{4}/π
⇒ F_{12} = 1.243 x 10^{4}
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.
A_{1} = 2 cm^{2}
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?
Also by definition of black surface (diffuse characteristics)
I_{c1} = σT_{1}^{4}/π
274 x 10^{6} W = 5.67 x 10^{1} W/m^{2} K^{4} x T_{1}^{4} x ( 2cm^{2} x 6 cm^{2}) 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}/ 10^{4} x 1/2 x ✓3/4 x T_{1}^{4}
274 x 10^{6} x 10^{8} x 4 / 5.67 x 12 x ✓3 = T_{1}^{4}
∴ T_{1}^{4} = 9.30 x 10^{10}
T_{1} = 552.23 K
A small stationary surface of area A_{1} = 10^{4}m^{2} 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
ፀ_{2} = dፀ_{2}/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^{6}j
(B) 4 x 10^{6}j
(C) 8.0 x 10^{6}j
(D) 16 x 10^{6}j
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 = dፀ_{2}/θ^{2}
This face of the surface 2 is in front of surface
1 for θ_{2} = 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_{c1}A_{1}A_{2} / r^{2}ፀ_{2}
= 4 x 100 x 10^{4} x 10^{4}/(0.5)^{2} x 2
= 8 x 10^{6} j
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 I_{n} = 80 W/m^{2}is 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?
By definition of intensity of radiation
dQ_{1} = I(dA_{1} cos ፀ)ω_{dA2➝dA1} where dA_{2} is a small area on hemispherical surface.
∴ ω_{dA2➝dA1} = (rsin ፀ dф)(rdፀ)/r^{2}
= sin ፀ dፀ dф
∴ dQ_{1}  IdA_{1} sin θ dθ dф
dQ_{1}  I_{n}cos ө dA_{1} sin θ dθ dф
dQ_{1}  I_{n}dA_{1} sin 2θ/2 x dθ dф
G = Q_{i}/dA_{1} =I_{n} x π
∴ G = 80 x π
G = 251.32 W/m^{2}
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?
Total irradiation on the surface is
i.e. the area under the curve G_{λ} vs λ.
G = 0+1/2 x 500 x (62) + 500 x (126) + 1/2 x 500 x (1612) + 0
G = 0 + 1000 + 3000 + 1000 + 0
G = 5000 W/m^{2}
Similarly radiation absorbed (in W/m^{2} ) is given as
G_{a} = 1000+800+2000+100
G_{a } = 4800 W/m^{2}
∴ α = G_{a}/G
α = 4800/5000
α = 0.96
Quesiton_type: 5
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