Fluid Forces: Chapters 9 and 11: Section 1--Overview. A cylinder is falling through a fluid at Velocity V. What are the Forces Acting?. Buoyancy. Would you expect the force Acting to be a function of Velocity?. Weight. V. There is a pressure Difference between back and - PowerPoint PPT Presentation
der is falling through a fluid at Velocity V Forces: Chapters 9 and 11: Section 1--Overview What are the Forces Acting? Weight mg F W Buoyancy Vol F B V Would you expect the force Acting to be a function of Velocity? There is a pressure Difference between back and front –leads to an additional force called DRAG Net Force: D W B F F F F Drag Force—function of Velocity and “projected P D D A V C F 2 2 A Common Form V D C A dimensionless Coef, free stream velocity— vel. re In same Area projected in flow direction (a circle here) P A
Transcript
A cylinder is falling through a fluid at Velocity V
Fluid Forces: Chapters 9 and 11: Section 1--Overview
What are the Forces Acting?
Weight mgFW
Buoyancy VolFB
V
Would you expect the forceActing to be a function of Velocity?
There is a pressure Difference between back and front –leads to an additional force called DRAG
Net Force: DWB FFFF Drag Force—function of Velocity and “projected Area”
PDD AV
CF2
2
A Common Form
VDC A dimensionless Coef, free stream velocity—
vel. relative to object In same direction as Drag
Area projected in flow direction (a circle here)PA
So Drag is a result of fluid moving over a body—Are there other forces that can arise in this way
Weight mgFW
Buoyancy VolFB Flow over a thin plate—Drag not important
V
Small projected area
But fluid will impose a shear stress—recall pipe friction—on Surface of plate
Resulting in a Surface Resistance Force in direction of free stream velocity
SfS AV
CF2
2
fC Dimensionless coef. sA Surface area (parallel to flow)
SWB FFFF Net Force:
What Area here?
Also Lift: Consider rotating Cylinder in Free stream Flow V
V
High Vel, Low p
Low Vel, Hight p
Why?
Pressure Difference will lead to a Force— The Lift Force —NORMAL to free streamDirection—Engineers calculate this as
PlanLL AV
CF2
2
LC Dimensionless coef. planA plan area (looking down on flow)
The Magnus Effect
So—there are three forces that can occur due to flow over a body
PDD AV
CF2
2
SfS AV
CF2
2
PlanLL A
VCF
2
2
In flow direction ------- normal to flow
Trick is finding C
Drag Surface Resistance Lift
For given BodiesLook up in tables Correlated to Account for Dragand shear resistancesee next slide and later notes
Can be calculated for plateSee section 2
Also in tables
Relation between Stress Distribution and Flow Forces (See Fig 11.2)
Free-Stream Velocity V will give riseto pressure acting normal tobody and shear acting along body
• Far from the surface, the fluid velocity is unaffected.
• In a thin region near the surface, the velocity is reduced
• Which is the “most correct” velocity profile?
…this is a good approximation near the “front” of the plate
Boundary layer growth
• The free stream velocity is u0, but next to the plate, the flow is reduced by drag
• Farther along the plate, the affect of the drag is felt by more of the stream, and because of this the boundary layer grows
• Fluid friction on the surface is associated with velocity reduction throughout the boundary layer
0y
o dy
du
Local stress & total force, skin friction
•Not immediately straightforward (unlike approximations we made with thin films): •du/dy decreases with x & y
•We need to find then
•And there is more trouble…
L
x
s dxBF0
0y
o dy
du
)x(yo 0
Breadth
Boundary layer transition to turbulenceAt a certain distance along a plate, viscous forces become to small relative to inertial forces to damp fluctuations
figure_09_04
Picture of boundary layer from text
Boundary layer transition• How can we solve problems for such a complex system?
• We can think about key parameters and possible dimensionless numbers
• Important parameters:– Viscosity μ, density ρ– Distance, x– Velocity uO
• Reynolds number combines these into one number
xuxu
Re OOx
δ(x)0y
B L thickness in laminar region
xUxU
Re OOx
Oxu
x5
Re
x5
Self-similar shape
Boundary layer questions• How can we solve problems for such a complex system?
• We can think about key parameters and possible dimensionless numbers
• What about stress?
• We talk about (local) stress and (total) force on a boundary in terms of local cf and average CF stress coefficients:
xUxU
Re OOx
2Uc
2o
of
2U*Area
FC
2
O
SF
δ(x)0y
Average shear-stress coefficientOn Plate of Length L
figure_09_12
Ignore this part just for a moment
Note New Reynolds No
Example 9.6 from text• A plate is 3 m long x 1 m wide• Air at 20°C and atmospheric pressure flows past this plate with a
velocity of 30 m/s • A boundary layer over a smooth, flat plate is laminar at first and
then becomes turbulent. The turbulent forms of drag, etc., are reasonable above Re = 5 x 105.
