Heat 4e SM Chap12

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preparation. If you are a student using this Manual, you are using it without permission.

Solutions Manual

for

Heat and Mass Transfer: Fundamentals & Applications

Fourth Edition

Yunus A. Cengel & Afshin J. Ghajar

McGraw-Hill, 2011

Chapter 12 FUNDAMENTALS OF THERMAL RADIATION

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Electromagnetic and Thermal Radiation

12-1C Electromagnetic waves are characterized by their frequency v and wavelength λ . These two properties in a medium are related by λ=c /v where c is the speed of light in that medium.

12-2C Visible light is a kind of electromagnetic wave whose wavelength is between 0.40 and 0.76 µm. It differs from the

other forms of electromagnetic radiation in that it triggers the sensation of seeing in the human eye.

12-3C Microwaves in the range of are very suitable for use in cooking as they are reflected by metals,

transmitted by glass and plastics and absorbed by food (especially water) molecules. Thus the electric energy converted to radiation in a microwave oven eventually becomes part of the internal energy of the food with no conduction and convection thermal resistances involved. In conventional cooking, on the other hand, conduction and convection thermal resistances slow down the heat transfer, and thus the heating process.

m 10 to

102 5_µ

12-4C Thermal radiation is the radiation emitted as a result of vibrational and rotational motions of molecules, atoms and

electrons of a substance, and it extends from about 0.1 to 100 µm in wavelength. Unlike the other forms of electromagnetic radiation, thermal radiation is emitted by bodies because of their temperature.

12-5C Light (or visible) radiation consists of narrow bands of colors from violet to red. The color of a surface depends on its

ability to reflect certain wavelength. For example, a surface that reflects radiation in the wavelength range 0.63-0.76 µm while absorbing the rest appears red to the eye. A surface that reflects all the light appears white while a surface that absorbs the entire light incident on it appears black. The color of a surface at room temperature is not related to the radiation it emits.

12-6C Radiation in opaque solids is considered surface phenomena since only radiation emitted by the molecules in a very

thin layer of a body at the surface can escape the solid.

12-7C Because the snow reflects almost all of the visible and ultraviolet radiation, and the skin is exposed to radiation both

from the sun and from the snow.

12-8C Infrared radiation lies between 0.76 and 100 µm whereas ultraviolet radiation lies between the wavelengths 0.01 and

0.40 µm. The human body does not emit any radiation in the ultraviolet region since bodies at room temperature emit radiation in the infrared region only.

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12-9 A cordless telephone operates at a frequency of 8.5×108 Hz. The wavelength of these telephone waves is to be

determined.

Analysis The wavelength of the telephone waves is

mm 353 = = × × = = 0.353m Hz(1/s) 10 5 . 8 m/s 10 998 . 2 8 8 v c λ

12-10 Electricity is generated and transmitted in power lines at a frequency of 50 Hz. The wavelength of the electromagnetic

waves is to be determined.

Analysis The wavelength of the electromagnetic waves is

Power lines m 10 5.996× 6 = × = = Hz(1/s) 50 m/s 10 998 . 2 8 v c λ

12-11 A microwave oven operates at a frequency of 2.2×109 Hz. The wavelength of these microwaves and the energy of

each microwave are to be determined.

Analysis The wavelength of these microwaves is

Microwave oven mm 136 = = × × = = 0.136m Hz(1/s) 10 2 . 2 m/s 10 998 . 2 9 8 v c λ

Then the energy of each microwave becomes

J 10 1.46 −24 − × = × × = = = m 136 . 0 ) m/s 10 998 . 2 )( Js 10 625 . 6 ( 34 8 λ hc hv e

12-12 A radio station is broadcasting radiowaves at a wavelength of 150 m. The frequency of these waves is to be

determined.

Analysis The frequency of the waves is determined from

Hz 10 2.00× 6 = × = = ⎯→ ⎯ = m 150 m/s 10 998 . 2 8 λ λ v c v c

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12-13 The speeds of light in air, water, and glass are to be determined.

Analysis The speeds of light in air, water and glass are

Air: = = × =3.0×108m/s 1 m/s 10 0 . 3 8 0 n c c Water: = = × =2.26×108m/s 33 . 1 m/s 10 0 . 3 8 0 n c c Glass: = = × =2.0×108m/s 5 . 1 m/s 10 0 . 3 8 0 n c c

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Blackbody Radiation

12-14C A blackbody is a perfect emitter and absorber of radiation. A blackbody does not actually exist. It is an idealized

body that emits the maximum amount of radiation that can be emitted by a surface at a given temperature.

12-15C We defined the blackbody radiation function because the integration cannot be performed. The

blackbody radiation function represents the fraction of radiation emitted from a blackbody at temperature T in the wavelength range from

λ f

∫

∞ 0 ) ( λ λ T d E_b λ f

λ = 0 to λ . This function is used to determine the fraction of radiation in a wavelength range between λ₁andλ₂.

12-16C Spectral blackbody emissive power is the amount of radiation energy emitted by a blackbody at an absolute

temperature T per unit time, per unit surface area and per unit wavelength about wavelength λ . The integration of the spectral blackbody emissive power over the entire wavelength spectrum gives the total blackbody emissive power,

∫

∞ = = 0 4 ) ( ) (T E T d T E_b _b_λ λ σ

The spectral blackbody emissive power varies with wavelength, the total blackbody emissive power does not.

12-17C The larger the temperature of a body, the larger the fraction of the radiation emitted in shorter wavelengths.

Therefore, the body at 1500 K will emit more radiation in the shorter wavelength region. The body at 1000 K emits more radiation at 20 µm than the body at 1500 K since λT =constant.

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12-18 The temperature of the filament of an incandescent light bulb is given. The fraction of visible radiation emitted by the

filament and the wavelength at which the emission peaks are to be determined.

Assumptions The filament behaves as a black body.

Analysis The visible range of the electromagnetic spectrum extends from λ₁=0.40µm toλ₂ =0.76µm . Noting that T =

2500 K, the blackbody radiation functions corresponding to λ₁Tandλ₂T are determined from Table 12-2 to be

053035 . 0 mK 1900 = K) m)(2500 76 . 0 ( 000321 . 0 mK 1000 = K) m)(2500 40 . 0 ( 2 1 2 1 = ⎯→ ⎯ = = ⎯→ ⎯ = λ λ f T f T µ µ λ µ µ λ T = 2500 K

Then the fraction of radiation emitted between these two wavelengths becomes

0.052714 = − = − ₁ 0.053035 0.000321 2 λ λ f f (or 5.2%)

The wavelength at which the emission of radiation from the filament is maximum is

m 1.16µ µ λ µ λ = ⋅ ⎯⎯→ = ⋅ = K 2500 K m 8 . 2897 K m 8 . 2897 )

( T _max_power _max_power

Discussion Note that the radiation emitted from the filament peaks in the infrared region.

