Chapter 4
4-101
cr – 412.5
C2
– 412.5
Node
I
2
A”**.
23.23
23.23
The
Equations
ABc
<(0.2t( (2.€l *O Os
arcr-r: r r+n zfu7ffiu.e^G r
(\ rzAA r -r r
ITl=
212= <(0.2.(Ct{2}+0 zr
3T3=
lq+
Chapter
4
11{
Chapter
4
4-702
Cl=87’4
C2
– l’14’8
C3
= 174’8
C4
= 174’8
Node
I
2
3
4
Afrn”*
4.263
4.263
4.263
4.37
llL
Chapter
4
4-103
k-43
p=7800 C-474 &-5 cm h=35 y
mt.oc
tt1
Chapter
4
Excel
solution
for Lr = 25
sec
and
75
sec
shown
below
rz – 190oC
occurs
at
six,
25
sec
time
increments.
Time – (25)(6)
– 150
sec
Steady
state
reached
at about
(30X75)
= 2250
sec
I?E
Chapter
4
The
Solution
A,r
– 75
sec
FGHII
ITI: T2= T3: T4= NO. time
2250 250 2so 2sa 0
3t99.6ms6t7 r99.607 r99.607 r93.22s t
4r7t.482?l
t5 160.756 l5t.5l
rr55.591 2
t?c
Chapter
4
4-105
aAr
–=-
(Lx)” 4
Ct=CZ=C3=C4=78,488
(0.15)2
Af*.* =# -436sec
f
4:o
4-106
Cr
=
350
Node
I
A?rn r,
16.93
l4t
Chapter
4
The
Solution l0 sec
GHIJKLM
IIIF 0.25 TI: T2: T5: TF
2Time
3incr 50 50 50 50
4I50.14286
50.07143 50 49.9E214
l4’r-
Chapter
4
The
Solution I min
G H IJKLM
IDt= 5TI: D: T5: T6=
2Time
l*s
Chapter
4
Chapter
4
4.TT2
Wall thickness
= 0.25
mT* = 600″C
x
=2.O
cm
h-100 y
4-ll4
ft
=
l.W a
– 5.4x
l0-7 Ti
-20″C
t45
at
x=2cm r=139A
4-tt7
Chapter
4
3.2
hro (350)(0.075)
= Q. 122
(13
x
10-7)@X60)
=0.291
at
)
fo- (0.a75)2
| *j,
Chapter
4
4-127
!=4=(20×9’ol)-20 o=L=5xlo{
R Lx 0.01 pc
| +’?
Chapter
4
4-134
Plate
hY-
=o(?y)
=zlL
<
o.l hL
<
o.o5
latakk
From
figure
+= 0.98
for
+ =
20
and
r =
1.0
worst
case
0shLL
l*a
Chapter
4
4-135
Aluminum k=204 d=8.4×10-5 m2/s 4=2o0″C
4-136
!=oforh+oo
hL
4-138
l4?
Chapter 4
4-t39
h=23 k
=1.37 a
=7.5x
l0-7 L
=
lg cm
4 =30″C L =OoC 7x=o
=5oC
a” Back
side insulated
4-l4A
T1=]O”F L =
350oF h=2.5 k* =0.395 p
= 59.6
4-145
r0
=
1.5 in
=
3.81 cm ft
=
0.585 a =1.4x
l0-7 p
=999
c
= 4195
l5o
4-147
4-749
c-2100 rl,-1.64×10-7
c
– 1300 u – 1.3
1x 10-7
Polyethylene k – 0.33
Particleboard
k-0.17
p-960
p – 1000
t5l
Chaptar
4
4-150
4 = surface
area
for radiation
4 = surface
area
forconvection
lrt