Chapter
I
Outer
Surface:
coolins
_ 91,051-
(}f<0l
_ r.238 +e3.55)rc(0.6×1323
_643)__98,124
Wlm
8-E1
Eur=37,194 Eh=14,513 9=g.tSZ
| +=0.3142
LL
+ =e.548 +=2.122 f- =42.44
etAt ez\, 4ryses
– -”
8-82
Assume
large
number
of pins
so
that
array
behaves
like single
pin
— rAt
i1 A2
7zt
Chapter
I
g-93
g-94
– 490.8’l
**2
x = depth
4 =
n(3)2
+ n(3
+ 12.5)lx2
+
(12.5
– 312
f r z
= nI9+
I 5. 5(r2
+ go.zs)t
|
2
|
8-85
3
– 1L.537″ 4 – lc])z
=28-274
il + ei($ – 4) 28.n4
+
Q.5)(2770
7tl
Chapter
I
8-96
(f),t
-€i)+k
(*),t -t;)+k
8-97
AlFrz(G pz,s)
Jr J2′
1-“t
etAt
pzD
e2A2(l h5)
1\r
Ar(Fn
+ pzsFrtzlr)
A2F3Q h,s)
J3
7zr
SM:
Heat
Transfer,
9e
8-88
Chapter
I
Jw r2(l Pzs)
I prs /12 Euz
A1
g-E9
|
Jw
Eb, I prs Jz Euz
q= >n -r
_ 4o(I- ni?f -rz4\
++1
Same
answer
with diffuse
PtD * ,,1
mrZrfl-)rs)
o(Tra
3lc
Chapter
I
g-90
l”
lc
r(2)
\=643K
Pzs
= 0′-l
FZlt
= 1.0
€l = 0’6
Pzo
=
o-1
FZln
=
0.75
Ar=(0.3X0.6)
=
S.
18
m2 T\=363K
E2=O-2 4z
– F\ =
0’25 4l =0’75
4rzll
= Ftz
=
a.25 1- et – 3’704
eth
h=’412K-139oC
6234
– 984
1859
– Ebz
_
74.07
+ 14.84
9.259
Jz, –
rsse
l- Pzs
Ek-1640
=dlr4
1859
-984 37.038
1L1
8-91
H
\=533K \=0.6
Chapter
I
l5cm
Pzs=0.1 T\=293K Pzo=0’1
=4575
– 418
n – 660.32
h.Fzzr(l
-12- Pzs)
Ito
Chapter
I
8-95
8-96
4z=r.o hr=! 4r
=o 4z=v!
A2k
/r(l – 0)
– 0 – e)(\zl) =
erEh
,,1,
-[t –
+],t –
“,.|
–
(r
–
e)(FzJ)
=
ezEk
-L\42) r
For A2
+ oo,
FZZ + 1.0, and
4t+ 0, and
Jr
-0 – e)Jz
—
erEh Jz$-(l-€2)l-(1- ezx0)/r
=
ezEh
Tlrs, Jz= Eu,
“/1
=
(1-
e)Eaz+
e1E5,
e
=
fttEh -r,)
=
ffi [Eu,
–
eEh
–
Q-
e1)E4l=
e1A1(E4
–
Eu)
This agrees
with Eq.
