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or 94.9 C
T
367.925 K
=
T
exp
h
−
A
s
⋅
ρ
V
⋅
c
p
⋅
t
⋅
⎛
⎜
⎝
⎞
⎠
T
i
T
air
−
()
⋅
⎡
⎢
⎣
⎤
⎥
⎦
T
air
+
:=
V
A
s
3.333
10
3
−
×
m
=
V
4.189
10
6
−
×
m
3
=
V
4
3
π
⋅
r
3
⋅
:=
A
s
1.257
10
3
−
×
m
2
=
A
s
4
π
⋅
r
2
⋅
:=
Bi << 0.1, so LC method is valid, and Temperature gradient in solid is
negligible.
Bi
6.65
10
4
−
×
=
Bi
hL
c
⋅
k
:=
L
c
3.333
10
3
−
×
m
=
L
c
r
3
:=
for sphere
Calculate Biot number
Assume
1. Unsteady state cooling
2. Check to see if Lumped Capacitance is valid
3. Neglect radiation exchange with surroundings
4. Constant properties
h8
0
W
m
2
K
:=
r
0.01 m
=
r
D
2
:=
D2
c
m
:=
t
2min
:=
T
air
303K
:=
T
i
453 K
=
T
i
180
273
+
()
K
:=
α
1.166
10
4
−
×
m
2
s
=
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This note was uploaded on 03/11/2011 for the course CHEME 333 taught by Professor Anthony during the Spring '11 term at UNL.
 Spring '11
 Anthony

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