Download Chapter 9, Flow Over Immersed Body (External Flow)

Chapter 9, Flow Over Immersed Body (External
Flow)
1. Lift and drag concepts
dFx = pdA cos + wdA sin
dFy = -pdA sin + wdA cos
Drag force, D =  dFx
Lift force, L =  dFy
Drag coefficient, CD = D/[(1/2)V2AD]
Lift coefficient, CL = L/[(1/2)V2AL]
2. Boundary Layer Characteristics
i) Flow over a flat plate
Boundary layer thickness
y=
when u/U = 0.99
/L << 1,
(Since v/u << 1)
Displacement thickness, *
bU* = b0 (U - u)dy
* = 0 [1 - (u/U)]dy
Momentum thickness, 
bU2 = b0 u(U - u)dy
 = 0 (u/U)[1 - (u/U)]dy
Using the liniear momentum equation
Fx = cs uVndA
-b0 w dx = b0 [u2 -uU]dy
0 w dx = U20 (u/U)[1 - (u/U)]dy
w = U2d/dx
eq.
Integrated linear momentum
Skin friction drag, D = 0 w dx
Friction coeffricient, Cf = w/[(1/2)U2] = 2d/dx
Drag coeficient, CD = D/[(1/2)U2] = (1/)0 Cf dx
If the velocity profile is known (Table 9.2), the shear
stress at the wall can be found
w = µdu/dy| y=0
Blasius’ solution is to solve the differential equation
for u.
However, if an approximate velocity profile is
assumed, the momentum thickness, , can be found
as a function of , the shear stress can be found from
the integrated linear momentum equation:
Assume velocity profile, u/U = 2(y/) - (y/)2
* = 0 [1 - (u/U)]dy
*/ = 01 [1 - 2(y/) + (y/)2]d(y/) = 1/3
 = 0 (u/U)[1 - (u/U)]dy
/ = 01 [2(y/) - (y/)2][1 - 2(y/) + (y/)2]d(y/) =
2/15
w = µdu/dy| y=0 = 2µU/
Cf = w/[(1/2)U2] = [2µU/]/[(1/2)U2] = 2d/dx
= 2d(2/15)/dx
d = 15µdx/(U)
/x = 5.48/(Rx)1/2,
*/x = 1.83/(Rx)1/2, Table
9.2
/x = Cf = 0.73/(Rx)1/2,
CD = 1.46/(R)1/2
Laminar flow, Rx = Ux/µ < 5 x 105
Shear at the wall (Blasius’ solution)
w = µdu/dy|y=0 = 0.332 U2/3 [µ/x]
Friction coefficient
Cf = w /[(1/2)U2]
Friction drag coefficient
(CD)f = (1/)o Cf dx = 1.328/R1/2
Turbulent flow, Rx = Ux/µ > 5 x 105
w = 0.0228U2/Rx1/5
Emperical
Cf = w /[(1/2)U2] = 0.0576/Rx1/5
(CD)f = (1/)o Cf dx = 0.72/R1/5
ii) Effect of pressure gradient
In addition to friction drag, there is pressure drag
which can only be determined experimentally
3. Drag
CD = (CD)f + (CD)p
depending in Reynolds number and shape of the
body
A. Skin Friction Drag
Shear stress at wall
w = µdu/dy| y=0
Friction coefficient
Cf = w/[(1/2)U2]
Drag coefficient

Cf dx
CD = D/[(1/2)U2] = (1/)0
Boundary Layer Theory, solution for velocity
a.
Differential Method
Continuity
u/x + v/y = 0
Momentum uu/x + vu/y = -(1/)dp/dx +
µ2u/x2
i) Flat Plate, dp/dx = 0
Blasius solution
ii) Turbine Blade, dp/dx  0
Falkner and Skan solution
b) Integral Method
Displacement thickness
* = 0 [1 - (u/U)]dy
Momentum thickness
 = 0 (u/U)[1 - (u/U)]dy
Shear stress at wall
w = U2d/dx
Friction coefficient
Cf = w/[(1/2)U2]
Drag coefficient

Cf dx
CD = D/[(1/2)U2] = (1/)0
Assuming a velovity profile
u/U = a(y/) + b(y/)2
B. Pressure Drag
Experimentally determined data are in
Figs. 9-19, -20, -22, -23, -24
Ex.9-10
Fz = d(mV)/dt
FB + D = W
W = V
FB = waterV
D = (1/2)U2ACD
CD is from Table 9-4,
solve for U and ckeck Reynolds number
Composite body drag
CD = 0.16 + 0.0095Ni
Ex. Autobomile, Table 9-5 Ni
Applications
Drag force, D = (1/2)U2ACD
Moment (torque),
Mo = 0
Power,  = M
Ex. 9-13
M = Ds(b + ds/2) + Dc(b/2)
Ds = (1/2)U2(/4)ds2 CDs
Dc = (1/2)U2bds CDc
Rs = Uds/ = 2.24 x 107
CDs = 0.3
Rc = Udc/ = 8.41 x 106
CDc = 0.7
M = 3.64 x 105 ft-lb