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Problem 4.204
[Difficulty: 3]
Given:
Compressed air bottle
Find:
Rate of temperature change
Solution:
Basic equations: Continuity; First Law of Thermodynamics for a CV
Assumptions:
1) Adiabatic
2) No work 3) Neglect KE 4) Uniform properties at exit 5) Ideal gas
Given data
p
500 kPa
⋅
=
T2
0
°
C
=
T
293K
=
V
100 L
⋅
=
m
exit
0.01
kg
s
⋅
=
Also
R
air
286.9N
⋅
m
⋅
kg K
⋅
=
c
v
717.4
Nm
⋅
kg K
⋅
⋅
=
From continuity
t
M
CV
∂
∂
m
exit
+
0
=
where m
exit
is the mass flow rate at the exit (Note: Software does not allow a dot!)
t
M
CV
∂
∂
m
exit
−
=
From the 1st law
0
t
M
u
⌠
⎮
⌡
d
∂
∂
u
p
ρ
+
⎛
⎝
⎞
⎠
m
exit
⋅
+
=
u
t
M
∂
∂
⎛
⎜
⎝
⎞
⎠
⋅
M
t
u
∂
∂
⎛
⎜
⎝
⎞
⎠
⋅
+
u
p
ρ
+
⎛
⎝
⎞
⎠
m
exit
⋅
+
=
Hence
um
exit
−
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Unformatted text preview: ⋅ M c v ⋅ dT dt ⋅ + u m exit ⋅ + p ρ m exit ⋅ + = dT dt m exit p ⋅ M c v ⋅ ρ ⋅ − = But M ρ V ⋅ = (where V is volume) so dT dt m exit p ⋅ V c v ⋅ ρ 2 ⋅ − = For air ρ p R air T ⋅ = ρ 500 10 3 × N m 2 ⋅ kg K ⋅ 286.9 N ⋅ m ⋅ × 1 20 273 + ( ) K ⋅ × = ρ 5.95 kg m 3 = Hence dT dt 0.01 − kg s ⋅ 500 × 10 3 × N m 2 ⋅ 1 100 L ⋅ × L 10 3 − m 3 ⋅ × kg K ⋅ 717.4 N ⋅ m ⋅ × m 3 5.95 kg ⋅ ⎛ ⎜ ⎝ ⎞ ⎠ 2 × = 1.97 − K s ⋅ = 1.97 − C s ⋅ =...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Dynamics, Fluid Mechanics

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