Problem 1.10 [Difficulty: 4]
NOTE : Drag f ormul a is in erro r: It shou ld be:
FD3π⋅V⋅d⋅=
Mg
FD = 3
π
Vd
a = dV/dt
Given: Dat a on sp here and for mula fo r dra g.
Find: Diameter of gasoline droplets that take 1 second to fall 10 in.
Solution: Use given data and data in Appendices; integrate equation of
motion by separating variables.
The data provided, or available in the Appendices, are:
μ4.48 10 7−
×lbf s⋅
ft2
⋅= ρw1.94 slug
ft3
⋅= SGgas 0.72=ρgas SGgas ρw
⋅= ρgas 1.40 slug
ft3
⋅=
New ton's 2 nd l aw for the sphere (mass M) is (ignor ing bu oyancy e ffec ts) MdV
dt
⋅Mg⋅3π⋅μ⋅V⋅d⋅−=
dV
g3π⋅μ⋅d⋅
MV⋅−
dt=
so
Integrating twice and using limits Vt() Mg⋅
3π⋅μ⋅d⋅1e
3−π⋅μ⋅d⋅
M
t⋅
−
⎛
⎜
⎝
⎞
⎠
⋅= xt() Mg⋅
3π⋅μ⋅d⋅tM
3π⋅μ⋅d⋅e
3−π⋅μ⋅d⋅
M
t⋅
1−
⎛
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅=
Replacing M with an expression involving diameter d Mρgas
πd3
⋅
6
⋅= xt()
ρgas d2
⋅g⋅
18 μ⋅t
ρgas d2
⋅
18 μ⋅e
18−μ⋅
ρgas d2
⋅
t⋅
1−
⎛
⎜
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
⋅=
This equation must be solved for d so that x1s⋅( ) 10 in⋅= . The answer can be obtained from manual iteration, or by using
Excel's Goal Seek.
d 4.30 10 3−
×in⋅=
0 0.025 0.05 0.075 0.1
0.25
0.5
0.75
1
t (s)
x (in)
0 0.25 0.5 0.75 1
2.5
5
7.5
10
t (s)
x (in)
Note That the particle quickly reaches terminal speed, so that a simpler approximate solution would be to solve Mg = 3πµVd for d,
with V = 0.2 5 m/s (a llowing fo r the fact that M is a function of d)!