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Discussion The drag force on a body is proportional to both the drag coefficient and the frontal area. Thus, one is able to reduce drag by reducing the drag coefficient or the frontal area (or both). 11-4C Solution We are to define the planform area of a body and discuss its applications.

Chapter 11 External Flow Drag and Lift

Drag Lift and Drag Coefficients

Solution We are to compare the speed of two bicyclists

Analysis The bicyclist who leans down and brings his body closer to his knees goes faster since the frontal area

and thus the drag force is less in that position The drag coefficient also goes down somewhat but this is a secondary effect

Discussion This is easily experienced when riding a bicycle down a long hill

Solution We are to discuss how the local skin friction coefficient changes with position along a flat plate in laminar

Analysis The local friction coefficient decreases with downstream distance in laminar flow over a flat plate

Discussion At the front of the plate the boundary layer is very thin and thus the shear stress at the wall is large As the

boundary layer grows downstream however the boundary layer grows in size decreasing the wall shear stress

Solution We are to define the frontal area of a body and discuss its applications

Analysis The frontal area of a body is the area seen by a person when looking from upstream the area projected

on a plane normal to the direction of flow of the body The frontal area is appropriate to use in drag and lift calculations for

blunt bodies such as cars cylinders and spheres

Discussion The drag force on a body is proportional to both the drag coefficient and the frontal area Thus one is able

to reduce drag by reducing the drag coefficient or the frontal area or both

Solution We are to define the planform area of a body and discuss its applications

Analysis The planform area of a body is the area that would be seen by a person looking at the body from above

in a direction normal to flow The planform area is the area projected on a plane parallel to the direction of flow and

normal to the lift force The planform area is appropriate to use in drag and lift calculations for slender bodies such as flat

plate and airfoils when the frontal area is very small

Discussion Consider for example an extremely thin flat plate aligned with the flow The frontal area is nearly zero and

is therefore not appropriate to use for calculation of drag or lift coefficient

Solution We are to explain when a flow is 2 D 3 D and axisymmetric

Analysis The flow over a body is said to be two dimensional when the body is very long and of constant cross

section and the flow is normal to the body such as the wind blowing over a long pipe perpendicular to its axis There is

no significant flow along the axis of the body The flow along a body that possesses symmetry along an axis in the flow

direction is said to be axisymmetric such as a bullet piercing through air Flow over a body that cannot be modeled as

two dimensional or axisymmetric is three dimensional The flow over a car is three dimensional

Discussion As you might expect 3 D flows are much more difficult to analyze than 2 D or axisymmetric flows

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Chapter 11 External Flow Drag and Lift

Solution We are to discuss the difference between upstream and free stream velocity

Analysis The velocity of the fluid relative to the immersed solid body sufficiently far away from a body is called

the free stream velocity V The upstream or approach velocity V is the velocity of the approaching fluid far ahead of

the body These two velocities are equal if the flow is uniform and the body is small relative to the scale of the free stream

Discussion This is a subtle difference and the two terms are often used interchangeably

Solution We are to discuss the difference between streamlined and blunt bodies

Analysis A body is said to be streamlined if a conscious effort is made to align its shape with the anticipated

streamlines in the flow Otherwise a body tends to block the flow and is said to be blunt A tennis ball is a blunt body

unless the velocity is very low and we have creeping flow

Discussion In creeping flow the streamlines align themselves with the shape of any body this is a much different

regime than our normal experiences with flows in air and water A low drag body shape in creeping flow looks much

different than a low drag shape in high Reynolds number flow

Solution We are to discuss applications in which a large drag is desired

Analysis Some applications in which a large drag is desirable parachuting sailing and the transport of pollens

Discussion When sailing efficiently however the lift force on the sail is more important than the drag force in

propelling the boat

Solution We are to define drag and discuss why we usually try to minimize it

Analysis The force a flowing fluid exerts on a body in the flow direction is called drag Drag is caused by

friction between the fluid and the solid surface and the pressure difference between the front and back of the body

We try to minimize drag in order to reduce fuel consumption in vehicles improve safety and durability of structures

subjected to high winds and to reduce noise and vibration

Discussion In some applications such as parachuting high drag rather than low drag is desired

Solution We are to define lift and discuss its cause and the contribution of wall shear to lift

Analysis The force a flowing fluid exerts on a body in the normal direction to flow that tends to move the body

in that direction is called lift It is caused by the components of the pressure and wall shear forces in the direction

normal to the flow The wall shear contributes to lift unless the body is very slim but its contribution is usually small

