Chapter
I
Er, – 4575 Eu,
7o,
Chapter
I
Node
Jr
4575-4
*2-Jr-0-,rr _0
2256 2t2r 2t22
g-46
Af tz
: Er- :0
“Dz
1- “t
A:’et
0 .l .2 .3 .4 .5 .6 .7 .g ,g 1.0
SURFACE
EMISSIVITY
E^
-0
ul
0
lea
Chapter
I
8-47
le
Eur A’rn eAl
I
rtt Qt
tappl – ^
r r \Eu,
tuppl = tupp| =rT:X*)*ffi
4z is
obtained
from Fig. 8-13.
Note
that €app
approaches
1.0 ur { -.”.
d
.x
As
–+0 euoo +0
-0 t2
8-48
11=0.5 L-1.0 ry,=1.0 eZ
Ti-300K s1
=0.65 \-0.5
ry,
-o \=8ooK
T
u – 1.0 4t – 0.35
ry
– l-0.35- 0.23
J3 Eut
7o;
Chapter
I
g-49
L – 0.
1
m hr
-(t.42)(100/0.1)t,o:7.99
7t9
8-50
Tr=gI3K E4=
50,811
Wl^’
(2) Surfac
e = side
walls
(3) room \= 303
h
€r
=
0’8 4=(2)(3)
-6^2
Az=
(10)
(2)=20
^2
Eh=478
Wl*’
J2
L =l.o Y
=1.5
DD 6,
Fn=;(0.75) =0.225
l- tr =
{.1666
x
10a
4Ar
AzFzt
: Euz
4t =0.25 \Z=A.75
Fzl=o’225
At4rt
–0.2222
Ar4rz
– 0.2222
h.Fzl
cl= -1’633x10s
W
T.ffi-ffi
Afn
lol
Chapter
I
8-51
\- 1273K Eur-
I
.489x 10s
wl^’
Surfac
e
(2) is sphere
in radiant
balance.
€l=0.6 T3=303K
Eh=478
wl^’
235.6
+
353 .4
+
(2)(91
.86)
+39.37
E- r,R
8-52
– 3.4’72= 1
1 – l3.EB9
Tt=873 €r=0.65 qz=o 7i=303t< 4z=O.2
4z= 4t–0.8 4,=
4,=(0.6)2
=0.36 l–q =!.496
4Ar
– 478
Eh =32’928
L.496
13.889 3.472
J2 : Euz
J”t
Chapter
I
8-53
11-Jcm ry,-10cm
€1
=
0.6 S2
=
0.7
L- l0
cm \-973K
ez-o n
=0.5
12
T\-303K
I
u =1.0
ry,
8-54
4-4:25^2
EU,
= 465.3
x =Y -L.25
DD
’42=
(4X5)(4):80
*2
Eh- 417.8 Wl^’
4t =
0-35 4z
=
0-65
\ =
301
K T\
-293K
tl =
0.62 €3
=
0.75
I
– at
=o.o24sz
?o
g
wl^z J3=424.3
8-55
Chapter
I
€1
=0.7 4t =
o’5
\-1000K
4l =
Ftz
=
o’5
@
m
d_50c
7r2-300K
=0,[
.3
183
8-56
Eur=o(6n)4-11630
Eh = o(8:l3J)4
=32,928
Wl^’
?ro
Chapter
I
8-57
Al=Ae=1.0
EU,
= 23,216
4z=o’2
Eur
g-59
d
– 50 =4.0
x 12.5
wl^2
4-)oo
Eh
= 459
wl^z
4t -0.8- F)3roo 4lt, =
l’o
Jzr Eoz
4z:0’59
Eot: J3
– 8.634
At4z
Eur
J3: Eh
56,690-4
*Jz-Jr _459-4 _0
I.274 9.634 12.425
4l=Fzl=0’41
– 12.425
/e4lt
J2
Ar4t
7u
Chapter
I
8-59
Disk=(l) Shield=(2) Room=(3) d=2=Z.O 4s=O.ZB
x25
4z=O.72 Tt=873K €r=0.55 Ez=O.l 4=30oC
Eq
=32,928
wl^’ Eh -478 Wl^’ A1
=Ic(0.25)2
=S.
1963m2
4 =
n(O.5X0.25)
=
0.3927 m2 4r =
4z* =
a.lO=
S3
(inside)
‘ “42
– 4.16g l- ez
^-Eur-Eh
n-r
lll=
-+-
R2 Rr +
7 .O25 18.194
18.194
I
Rr (22.918X2)
+2.546
1
+-
7.074
Rz
= 7 .648 IR-lt.8l7
n _ 32,928
– 479 n1 Ar rr f _ 32,929
– J1
q—‘– . =2746W=-
‘ I 1.817 4.169
21,482-Jzi
_Jrr-478 Jzr
– 10,264
7.075 6.17 L6
Jzi – Jzo _ Jzi – 478 D
(2)(22.gtg) 4g.3n -b2,
fz = 561 K = 288″C
IR – 4.168+
R2 Rl = S.
