Problem 1

12ft

7 100

V

Given: Truck traveling down 7% grade Width = 10ft m = 25 tons = 50,000 lb Rolling resistance on concrete = 1.2% of weight CD = 0.96 without air deflector CD = 0.70 with air deflector

Drag Weight F

Find: Velocity of truck for both drag situations θ Equations: For force balance analysis: ∑ ∑ ∑ === zzyyxx maFmaFmaF Lift and Drag:

AV

FCdragoftCoefficien D

D2

21 ρ

≡=

AV

FCliftoftCoefficien LL

2

21 ρ

≡=

Solution: 1. Use force balance to find equation for motion of truck. 2. Calculate drag due to rolling resistance by 0.012*weight. 3. Calculate aerodynamic drag using both given CD (separately) and eqn. for drag. 4. Plug information into original force balance eqn. to find truck velocity. Numbers:

Problem 2

U = 20ft/sGiven: Box kite blowing in wind Vwind = 20 ft/s W = 0.9 lb θ Tension in string = 3.0 lb Angle to ground = 30o

Frontal area = 6.0 ft2

Find: a) Determine the lift and drag coefficients b) If Vwind = 30 ft/s and CD and CL are as found, will the kite rise or fall?

T W

D

L

x

y

Assumptions: Equations: For force balance analysis: ∑ ∑ ∑ === zzyyxx maFmaFmaF Lift and Drag:

AV

FCdragoftCoefficien D

D2

21 ρ

≡=

AV

FCliftoftCoefficien LL

2

21 ρ

≡=

Solution: a) 1. Find the density of air from Table A.9. 2. Use force balance in both x and y to find equations for drag and lift respectively. 3. Using equations for lift and drag calculate CD and CL. b) 1. Plug new velocity into equations for lift and drag and recalculate θ .

Numbers:

Problem 3 Given: Old airplane with wires used to strengthen design, with: U = 70mph, Awing = 148 ft2, CD wing = 0.020, Lwire = 160 ft, Dwire = 0.05 in Find: Ratio of drag from wire bracing to that from the wings Assumptions: (1) Wires modeled as smooth cylinders Equations: Lift and Drag:

AV

FCdragoftCoefficien D

D2

21 ρ

≡=

AV

FCliftoftCoefficien LL

2

21 ρ

≡=

Solution: 1. Calculate the Reynolds number of the wire braces. 2. Look up CD for wires in Fig 9.13. 3. Calculate Dwire / Dwing. Numbers:

Problem 4 Given: Old airplane with wires used to strengthen design, with: U = 150 km/h, Lwire = 60 m, Dwire = 6 mm Find: Power required to move the guy wires Assumptions: (1) Wires modeled as smooth cylinders Equations: Lift and Drag:

AV

FCdragoftCoefficien D

D2

21 ρ

≡=

AV

FCliftoftCoefficien LL

2

21 ρ

≡=

Solution: 1. Look up the kinematic viscosity and density of air at 15oC in Table A.10.

325 /23.1,/1045.1 mkgsmx == − ρν 2. Calculate the Reynolds number of the wire braces.

241,17/1045.1

100016

36001

11000150

Re 25 === − smxmm

mmms

hrkm

mhrkm

UDν

3. Look up CD(Re) for wires in Fig 9.13. CD ~ 1.2 4. Calculate the force of drag using the equation for the coefficient of drag.

NF

mmshrkmmhrkmmkgF

AV

FC

D

D

DD

25.461

)60)(006.0()3600/1*/1000*/150)(/23.1)(5.0(2.1

21

23

2

=

=

≡ρ

5. Now calculate power necessary, knowing P = FV = (461.25 N)(41.7m/s) = 19.2 kW.

Problem 5 Given: Optimally faired wires of the same length and other properties as in problem 4 Find: Percent power saved by fairing guy wires Equations: Lift and Drag:

AV

FCdragoftCoefficien D

D2

21 ρ

≡=

AV

FCliftoftCoefficien LL

2

21 ρ

≡=

Solution: 1. Look up the kinematic viscosity and density of air at 15oC in Table A.10.

325 /23.1,/1045.1 mkgsmx == − ρν 2. Calculate the Reynolds number of the wire braces.

241,17/1045.1

100016

36001

11000150

Re 25 === − smxmm

mmms

hrkm

mhrkm

UDν

3. Look up CD(Re) for streamlined strut in Fig 9.14. CD ~ 0.06 4. Calculate the force of drag using the equation for the coefficient of drag.

NF

mmshrkmmhrkmmkgF

AV

FC

D

D

DD

23

)60)(006.0()3600/1*/1000*/150)(/23.1)(5.0(06.0

21

23

2

=

=

≡ρ

5. Now calculate power necessary, knowing P = FV = (23 N)(41.7m/s) = 960 W. 6. Comparing this to the power in problem 4:

%95%100200,19

960200,19%100 =−

=−

xW

WWxP

PP streamline less power necessary!!!