Discussion Typically the nonsymmetrical shape of the body is what causes the lift force to be produced

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Chapter 11 External Flow Drag and Lift

Solution We are to explain how to calculate the drag coefficient and discuss the appropriate area

Analysis When the drag force FD the upstream velocity V and the fluid density are measured during flow over a

body the drag coefficient is determined from

where A is ordinarily the frontal area the area projected on a plane normal to the direction of flow of the body

Discussion In some cases however such as flat plates aligned with the flow or airplane wings the planform area is

used instead of the frontal area Planform area is the area projected on a plane parallel to the direction of flow and normal to

the lift force

Solution We are to explain how to calculate the lift coefficient and discuss the appropriate area

Analysis When the lift force FL the upstream velocity V and the fluid density are measured during flow over a

body the lift coefficient can be determined from

where A is ordinarily the planform area which is the area that would be seen by a person looking at the body from above

in a direction normal to the body

Discussion In some cases however such as flat plates aligned with the flow or airplane wings the planform area is

used instead of the frontal area Planform area is the area projected on a plane parallel to the direction of flow and normal to

the lift force

Solution We are to define and discuss terminal velocity

Analysis The maximum velocity a free falling body can attain is called the terminal velocity It is determined by

setting the weight of the body equal to the drag and buoyancy forces W FD FB

Discussion When discussing the settling of small dust particles terminal velocity is also called terminal settling speed

or settling velocity

Solution We are to discuss the difference between skin friction drag and pressure drag and which is more significant

for slender bodies

Analysis The part of drag that is due directly to wall shear stress w is called the skin friction drag FD friction since

it is caused by frictional effects and the part that is due directly to pressure P and depends strongly on the shape of

the body is called the pressure drag FD pressure For slender bodies such as airfoils the friction drag is usually more

significant

Discussion For blunt bodies on the other hand pressure drag is usually more significant than skin friction drag

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Chapter 11 External Flow Drag and Lift

Solution We are to discuss the effect of surface roughness on drag coefficient

Analysis The friction drag coefficient is independent of surface roughness in laminar flow but is a strong function

of surface roughness in turbulent flow due to surface roughness elements protruding farther into the viscous sublayer

Discussion If the roughness is very large however the drag on bodies is increased even for laminar flow due to

pressure effects on the roughness elements

Solution We are to discuss the effect of streamlining and its effect on friction drag and pressure drag

Analysis As a result of streamlining a friction drag increases b pressure drag decreases and c total drag

decreases at high Reynolds numbers the general case but increases at very low Reynolds numbers creeping flow since

the friction drag dominates at low Reynolds numbers

Discussion Streamlining can significantly reduce the overall drag on a body at high Reynolds number

Solution We are to define and discuss flow separation

Analysis At sufficiently high velocities the fluid stream detaches itself from the surface of the body This is

called separation It is caused by a fluid flowing over a curved surface at a high velocity or technically by adverse

pressure gradient Separation increases the drag coefficient drastically

Discussion A boundary layer has a hard time resisting an adverse pressure gradient and is likely to separate A

turbulent boundary layer is in general more resilient to flow separation than a laminar flow

Solution We are to define and discuss drafting

Analysis Drafting is when a moving body follows another moving body by staying close behind in order to

reduce drag It reduces the pressure drag and thus the drag coefficient for the drafted body by taking advantage of the

low pressure wake region of the moving body in front

Discussion We often see drafting in automobile and bicycle racing

Solution We are to discuss how drag coefficient varies with Reynolds number

Analysis a In general the drag coefficient decreases with Reynolds number at low and moderate Reynolds

numbers b The drag coefficient is nearly independent of Reynolds number at high Reynolds numbers Re 104

Discussion When the drag coefficient is independent of Re at high values of Re we call this Reynolds number

independence

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Chapter 11 External Flow Drag and Lift

Solution We are to discuss the effect of adding a fairing to a circular cylinder

Analysis As a result of attaching fairings to the front and back of a cylindrical body at high Reynolds numbers a

friction drag increases b pressure drag decreases and c total drag decreases

Discussion In creeping flow very low Reynolds numbers however adding a fairing like this would actually increase

the overall drag since the surface area and therefore the skin friction drag would increase significantly