L716
Jr=21,482
Wl^’
wl*’ Jzo
-993
– Jzi
* Jzo
= 562g
W
l^, – dTr4
2
Vtt
8-62
(a)
Chapter
I
8-61
1500c
= edT4 t=&
Dark
Side
6.45
&–
alrt t,
R t lt t I
r v va- g
Y
Eu,
– ,rs
n ( 479.3 )l/4
T*
=l-i =303.2K=30.2oC
-s [5.669
x rc-9
)
at
equilibrium Js = Eb,
Ebro
. , 6.45
slnO-
‘ 6.95
0
=’lo.3o
cos
0
– 0.3367
solid
angle
= 1-:os0 =
0.3316
=
F*
24, =
0.6684 Frro
=
1.0
4 = rc(Dz
– n ^2 Eu,
– (5
.66gx
l0-8xzsg)4
=390
wl^z
Ebro
– 0#*.’? ,J*,
-o /,
=
rzg.3=
Er-
dTr4
v0.33t6
v0.6694
4=218.5K=-54.5oC
4i.* =
o(
+)t =
+ for
sun
\2) 4
For
polished
Aluminum
t – 0.048
q
absorbed
=
(0.M8/+’)(1400)
= sz.t79 w
‘\4/’
T-403.3K-130.30c
4,F,,
/s* Olt’ -0
Itt
8-63
d-0.02; x-0.03
Ar: rct0.02×0.03)
+
(0-02)’ru7
– rr(0
8-64
Btack
shiny
lacquer’
t1
nv
0’88’ ta: 0’99
7t+
g-65
At=
4,=0.09
^2
T\
-2e3K
FZln
= 1.0
\=423K
h-r.42(
{)r/4
‘J-r.–\T
)
Chapter
I
S1
= t2 = 0.8
4Z =
0.55 4l =
FZlt
=
0.45
q2
7tr
\ – 1200
K T\
=300
K Eh – 1.1755
x
105 Eh – 459
s1
=
0.2 t2 =
0.5 t3
=0.8 4z =
Fzl
=
1.0 l-tl =4.0
t1
Chapter
I
E-66
Zg”‘-
P
8-67
xlu
Chapter I
Eq =3’l’194 Eh- 14sr
wl^z
8-69
\=900 K €l
=0.3
T\-400K t3=0.5 t2
=
o.o5
8-69
c@
EU,
\z=a-2
Ja: Eu+
4+=FZ+=0.8
3tt
Chapter
8
Node
{
u’1?-
tt
*J?
,
J]
*s!- Jr-o
1.042 3.472 0.g6g
l
Node
/2
h-J2,544-J2,6lll-J
-+-.-+ , -2-0
3.472 0.9691 0.4630
64,552
– 26,79r
=
36,239W
J1
=26,791
Wl^2
J2- 5984
wl*2
611I
– 5984
Qz= 0.463
Ql= =274W
1.042
ltE
Chapter
I
8-70
@
\=773K
@
\-298K €1
=0.8 €2=02 L=0.07854
L
f2=533K-260″C
tj = o’4
4571
8-71
\=1073
K T\=373K \=0.8
4-4cm d3=8cm d2-6cm €2=0.3
Eh
=75,146
wl^’ Eh- 1og7
wl^’
tt9
Chapter
I
g-73
Eu,
=
o(l 0l-
i)4
=75,133 Eur=
o(6:6)4
=
l.
1,62g
Eh
=o(2r4
– o
g-75
d
– 0.03;
x:0.06
– 0.021,
)za
…….-.’
l.– .- ” -“1 “- –T – .4′” -‘ – | | ‘ | –*
8-77
dl
=
o’3
h=643K
dz
=0.6 €l
=
0’5
AI=n(0’3)- 09425
tz=0.3 \= 1033
K
t8=o’7
\z =
1’o
–^ 64,552J- J2-9691
15,773 1.23g
4’7
,8
16
– Eun 3
-537
“O Ae
{7,816
-29,218 3.537
+
I ‘768
Ts=889(=6l6oc
h= 47’816
Eus=35’4L6=dfro
7zt
8-79
Tt=1033K f2=643K
S1
=
0.5 t2 =
0.3 \Z =
1.0
&g=€g=0.3 Tg=0J
Consider
per
unit
area,
4, = Ae
= A2
= 1.0
Chapter
I
Eu’
=9690
F)s
=
1’o
EU,
= 64,550
\s =
1’o
Same
network as
8-61
m+{zQ-€g)
Mt@
_ 64,550
– 9690
1+2.333+*
iadffr+0.7
– 54’860
– 1
2.160
w
4.5097
q
– rz,t60
– 64’519-
h = Jz -9690 Jr
= 52,385
1.0 2.333 J2
=
38,063
g-90
Re
= =
53,330
x
lo-s
) =
7z-