Problem 6 Given: Spherical hailstones, D = 10mm, falling through air at STP Find: Terminal velocity Equations:

(1) AV

FCdragoft DD

2

21 ρ

≡=Coefficien

Solution:

1. Look up properties for air in Table A.10: 325 /23.1,/1045.1 mkgsmx == − ρν 2. Look up the specific gravity of ice in Table A.1 and calculate the density of ice.

330 /917)/1000)(917.0(917.0

2mkgmkgSGSG Hiceiceice ===→= ρρ

3. Since the sum of the forces in y is zero when terminal velocity is reached, sum

forces: ∑ −== Dy FmgF 0

4. Rearrange equation (1) for the drag force and plug it into the force balance

equation just found to find Vt. ACmgVAVCmgD

ttD ρρ 2)5.0( 2 =→=

5. Calculate the mass of the hail: kgxmmkgVm 43

3 108.46

)01.0(/917 −===πρ

6. Calculate the area: 2522

1085.7401.0

4mxDA −===

ππ

7. Now calculate Vt(CD): DD

t Csm

mxmkgCsmkgxV /87.9

)1085.7)(/23.1()/81.9)(108.4(2

253

24

== −

−

8. Assume CD is in the flat region of Figure 9.11 so that CD ~ 0.47.

9. Now calculate: smsmVt /4.1447.0

/87.9==

10. For this velocity calculate: 9931/1045.1

)01.0)(/4.14(Re 25 === − smxmsmDVt

ν

11. Now go back to Figure 9.11 and look up CD for this Reynolds number: CD ~ 0.41

12. Recalculate: smsmVt /4.1541.0

/87.9==

* Although hailstones are not perfectly spherical this is probably a good estimate

Problem 7 Given: Bicycle with frontal A = 3.9 ft2, and CD = 0.88, and peddling at 15 mph in still air Find: Extra power necessary to peddle against a 20 mph head wind Equations:

(1) AV

FCdragoft D

D2

21 ρ

≡=Coefficien

Solution:

1. Look up density of air in Table A.10: 3/00238.0 ftslug=ρ 2. Change units of bike velocity: Vbike = 15 mph (88ft/s)/(60mph) = 22 ft/s 3. Wind speed relative to bike: Vrel = (15+20)mph(88ft/s)/(60mph)=51.3 ft/s 4. Rearrange equation (1) into an equation for drag:

242232 1008.4)039)(/00238.0)(88.0)(5.0(5.0 relrelDD VxVftftslugAVCF −=== ρ

5. Get equation for power necessary to peddle bike:

223 0898.0)1008.4)(/22( relrelDbike VVxsftFVP === −

6. Calculate power for peddling in still air:

slbftsftVP rel /5.43)/22)(0898.0(0898.0 22 −===

7. Calculate power for peddling in 20mph headwind:

slbftsftVP relrel /236)/3.51)(0898.0(0898.0 22 −===

8. Additional power necessary: Prel –P = (236-43.5)(ft-lb/s)(1hp/550ft/lb/s) = 0.35 hp

Problem 8 Given: Power to overcome aerodynamic drag: P ~ Un

Find: n Equations: (1) Power = UFDrag

(2) AV

FCdragoftCoefficien D

D2

21 ρ

≡=

Solution:

1. Combining equations (1) and (2) for power: AUUCP D25.0 ρ=

2. For many vehicles the drag coefficient is essentially independent of Reynolds

number, thus CD is not a function of U so that . So n = 3. AUCP D35.0 ρ=

Problem 9 Given: Cork ball in water with: d = 0.3m, SG = 0.12, angle = 30o

U Find: Current speed, U

θ Equations: (1) For force balance for stationary object:

Tension = TW = mg

Buoyancy force = FB

FD ∑ ∑ ∑ === 000 zyx FFF

(2) AV

FCdragoftCoefficien D

D2

21 ρ

≡=

Solution:

1. Do some geometry and use equations (1) to find in the x-direction: FD = Tcos30o, and in the y-direction: FB = W + Tsin30o.

2. Calculate the buoyancy force and weight of the ball: FB = ρgVol = (9.80

kN/m3)(4π/3)(0.3/2 m)3 = 0.1385 kN, and W = SG FB = 0.21 (0.1385 kN) = 0.0291 kN

3. Now balancing forces in the y-direction and using equation (2): 0.1385 kN =

0.0291 kN + D tan30o , Or D = 0.189 kN, where D = CD ½ U2A = CD U2 (1/2) (999 kg/m3) (π(0.3m)2/4) = 35.3CDU2N, where U ~ m/s. HENCE: (eq. A) 5.3 CDU2 = 189 (in N) or CDU2 = 5.35

(eq. B) Re = UD/ν = 0.3m U/ 1.12x10-6m2/s =2.68 x 105 U

4. Now the strategy is to use trial and error to determine U. Assume CD, then

calculate U from Eq. (A) and Re from Eq. (B); check CD from Fig. 9.11, and

iterate until CD’s converge. See below.