Solution The drag force acting on a car is measured in a wind tunnel The drag coefficient of the car at the test

conditions is to be determined

Assumptions 1 The flow of air is steady and incompressible 2 The cross section of the tunnel is large enough to simulate

free flow over the car 3 The bottom of the tunnel is also moving at the speed of air to approximate actual driving

conditions or this effect is negligible 4 Air is an ideal gas

Wind tunnel

Properties The density of air at 1 atm and 25 C is 90 km h

1 164 kg m3

Analysis The drag force acting on a body and the drag

coefficient are given by

FD C D A and CD FD

where A is the frontal area Substituting and noting that 1 m s 3 6 km h the drag coefficient of the car is determined to be

2 220 N 1 kg m s 2

1 164 kg m 1 25 1 65 m 90 3 6 m s

Discussion Note that the drag coefficient depends on the design conditions and its value will be different at different

conditions Therefore the published drag coefficients of different vehicles can be compared meaningfully only if they are

determined under identical conditions This shows the importance of developing standard testing procedures in industry

Solution The resultant of the pressure and wall shear forces acting on a body is given The drag and the lift forces

acting on the body are to be determined

Analysis The drag and lift forces are determined by decomposing

the resultant force into its components in the flow direction and the FR 580 N

normal direction to flow

Drag force FD FR cos 580 N cos 35 475 N

Lift force FL FR sin 580 N sin 35 333 N

Discussion Note that the greater the angle between the resultant force and the flow direction the greater the lift

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Chapter 11 External Flow Drag and Lift

Solution The total drag force acting on a spherical body is measured and the pressure drag acting on the body is

calculated by integrating the pressure distribution The friction drag coefficient is to be determined

Assumptions 1 The flow of air is steady and incompressible 2 The surface of the sphere is smooth 3 The flow over the

sphere is turbulent to be verified

Properties The density and kinematic viscosity of air at 1 atm and 5 C are 1 269 kg m3 and 1 382 10 5 m2 s

The drag coefficient of sphere in turbulent flow is CD 0 2 and its frontal area is A D2 4 Table 11 2

Analysis The total drag force is the sum of the friction and pressure drag forces Therefore

FD friction FD FD pressure 5 2 4 9 0 3 N Air

V 2 V 2 D 12 cm

where FD C D A and FD friction C D friction A

Taking the ratio of the two relations above gives

FD friction 0 3 N

C D friction CD 0 2 0 0115

Now we need to verify that the flow is turbulent This is done by calculating the flow velocity

from the drag force relation and then the Reynolds number

V 2 2 FD 2 5 2 N 1 kg m s 2

FD C D A V 60 2 m s

2 C D A 1 269 kg m 3 0 2 0 12 m 2 4 1N

VD 60 2 m s 0 12 m

1 382 10 m s

which is greater than 2 105 Therefore the flow is turbulent as assumed

Discussion Note that knowing the flow regime is important in the solution of this problem since the total drag

coefficient for a sphere is 0 5 in laminar flow and 0 2 in turbulent flow

Solution A car is moving at a constant velocity The upstream velocity to be used in fluid flow analysis is to be

determined for the cases of calm air wind blowing against the direction of motion of the car and wind blowing in the same

direction of motion of the car

Analysis In fluid flow analysis the velocity used is the Wind

relative velocity between the fluid and the solid body 110 km h

a Calm air V Vcar 110 km h

b Wind blowing against the direction of motion

V Vcar Vwind 110 30 140 km h

c Wind blowing in the same direction of motion

V Vcar Vwind 110 30 80 km h

Discussion Note that the wind and car velocities are added when they are in opposite directions and subtracted when

they are in the same direction

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Chapter 11 External Flow Drag and Lift

Solution The frontal area of a car is reduced by redesigning The amount of fuel and money saved per year as a result

are to be determined

Assumptions 1 The car is driven 12 000 miles a year at an average speed of 55

km h 2 The effect of reduction of the frontal area on the drag coefficient is

negligible

Properties The densities of air and gasoline are given to be 0 075 lbm ft3

and 50 lbm ft respectively The heating value of gasoline is given to be 20 000

Btu lbm The drag coefficient is CD 0 3 for a passenger car Table 11 2

Analysis The drag force acting on a body is determined from

where A is the frontal area of the body The drag force acting on the car before redesigning is

