Estimate drag force on car. Let the car’s principle dimension be R, which is to say “consider a spherical car”, [ R ] = [ L ] = [ characteristic length of body]; [η ] = [ML−1T −1 ] = [fluid viscosity]; (1.1) [v ] = [ LT −1 ] = [ velocity of body]; [ ρ ] = [ ML−3 ] = [fluid-density];
Force: estimate the force by considering continuity, J p ≡ η = [ ML−1T −2 ] = [ momentum current ] ~ P = [ MT −2 L−1 ] = [pressure]; J p ~ P; Jp
= η ∇v
(1.2)
~ η v / R; J p ~ P → P ~ η v / R
(1.3)
Surface area, S: let the body be approximately spherical. Then, we can immediately compute resistive force 2 S ~ 4π R
→F
~ PS
2 R) ⋅ ( 4π R ) = 4πη vR ; Compare: exact F for sphere is: F
= (η v /
= 6πη v R ;
(1.4)
Reynold’s number: Using a power-law-ansatz for a dimensionless quantity, we form the Reynold’s number as, Re = [ Reynold's number ] = v a Rbη c ρ d
Let
Ν ≡ η / ρ ,
=
a
a
b
c
c
c
d
3d
L T − L M L− T − M L−
the “dyn “dynami amicc visc viscosi osity” ty”.. We know know v = 60 mph mph
→ ( a, b, c, d ) = (1,1, −1,1) →
= 27 2
m s
Re = vRρ / η ; (1.5)
, and R could be about 2 meters. Thus, it
−1
remains to estimate the dynamic viscosity, which has units of L T , the same as diffusivity. Using DA again, [ Ν ] = L2 T −1 ~ ℓv RMS = [ mean free path ][ RMS thermal velocity] (1.6) The RMS-velocity is well-known to be T m ≈ 0.3mO2
+ 0.7 mN
2
, and mO2
= 5.3 × 10
= U therm →
−26
kg
≈
16 14
1 2
mv
2
=
3 2
k BT
↔
v=
3k B T m
≡
vRMS , in which
mN 2 . The mean free path can be estimated by considering a
typical atom of air, saying that it sweeps out a cylinder of space of volume V ~ nV −1 , where nV is the numberdensity of atoms, meaning V
2
=πr ⋅ ℓ =
nV
↔ ℓ = (π r
2
− nV ) 1 . The cross section of the molecule r could be
=
estimated by considering the n = 2 level of atomic states, E2 which E 1 = −13.6 eV , and V r ~
= V (r ) =
kZe
2
=
E1 = 2 V
=
2 T (c.f., virial theorem) in
kZe 2 / r , yielding,
(9 × 109 )( 7 2+8 )(1.6 × 10−19 )2 1
V
1 22
2
2+1
(13.6)(1.6 × 10−19 )
Meanwhile: the ideal gas law gives us nV as: PV
=
Nk BT
↔
= 6.4
nV
=
nm = 12 128 Bohr radii
N /V
=
P / ( kB T ) . So, we finally estimate the
dynamic viscosity as, Ν
~
3k BT / m πr
2
nV
=
3kBT / m π ( kZV e
2
)2
P k BT
=
3 (0.026 ×1.6 ×10 −19 J ) 3/ 2 π
(6.4×10−9 m) 2 (1 (1×10 5
N 2 m
)
1 0.3 +
14 16
0.7 5.3 ×10
= −26
1.64 × 10
kg
We anticipate a very high reynold’s number, vR ( 27 ms )( 2m ) 9 e) × RvmnV ; Re = = = 3.3 × 10 → F ~ πη R × (Re) × v = πΝ (Re) × Rv ρ = πΝ (R −7 1.64 × 10 Ν
−7
m2 s
(1.7)
(1.8)
