Radiation Examples

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Example 1

Consider a 5mm square diffuse surface ∆Ao having a total emissive power Eo = 4000W/m2 . Assumptions: • diffuse emitter • non-participating medium above ∆Ao

a) What is the rate at which radiant energy us enutted by ∆Ao (qemit )?

 qemit = Eo ∆Ao = 4000

 W (5 × 10−3 m)(5 × 10−3 m) = 0.1W m2

(1)

b) What is the intensity, Io,e , of the radiation field emitted from the surface ∆Ao ? Since the radiation is diffuse, equation 12.14 (from the text) applies Io,e = Eo /π = 1273W/(m2 Sr)

(2)

This intensity is constant at any point in our hemisphere since the radiation is diffuse (independant of direction) and the medium is non-participating (does not absorb or emit radiation).

c) Beginning with 12.10 (text) and presuming knowledge of Io,e , obtain an expression for qemit .

Z qemit

=

Io,e ∆Ao cos θ sin θdθdφ # Z 2π "Z π/2 = Io,e ∆Ao cos θ sin θdθ dφ hemisphere

φ=0

Z

θ=0

2π

= Io,e ∆Ao

[1/2] dφ φ=0

= Io,e ∆Ao π = 1273π(5 × 10−3 m)(5 × 10−3 m) = 0.1W 1

The integration over the hemisphere is based solely on geometry when the radiation is diffuse. This can get ugly if Ie depends on directions.

d) Consider a hemispherical surface located at r=R1 = 0.5 m from the surface. Using conservation of energy, determine the rate at which energy is incident on this surface due to emission from ∆Ao . Defining a control surface above ∆Ao and on the hemisphere (A1 ), the radiant energy leaving ∆Ao must pass through A1 since the media in between is nonparticipating. q1,incident = Eo ∆Ao = 0.1W

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The average irradiation though will depend on the area (the energy is spreading out). Eo ∆Ao = 0.0637W/m2 2πR12

G1 = q1, incident/A1 =

(4)

e) Using 12.5 (text) determine the rate at which radiant energy leaving ∆Ao is intercepted by a small area ∆A2 (4 × 10−6 m2 ) located in the direction (45o , φ) on the hemispherical surface. What is the irradiation on ∆A2 ? The radiant power leaving ∆Ao intercepted by ∆A2 when ∆A2 = 4 × 10−6 m2 , located on the hemisphere in the direction (45o , φ) is qδAo −∆A2 = Io,e ∆Ao cos θ∆ω2−0

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and, from equation 12.2 (text)

∆ω2−0

= = =

∆A2 r2 4 × 10−6 0.52 1.6 × 10−5 Sr

Therefor, 

qδAo −∆A2

=

 W 1237 (2.5 × 10−6 [m2 ]) cos 45o (1.6 × 10−5 [Sr]) m2 Sr

=

3.6 × 10−7 W

The irradiation is simply G2 = q/∆A2 = (3.6−7 [W ])/(4 × 10−6 [m2 ]) = 90mW/m2 2

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f) Repeat part e) for the direction (0, φ). Are the irradiations equal? The solid angle is the same as above, but the surface normal is different.   W qδAo −∆A2 = 1237 (2.5 × 10−6 [m2 ]) cos 0o (1.6 × 10−5 [Sr]) m2 Sr 5.09 × 10−7 W

=

G2 = q/∆A2 = (5.09 × 10−7 [W ])/(4 × 10−6 [m2 ]) = 127mW/m2

(7)

You can see the entire surface area at 0o

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Example 2

An opaque grey surface at 27o C is exposed to irradiation of 1000 W/m2 and 800 W/m2 is reflected. Air at 17o C flows over the surface and the heat transfer coefficient is 15W/m2 K. Determine the net heat flux from the surface. Assumptions: • Surface is opaque and grey • Surface is diffuse • Surroundings are already accounted for For a diffuse gray opaque surface,  = α = 1 − ρ =α=1−

Gr ef = 1 − 800/1000 = 0.2 G

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Energy balance: q”net

= E˙ out − E˙ in = q”conv + E + Gref − G = h(Ts − T∞ ) + σT s4 + Gref − G     W W −8 = 15 (27 − 17)[K] + 0.2(5.67 × 10 ) (27 + 273)4 [K 4 ] m2 m2 K 4 +800[W/m2 ] − 1000[W/m2 ] = 42[W/m2 ]

The radiosity is J

= Gref + E = 800 + σTs4 = 892W/m2 3

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Example 3

A sphere (k = 185W/m2 , αt = 72.5m2 /s ) of diameter 30 mm. The surface of the sphere is diffuse and grey, with  = 0.8. It is placed in an oven whos walls are at To = 600K. The air in the oven is at T∞ = 400K and the convection coefficient between the sphere and the air is 15 W/m2 K. a) Find the heat transfer to the sphere when its temperature is 300K? b) What is the steady state temperature of the sphere? c) How long will it take for the sphere initally at 300K to come within 20K of the steady state T? Assumptions: • surface is diffuse and grey • surface is much smaller than the wall area • surfaces are isothermal

a) Find the heat transfer to the sphere when its temperature is 300K? Surface energy Balance: qnet

= qin − qout = =

αGAs + qc onv − EAs ασTo4 As + hAs (T∞ − Ts ) − σTs4 As

The irradiation to the sphere is the emissive power of a blackbody at the temperature of the oven walls. α =  since the surface is grey and dsiffuse.

qnet

 W (600[K])4 + 15[W/m2 K](400 − 300)K = [0.8(5.67 × 10 ) m2 K 4   2 W 4 πD −8 (300[K]) ] −0.8(5.67 × 10 ) m2 K 4 4 = 19.8W −8



b) What is the steady state temperature of the sphere? At steady state, qnet = 0. 0 Tss

4 = ασTo4 As + hAs (T∞ − Tss ) − σTss As = 538.2[K]

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c) How long will it take for the sphere initally at 300K to come within 20K of the steady state T? This is a basic lumped capacitance problem with the additional fluxes due to radiation complicating things with fourth order dependancies on T. The time ti reach (538.2 - 20) = 518.2K is found to be 14.3 minutes.

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More examples

Do not forget to look at the examples in your text where the surface properties are given differently for different wavelengths.

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