8-1
(a) T
– 800’C
=
1073
K
br =
(aXIAn)=
4292
Chapter
I
L{ =(0.2X1073)
– 214.6
Euo-h,
=
oEUo-f,
=
0.53131
E6
– (5.66,0
x
to-811t otl)a
=
7.515
x
lOa
w
I
^’
4trans
=
(0.53131X0.9X7.515 x
t04)
=35,933
Wl^’
8-.2
(a) hf – (5.5X1073)
=
5901.5 Euo-rr,
=
O-‘72921
qtrans
– (0.7
2g21X0.85X75,150)
= 46,580
W
l^’
zt?
8-3
(a)
Chapter
I
hf – 62)(1073)
=
55,796 Euo-h
=
1.0
qtrans
=
(l
.0X0.92)(7
5,150)
= 69,138
W
I
^2
8,-4
L{ =
(4X560)
=2240
1t.oR
T-300K
127′
– (15X560)
=
8400
Ir.oR
T-1000K
8-6
0 0.05
0
0
0
0
T-5000K
AT
00
z.81
v7
T-5795K
h
0
8-10
T-1073K
h
I
2
3
4
5
6
LT
0
M(p-k)
1073
2r46
3219
4292
5365
6348
T-300K
0.155
Frac
fraction 0
– 1T
7.74×10{
g.l3
x 101
0.32233
0.53
l3l
0.6765r
0.76538
T-1000K
0.37
r
2″
w.2
fraction
h{ – hf
8.35
x 10-3
o.23103
0.20898
0.1452
0.08887
T-50mK
0.164
Frac
contained
0.0015
Chapter
I
zEr
Chapter
I
8-11
T-t373K
fraction
4.25
8-12
AT
2897
1(tt)
2.ll
Eb
=
o(1
4Aq4
=).I77
x
105
Wl^z
If Fraction
8-13
enEutdh T-2000K Eb:g.O7x10s
€7 Eut,
M
E-J;
L
T5
=
3.2x
1016
e
lEun”
0
0
0
0
0
7Er’
8-14
(a)
Find 4-z
Y
=3 -0.6 z -!=0.8
Chapter
I
4_z
284
Chapter
I
(c)
Azry-l
= At[-z 4-z
=
0.06
2=0.25
L
L
=2.0
11
8-15
n=
5 cm ry,-2.5
cm
L- l0
cm
zEg
Chapter
I
)t
8-16
\
ocm @
– Q)n(L2.5)2
= 98
= /Cr2
= n(5)2
:78.5
Fn= FZz=0 4g
8-17
11= 5.0
cm rz=12.5
cm L=7 cm
open
ends
=3 & 4
L =
0.56 L =
0.4 4z
=
o.t8 Fzr
=
o.tg
12 12
– 5.02)
= 412.3 cm2
7.’ 1
– 0.
19
– 0.18_
=
0.3
15
r23=T
4z
=o.re(#)=
o.ns
\z + 4t+
4+
=
1.0 4z
=
4+ nt
=ff =
o.263
ry,
– (o
26r{t)
=
ro.zot{4)
=
o
too
Fzz
=,o.rtr,[f)
=
ro
rtr{#) =
o.42r
4t+ Ftz+
4c =
1.0
Ft+
=
1
– 0. 140
– 0.421
=
O.439
Le0
Chapter
I
8-19
4-20 d2-d3=50 L-t}
g -+ =
8-20
(a) 4t =
1.0
4z_+ =4-_2 =0.637
4rcfn
4r
-t
-+-t -?-0.363
Ar 7c
a9o
Chapter I
?fr
Chapter
I
4r
+
fi2* Fn:1.0 4r – 1-0.