0 075 lbm ft 3 55 mph 2

2 1 4667 ft s 1 lbf

FD 0 3 18 ft 40 9 lbf

1 mph 32 2 lbm ft s

Noting that work is force times distance the amount of work done to overcome this drag force and the required energy

input for a distance of 12 000 miles are

5280 ft 1 Btu 6

Wdrag FD L 40 9 lbf 12 000 miles year 3 330 10 Btu year

1 mile 778 169 lbf ft

3 330 10 6 Btu year

E in 1 110 10 7 Btu year

Then the amount and costs of the fuel that supplies this much energy are

mfuel E in HV 1 110 10 7 Btu year 20 000 Btu lbm

Amont of fuel 11 10 ft 3 year

fuel fuel 50 lbm ft 3

7 4804 gal

Cost Amount of fuel Unit cost 11 10 ft 3 year 3 10 gal 257 4 year

That is the car uses 11 10 ft3 83 03 gallons of gasoline at a cost of 257 4 per year to overcome the drag

The drag force and the work done to overcome it are directly proportional to the frontal area Then the percent

reduction in the fuel consumption due to reducing frontal area is equal to the percent reduction in the frontal area

A Anew 18 15

Reduction ratio 0 1667

Amount reduction Reduction ratio Amount

0 1667 83 03 gal year 13 8 gal year

Cost reduction Reduction ratio Cost 0 1667 257 4 year 42 9 year

Therefore reducing the frontal area reduces the fuel consumption due to drag by 16 7

Discussion Note from this example that significant reductions in drag and fuel consumption can be achieved by

reducing the frontal area of a vehicle

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Chapter 11 External Flow Drag and Lift

Solution The previous problem is reconsidered The effect of frontal area on the annual fuel consumption of the car

as the frontal area varied from 10 to 30 ft2 in increments of 2 ft2 are to be investigated

Analysis The EES Equations window is printed below along with the tabulated and plotted results

rho 0 075 lbm ft3

V 55 1 4667 ft s

Price 2 20 gal

efuel 20000 Btu lbm

rho gas 50 lbm ft3

L 12000 5280 ft

FD CD A rho V 2 2 32 2 lbf

Wdrag FD L 778 169 Btu

Ein Wdrag Eff

m Ein efuel lbm

Vol m rho gas 7 4804 gal

Cost Vol Price

A ft2 Fdrag lbf Amount gal Cost

10 22 74 43 27 95 2

12 27 28 51 93 114 2

14 31 83 60 58 133 3

16 36 38 69 24 152 3

18 40 92 77 89 171 4

20 45 47 86 55 190 4

22 50 02 95 2 209 4

24 54 57 103 9 228 5

26 59 11 112 5 247 5

28 63 66 121 2 266 6

30 68 21 129 8 285 6

Discussion As you might expect the cost goes up linearly with area since drag force goes up linearly with area

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Chapter 11 External Flow Drag and Lift

Solution A circular sign is subjected to high winds The drag force acting on the sign and the bending moment at the

bottom of its pole are to be determined

Assumptions 1 The flow of air is steady and incompressible 2 The drag force on the pole is negligible 3 The flow is

turbulent so that the tabulated value of the drag coefficient can be used

Properties The drag coefficient for a thin circular disk is CD 1 1 Table 11

2 The density of air at 100 kPa and 10 C 283 K is

P 100 kPa 150 km h

1 231 kg m 3 50 cm

RT 0 287 kPa m 3 kg K 283 K

Analysis The frontal area of a circular plate subjected to normal flow is A

D2 4 Then the drag force acting on the sign is

1 231 kg m 3 150 3 6 m s 2 1N

1 1 0 5 m 2 4 231 N

2 1 kg m s 2

Noting that the resultant force passes through the center of the stop sign the

bending moment at the bottom of the pole becomes

M bottom FD L 231 N 1 5 0 25 m 404 Nm

Discussion Note that the drag force is equivalent to the weight of over 23 kg of mass Therefore the pole must be

strong enough to withstand the weight of 23 kg hanged at one of its end when it is held from the other end horizontally

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Chapter 11 External Flow Drag and Lift

Solution We are to estimate how much money is wasted by driving with a pizza sign on a car roof

Properties fuel 50 2 lbm ft3 HVfuel 1 53 107 ft lbf lbm air 0 07518 lbm ft3 air 1 227 10 5 lbm ft s

Analysis First some conversions V 45 mph 66 0 ft s and the total distance traveled in one year L 10 000

miles 5 280 107 ft The additional drag force due to the sign is

FD 12 airV 2C D A

where A is the frontal area The work required to overcome this additional drag is force times distance So letting L be the

total distance driven in a year

Work drag FD L 12 airV 2C D AL

The energy required to perform this work is much greater than this due to overall efficiency of the car engine transmission

