Radiation

1 . حطس یامد اب هایس مسج کی ار دیشروخ 10400

نودا م ی شبات یژرنا خرن .دیریگب رظن رد

R

جوم لوط هزاب رد ار حطس زا هدش لیسگ زمرق λ=0.76−100 μm

.دیروآ تسدب

Analysis T = 10400 R = 5778 K, the blackbody radiation functions corresponding to are determined from Table 11-2 to be

λ

₁

T =( 0.76 μm )( 5778 K ) =4391.3 μ mK ⃗ f

_λ

1

= 0.547370 λ

₂

T =( 100 μm )( 5778 K ) =577,800 μ mK ⃗ f

_λ

2

= 1.0

Then the fraction of radiation emitted between these two wavelengths becomes f_λ

2−f_λ

1=1.0−0 .547=0 . 453

(or 45.3%)

The total blackbody emissive power of the sun is determined from Stefan-Boltzman Law to be E_b=σT⁴=(0.1714×10⁻⁸ Btu/h .ft².R⁴)(10,400 R)⁴=2.005×10⁷ Btu/h .ft²

Then,

E_infrared=(0. 451)E_b=(0. 453)(2.005×10⁷ Btu/h .ft²)=9.08×10⁶ Btu/h.ft²

2 . جو م لو ط رد ی شبات یژر نا رثکاد ح ه ک تسا یتروص هب پملا کی یحارط λ=0.47 μm

جو م لو ط هزاب رد ار هدش عطاس یشبات یژنا تبسن و پملا نیا یامد .دوش یم عطاس یئرم λ=0.40−0.76 μm .دیروآ تسدب

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

( λT )

_{max power}

=2897 .8 μm ⋅K ⃗ T = 2897 .8 μm ⋅ K

0. 47 μm =6166 K

The visible range of the electromagnetic spectrum extends fromλ₁=0. 40μm to λ₂=0.76μm . Noting that

=6166 K, the blackbody radiation functions corresponding to are determined from Table 11-2 to be

λ

₁

T =( 0. 40 μm )( 6166 K ) =2466 μ mK ⃗ f

_λ

1

= 0.15444 λ

₂

T =( 0.76 μm )( 6166 K ) =4686 μ mK ⃗ f

_λ

2

= 0.59141

Then the fraction of radiation emitted between these two wavelengths becomes f_λ

2−f_λ

1=0.59141−0.15444≃0.437

(or 43.7%)

3 . هرجنپ کی یحارط 2

m*2 m

هک تسا یا هنوگ هب

90 جوم لوط هزاب رد یشبات یژرنا %

λ=0.3−3.0μm

مسج کی دننام هشیش اه جوم لوط ریاس یازاب و دنک یم روبع هشیش زا

SUN T = 10,400 R

یا مد ود رد هایس مسج کی دیشروخ رگا .دنک یم راتفر تام لاماک 5800

و

K

1000

K

.دیروآ تسدب ار هرجنپ زا یروبع یشبات یژرنا خرن ،دوش هتفرگ رظن رد

Analysis The surface area of the glass window is A_s=4 m²

(a) For a blackbody source at 5800 K, the total blackbody radiation emission is E_b(T)=σT⁴A_s=(5.67×10⁻⁸ kW/m².K)⁴(5800 K)⁴(4 m²)=2.567×10⁵ kW The fraction of radiation in the range of 0.3 to 3.0 m is

λ

₁

T =( 0.30 μm )( 5800 K ) =1740 μ mK ⃗ f

_λ

1

= 0.03345 λ

₂

T =( 3.0 μm )( 5800 K ) =17,400 μ mK ⃗ f

_λ

2

= 0.97875

Δf=f_λ

2−f_λ

1=0 . 97875−0 .03345=0 . 9453

Noting that 90% of the total radiation is transmitted through the window,

E

_transmit

= 0.90 Δ fE

_b

( T )

¿(

0.90 )( 0.9453 )( 2.567 × 10

⁵

kW )= 2. 184 × 10

⁵

kW

(b) For a blackbody source at 1000 K, the total blackbody emissive power is E_b(T)=σT⁴A_s=(5.67×10⁻⁸ W/m².K⁴)(1000 K)⁴(4 m²)=226 .8 kW The fraction of radiation in the visible range of 0.3 to 3.0 m is

λ

₁

T =( 0.30 μm )( 1000 K ) =300 μ mK ⃗ f

_λ

1

= 0.0000 λ

₂

T =( 3.0 μm )( 1000 K ) =3000 μ mK ⃗ f

_λ

2

= 0.273232

Δf=f_λ

2−f_λ

1=0 .273232−0 and

E_transmit=0.90Δ fE_b(T)=(0.90)(0.273232)(226.8 kW)=55.8 kW SUN

L = 2 m

Glass

 = 0.9

4 . رتمک یاه جوم لوط یازاب سم تافلوس اب هدش هدیشوپ یموینیمولآ هحفص کی ساکعنا بیرض

زا

3μm

ربارب 0.35 و زا رتشیب یاه جوم لوط یازاب

3μm

ربارب 0.95 یا مد ود رد .تسا

5800 و

K

300 ( حط س ط سوتم سا کعنا بیر ض

K

reflectivity

( لی سگ بیر ض ،)

emissivity

( بذج و )

absorptivity

.دیروآ تسدب ار )

Analysis The average reflectivity of this surface for solar radiation ( = 5800 K) is determined to be

Noting that this is an opaque surface,

At = 5800 K:

α + ρ = 1 ⃗ α = 1 − ρ = 1 − 0.362 = 0. 638

Repeating calculations for radiation coming from surfaces at = 300 K,

ρ( T )=( 0.35 )(0.0001685 )+(0.95 )( 1−0.0001685)=0. 95

At = 300 K:

α + ρ = 1⃗ α = 1 − ρ = 1 − 0.95 = 0. 05

and

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,

5 .