12-19E The radiation energy emitted by the black surface per unit area (at 2060 °F) for λ ≥ 4 µm is to be determined.

Assumptions 1 The surface behaves as a black body.

Analysis The blackbody radiation function corresponding to λ1 = 4 µm is determined from Table 12-2 to be K m 5600 K ) 8 . 1 / 1 )( 460 2060 )( m 0 . 4 ( 1 = µ + = µ ⋅ λT → f_λ₁ =0.701046

Then, the radiation energy emitted is determined using

2 ft Btu/h 20,700 ⋅ = − ⋅ ⋅ × = − = = = − ∞ ∞ − ∞ ∞ −

∫

) 701046 . 0 1 ( ) R 2520 )( R ft Btu/h 10 1714 . 0 ( )] ( ) ( [ ) ( ) , ( ) ( 4 4 2 8 1 4 4 , 1 1 1 T f T f T T f T T E T Eb b λ λ λ λ λ σ σ λ

Discussion The total radiation energy emitted by this black surface is simply

2 4 ₆₉_,₁₀₀_Btu/h _ft )

(T = T = ⋅

E_b σ

12-20 The maximum thermal radiation that can be emitted by a surface is to be determined.

Analysis The maximum thermal radiation that can be emitted by a surface is determined from Stefan-Boltzman law to be

2 W/m 56,700 = × = = 4 ₍₅_.₆₇ ₁₀−8_W/m2_._K4₎₍₁₀₀₀_K)4 ) (T T E_b σ

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12-21 An incandescent light bulb emits 15% of its energy at wavelengths shorter than 0.8

µm. The temperature of the filament is to be determined.

T = ? Analysis From the Table 12-2 for the fraction of the radiation, we read

mK 2445 15 . 0 λ µ λ = ⎯⎯→ T = f

For the wavelength range of λ1 =0.0µm toλ2 =0.8µm

K 3056 = ⎯→ ⎯ = ⎯→ ⎯ T 2445 mK T m 0.8 = µ λ µ λ

12-22E The sun is at an effective surface temperature of 10,400 R. The rate of infrared radiation energy emitted by the sun is

to be determined.

Assumptions The sun behaves as a black body.

Analysis Noting that T = 10,400 R = 5778 K, the blackbody radiation functions corresponding to

T

T ₂

1 andλ

λ are determined from Table 12-2 to be

SUN T = 10,400 R 0 . 1 mK 577,800 = K) m)(5778 100 ( 547370 . 0 mK 4391.3 = K) m)(5778 76 . 0 ( 2 1 2 1 = ⎯→ ⎯ = = ⎯→ ⎯ = λ λ f T f T µ µ λ µ µ λ

Then the fraction of radiation emitted between these two wavelengths becomes (or 45.3%) 453 . 0 547 . 0 0 . 1 1 2− λ = − = λ f f

The total blackbody emissive power of the sun is determined from Stefan-Boltzman Law to be 2 7 4 4 2 8 4 ₌₍₀_.₁₇₁₄_×₁₀ _Btu/h.ft _._R ₎₍₁₀_,₄₀₀_R) ₌₂_.₀₀₅_×₁₀ _Btu/h.ft = _T − E_b σ Then, 6 2 Btu/h.ft 10 9.08× = × = =(0.453) (0.453)(2.005 107Btu/h.ft2) infrared Eb E

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12-23 The spectral blackbody emissive power of the sun versus wavelength in the range of 0.01 µm to 1000 µm is to

be plotted.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" T=5780 [K] lambda=0.01[micrometer] "ANALYSIS" E_b_lambda=C_1/(lambda^5*(exp(C_2/(lambda*T))-1)) C_1=3.742E8 [W-micrometer^4/m^2] C_2=1.439E4 [micrometer-K] λ [µm] E[W/mb, λ 2_-µm] 0.01 0 10.11 12684 20.21 846.3 30.31 170.8 40.41 54.63 50.51 22.52 60.62 10.91 70.72 5.905 80.82 3.469 90.92 2.17 … … … … 909.1 0.0002198 919.2 0.0002103 929.3 0.0002013 939.4 0.0001928 949.5 0.0001847 959.6 0.000177 969.7 0.0001698 979.8 0.0001629 989.9 0.0001563 1000 0.0001501 0.01 0.1 1 10 100 1000 10000 0.0001 0.001 0.01 0.1 1 10 100 1000 10000 100000

λ [µm]

Eb

λ [

W

/m

2

- µ

m]

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12-24 The temperature of the filament of an incandescent light bulb is given. The fraction of visible radiation emitted by the

filament and the wavelength at which the emission peaks are to be determined.

Analysis The visible range of the electromagnetic spectrum extends from λ₁=0.40µm toλ₂ =0.76µm . Noting that T =

2800 K, the blackbody radiation functions corresponding to λ₁Tandλ₂T are determined from Table 12-2 to be

088590 . 0 mK 2128 = K) m)(2800 76 . 0 ( 0014088 . 0 mK 1120 = K) m)(2800 40 . 0 ( 2 1 2 1 = ⎯→ ⎯ = = ⎯→ ⎯ = λ λ f T f T µ µ λ µ µ λ T = 2800 K

Then the fraction of radiation emitted between these two wavelengths becomes

0.08718 = − = − ₁ 0.088590 0.0014088 2 λ λ f f (or 8.7%)

The wavelength at which the emission of radiation from the filament is maximum is

m 1.035µ µ λ µ λ = ⋅ ⎯⎯→ = ⋅ = K 2800 K m 8 . 2897 K m 8 . 2897 )

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12-25 Prob. 12-24 is reconsidered. The effect of temperature on the fraction of radiation emitted in the visible range is

to be investigated.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" T=2800 [K] lambda_1=0.40 [micrometer] lambda_2=0.76 [micrometer] "ANALYSIS" E_b_lambda=C_1/(lambda^5*(exp(C_2/(lambda*T))-1)) C_1=3.742E8 [W-micrometer^4/m^2] C_2=1.439E4 [micrometer-K]

f_lambda=integral(E_b_lambda, lambda, lambda_1, lambda_2)/E_b E_b=sigma*T^4

sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant"

T [K] fλ 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000 0.000007353 0.0001032 0.0006403 0.002405 0.006505 0.01404 0.02576 0.04198 0.06248 0.08671 0.1139 0.143 0.1732 0.2036 0.2336 0.2623 1000 1500 2000 2500 3000 3500 4000 0 0.05 0.1 0.15 0.2 0.25 0.3

T [K]

f λ

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12-26 An isothermal cubical body is suspended in the air. The rate at which the cube emits radiation energy and the spectral

blackbody emissive power are to be determined.

Assumptions The body behaves as a black body.

