Transcript CE319F Elementary Fluid Mechanics

CE 319 F
Daene McKinney
Elementary Mechanics of Fluids
Resistance
Flow Past a Flat plate
•
•
•
Boundary layer: Region next to an object
where fluid has its velocity changed due to
shear resistance of the boundary.
Velocity gradient exists between free stream
and object, thus shear stress exists at surface
which retards the flow.
Boundary layer grows in downstream
direction, until the onset of turbulence
VD
,

Vx
Plates Re 
,

Pipes Re 
•
Transition Re  2,000
Transition Re  500,000
Shear stress is high at leading edge and
decreases until transition and then increases
again
Turbulent region
Laminar region
Stress :   0.5U o2
Uo
Re x
x
Resistance : F  0.664BU o Re L
Stress :   0.332
Coefficien t :
CF 
Fs
BLU o2

/2
0.058
Re x 1 / 5
Resistance : F  0.5U o2 BL
1.33
Re L
Coefficien t :
CF 
F
0.072
Re L 1 / 5
0.523
1520


AU o2 ln 2 0.06 * Re L  Re L
Shear Stress Coefficients
•
Shear stress coefficient = ratio of shear stress
at wall to dynamic pressure of free stream
cf 
•
0
U o2 / 2
Total shearing force
Fs    0dA
A

Cf 
U o2
 c f dA
2 A
Fs
U o2 / 2
Ex (9.77)
Given: Ship prototype 500 ft long, wetted area of 25,000
ft2, and velocity of 30 ft/s in sea water at 10oC.
Model is in fresh water at 60oF, model:prototype
scale = 1/100, and Froude numbers are matched.
Drag is calculated as flat plate with model wetted
area and length. A drag of 0.1 lbf is measured in the
model tests.
Find: Drag on the ship.
Froude No. similarity
Frm  Frp
Vp
Vm

gLm
gL p
Vm

Vp
Lm
1
1


Lp
100 10
Vm  (1 / 10) * 30  3 ft / s
Shear resistance on model
Re L 
VL


3 * 5L
5
 1.23x106
1.22 x10
0.523
1520
Cf  2

 0.00293
ln (0.06 Re L ) Re L
 V 2 A 
  (0.00293) * 0.5 *1.94 * 32 * 2.5
Fm  C f 
 2 


Fm  0.0639 lbf
Fwave  Ftotal  Fshear  0.1  0.0639  0.0361lbf
Ex (9.77)
Scale up of wave drag
C p m  C p p
 p 
 p 

 

 V 2 / 2 
 V 2 / 2 

m 
p
pm  mVm2

p p  pV p2
Force on Prototype
3
3
Fm pm Am  mVm2 Am  m  Lm  1.94  1 






F p p p Ap  pV p2 Ap  p  L p  1.99  100 
1.99
1003  0.0361* 1.99 *106
F p  Fm
1.94
1.94
F p  3.7 x104 lbf
Lift of an Airfoil
•
•
Forces acting on airfoil: Velocity over the top
of foil is greater than free stream velocity,
pressure there is less than freestream.
Similarly, the pressure on the bottom of the
foil is greater than freestream pressure. This
difference in pressure contributes to the lift of
the foil.
Shear stress along the foil acts to drag on the
foil.
Drag of a Thin Plate
•
For a plate parallel to the flow, shear forces
are the only ones acting
Vo2
FD  2C f BL
2
•
For a plate normal to the flow, shear and
pressure forces act
Vo2
FD  (0.8  1.2) BL
2
•
For a more general object
Vo2
FD  C D Ap 
2
Drag Coefficients
•
Coefficient of Drag
CD 
FD
Vo2
Ap 
2
HW (11.4, 11.7, 11.56)
• 11.4
Ex (11.3)
•
•
•
Given: Pressure distribution is shown, flow
is left to right.
Find: Find CD
Solution: CD is based on the projected area of
the block from the direction of flow. Force
on downstream face is:
FD  Drag  C p Ap V 2 / 2  0.5 Ap V 2 / 2
The total force on each side face is: FS  C p Ap V / 2  0.5 Ap V / 2
2
The drag force on one face is:
The total drag force is:
2
FS  Drag  FS sin   0.5 Ap V 2 / 2 * 0.5
FDrag  2( FS ) Drag  ( FD ) Drag
 2 * (0.5 Ap V 2 / 2 * 0.5)  0.5 Ap V 2 / 2  C D Ap V 2 / 2
Coefficient of Drag is: CD=1
Ex (11.8)
•
•
•
Given: Flag pole, 35 m high, 10 cm diameter,
in 25-m/s wind, Patm = 100 kPa, T=20oC
Find: Moment at bottom of flag pole
Solution:
  1.51x10
Re 
VD


5
m / s,   1.20 kg / m
2
25 * 0.1
5

1
.
66
x
10
1.51x105
C D  0.95 (figure 11  5)
3
Vo2 H
H
M  FD  C D Ap 
2
2 2
252 35
 0.95 * 0.10 * 35 *1.2 *
*
2
2
 21.8 kN
FB
Ex (11.57)
•
•
•
Given: Spherical balloon 2-m diameter, filled
with helium @std conditions. Empty weight
= 3 N.
Find: Velocity of ascent.
Solution:
  air
6
D  FD  3   He
3
FD  3  ( air   He )

D3
6
287  3
 3   air (1 
) 2
2,077 6
 1.422 N
WB
FD  C D Ap
Vo2


6
D
3
2
2 FD
2 *1.422
0.739


2
C D Ap 
CD
C D ( / 4) * 2 *1.225
Vo 
Vo 
0.739
 1.36 m / s
0.4
Check Re
Re 
VD


1.36 * 2
1.46 x10
5
 1.86 x105
Iteration: CD=0.42
Vo 
WH
e
Iteration: Guess CD=0.4
 Fy  0  FB  FD  WB  WHe

FD
0.739
 1.33 m / s
0.42