؟دید یایاوز

0.95

3 , m



Assumptions The surfaces are diffuse emitters and reflectors.

Analysis From Figure:

L 3 W = 1

2 = 0.5 ¿ } ¿¿ F 31 = 0.24 ¿

and

L 3 W = 1

2 = 0.5 ¿ } ¿¿ F 3 →( 1 + 2 ) = 0.29 ¿

We note that A1 = A3. Then the reciprocity and superposition rules give

A

₁

F

₁₃

= A

₃

F

₃₁

⃗ F

₁₃

= F

₃₁

=0. 24

F

_3→(1+2)

= F

₃₁

+ F

₃₂

⃗ 0.29 = 0.24 + F

₃₂

⃗ F

₃₂

= 0.05

Finally,

6 . دید یایاوز F

₁₂

∧ F

₂₁

:)لیوط( تیاهن یب همین یاه تکاد یارب

(1) A2

A1 A3 (3) L3 = 1 m

L2 = 1 m (2) W = 2 m

Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected.

Analysis (a) Surface (1) is flat, and thus . summation rule: F₁₁+F₁₂=1→F₁₂=1

reciprocity rule: A₁F₁₂=A₂F₂₁⃗F₂₁=A₁

A₂F₁₂= Ds

(

^πD2

)

^s(1)=

2 π=0 .64

(b) Noting that surfaces 2 and 3 are symmetrical and thus , the summation rule gives

F

₁₁

+ F

₁₂

+ F

₁₃

= 1 ⃗ 0 + F

₁₂

+ F

₁₃

= 1 ⃗ F

₁₂

= 0.5

Also by using the equation obtained in Example 12-4 (we solved it in our class),

reciprocity rule: A₁F₁₂=A₂F₂₁⃗F₂₁=A₁

A₂F₁₂=a

b

(

¹²

)

^=2ba

D (2)

(1)

a (3) (2)

(1)

L1 = a L2 = a

L3 = b L4 = b

(c) Applying the crossed-string method gives

7 .

؟دید یایاوز

From Fig.12-5,

L 2 D = 2

2 ¿ } ¿¿ F ( 2 + 4 )→( 1 + 3 ) = 0.20 ¿

^and

L D 2 = 2 2 ¿ } ¿¿ F 14 = 0.12 ¿

superposition rule: F

_{(2+4)→(1+3)}

= F

_(2+4)→1

+ F

_(2+4)→3

symmetry rule: F

_(2+4)→1

= F

_(2+4)→3

Substituting symmetry rule gives

reciprocity rule: A

₁

F

_1→(2+4)

= A

₍₂₊₄₎

F

_(2+4)→1

⃗( 2 ) F

_1→(2+4)

=( 4 )( 0.10 )⃗ F

_1→(2+4)

= 0.20 superposition rule : F

_1→(2+4)

= F

₁₂

+ F

₁₄

⃗ 0.20 = F

₁₂

+ 0.12 ⃗ F

₁₂

= 0.20 − 0.12 = 0.08

8 . یا ه ر طق ه ب ز کرم مه هر ک ود D

₁

=0.3 m∧ D

₂

=0.8 m

تبا ث یا هامد رد

T₁=700K∧T2=400K

هر ک ود نیا لی سگ بیارض .دنا هدش هتشاد هگن

ε₁=0.5∧ε2=0.7

ر گا نین چمه .د یروآ ت سدب ار هرک ود نایم یشبات صلاخ ترارح لاقتنا خرن .دنشاب یم فارطا طیحم یامد

30℃

هر ک یجرا خ حطس رد ار ییاجباج ترارح لاقتنا بیرض دشاب

ربارب هرک یجراخ حطس لیسگ بیرض( .دیروآ تسدب اه 0.35

).دوش هتفرگ رظن رد

Two concentric spheres are maintained at uniform temperatures. The net rate of radiation heat transfer between the two spheres and the convection heat transfer coefficient at the outer surface are to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray.

Analysis The net rate of radiation heat transfer between the two spheres is

Radiation heat transfer rate from the outer sphere to the surrounding surfaces are Q˙_rad=ε FA₂σ(T

24−T

surr4)

=(0.35)(1)[π(0.8 m)²](5.67×10⁻⁸ W/m²⋅K⁴)[(400 K)⁴−(30+273 K)⁴]=685W

The convection heat transfer rate at the outer surface of the cylinder is determined from requirement that heat transferred from the inner sphere to the outer sphere must be equal to the heat transfer from the outer surface of the outer sphere to the environment by convection and radiation. That is,

Q ˙

_conv

= ˙ Q

₁₂

− ˙ Q

_rad

= 1669 − 685 = 984 W

Then the convection heat transfer coefficient becomes

˙

Q_conv.=hA₂

(

^T2−T∞

)

984 W=h

[

^{π(0.8 m)2}

]

⁽400 K-303 K)⃗h=5.04 W/m²⋅° C

= 0.35

Tsurr = 30C

T = 30C D1 = 0.3 m T1 = 700 K D2 = 0.8 m

T2 = 400 K