Analysis (a) The total blackbody emissive power is determined from Stefan-Boltzman Law to be

2 2 2 ₆₍₀_.₂ ₎ ₀_.₂₄_m 6 = = = a As 20 cm T = 900 K W 8928 = × = = (5.67 10− W/m .K )(900K) (0.24m ) ) ( 4 8 2 4 4 2 s b T T A E σ 20 cm (b) The spectral blackbody emissive power at a wavelength of 4 µm is

determined from Plank's distribution law,

µm kW/m 6.84 2⋅ = ⋅ = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × ⋅ ⋅ × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = µm W/m 6841 1 K) m)(900 4 ( K m 10 43878 . 1 exp m) 4 ( /m m W 10 74177 . 3 1 exp 2 4 5 2 4 8 2 5 1 µ µ µ µ λ λ λ T C C Eb 20 cm

12-27 Radiation emitted by a light source is maximum in the blue range. The temperature of this light source and the fraction

of radiation it emits in the visible range are to be determined.

Assumptions The light source behaves as a black body. Analysis The temperature of this light source is

K 6166 = ⋅ = ⎯→ ⎯ ⋅ = m 47 . 0 K m 8 . 2897 K m 8 . 2897 ) (λT max power µ T µ_µ T = ?

The visible range of the electromagnetic spectrum extends from

m 76 . 0 to m 40 . 0 ₂ 1 µ λ µ

λ = = . Noting that T = 6166 K, the blackbody radiation

functions corresponding to λ₁Tandλ₂T are determined from Table 12-2 to be

59144 . 0 mK 4686 = K) m)(6166 76 . 0 ( 15440 . 0 mK 2466 = K) m)(6166 40 . 0 ( 2 1 2 1 = ⎯→ ⎯ = = ⎯→ ⎯ = λ λ f T f T µ µ λ µ µ λ

Then the fraction of radiation emitted between these two wavelengths becomes (or 43.7%) 0.437 ≅ − = − 0.59144 0.15440 1 2 λ λ f f

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12-28 A glass window transmits 90% of the radiation in a specified wavelength range and is opaque for radiation at other

wavelengths. The rate of radiation transmitted through this window is to be determined for two cases.

Assumptions The sources behave as a black body. Analysis The surface area of the glass window is

SUN 2 m 9 = s A

(a) For a blackbody source at 5800 K, the total blackbody radiation emission is W 10 775 . 5 ) m 9 ( K) 5800 ( ) K . W/m 10 67 . 5 ( ) ( = 4 = × −8 2 4 4 2 = × 8 s b T T A E σ

The fraction of radiation in the range of 0.3 to 3.0 µm is

L = 3 m Glass τ = 0.9 97875 . 0 mK 17,400 = K) m)(5800 0 . 3 ( 03345 . 0 mK 1740 = K) m)(5800 30 . 0 ( 2 1 2 1 = ⎯→ ⎯ = = ⎯→ ⎯ = λ λ f T f T µ µ λ µ µ λ ∆f = f_λ₂− f_λ₁ =0.97875−0.03345=0.9453

Noting that 90% of the total radiation is transmitted through the window, kW 491,300 = × = ∆ = ) kW 10 775 . 5 )( 9453 . 0 )( 90 . 0 ( ) ( 90 . 0 5 transmit fE T E _b

(b) For a blackbody source at 1000 K, the total blackbody emissive power is W 300 , 510 ) m 9 ( K) 1000 )( K . W/m 10 67 . 5 ( ) ( = 4 = × −8 2 4 4 2 = s b T T A E σ

The fraction of radiation in the visible range of 0.3 to 3.0 µm is

273232 . 0 mK 3000 = K) m)(1000 0 . 3 ( 0000 . 0 mK 300 = K) m)(1000 30 . 0 ( 2 1 2 1 = ⎯→ ⎯ = = ⎯→ ⎯ = λ λ f T f T µ µ λ µ µ λ 0 273232 . 0 1 2− = − = ∆f f_λ f_λ and kW 125.5 = = ∆ =0.90 ( ) (0.90)(0.273232)(510.3kW) transmit fE T E _b

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12-29 The radiation energy emitted within the visible light region by daylight and candlelight is to be determined.

Assumptions 1 The sun and the candlelight behave as black bodies.

Analysis The visible range of the electromagnetic spectrum extends from λ1 = 0.40 µm to λ2 = 0.76 µm. For daylight (T =

5800 K), the blackbody radiation functions corresponding to λ1T and λ2T are determined from Table 12-2 to be K m 2320 ) K 5800 )( m 40 . 0 ( 1 = µ = µ ⋅ λT → f_λ₁_,_daylight=0.124509 K m 4408 ) K 5800 )( m 76 . 0 ( 2 = µ = µ ⋅ λ T → fλ2 ,daylight=0.550015

Then the fraction of radiation emitted between these two wavelengths (for daylight) becomes 4255 . 0 124509 . 0 550015 . 0 daylight , 2 1−λ = − = λ f

Hence, the radiation energy emitted (for daylight) is determined using

2 7 W/m 10 2.73× = ⋅ × = = = − − − ( )

∫

( , ) 4 ,daylight (5.67 10 8 W/m2 K4)(5800K)4(0.4255) , 1 2 2 1 2 1 λ λ λ λ λ λ λ T E λ T σT f E_b _b

For candlelight (T = 1800 K), the blackbody radiation functions corresponding to λ1T and λ2T are determined from Table

12-2 to be K m 720 ) K 1800 )( m 40 . 0 ( 1 = µ = µ ⋅ λT → f_λ₁_,_candle=0.0000096 K m 1368 ) K 1800 )( m 76 . 0 ( 2 = µ = µ ⋅ λ T → fλ2 ,candle=0.006885

Then the fraction of radiation emitted between these two wavelengths (for candlelight) becomes 006875 . 0 0000096 . 0 006885 . 0 candle , 2 1−λ = − = λ f

Hence, the radiation energy emitted (for candlelight) is determined using

2 W/m 4090 = ⋅ × = = = − − − ( )

∫

( , ) 4 ,daylight (5.67 10 8 W/m2 K4)(1800K)4(0.006875) , 1 2 2 1 2 1 λ λ λ λ λ λ λ T E λ T σT f E_b _b

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12-30 The peak spectral blackbody emissive power for a match flame and moonlight is to be determined.

Assumptions 1 The match and the moon behave as black bodies.