(843a)
?t)
ChapbrS
8-100
dt=dz=30cm x=5cm !=O 4z=4t=O.7
x
4s=4t=0.3 Ts=293K 4=txt05flm2
A
8-101
rAz
/\
3
4 =900
K €i
=06, 4 =
15fi)
K €z
=O.8
4 =looo
flm2 et–o.1
Alt
=
Fgz
=0.5 Eq
=37J9[ Eu,
=2.87
x
lOs
7to
Chapter
I
t-102
At=h.=lm2 \=573N sr.=0.5 Eh=6lll
Qz=O ez=0.7 7i=303tr E4=478
Ftz=O
– O.5)(0.2J2+
=
8-103
dt=
dz=
60
cm x
=
15
cm 4
=
813
K Tz=573K 4=O.7
sz
=
0.5 ?i
=
303
t< At
=
h, =
O.2827
m2 Ftz
= 0’6
= Fzr
0.4 Eh
=24,767
wl^’ Eut
=6ltl
?7(
Chapter
I
8-104
dt=dZ=lm
4t=FZl=0.4
Tt-303K
x=0.25 !-+
x
4- Az=n(0.5)z
=0.7854
\:0.5 Eu,
– 61 1l
4z= Fh
=
0’6
\=573K
Eu,
– 478
L’273*#
3.183
‘ 2.122+3.183
8-105
dr=dz=50cm x=10cm 4z=4t=0.65 4z=41=0.35
7i = 350
K €1= t2 – $.$ Ebt
= 850.7
+l=lo,ooowl^’ 4=Q Fr
=hz=o Q3+l.o
Alz Alt rr
q-
7tu
Chapter
I
8-106
\=900K €1
=0.5
Az=2n(2)2
–25.13
m2
Qg=0 A3
= tc(22
EU,
=
37,194 Ar: n(0.5)z
=
0.7854
fz=300K t2=0.8 Eh=459
– 0.52)
=
I I.78
I ^2 4z = Flz=
1.0
8-107
d=l .0
x
4z =
0.17 (l) disk (Z)
room (3)
shield
=
7r,
.s6’n
x
r 0-3
X20,
z4r
8-108
!-l.o
x
4z =
0.11 (1)
disk (2) room
Chapter
I
\=1273K
Euz
q- 148,900
– 418
J3
– 11,532 W
4.168+ = 1 =
zs*r’*2,6.b
74.26 74.26 31.93
749
lt+
Chapter
8
8-109
Er,
=644 =23,2A0
Wl^z Eu, – 459
8-110
Bottom
– (l) Sides
= (2) Room
= (3)
Az
= rc(a.005
+
0.01X(0.0
L2
-0.005′)
*0.032
ft2 =
0.00
1433
Al = 7-854xl0-5
Eur=478-
J3 T\=
303
K
\=773K Eu,
=20,241
€l =
0-6
7a4e-,
o
@
\\
I
J
I
I
I
I
I
I
I
J
O
o
–6-
@
Chapter
I
8-lll
Tt=373K Tz.=3131< i=0.7 Ez=0.5
€3=t4=t5=€O=0.6 TZ=T+=343K TS=TO=60oC=333K
8-112
rr
=
l0 cm 12
=
5 cm L
=
l0
cm |=O.t t=r.O
=0.11 (3)room 4f =l-0.11=0.89 fi3oo,,o,n
=1.0
7tr
l9l l9l r27
Eug
Chapter
I
8-lt3
4=873f €r=0.63 rr=4=25cm ez=g0UWf^2
ez=0.75 d
=125
cm ?i
=
303
K Eq
=32,g}g
Wl^,
77c
Heater
surface
(3) Room is
(4) dr
=
l0 cm €r
=
0.4 dz
=2O
cm
€z=0.6 Z=l0cm €3=0.8 4=90kVm, To=303K
Als
Chapter
I
8-115
z(0.1)2
=
0.03142
1e2
l-et
etAt
efz
Jzr I e2
l-“rJu
etAt
Eur
AtFt+o
AtFzqr
7?1
Chapter
8
1–t =47.74 += lo.6r I =
176.8
€rAt ezb, At4z
+ =7o.74 +-=176.8 + =77.06 I =4e.73
hrzs Ar4qo Ar4+i Ar4z —
:-!=106.1 +=7O.74 I =15.92
4rt+ bFz+t /p4z+o
/\
qz
=[
4)or= (90,000)(0.023s6) = 2r2o w
\A/ –
8-ll6
77t
Chapter
I
8-117,
T* = -‘lO”C
= 203
K dsolar
= 0.46 dtow
temp
= 0.95
(1070X0.46)
=
(0.95X5
.669
x1g-a;17a
-2cB4) T
=322.6
K =
49.6oC
8-,119
as(950)=
oe(74
-3000)* l2(T-300)
=
-902.5+
T
&I19
8-120
hA(T*
-f) +
o€A(Tr4
-f,4) =
0
T*
=
l05oc
:378 K f,
=
98oC
– 37 1K
Solving
Tr=305K-32oC
8-121
8-122
Tr=-40oC
-233K €- 0.8
hA(Tw
-To*)lurn
+hA(Tw
-To*)turb
– -oAt(T*a -Tro)
– 56.26 xc
– 584 K
,7I
Chapter
I
8-123
(28)A(Ta
-273) = (s.669
x rO-8)a(t
.O)(Zn4
– 2$4.)
Ta=28O.8K=7.8t
E-124
h
=
n2( +)”o Tf
= 6sooc
= 923
K ?a
=
560oc
=
833
K
8-125
T*
=2O”C
=293K T,
=32″C=
4 =291K=24oC
8-t26
7{o
Chapter
I
8-127
442=kFzt 4, =(0.86{i)=0.* Fzz=1-0.43
-0.24=0.33
Jr
– (l
– e)(4zJz;
+
\384) = erE4
8-12E
8-129
-}= Tn
=363K e
=O-94
74r