Transcript L20

Flow Around A Corner
• A particle at (x0,y) will descend at
speed U until y=x0.
– Ignore gravity
• Then it finds itself in a horizontal
fluid flow.
• Drag force on particle takes more
general form:
– FD = -(v-u)/B
• Solve for particle trajectory in x
and y directions.
t
x  x 0   vdt '
0


 t 
 t 
x  x 0  mBv 0 1exp  
   ut  umB1  exp  
 
 mB  
 mB  


dv
v
u


dt
mB mB

v
u
u 
 v
 t 

  0 
 exp  

mB mB  mB mB 
 mB 
 t 
v  u  v 0  u  exp  

 mB 
 t  
 t 
v  v 0 exp  
  u1  exp  
 
 mB  
 mB  
Approaching Drift Velocity
Here we consider the trajectory of a charged particle in a constant electric field
• The force on a charged particle is the charge on the particle times the electric field at its location
• e is the elementary unit of charge, and –e is the charge on a single electron. Assume the aerosol
particle has a single extra electron.
• The electric field is calculated as E = -V, where V is the electric potential (voltage)
V2
Parallel plates at different
voltages produce a nearly constant
field between them. Let V2 > V1.
The distance between them is h.
E
V2  V1 ˆ
k
h
FE  eE
V1
dv
m
 FD  FE
Now consider the generalized force equation for the particle,
dt
dv
v
m
   eE
dt
B
Equations of this form have the solution,
v
e
e 
 v
 t 

 E    0  E  exp  

mB m
 mB 
 mB m 
V V
 t 
v  v 0 exp  
  eB 1 2
h
 mB 
Memory of original velocity decays away

 t 
1

exp



mB



A “drift velocity” takes over on same timescale
Approaching Drift Velocity
Now that we have solved for velocity, we need to integrate to get trajectory

 t 
 t 
v  v 0 exp  
  v D 1  exp  

mB
 mB 



Assume particle starts at rest, vD = 1 cm/s, t = 10s
t
z  z 0   vdt '
0

 t 
z  z 0  v Dt  v D  v 0 mB1  exp  

 mB 

zoffset
The result is that the particle initially
accelerates until it approaches a path parallel
to the constant drift path.
The offset between these paths asymptotes to
z offset  v D  v 0 mB
z=vDt
z