Analysis Using the combination of Planck’s law and Wien’s displacement law, the peak spectral blackbody emissive power

can be determined: m W/m ] 1 ) / 10 43878 . 1 [exp( 10 74177 . 3 ] 1 ) / [exp( ) , ( ₅ ₄8 2 5 1 1 _λ _λ _λ _λ µ λ λ ⋅ − × × = − = T T C C T E_b and K m 8 2897 ) (λT _max_power= . µ ⋅ Then, m W/m 10 278 . 1 m W/m ] 1 ) 8 2897 / 10 43878 . 1 [exp( ) 8 2897 ( 10 74177 . 3 ) ( 5 11 4 5 5 8 max µ µ λ ⋅ × = ⋅ − × × = − _T . . T T E_b

For a match flame (T = 1700 K), the peak spectral blackbody emissive power is

µm W/m 10 1.81× 5 ⋅ = ⋅ × =₁_.₂₇₈ ₁₀− ₍₁₇₀₀₎ _W/m _m ) 1700 ( 11 5 max µ λ b E

For moonlight (T = 4000 K), the peak spectral blackbody emissive power is

µm W/m 10 1.31× 7 ⋅ = ⋅ × =₁_.₂₇₈ ₁₀− ₍₄₀₀₀₎ _W/m _m ) 4000 ( 11 5 max µ λ b E

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Radiation Intensity

12-31C A solid angle represents an opening in space, whereas a plain angle represents an opening in a plane. For a sphere of

unit radius, the solid angle about the origin subtended by a given surface on the sphere is equal to the area of the surface. For a cicle of unit radius, the plain angle about the origin subtended by a given arc is equal to the length of the arc. The value of a solid angle associated with a sphere is 4π.

12-32C When the variation of a spectral radiation quantity with wavelength is known, the correcponding total quantity is

determined by integrating that quantity with respect to wavelength from λ = 0 to λ = ∞.

12-33C Radiosity J is the rate at which radiation energy leaves a unit area of a surface by emission and reflection in all

directions.. For a diffusely emitting and reflecting surface, radiosity is related to the intensity of emitted and reflected radiation by J=πI_e₊_r (or Jλ =πIλ,e+r for spectral quantities).

12-34C The intensity of emitted radiation Ie(θ, φ) is defined as the rate at which radiation energy is emitted in the (θ,

φ) direction per unit area normal to this direction and per unit solid angle about this direction. For a diffusely emitting surface, the emissive power is related to the intensity of emitted radiation by

e

Q d &

e

I

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12-35 A surface (A2) is subjected to radiation emitted by another surface (A1). The rate at which emitted radiation is received

and the irradiation on A2 are to be determined.

Assumptions 1 Surface A1 emits diffusely as a blackbody. 2 Both A1 and A2 can be approximated as differential surfaces

since both are very small compared to the square of the distance between them.

Analysis Approximating both A1 and A2 as differential surfaces, the solid angle subtended by A2 when viewed from A1 can be

determined from sr 10 51 . 24 ) cm 50 ( 40 cos ) cm 8 ( cos ₄ 2 2 2 2 2 2 2 , 1 2− ≅ = = °= × − r A r A_n θ ω

since the normal of A2 makes 40° with the direction of viewing. Note

that solid angle subtended by A2 would be maximum if A2 were

positioned normal to the direction of viewing. Also, the point of viewing on A1 is taken to be a point in the middle, but it can be any

point since A1 is assumed to be very small.

The radiation emitted by A1 that strikes A2 is equivalent to the

radiation emitted by A1 through the solid angle ω2-1. The intensity of

the radiation emitted by A1 is

sr W/m 18048 sr W/m 10 67 . 5 ) ( 4 2 ₂ 1 1= _π = ×_π = ⋅ T E I b

This value of intensity is the same in all directions since a blackbody is a diffuse emitter. Intensity represents the rate of radiation emission per unit area normal to the direction of emission per unit solid angle. Therefore, the rate of radiation energy emitted by A1 in the direction of θ1 through the solid angle ω2-1 is determined by multiplying I1 by the area of A1

normal to θ1 and the solid angle ω2-1. That is,

W 10 76.1 −4 − − − − × = × ° × ⋅ = = ) sr 10 51 . 24 )( 55 )(cos m 10 3 )( sr W/m 18048 ( ) cos ( 4 2 4 2 1 2 1 1 1 2 1 I A θ ω Q&

Irradiation is the rate at which radiation is incident upon the surface per unit surface area. Hence, the irradiation on A2.is

2 W/m 9.51 = × × = = − − 2 4 -4 2 2 1 2 m 10 8 W 10 1 76. A Q G &

Discussion The total rate of radiation emission from surface A1 is = = 17.01 W. Therefore, the fraction of

emitted radiation that strikes A

e

Q&

A₁E_b(T₁)

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12-36 A surface (A2) is subjected to radiation emitted by another surface (A1). The intensity of the radiation emitted by A1 is

to be determined.

Analysis Approximating both A1 and A2 as differential

surfaces, the solid angle subtended by A2 when viewed from A1

can be determined from

sr 10 89 . 10 ) cm 75 ( 40 cos ) cm 8 ( cos 4 2 2 2 2 2 2 2 , 1 2− ≅ = = °= × − r A r A_n θ ω

The rate of radiation energy emitted by A1 in the direction of

θ1 through the solid angle ω2-1 is given as

1 2 1 1 1 2 1− =I (A cosθ )ω − Q&

Hence, the intensity of the radiation emitted by A1 can be

determined with sr W/m 1460 2⋅ = × ° × × = = ₋ − ₋ − − ) sr 10 89 . 10 )( 55 )(cos m 10 3 ( W 10 274 ) cos ( 4 2 4 6 2 1 1 1 2 1 1 _A _θ _ω Q I &

The temperature of A1 is determined using

π σ π 4 1 1 1 ) (T T E I = b = → 4 / 1 1 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = σ π I T K 533 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⋅ × ⋅ = ₋ 4 / 1 4 2 8 2 1 K W/m 10 67 . 5 ) sr )( sr W/m 1460 ( π T

Discussion If A2 were directly above A1 at a distance 75 cm, then θ1 = 0° and θ2 = 90°. That means the rate of radiation

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12-37A small surface emits radiation. The rate of radiation energy emitted through a band is to be determined.

Assumptions Surface A emits diffusely as a blackbody.

Analysis The rate of radiation emission from a surface per unit surface area

in the direction (θ,φ) is given as

45° 60° φ θ θ θ φ θ d d I dA Q d

dE= &e = _e( , )cos sin

The total rate of radiation emission through the band between 60° and 45° can be expressed as 4 4 4 sin cos ) , ( 4 4 2 0 60 45 T T I d d I E _e _b π σ π σ π φ θ θ θ φ θ π φ θ = = = =

∫ ∫

= = A = 1 cm 2 T = 2100 K since the blackbody radiation intensity is constant (Ib = constant), and

4 / ) 45 sin 60 (sin sin cos 2 sin cos 60 2 2 45 2 0 60 45 θ θ θ φ π θ θ θ θ π π π φ θ =

∫

= − =

∫ ∫

= = d d = d

Approximating a small area as a differential area, the rate of radiation energy emitted from an area of 1 cm2_{in the specified}

band becomes W 27.6 = × ⋅ × = = = − (1 10− m ) 4 K) 2100 )( K W/m 10 67 . 5 ( 4 2 4 4 4 2 8 4 dA T EdA Q&_e σ

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12-38 Radiation is emitted from a small circular surface located at the center of a sphere. Radiation energy streaming through

a hole located on top of the sphere and the side of sphere are to be determined.