Work drag 1

2 airV 2C D AL

overall overall

But the required energy is also equal to the heating value of the fuel HV times the mass of fuel required In terms of

required fuel volume volume mass density Thus

mfuel required Erequired HV 1

2 airV 2 CD AL

Vfuel required

fuel fuel fuel overall HV

The above is our answer in variable form Finally we plug in the given values and properties to obtain the numerical

Vfuel required

2 0 07518 lbm ft 66 0 ft s 0 94 0 612 ft 5 280 10 ft

50 2 lbm ft 0 332 1 53 10 ft lbf lbm

32 174 lbm ft s

which yields Vfuel required 0 6062 ft3 which is equivalent to 4 535 gallons per year At 3 50 per gallon the total cost

is about 15 87 per year or rounding to two significant digits the total cost is about 16 per year

Discussion Any more than 2 significant digits in the final answer cannot be justified Bill would be wise to lobby for a

more aerodynamic pizza sign

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Chapter 11 External Flow Drag and Lift

Solution An advertisement sign in the form of a rectangular block that has the same frontal area from all four sides is

mounted on top of a taxicab The increase in the annual fuel cost due to this sign is to be determined

Assumptions 1 The flow of air is steady and incompressible 2 The car is

driven 60 000 km a year at an average speed of 50 km h 3 The overall

efficiency of the engine is 28 4 The effect of the sign and the taxicab on

the drag coefficient of each other is negligible no interference and the

edge effects of the sign are negligible a crude approximation 5 The flow AD

is turbulent so that the tabulated value of the drag coefficient can be used

Properties The densities of air and gasoline are given to be 1 25 kg m3

and 0 72 kg L respectively The heating value of gasoline is given to be TAXI

42 000 kJ kg The drag coefficient for a square rod for normal flow is CD

2 2 Table 11 1

Analysis Noting that 1 m s 3 6 km h the drag force acting on the sign is

V 2 1 25 kg m 3 50 3 6 m s 2 1N

FD C D A 2 2 0 9 0 3 m 2 71 61 N

Noting that work is force times distance the amount of work done to overcome this drag force and the required energy input

for a distance of 60 000 km are

Wdrag FD L 71 61 N 60 000 km year 4 30 10 6 kJ year

W drag 4 30 10 6 kJ year

E in 1 54 10 7 kJ year

Then the amount and cost of the fuel that supplies this much energy are

mfuel Ein HV 1 54 10 7 kJ year 42 000 kJ kg

Amont of fuel 509 L year

fuel fuel 0 72 kg L

Cost Amount of fuel Unit cost 509 L year 1 10 L 560 year

That is the taxicab will use 509 L of gasoline at a cost of 560 per year to overcome the drag generated by the

advertisement sign

Discussion Note that the advertisement sign increases the fuel cost of the taxicab significantly The taxicab operator

may end up losing money by installing the sign if he she is not aware of the major increase in the fuel cost and negotiate

accordingly

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Chapter 11 External Flow Drag and Lift

Solution A person who normally drives at 55 mph now starts driving at 75 mph The percentage increase in fuel

consumption of the car is to be determined

Assumptions 1 The fuel consumption is approximately proportional to the drag force on a level road as stated 2 The

drag coefficient remains the same

Analysis The drag force is proportional to the square of the velocity and power is force times velocity Therefore

FD CD A and W drag FDV CD A

Then the ratio of the power used to overcome drag force at V2 75 V

mph to that at V1 55 mph becomes

W drag2 V23 75 3

W drag1 V13 553

Therefore the power to overcome the drag force and thus fuel consumption per unit time more than doubles as a result of

increasing the velocity from 55 to 75 mph

Discussion This increase appears to be large This is because all the engine power is assumed to be used entirely to

overcome drag Still the simple analysis above shows the strong dependence of the fuel consumption on the cruising speed

of a vehicle A better measure of fuel consumption is the amount of fuel used per unit distance rather than per unit time A

car cruising at 55 mph will travel a distance of 55 miles in 1 hour But a car cruising at 75 mph will travel the same distance

at 55 75 0 733 h or 73 3 of the time Therefore for a given distance the increase in fuel consumption is 2 54 0 733

1 86 an increase of 86 This is large especially with the high cost of gasoline these days

Solution A submarine is treated as an ellipsoid at a specified length and diameter The powers required for this

submarine to cruise horizontally in seawater and to tow it in air are to be determined