• What is the average resistance coefficient Cf for the plate? • Also, what will be the total shearing resistance force of one side
of the plate?• What will be the resistance due to the turbulent part and the
laminar part of the boundary layer?
smVAsmmkgair /30331/1051.1/2.1 252
SfS AV
CF2
2
Find , shearing resistance on one side of plate, and resistance due to laminar flow fC
Here it is !
smVAsmmkgair /30331/1051.1/2.1 252
Find , shearing resistance on one side of plate, and resistance due to laminar flow fC
65
1096.51051.1
330
VL
ReL
1st calculate plat Reynolds number
Mixed laminar-Turbulent
00294.01520
)06.0(ln
523.02
eleL
f RRC
smVAsmmkgair /30331/1051.1/2.1 252
Find , shearing resistance on one side of plate, and resistance due to laminar flow fC
65
1096.51051.1
330
VL
ReL
00294.01520
)06.0(ln
523.02
eleL
f RRC
NAV
CF SfS 76.432
)30(2.100294.0
2
22
smVAsmmkgair /30331/1051.1/2.1 252
Find , shearing resistance on one side of plate, and resistance due to laminar flow fC
65
1096.51051.1
330
VL
ReL
00294.01520
)06.0(ln
523.02
eleL
f RRC
NAV
CF SfS 76.432
)30(2.100294.0
2
22
Now Calculate Transition point mVxVx
tt 252.0/000,500000,500
smVAsmmkgair /30331/1051.1/2.1 252
Find , shearing resistance on one side of plate, and resistance due to laminar flow fC
65
1096.51051.1
330
VL
ReL
00294.01520
)06.0(ln
523.02
eleL
f RRC
NAV
CF SfS 76.432
)30(2.100294.0
2
22
Now Calculate Transition point mVxVx
tt 252.0/000,500000,500
So laminar layer Coefficient is
smVAsmmkgair /30331/1051.1/2.1 252
Find , shearing resistance on one side of plate, and resistance due to laminar flow fC
65
1096.51051.1
330
VL
ReL
00294.01520
)06.0(ln
523.02
eleL
f RRC
NAV
CF SfS 76.432
)30(2.100294.0
2
22
Now Calculate Transition point mVxVx
tt 252.0/000,500000,500
So laminar layer Coefficient is
And laminar force is
Average shear-stress coefficient
figure_09_12
Laminar Turbulent Induced
δ(x)
cf
FS
Cf
Laminar, Turbulence, Induced Turbulence
x0
L
0x
dxB
2UBL
F2
O
S
2U 2
O
O
7/1XRe
x16.0
XRe
x5
f
2O c
2
U
7/1
XRe
x16.0
7/1XRe
027.0
X
2 Re06.0ln
455.0
XRe
664.0
f
2o
2
U*Area C
7/1LRe
032.0 LL
2 Re
1520
Re06.0ln
523.0
LRe
33.1
xUxU
Re OOx
LURe O
L
BACK TO DRAG ON SUBMERGED OBJECTS…
Drag on a surface – 2 types• Pressure stress / form drag
• Shear stress / skin friction drag
• A boundary layer forms due to skin friction• For shapes more complex than a plane, these result
in total drag forces which are usually hard to solve analytically
Shortcuts for total drag• For less precise design and/or well-known /
well-studied (simple) objects, we rely on charts for an average coefficient of drag
2AVCF 2DD
Frontal Area
+ shear
Drag coefficients for
2d or infinitely long objects for 3d bodies
figure_11_04
figure_11_07
Vbor
VdRe
P 11.18, 9th edition
Compute the overturning moment exerted by a 35 m/s wind on a smokestack that has a diameter of 2.5 m and a height of 75 m. Assume that the air temperature is 20° C and that the atmospheric pressure is 99 kPa absolute.
V= 35m/s
2.5 m
75m
smmkg /1051.1,/17.1 253
Object is ~an infinite cylinder
61079.5Re Vd
V= 35m/s
d=2.5 m
75m
smmkg /1051.1,/17.1 253
Object is ~an infinite cylinder
61079.5Re Vd
62.0 DC
V= 35m/s
d=2.5 m
75m
smmkg /1051.1,/17.1 253
Object is ~an infinite cylinder
61079.5Re Vd
62.0 DC
kNAVCF DD 31.83)755.2(2
)35(17.162.02
22
Then turning moment
mMNFh
M D .12.331.835.372
Lift
Total lift• Similar to our calculations of total drag, we rely on
charts for an average coefficient of lift
• A is a reference area, usually “planform area”
22AVCF LL
Example 11.6 Lift on a Rotating SphereA ping-pong ball is moving at 10 m/s in air and is spinning CW at 6000 rpm as shown. The ball diameter = 3 cm. Calculate the lift and drag forces and indicate the direction of each. Assume standard atmospheric pressure and a temperature of 20 C.
How does the answer change if the ball is spinning CCW?
Rotation parameterV
r
6000
0.03
3/2.1 mkg
Find Lift and Drag Forces on Ping-Pong
Rotation parameterV
r
6000
0.03
3/2.1 mkg
Rotation rate
sradrevradsrev /628)/(2)/(100
Rotation parameter
942.10
015.0628
V
r
64.0,26.0 DL CC NAVCF LL22 101.12
NAVCF DD22 1071.22
So—there are three forces that can occur due to flow over a body
PDD AV
CF2
2
SfS AV
CF2
2
PlanLL A
VCF
2
2
In flow direction ------- normal to flow
Trick is finding C
Drag Surface Resistance Lift
For given BodiesLook up in tables ch 11Correlated to Account for Dragand shear resistancesee next slide and later notes