Analysis (a) Approximating both A1 and A2 as differential surfaces, the solid

angle subtended by A2 when viewed from A1 can be determined from D2 = 1 cm

r = 1 m A1 = 2 cm2 T1 = 1000 K sr 10 .854 7 m) 1 ( m) 005 . 0 ( ₅ 2 2 2 2 2 2 , 1 2− ≅ = =π = × − ω r A r A_n

since A2 were positioned normal to the direction of viewing.

The radiation emitted by A1 that strikes A2 is equivalent to the radiation emitted by A1

through the solid angle ω2-1. The intensity of the radiation emitted by A1 is

sr W/m 048 , 18 K) 1000 )( K W/m 10 67 . 5 ( ) ( 4 8 2 4 4 ₂ 1 1 1 = ⋅ ⋅ × = = = − π π σ π T T E I b

normal to θ1 and the solid angle ω2-1. That is,

W 10 2.835 4 5 2 4 2 1 2 1 1 1 2 1 sr) 10 854 . 7 )( m 0 cos 10 2 )( sr W/m 048 , 18 ( ) cos ( − − − − − × = × ° × ⋅ = =I A θ ω Q& r = 1 m θ1 = 45° θ2 = 0° D2 = 1 cm A1 = 2 cm2 T1 = 1000 K

where θ1 = 0°. Therefore, the radiation emitted from surface A1 will strike

surface A2 at a rate of 2.835×10-4 W.

(b) In this orientation, θ1 = 45° and θ2 = 0°. Repeating the calculation we

obtain the rate of radiation to be W 10 2.005 4 5 2 4 2 1 2 1 1 1 2 1 sr) 10 854 . 7 )( m 45 cos 10 2 )( sr W/m 048 , 18 ( ) cos ( − − − − − × = × ° × ⋅ = =I A θ ω Q&

(20)

12-39 Radiation is emitted from a small circular surface located at the center of a sphere. Radiation energy streaming through

a hole located on top of the sphere and the side of sphere are to be determined.

Analysis (a) Approximating both A1 and A2 as differential surfaces, the solid angle subtended

by A2 when viewed from A1 can be determined from to be D2 = 1 cm

r = 2 m A1 = 2 cm2 T1 = 1000 K sr 10 .963 1 m) 2 ( m) 005 . 0 ( ₅ 2 2 2 2 2 2 , 1 2− ≅ = =π = × − ω r A r A_n

since A2 were positioned normal to the direction of viewing.

The radiation emitted by A1 that strikes A2 is equivalent to the radiation emitted by A1

through the solid angle ω2-1. The intensity of the radiation emitted by A1 is

( 1) 14 (5.67 10 8 W/m2 K4)(1000K)4 18,048 W/m2 sr 1 = ⋅ ⋅ × = = = − π π σ π T T E I b

W 10 7.086 5 5 2 4 2 1 2 1 1 1 2 1 sr) 10 963 . 1 )( m 0 cos 10 2 )( sr W/m 048 , 18 ( ) cos ( − − − − − × = × ° × ⋅ = =I A θ ω Q& r = 2 m θ1 = 45° θ2 = 0° D2 = 1 cm A1 = 2 cm2 T1 = 1000 K

where θ1 = 0°. Therefore, the radiation emitted from surface A1 will strike

surface A2 at a rate of 2.835×10-4 W.

(b) In this orientation, θ1 = 45° and θ2 = 0°. Repeating the calculation we obtain

the rate of radiation as W 10 5.010 5 5 2 4 2 1 2 1 1 1 2 1 sr) 10 963 . 1 )( m 45 cos 10 2 )( sr W/m 048 , 18 ( ) cos ( − − − − − × = × ° × ⋅ = =I A θ ω Q&

(21)

12-40 A surface is subjected to radiation emitted by another surface. The solid angle subtended and the rate at which emitted

radiation is received are to be determined.

Analysis Approximating both A1 and A2 as differential surfaces, the solid angle

subtended by A2 when viewed from A1 can be determined from

A2 = 7 cm2 θ2 = 60° r = 80 cm θ1 = 45° ₋ ≅ = = ₂ °=5.469×10−4 sr 2 2 2 2 2 2 , 1 2 cm) 80 ( 60 cos ) cm 7 ( cos r A r An θ ω

since the normal of A2 makes 60° with the direction of viewing. Note that solid

angle subtended by A2 would be maximum if A2 were positioned normal to the

direction of viewing. Also, the point of viewing on A1 is taken to be a point in the

middle, but it can be any point since A1 is assumed to be very small.

The radiation emitted by A1 that strikes A2 is equivalent to the radiation

emitted by A1 through the solid angle ω2-1. The intensity of the radiation emitted

by A1 is A1 = 7 cm2 T1 = 600 K sr W/m 7393 K) 800 )( K W/m 10 67 . 5 ( ) ( 4 8 2 4 4 ₂ 1 1 1 = ⋅ ⋅ × = = = − π π σ π T T E I b

W 10 2.001 3 4 2 4 2 1 2 1 1 1 2 1 sr) 10 469 . 5 )( m 45 cos 10 7 )( sr W/m 7393 ( ) cos ( − − − − − × = × ° × ⋅ = =I A θ ω Q&

Therefore, the radiation emitted from surface A1 will strike surface A2 at a rate of 2.001×10-3 W.

(22)

12-41 A small surface is subjected to uniform incident radiation. The rates of radiation emission through two specified bands

are to be determined.

Assumptions The intensity of incident radiation is constant.

45°

Analysis (a) The rate at which radiation is incident on a surface per

unit surface area in the direction (θ,φ) is given as φ θ θ θ φ θ d d I dA Q d dG= i = i( , )cos sin &

The total rate of radiation emission through the band between 0° and 45° can be expressed as 2 sin cos ) , ( 2 0 45 0 1 π φ θ θ θ φ θ π φ θ Ii d d Ii G =

∫ ∫

= = = A = 1 cm2

since the incident radiation is constant (Ii = constant), and

2 / ) 0 sin 45 (sin sin cos 2 sin cos 45 2 2 0 2 0 45 0 θ θ θ φ π θ θ θ θ π π π φ θ =

∫

= − =

∫ ∫

= = d d = d

Approximating a small area as a differential area, the rate of radiation energy emitted from an area of 1 cm2_{in the specified}

band becomes W 3.46 = × ⋅ × = = = ₀_.₅ ₀_.₅ ₍₂_.₂ ₁₀4_W/m2 _sr₎₍₁ ₁₀−4 _m2₎ 1 1 , GdA πI dA π Q&_i _i

(b) Similarly, the total rate of radiation emission through the band between 45° and 90° can be expressed as

A = 1 cm2 90° 45° 2 sin cos ) , ( 2 0 90 45 1 π φ θ θ θ φ θ π φ θ Ii d d Ii G =

∫ ∫

= = = since 2 / ) 45 sin 90 (sin sin cos 2 sin cos 2 2 90 45 2 0 90 45 π π θ θ θ π φ θ θ θ θ π φ θ = − = =

∫

∫ ∫

= = d d = d and W 3.46 = × ⋅ × = = = ₀_.₅ ₀_.₅ ₍₂_.₂ ₁₀4_W/m2 _sr₎₍₁ ₁₀−4_m2₎ 2 2 , G dA πI dA π Q&_i _i

Discussion Note that the viewing area for the band 0° - 45° is much smaller, but the radiation energy incident through it is

(23)

12-42 The intensity of solar radiation incident on earth’s surface is given. The peak value for the intensity of incident solar

radiation and the solar irradiation on earth’s surface are to be determined.