Assumptions 1 The submarine can be treated as an ellipsoid 2 The flow is turbulent 3 The drag of the towing rope is

negligible 4 The motion of submarine is steady and horizontal

Properties The drag coefficient for an ellipsoid with L D 40 km h

25 5 5 is CD 0 1 in turbulent flow Table 11 2 The Submarine

density of sea water is given to be 1025 kg m3 The density of

air is given to be 1 30 kg m3

Analysis Noting that 1 m s 3 6 km h the velocity of the submarine is equivalent to V 40 3 6 11 11 m s The

frontal area of an ellipsoid is A D2 4 Then the drag force acting on the submarine becomes

V 2 1025 kg m 3 11 11 m s 2 1 kN

In water FD C D A 0 1 5 m 2 4 124 2 kN

2 2 1000 kg m s 2

V 2 1 30 kg m 3 11 11 m s 2 1 kN

In air FD C D A 0 1 5 m 2 4 0 1575 kN

2 2 1000 kg m s 2

Noting that power is force times velocity the power needed to overcome this drag force is

In water W drag FDV 124 2 kN 11 11 m s 1380 kW

In air W drag FDV 0 1575 kN 11 11 m s 1 75 kW

Therefore the power required for this submarine to cruise horizontally in seawater is 1380 kW and the power required to

tow this submarine in air at the same velocity is 1 75 kW

Discussion Note that the power required to move the submarine in water is about 800 times the power required to move

it in air This is due to the higher density of water compared to air sea water is about 800 times denser than air

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Chapter 11 External Flow Drag and Lift

Solution A billboard is subjected to high winds The drag force acting on the billboard is to be determined

Assumptions 1 The flow of air is steady and incompressible 2 The drag force on the supporting poles are negligible 3

The flow is turbulent so that the tabulated value of the drag coefficient can be used 4 The 3 D effects are negligible and

thus the billboard is approximated as a very long flat plate two dimensioal

Properties If the plate were two dimensional the drag coefficient would be CD 1 9 Table 11 1 However for this

3 D case Table 11 2 yields CD 1 10 0 02 20 8 8 20 1 158 which is a better approximation of the drag coefficient

We calculate the gas constant for air Table A 1E as

Ru 10 732 psia ft 3 lbmol R 20 ft

Rair 0 37045 psia ft 3 lbm R

M 28 97 lbm lbmol 55 mph 12ft

The density of air at 14 3 psia and 40 F 40 459 67 499 67 R is thus

P 14 3 psia

0 077254 lbm ft 3

RT 0 37045 psia ft 3 lbm R 499 67 R

Analysis The drag force acting on the billboard is determined from

V 2 0 077254 lbm ft 3 55 mi h 2 1 lbf 1 4667 ft s

FD C D A 1 158 12 20 ft 2 2

2 2 32 174 lbm ft s 1 mi h

2171 lbf 2170lbf

Discussion Note that the drag force is equivalent to the weight of 2170 lbm of mass Therefore the support bars must

be strong enough to withstand the weight of 2170 lbm hanging on one of their ends when they are held from the other end

horizontally which causes a fairly large moment or torque on the support bars

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Chapter 11 External Flow Drag and Lift

Solution A semi truck is exposed to winds from its side surface The wind velocity that will tip the truck over to its

side is to be determined

Assumptions 1 The flow of air in the wind is steady and incompressible 2 The edge effects on the semi truck are

negligible a crude approximation and the resultant drag force acts through the center of the side surface 3 The flow is

turbulent so that the tabulated value of the drag coefficient can be used 4 The distance between the wheels on the same

axle is also 2 m 5 The semi truck is loaded uniformly so that its weight acts through its center

Properties The density of air is given to be 1 10 kg m3 The drag coefficient for a body of rectangular cross section

corresponding to L D 2 2 1 is CD 2 2 when the wind is normal to the side surface Table 11 2

Analysis When the truck is first tipped the wheels on the wind loaded side of the truck will be off the ground and

thus all the reaction forces from the ground will act on wheels on the other side Taking the moment about an axis passing

through these wheels and setting it equal to zero gives the required drag force to be

M wheels 0 FD 2 m W 1 m 0 FD W 2 9m 2m

Substituting the required drag force is determined to be V 5 000 kg 2 5 m

FD mg 2 5000 kg 9 81 m s 2 2 24 525 N

The wind velocity that will cause this drag force is determined to be 0 75 m

V 2 1 10 kg m 3 V 2 1 N

FD CD A 24 525 N 2 2 2 5 9 m 2 V 31 48 m s

2 2 1 kg m s

which is equivalent to a wind velocity of V 31 48 3 6 113 km h

Discussion This is very high velocity and it can be verified easily by calculating the Reynolds number that the flow is

turbulent as assumed

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