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16.9:
Solving Eq. (16.10) for the temperature,
(
29
(
29
K,
191
K
mol
J
3145
.
8
40
.
1
hr
km
6
.
3
s
m
1
0.85
h
km
850
mol)
kg
10
8
.
28
(
2
3
2
=
⋅
×
=
=

R
Mv
T
γ
or
C.
82
°

b) See the results of Problem 18.88, the variation of atmospheric pressure
with altitude, assuming a nonconstant temperature. If we know the altitude we can use
the result of Problem 18.88,
.
1
0
0
α
α

=
R
Mg
T
y
p
p
Since
,
y
T
T
o
α

=
m,
C
10
6
.
K,
191
for
2
°
×
=
α
=

T
and
).
ft.
840
,
44
(
m
667
,
13
K,
273
0
=
=
y
T
Although a
very high altitude for commercial aircraft, some military aircraft fly this high. This result
assumes a uniform decrease in temperature that is solely due to the increasing altitude.
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Unformatted text preview: Then, if we use this altitude, the pressure can be found: , K 273 m) (13,667 ) m C 10 6 (. 1 p m) C 10 K)(.6 mol J 315 . 8 ( ) s m mol)(9.8 kg 10 8 . 28 ( 2 o 2 2 3 ° × ⋅ × ° ×= p and , p 13 . ) 70 (. p o 66 . 5 o = = p or about .13 atm. Using an altitude of 13,667 m in the equation derived in Example 18.4 gives , p 18 . o = p which overestimates the pressure due to the assumption of an isothermal atmosphere....
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