1l1l
.?,
, l-
Chapter
8
8’21
=0
=O-2
s-22
E.23
Tt=Zffi”C=533K E4=a575Wfm2
S24
=l’441112 li =800x Q
=500
K
er2=
8-2s
4 =
540oC
=
813
K Tz=30O”C=
573
K 4=O]
tz
=O.5 ?i
=
100″C
=373K At=
4=O’2827
m2
w1^’ E4=6tll
a7,
Chapter I
8-26
J2D
=
ezEh=
3.056 pzz
=
0.5 I =11.79
ArIr20- P::,)
– Jr* #-,rt +478- 4 – o
8-27
Ea.,=a59
Wf m2 Eur=aM wf
m2 4z=0.25 A1–A2–)
€t
=
0.8 €Z
= 0.8
8-28
4 =8oooC=1073
K 6br=dT4
-75,146
wl^’
€r
=0.8 tz=0.1 4 =300
K Eur=d\a
=459
7_g+
p – 1.7
atm
Pr
= A.697
Chapter
I
f,-1.0
8-29
Tf=350K
20.76x
10-6
v-
Tr:22″C:295K
k
– 0.03003
1.7
27
-zTQ.5)3
(0.697)(r.72 l
Qtotat=
8-30
Ar=h.=1t.5;2
=2.25m2 4=800oC=1073K Tz=553I<
sr
=0.5 rz
=0.8 Tr
=0
K 4z=0.7 \z= Fzl=0.3
= szoz
z1
f
Chapter
I
8-32
!l-
=
6.eosx
t0-8)z(0.05X0.6X36
6a
-zgza)=
56’5
Wm =
17
’22
w lft
s?c’
8-33
AT
– 1
A0
– 27
:73oC:
k
: 0.04;
W
– 0.05′,
r
:0’0125 L – 2’A
Tabre
3-1; s
: zTE(zllnl0.54(0,05)/0.01
251:
l’632
– 15’318
8-34
q
=
oAe(Trn
-rr\= (5.669
x to-8Xo.OX0.3X0.sX36ga
-zgla,l= 89.55 W
a11
Chapter
I
8-3s
Iro=l50oc=423K T*=20″C=293K ?r=38oC=3llK
d=0.125m L=6m
8-36
4 =
heater ,t =
side walls 4 = room
->
oo Af =
belt
Jz= Eh 4 =698
t< Tt=393rc T+=298K
Eh
=13,456
Wl^z Eu,
=1352 E6o
= 447
=0.9 h= 4=3m2 h=l.gmz
20.5g0
w
Q’=
z9g
Chapter
I
8-37
f2
– 1200
K €2
=
0.75 \ =
0.3 Ql
=0
Eu,
=
dTz4
– l.l76x los Er,
– drt4
– 417.g
I
AzFzt AtFtt
Eut
Tt
-293K
167.53
+
361
.84
rri:.53
4z: l -o
T3
= 1052
K
8-38
Ar
= tr(0.A2)
=
0.06283 Az
=
n(0.06)
=
Q.
1884
Eq – d44
=32,922
wl*z Eh – dr\4 – 4sg
AzFzt
Tgg
t-39
L
=
30 cm i= 4
cm 4z =
0.8 r2F21= r1F12
Iteration
Solution
g-40
Ebr
: I
17500
Euz
– 3
543
ez
Qr+
Qz*
8-41
(Eur
– Euz)
(0.52)
– 0.48XEb
z
-Eur)
(1.0(Eb
z
-Eur)
8-42
d-75cm
s2
=
o’6
x- 50
cm
Tz=400K
d
=1.5 gl =
Tooo
x Alr
4z=Fzt=0’3
€l
=
0’8
4l = Fzl
=a’7
7ot
Chapter
I
8-43
f2=288K 4l=0.5
4z
=
o’5
Eu,
– 139,736
Wl^’
t39,736
– 390
q- 1.061 + sfseq6{p
q wlo reflector = ttAt(Eu,
Eb2
=
390
wl^’
= 23,758
IV/m
– Eu) = 26,264
Wm
\ -12s3K
7oz