# 978-0123869449 Chapter 6 Part 3

Document Type

Homework Help

Book Title

Radiative Heat Transfer 3rd Edition

Authors

Michael F. Modest

For full access to CoursePaper, a membership subscription is required.

CHAPTER 6 201

0.5−1

0.5−1×0.5199

202 RADIATIVE HEAT TRANSFER

6.26

w

w

w

w

T,

∋

S´

S´

(2)

A2

A1

An infinitely long corner of characteristic length w=1 m is

a gray, diﬀuse emitter and purely specular reflector with ǫ=

ρs=1

2. The entire corner is kept at a constant temperature

T=500 K, and is irradiated externally by a line source of

strength S′=20 kW/m, located a distance waway from both

sides of the corner, as shown in the sketch. What is the total

heat flux Q′(per m length) to be supplied or extracted from

the corner to keep the temperature at 500 K?

8. It can also be found from Configuration D-37 (with b2=0, b1=a=w,

and rsmall) as

B1=1,B2=0,Fs′−1=1

2πtan−11−tan−10=1

2π×π

4=1

8.

Similarly, Fs′(2)−1is found also from Appendix D Configuration 37 (with b2=a=w,b1=2w) as

q=0.5×h(1−0.50.2929) 5.670 ×10−8×5004−3012i=6.3 W/m2,

Q′=q×2w=12.6 W/m.

CHAPTER 6 203

6.27

60°

A2

A1

A3

60°

L = 1 m

ground

qsun

A long greenhouse has the cross-section of an equilateral triangle as

shown. The side exposed to the sun consists of a thin sheet of glass

(A1) with reflectance ρ1=0.1. The glass may be assumed perfectly

transparent to solar radiation, and totally opaque to radiation emitted

inside the greenhouse. The other side wall (A2) is opaque with emittance

ǫ2=0.2, while the floor (A3) has ǫ3=0.8. Both walls (A1and A2) are

specular reflectors, while the floor reflects diﬀusely. For simplicity, you

may assume surfaces A1and A2to be perfectly insulated, while the floor

loses heat to the ground according to

q3,conduction =U(T3−T∞)

where T∞=280 K is the temperature of the ground, and U=19.5W/m2K

is an overall heat transfer coeﬃcient.

Determine the temperatures of all three surfaces for the case that the sun shines onto the greenhouse with

strength qsun =1000 W/m2in a direction parallel to surface A2.

If Eb1=Eb2, then with 1 −ǫ1Fs

1−1−ǫ2Fs

1−2=Fs

1−3and 1 −ǫ1Fs

2−1−ǫ2Fs

2−2=Fs

2−3we get

=789.4 W/m2

or

T1=T2=789.4/5.67 ×10−81/4=343.5 K.

T1=T2=343.5 K,T3=320 K.

CHAPTER 6 205

6.29 Consider the solar collector shown. The collec-

tor plate is gray and diﬀuse, while the insulated

guard plates are gray and specularly reflecting.

Sun strikes the cavity at an angle α(α < 45◦).

How much heat is collected? Compare with a

collector without guard plates. For what val-

ues of αis your theory valid?

3 = 0.2

1 = 0.9

T1 = 400 K

2 = 0.2

3 m

qsun = 1250 W/m2

Insulated

∋

∋

∋

F2i−1=diagonals −sides

2×base

Fs

X

(ρs)i√16+9i2+3(i+1) −h3i+p16+9(i+1)2i

=0.3860,

A3(23)

A3(2)

A2(32)

and

Fs

1−2=L

hFs

2−1=4

3×0.3860 =0.5147.

Similarly,

Fs

2−2=ρsF2(3)−2+(ρs)3F2(323)−2+···

X

X

(ieven)

=0.5+0.82×0.21221 +0.84×0.13104 +··· =0.7412.

One of these could also have been found using the summation rule, equation (6.21),

2Fs

2−1+ǫ2Fs

2−2+ǫ3Fs

2−3=20.3860+0.2(0.3984+0.7412) =0.9999.

h(1−ρs

2)#qscos α=1−ǫ2

htan αqscos α,

α

since d=Ltan α.

0.8−1×0.3860q1,

or

Eb1−0.1029Eb2−0.1029Eb3=1.1111q1+(1−0.2667 tan α)qscos α,

or

=0.8158 ×5.670 ×10−8×4004−(1.0305 cos α−0.2425 sin α)×1250

q1=(−1288 cos α+303 sin α+1184) W/m2,

q1,max =q1(α=0◦)=1288 +1184 =−104 W/m2.

CHAPTER 6 207

208 RADIATIVE HEAT TRANSFER

6.30

w1 = 4 cm

qs

A2

A1

w2 = 3 cm

φ

A rectangular cavity as shown is irradiated by a parallel-light source

of strength qs=1000 W/m2. The entire cavity is held at constant

temperature T=300 K and is coated with a gray material whose

reflectance may be idealized to consist of purely diﬀuse and specular

components, such that ǫ=ρd=ρs=1

3. How must the cavity be

oriented toward the light source (i.e., what is φ?) so that there is no

net heat flux on surface A1?

0.459

φ=17.5◦−3.5◦=14.0◦.

210 RADIATIVE HEAT TRANSFER

6.31

h = 1 m

w = 2 m

Tw, w = 1

0 K

= 0.1

∋

∋

0 K

N shields,

Reconsider the spacecraft of Problem 6.10. To decrease the heat

loss from Surface 2 the specularly reflecting shield 1 is replaced

by an array of Nshields (parallel to each other and very closely

spaced), of the same dimensions as the black surface and made

of the original, specularly reflecting shield material with emittance

ǫ=0.1. Determine the net heat loss from the black plate as a function

of shield number N.

212 RADIATIVE HEAT TRANSFER

6.34

h

w

A2: T2,

A1: T1,

0K

∋

∋

An infinitely long corner piece as shown is coated with a material of (diﬀuse

and gray) emittance ǫ, and purely specular reflectance. Calculate the variation

of heat flux along the surfaces per unit area. Both surfaces are isothermal at T1

and T2, respectively.

CHAPTER 6 213

6.35

L

L

0 K

1, T1

∋

2

∋

An infinitely long cavity as shown is coated with gray, specular ma-

terials ǫ1and ǫ2(but the materials are diﬀuse emitters). The vertical

surface is insulated, while the horizontal surface is at constant tem-

perature T1. The surroundings may be assumed to be black at 0 K.

Specify the variation of the temperature along the vertical plate.

1)Eb1dFdy−dx =ǫ1Eb1ZA1

Fdy−1is easily evaluated, or can be found by setting φ=π

2in Configuration 23 in Appendix D:

## Trusted by Thousands of

Students

Here are what students say about us.

###### Resources

###### Company

Copyright ©2021 All rights reserved. | CoursePaper is not sponsored or endorsed by any college or university.