Assumptions 1 The intensity is not dependent of the azimuth angle φ.

Analysis The intensity of solar radiation incident on earth’s surface peaks at the zenith angle of θ = 0°, hence

sr W/m 100 2⋅ = ° =100cos0 max , i I

The solar irradiation on earth’s surface is

2 W/m 209 = = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − = = = = =

∫

∫ ∫

∫

3 200 3 ) (cos 200 sin cos ) 2 ( 100 sin cos 100 sin cos ) ( 2 / 0 3 2 2 / 0 2 0 2 2 / 0 2 0 2 / 0 hemisphere π θ π θ θ θ π φ θ θ θ φ θ θ θ θ π π π π π π d d d d d I dG G _i

Discussion The intensity of incident solar radiation is at minimum when the sun is at the horizon, where the zenith angle is

(24)

12-43 A radiation detector (A2) is placed normal to the direction of viewing from another surface (A1), and is measuring a

specified amount of irradiation. The distance between the radiation detector and the radiation emitting surface is to be determined.

Analysis Approximating both A1 and A2 as differential

surfaces, the solid angle subtended by A2 when viewed from A1

can be determined from 2 2 2 2 2 , 1 2 cos L A L A_n θ ω ₋ ≅ = (1)

The radiation emitted by A1 that strikes A2 is equivalent to the

radiation emitted by A1 through the solid angle ω2-1. The

intensity of the radiation emitted by A1 is

π σ π 4 1 1 1 ) (T T E I ₌ b ₌ ₍₂₎

Therefore, the rate of radiation energy emitted by A1 in the direction of θ1 through the solid angle ω2-1 is determined by

multiplying I1 by the area of A1 normal to θ1 and the solid angle ω2-1. That is,

(3) 1 2 1 1 1 2 1− =I (A cosθ )ω − Q&

Irradiation is the rate at which radiation is incident upon the surface per unit surface area. Hence, the irradiation measured by the radiation detector A2.is

2 2 1 2 _A Q G = &− (4)

Substituting Eqs. (1) to (3) into Eq. (4) yields

2 2 2 2 1 1 4 1 2 ) cos )( cos ( L A A A T G π θ θ σ = → 2 / 1 2 2 1 1 4 1 cos cos ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = G A T L π θ θ σ

Since the radiation detector is placed normal to the direction of viewing from A1, we have θ1 = θ2 = 0°. Hence the distance L

is m 0.233 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × ⋅ × = − − 2 / 1 2 2 4 4 4 2 8 ) W/m 100 ( ) m 10 3 ( ) K 1000 )( K W/m 10 67 . 5 ( π L

Discussion The solid angle subtended by A2 is at maximum, since the radiation detector is positioned normal to the direction

(25)

12-44 A radiation sensor is placed with a 30° tilt off the normal direction of viewing from an aperture through which

radiation is emitted as a blackbody. The distance between the sensor and the aperture is to be determined.

Assumptions 1 The aperture emits diffusely as a blackbody. 2 Both aperture and sensor can be approximated as differential

surfaces since both are very small compared to the square of the distance between them.

Analysis Approximating both aperture (A1) and radiation sensor (A2) as differential surfaces, the solid angle subtended by A2

when viewed from A1 can be determined from

2 2 2 2 2 , 1 2 cos L A L A_n θ ω ₋ ≅ = (1)

The radiation emitted by A1 that strikes A2 is equivalent to

the radiation emitted by A1 through the solid angle ω2-1.

The intensity of the radiation emitted by A1 is

πb

E

I₁= (2)

Therefore, the rate of radiation energy emitted by A1 in the direction of θ1 through the solid angle ω2-1 is determined by

multiplying I1 by the area of A1 normal to θ1 and the solid angle ω2-1. That is,

(3) 1 2 1 1 1 2 1− =I (A cosθ )ω − Q& where A₁ =πD₁2/4

Irradiation is the rate at which radiation is incident upon the surface per unit surface area. Hence, the irradiation measured by the radiation detector A2.is

2 2 1 2 _A Q G = &− (4)

Substituting Eqs. (1) to (3) into Eq. (4) yields

2 2 2 2 1 1 2 ( cos )( cos ) L A A A E G b π θ θ = → 2 / 1 2 2 1 2 1 4 cos cos ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = G D E L b π θ θ π

With θ1 = 0° and θ2 = 30°, the distance between the sensor and the aperture is

m 0.353 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × ° = 2 / 1 2 2 2 5 ) W/m 50 ( 4 30 cos ) m 01 . 0 )( W/m 10 87 . 2 ( L

Discussion If the radiation sensor is positioned normal to the direction of viewing (θ2 = 0°) with L = 0.353 m, it would

(26)

Radiation Properties

12-45C A body whose surface properties are independent of wavelength is said to be a graybody. The emissivity of a

blackbody is one for all wavelengths, the emissivity of a graybody is between zero and one.

12-46C The emissivity is the ratio of the radiation emitted by the surface to the radiation emitted by a blackbody at the

same temperature. The fraction of radiation absorbed by the surface is called the absorptivity α,

ε ) ( ) ( ) ( T E T E T b = ε and G G_abs = = radiation incident radiation absorbed α

When the surface temperature is equal to the temperature of the source of radiation, the total hemispherical emissivity of a surface at temperature T is equal to its total hemispherical absorptivity for radiation coming from a blackbody at the same temperature ε_λ(T)=α_λ(T).

12-47C The heating effect which is due to the non-gray characteristic of glass, clear plastic, or atmospheric gases is known

as the greenhouse effect since this effect is utilized primarily in greenhouses. The combustion gases such as CO2 and water

vapor in the atmosphere transmit the bulk of the solar radiation but absorb the infrared radiation emitted by the surface of the earth, acting like a heat trap. There is a concern that the energy trapped on earth will eventually cause global warming and thus drastic changes in weather patterns.

12-48C Glass has a transparent window in the wavelength range 0.3 to 3 µm and it is not transparent to the radiation which

has wavelength range greater than 3 µm. Therefore, because the microwaves are in the range of µm, the harmful microwave radiation cannot escape from the glass door.

5 2_to₁₀

(27)

12-49 A radiation sensor (A2) is placed normal to the direction of viewing from another surface (A1). An optical filter with

specified spectral transmissivity is placed in front of the sensor. The irradiation that is measured by the sensor is to be determined.