Assume CD = 0.5 then from Eq. (A) U = 3.27 m/s and Re = 8.76 x 105, and from Fig 9.11 CD = 0.15 which does not equal 0.5 so try again Assume CD = 0.15 then from Eq. (A) U = 5.97 m/s and Re = 1.6 x 106, and from Fig. 9.11 CD = 0.20 which does not equal 0.15 so try again Assume CD = 0.19 then from Eq. (A) U = 5.31 m/s and Re = 1.42 x 106, and from Fig. 9.11 CD = 0.19 so U = 5.31 m/s

Problem 10 Given: Flow around UN building where: w = 87.5m, h = 154m, CD = 1.3, U = 20 m/s Find: Drag Assumptions: (1) Air is at STP Equations:

(1) AV

FCdragoftCoefficien D

D2

21 ρ

≡=

Solution:

1. Look up density of air at STP in Table A.10. 3/23.1 mkg=ρ

2. Rearrange equation (1) to find: AVCF DD25.0 ρ=

3. Plug in the given conditions and calculate the drag:

NxmmsmmkgFD623 1031.4)154)(5.87()/20)(/23.1)(3.1)(5.0( ==

Problem 11 Given: Flow around UN building where: w = 87.5m, h = 154m, CD = 1.3, U = 20 m/s at y = h/2 = 77m Find: Drag Assumptions: (1) Air is at STP Equations: (1) Velocity profile given: u = Cy0.4

(2) AV

FCdragoftCoefficien D

D2

21 ρ

≡=

(3) ∫== DD dFFdragtotal Solution:

1. Look up density of air at STP in Table A.10. 3/23.1 mkg=ρ

2. Rearrange equation (1) to find C, knowing u(h/2): C = 20/(770.4) = 3.52

3. Rearrange equation (2) to find: AVCF DD25.0 ρ=

4. Plug this into equation (3) to find: ∫ ∫= dxdyuCF DD25.0 ρ

5. Now plug in equation (1) and known values and integrate from the ground to the

top of the building.

NxmmkgF

ywCCdyuwCF

D

DDD

68.1

23

154

0

8.12

154

0

2

1017.48.1

154)52.3)(5.87)(/23.1)(3.1)(5.0(

8.15.05.0

==

⎥⎦

⎤⎢⎣

⎡== ∫ ρρ

* Drag is just less than with a uniform flow profile

Problem 12 Given: Nuclear submarine cruising submerged

L = 107m

w = pi*D = 34.6m xy

V = 27 knots δ Find: a) Estimate percentage of hull length with laminar BL b) Calculate drag due to skin friction c) Estimate power consumed Assumptions: (1) Can treat hull as a flat plate with same wetted area (2) Neglect laminar boundary layer (if small percent of hull length) Equations: Boundary Layer:

000,500Re =transition

∫∫

∫∫

⎟⎠⎞

⎜⎝⎛ −≈⎟

⎠⎞

⎜⎝⎛ −==

⎟⎠⎞

⎜⎝⎛ −≈⎟

⎠⎞

⎜⎝⎛ −==

∞

∞

δ

δ

θ

δ

00

00

*

11

11

dyUu

Uudy

Uu

Uuthicknessmomentum

dyUudy

Uuthicknessntdisplaceme

Flow over flat plate parallel to the flow:

∫==surfaceplate wD dAFdragfriction τ

Turbulent Boundary Layer: 5 x 105 < ReL < 107 :

51

Re

0742.0

L

DC =

ReL < 109 :

58.2)Re(log455.0

LDC =

Lift and Drag:

AV

FCdragoftCoefficien D

D2

21 ρ

≡=

AV

FCliftoftCoefficien LL

2

21 ρ

≡=

Solution: a) 1. Look up value of viscosity of seawater in Table A.2. 2. Calculate ReL. 3. Calculate the ratio Rext / ReL which will give us the ratio xt / L. 4. Multiply this ratio by 100% to get the percent of hull length with BL (it is small). b) 1. Find CD using equation for turbulent boundary layer. 4. Plug CD into equation to find FD. c) 1. Power = FD V, so calculate the power.

Fox Solution:

Problem 13 Given: Paddle wheel is immersed in a river current. Find: a) Expression for force produced by the wheel b) Expression for torque produced by the wheel c) Expression for power produced by the wheel d) Find the optimum angular speed, ω Assumptions: (1) Neglect air resistance since waterair ρρ << (2) Use the velocity relative to the paddle (3) Assume the equivalent of one paddle is constantly immersed Equations: Lift and Drag:

AV

FCdragoftCoefficien DD

2

21 ρ

≡= AV

FCliftoftCoefficien LL

2

21 ρ

≡=

Other useful equations: Torque = FDR Power = Tω Solution: a)

1. Calculate the relative velocity, Vrel = V – U = V - Rω . 2. Plug into equation for drag force and find the expression. b)

1. Now plug this FD into the equation for torque and find the expression.

c)

1. Now plug this T into the equation for power and find the expression. d)

1. Differentiate P with respect to U, where U = Rω . 2. Set this equal to zero to find the minima and maxima. 3. Plug back in Rω = U and find the relationship ω = f (V,R).

Fox Solution:

Problem 14

Problem 15