ME 362
Drag
Drag (D):
wdA
psdA
dD

D  Dp  D f
  ps cos dA    w sin  dA
where D
Dp
Df
ps
w
dA
=
=
=
=
=
=
total drag
pressure drag (dominant for flow over a road vehicle)
friction drag (dominant for flow over an airplane)
surface pressure
wall shear stress
element surface area
Drag Coefficient (CD):
CD 
D
1
U  2 A
2
where A is the frontal area of the object
Note: A  dA
Pressure Coefficient (Cp):
Cp 
p s  p
1
U  2
2
Page 1 of 3
ME 362
Drag
Page 2 of 3
Flow Over a Circular Cylinder:
2
Dp 
p
s
cos dA   ( p s  p ) cos dA
0
C D, p 
2
2
(Note:
0
p

cos  dA  0 )
0
ps
Dp
dA = adb
1
U  2 A
2
U
where CD,p
a
b
dA
A
=
=
=
=
=
d
pressure drag coefficient
cylinder radius
cylinder width (into the page)
adb (ad = arc length)
2ab (frontal area)
a

Solution Procedure:
Step 1: Find stream function  or potential function  by superposition of
elementary flows
   uniform   doublet or
   uniform   doublet
Step 2: Find surface velocity V,s at cylinder surface (r = a)
V ,s  V
r a


r

r a
1 
r 
(Note: Vr ,s  Vr
r a
r a

1 
r 

r a

r
Step 3: Find surface pressure ps from Bernoulli equation
ps 
1
1
V2,s  p   U 2
2
2
Step 4: Find pressure coefficient Cp or pressure drag coefficient CD,p

Cp 

C D, p
p s  p
1
U  2
2
Dp

1
U  2 A
2
where D p 
2
2
0
0
 ps cos dA   ( ps  p ) cos dA
0)
r a
ME 362
Drag
Page 3 of 3
Example:
U

a
K
The stream function for potential flow past a circular cylinder (radius = a and width = b)
with circulation is given as:
a2
r
  U  sin  (r  )  K ln
r
a
The tangential velocity at the cylinder surface (r = a) is then calculated as:
v (r  a)  2U  sin  
K
a
K
 1 , show that two stagnation points (at which v  0 ) appear on the
Ua
surface at angles  = 30 and 150.
a) For
b) Assume that on the front surface ( -90    90 ), the velocity is given by the
K
 1 from which the surface pressure ps is computed
potential flow theory with
Ua
by Bernoulli’s equation. On the rear surface ( 90    270 ), the pressure is
assumed equal to its value at  = 210 to model the overall effect due to flow
separation.
Compute the pressure drag coefficient CD,p, where
C D, p 
Dp
1
U 2 (2ab)
2
Note that  sin n  cos d 
2
and
D p   ( p s  p ) cos (ad b)
0
sin n 1 
n 1