Analysis For T = 1000 K, the blackbody radiation functions corresponding to λ1T is determined from Table 12-2 to be K m 2000 ) K 1000 )( m 2 ( 1 = µ = µ ⋅ λT → f_λ₁ =0.066728

Hence, the transmissivity of the optical filter is

4666 . 0 ) 066728 . 0 1 ( 5 . 0 ) 066728 . 0 ( 0 ) 1 ( ) ( ₁ ₂ ₁ 1 + − = + − = =τ _λ τ _λ τ f f

The rate of radiation energy emitted by A1 in the direction of θ1 through the solid angle ω2-1 is determined by multiplying I1

by the area of A1 normal to θ1 and the solid angle ω2-1. That is,

τ ω θ1 21 1 1 2 1− =I (A cos ) − Q& where 2 2 2 2 2 , 1 2 cos L A L An θ ω ₋ ≅ = and π σ π 4 1 1 1 ) (T T E I ₌ b ₌

Since the radiation sensor is placed normal to the direction of viewing from A1, we have θ1 = θ2 = 0°, hence

τ π σ 2 2 1 4 1 2 1 L A A T Q&₋ =

Irradiation is the rate at which radiation is incident upon the surface per unit surface area. Hence, the irradiation measured by the radiation sensor A2 is

2 W/m 16.8 = × ⋅ × = = = = − − − ) 4666 . 0 ( ) m 5 . 0 ( ) m 10 5 ( ) K 1000 )( K W/m 10 67 . 5 ( 2 2 4 4 4 2 8 2 1 4 1 2 2 1 2 tr π τ π σ L A T A Q G G &

Discussion If the optical filter is removed, the irradiation measured by the radiation sensor would be

2 2 tr ₃₆_W/m 4666 . 0 W/m 8 . 16 4666 . 0 = = = G G

(28)

12-50 An opaque horizontal plate that is well insulated on the edges and the lower surface has a constant temperature of 500

K and α = 0.51, (a) the total hemispherical emissivity and (b) the radiosity of the plate surface are to be determined.

Assumptions 1 The plate has a uniform temperature. 2 The plate is well insulated on the edges and the lower surface. Properties The total hemispherical absorptivity of the plate is given to be 0.51.

Analysis (a) The average emissivity of the surface can be determined from

) 1 ( ) ( 1 1 1 1 1 1 2 1 2 0 1 2 0 1 λ λ λ λ λ λ λ λ ε ε ε ε λ ε λ ε ε f f f f E d E E d E T b b b b − + = + = + = ₋ ₋_∞ ∞

∫

For T = 500 K, the blackbody radiation functions corresponding to λ1T is determined from Table 12-2 to be K m 2000 ) K 500 )( m 4 ( 1 = µ = µ ⋅ λT → f_λ₁ =0.066728

Hence, the total hemispherical emissivity is

0.773 = − + =0.4(0.066728) 0.8(1 0.066728) ) (T ε

(b) The radiosity of the plate surface can be determined from the following expression:

G E G E

J= + _ref =ε _b+ρ

Since the plate is opaque (τ = 0), the reflectivity is then, ρ = 1 − α. Hence,

2 W/m 5480 = − + ⋅ × = − + = − _W/m _K ₎₍₅₀₀_K₎ ₍₁ ₀_.₅₁₎₍₅₆₀₀_W/m ₎ 10 67 . 5 )( 773 . 0 ( ) 1 ( 2 4 4 2 8 4 _G T J εσ α

Discussion Both emissive power (E) and reflected irradiation (Gref) contributed equal amount to the radiosity, with E = Gref =

(29)

12-51 The variation of emissivity of a surface at a specified temperature with wavelength is given. The average emissivity of

the surface and its emissive power are to be determined.

Analysis The average emissivity of the surface can be determined from

ελ ) 1 ( + ) ( + + + ) ( ) ( ) ( ) ( 2 1 2 1 2 2 1 1 2 2 1 1 3 2 1 -3 -2 -0 1 4 3 4 2 4 0 1 λ λ λ λ λ λ λ λ λ λ λ λ ε ε ε ε ε ε σ λ ε σ λ ε σ λ ε ε λ λ λ f f f f f f f T d T E T d T E T d T E T b b b − − = = + + = ∞ ∞

∫

0.7 0.4 0.3 where f_λ f are blackbody radiation functions corresponding to

1 and λ2 T T ₂ 1 andλ λ , determined from 3 6 _{λ, µm} 737818 . 0 mK 6000 = K) m)(1000 6 ( 273232 . 0 mK 3000 = K) m)(1000 3 ( 2 1 2 1 = ⎯→ ⎯ = = ⎯→ ⎯ = λ λ f T f T µ µ λ µ µ λ

f₀−λ₁ = fλ₁−f₀ = fλ₁since f₀ =0andf λ₂−∞= f∞−fλ₂ sincef∞ =1. and, 0.5132 = − + − + =(0.4)0.273232 (0.7)(0.737818 0.273232) (0.3)(1 0.737818) ε

Then the emissive power of the surface becomes

2 kW/m 29.1 = = × = = _T4 ₀_.₅₁₃₂₍₅_.₆₇ ₁₀−8_W/m2_._K4₎₍₁₀₀₀_K)4 ₂₉_,₁₀₀_W/m2 E εσ

12-52E A spherical ball emits radiation at a certain rate. The average emissivity

of the ball is to be determined at the given temperature.

D = 5 in

Ball

T=950 R

Analysis The surface area of the ball is

2 2

2 ₌ ₍₅_/₁₂_ft₎ ₌₀_.₅₄₅₄_ft

=πD π

A

Then the average emissivity of the ball at this temperature is determined to be

0.630 = ⋅ × = = ⎯→ ⎯ = 4 ₄ ₂ _-₈ ₂ ₄ ₄ R) 950 )( R Btu/h.ft 10 )(0.1714 ft (0.5454 Btu/h 480 T A E T A E σ ε σ ε

(30)

12-53 The variation of transmissivity of the glass window of a furnace at a specified temperature with wavelength is given.

The fraction and the rate of radiation coming from the furnace and transmitted through the window are to be determined.

Assumptions The window glass behaves as a black body.

λ, µm

3

τ

λ

Analysis The fraction of radiation at wavelengths smaller than 3 µm is 403607 . 0 mK 3600 = K) m)(1200 3 ( ⎯⎯→ = = f_λ T µ µ λ 0.7

The fraction of radiation coming from the furnace and transmitted through the window is

0.2825 = − + = − + = ) 403607 . 0 1 )( 0 ( ) 403607 . 0 )( 7 . 0 ( ) 1 ( ) ( τ₁ _λ τ₂ _λ τ T f f

Then the rate of radiation coming from the furnace and transmitted through the window becomes

W 5315 = × × = = _A _T4 ₀_.₂₈₂₅₍₀_.₄₀ ₀_.₄₀_m2₎₍₅_.₆₇ ₁₀−8_W/m2_._K4₎₍₁₂₀₀_K)4 G_tr τ σ

12-54 The variation of emissivity of a tungsten filament with wavelength is given. The average emissivity, absorptivity, and

reflectivity of the filament are to be determined for two temperatures.

