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Problem *3.98
[Difficulty: 3]
F
buoyancy
W
load
y
W
hot air
Given:
Data on hot air balloon
Find:
Maximum mass of balloon for neutral buoyancy; mass for initial acceleration of 2.5 ft/s
2
.
Assumptions:
Air is treated as static and incompressible, and an ideal gas
Solution:
Basic equation
F
B
ρ
atm
g
⋅
V
⋅
=
and
Σ
F
y
Ma
y
⋅
=
Hence
Σ
F
y
0
=
F
B
W
hotair
−
W
load
−
=
ρ
atm
g
⋅
V
⋅
ρ
hotair
g
⋅
V
⋅
−
Mg
⋅
−
=
for neutral buoyancy
MV
ρ
atm
ρ
hotair
−
()
⋅
=
Vp
atm
⋅
R
1
T
atm
1
T
hotair
−
⎛
⎜
⎝
⎞
⎟
⎠
⋅
=
M
320000 ft
3
⋅
14.7
×
lbf
in
2
⋅
12 in
⋅
ft
⎛
⎝
⎞
⎠
2
×
lbm R
⋅
53.33 ft
⋅
lbf
⋅
×
1
48
460
+
R
⋅
1
160
460
+
R
⋅
−
⎡
⎣
⎤
⎦
×
=
M
4517 lbm
⋅
=
Initial acceleration
Σ
F
y
F
B
W
hotair
−
W
load
−
=
ρ
atm
ρ
hotair
−
g
⋅
V
⋅
M
new
g
⋅
−
=
M
accel
a
⋅
=
M
new
2
ρ
hotair
⋅
V
⋅
+
a
⋅
=
Solving for M
new
ρ
atm
ρ
hotair
−
g
⋅
V
⋅
M
new
g
⋅
−
M
new
2
ρ
hotair
⋅
V
⋅
+
a
⋅
=
M
new
V
ρ
atm
ρ
hotair
−
g
⋅
2
ρ
hotair
⋅
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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
 Fluid Mechanics

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