Analysis (a) T = 1500 K 013754 . 0 mK 1500 = K) m)(1500 1 ( ₁ 1T = µ µ ⎯⎯→fλ = λ 0.5 0.15 ελ The average emissivity of this surface is

0.155 = − + = − + = ) 013754 . 0 1 )( 15 . 0 ( ) 013754 . 0 )( 5 . 0 ( ) 1 ( ) ( ε1 λ₁ ε2 λ₁ ε T f f

From Kirchhoff’s law,

0.155 = =α ε (at 1500 K) and α+ρ=1⎯⎯→ρ =1−α =1−0.155=0.845 λ, µm 1 (b) T = 2500 K 161688 . 0 mK 2500 = K) m)(2500 1 ( ₁ 1T = µ µ ⎯⎯→fλ = λ Then 0.207 = − + = − + = (1 ) (0.5)(0.161688) (0.15)(1 0.161688) ) ( ε₁ _λ₁ ε₂ _λ₁ ε T f f

From Kirchhoff’s law,

0.207

= =α

ε (at 2500 K)

(31)

12-55 The variations of emissivity of two surfaces are given. The average emissivity, absorptivity, and reflectivity of each

surface are to be determined at the given temperature.

Analysis For the first surface:

(3 m)(3000K)=9000 mK 0.890029 1 1T = µ µ ⎯⎯→fλ = λ _ε λ 0.8 0.2 0.1 0.9

The average emissivity of this surface is

0.28 = − + = − + = ) 890029 . 0 1 )( 9 . 0 ( ) 890029 . 0 )( 2 . 0 ( ) 1 ( ) ( ε₁ _λ₁ ε₂ _λ₁ ε T f f

The absorptivity and reflectivity are determined from Kirchhoff’s law 0.72 0.28 = − = − = ⎯→ ⎯ = + = = 28 . 0 1 1 1 K) 3000 (at α ρ ρ α α ε 3 λ, µm

For the second surface:

890029 . 0 mK 9000 = K) m)(3000 3 ( ₁ 1T = µ µ ⎯⎯→fλ = λ

The average emissivity of this surface is

0.72 = − + = − + = ) 890029 . 0 1 )( 1 . 0 ( ) 890029 . 0 )( 8 . 0 ( ) 1 ( ) ( ε₁ _λ₁ ε₂ _λ₁ ε T f f Then, 0.28 0.72 = − = − = → = + = = 72 . 0 1 1 1 K) 3000 (at α ρ ρ α α ε

Discussion The second surface is more suitable to serve as a solar absorber since its absorptivity for short wavelength

radiation (typical of radiation emitted by a high-temperature source such as the sun) is high, and its emissivity for long wavelength radiation (typical of emitted radiation from the absorber plate) is low.

(32)

12-56 The variation of emissivity of a surface with wavelength is given. The average emissivity and absorptivity of the

surface are to be determined for two temperatures.

Analysis (a) For T = 5800 K:

(5 m)(5800K)=29,000 mK 0.994715 1 1T = µ µ ⎯⎯→fλ = λ 0.9 0.15 ελ The average emissivity of this surface is

0.154 = − + = − + = ) 994715 . 0 1 )( 9 . 0 ( ) 994715 . 0 )( 15 . 0 ( ) 1 ( ) ( ε₁ _λ₁ ε₂ _λ₁ ε T f f (b) For T = 300 K: 013754 . 0 mK 1500 = K) m)(300 5 ( 1 1T = µ µ ⎯⎯→fλ = λ and 5 λ, µm 0.89 = − + = − + = ) 013754 . 0 1 )( 9 . 0 ( ) 013754 . 0 )( 15 . 0 ( ) 1 ( ) ( ε₁ _λ₁ ε₂ _λ₁ ε T f f

The absorptivities of this surface for radiation coming from sources at 5800 K and 300 K are, from Kirchhoff’s law,

0.154 = =ε α (at 5800 K) 0.89 = =ε α (at 300 K)

12-57 The variation of absorptivity of a surface with wavelength is given. The average absorptivity, reflectivity, and

emissivity of the surface are to be determined at given temperatures.

Analysis For T = 2500 K: 633747 . 0 mK 5000 = K) m)(2500 2 ( ₁ 1T = µ µ ⎯⎯→fλ = λ 0.8 0.3 λ, µm 2 αλ

The average absorptivity of this surface is

0.483 = − + = − + = ) 633747 . 0 1 )( 8 . 0 ( ) 633747 . 0 )( 3 . 0 ( ) 1 ( ) ( α₁ _λ₁ α₂ _λ₁ α T f f

Then the reflectivity of this surface becomes

0.517 = − = − = ⎯→ ⎯ = +ρ 1 ρ 1 α 1 0.483 α

Using Kirchhoff’s law, α =ε , the average emissivity of this surface at T = 3000 K is determined to be 737818 . 0 mK 6000 = K) m)(3000 2 ( ⎯⎯→ = = f_λ T µ µ λ 0.431 = − + = − + = ) 737818 . 0 1 )( 8 . 0 ( ) 737818 . 0 )( 3 . 0 ( ) 1 ( ) ( 1 1 2 1 λ ε λ ε ε T f f

(33)

12-58 The variation of reflectivity of a surface with wavelength is given. The average reflectivity, emissivity, and

absorptivity of the surface are to be determined for two source temperatures.

Analysis The average reflectivity of this surface for solar

radiation (T = 5800 K) is determined to be 0.95 0.35

ρ

978746 . 0 mK 17400 = K) m)(5800 3 ( → = = f_λ T µ µ λ 0.362 = − + = − + = + = ₋ ₋_∞ ) 978746 . 0 1 )( 95 . 0 ( ) 978746 . 0 )( 35 . 0 ( ) 1 ( ) ( ) ( ) ( 1 1 1 1 2 1 2 0 1 λ λ λ λ ρ ρ ρ ρ ρ f f T f T f T

Noting that this is an opaque surface, τ=0

3

At T = 5800 K: α+ρ=1⎯⎯→α =1−ρ=1−0.362=0.638 λ, µm

Repeating calculations for radiation coming from surfaces at T = 300 K, 0001685 . 0 mK 900 = K) m)(300 3 ( ⎯⎯→ ₁ = = f_λ T µ µ λ 0.95 = − + =(0.35)(0.0001685) (0.95)(1 0.0001685) ) (T ρ At T = 300 K: α+ρ=1⎯⎯→α =1−ρ=1−0.95=0.05 and ε =α =0.05

The temperature of the aluminum plate is close to room temperature, and thus emissivity of the plate will be equal to its absorptivity at room temperature. That is,

638 . 0 05 . 0 = = = = s room α α ε ε