Homework Solution 3

Problem 1
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
Find: Velocity of truck for both drag situations
Equations:
For force balance analysis:
F x = max
Fy
= ma y Fz
maz
∑ ∑ ∑ =
Lift and Drag:
Coefficient
Coefficient
Solution:
of
drag = C
D
FD
≡
1
ρV
2
FL
of lift = CL
≡
1
ρV
2
2
A
2
A
V
Drag
100
F
7
Weight
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:
12ft
θ

Problem 2
Given: Box kite blowing in wind
U = 20ft/s
Vwind = 20 ft/s
W = 0.9 lb θ
Tension in string = 3.0 lb
Angle to ground = 30 o
Frontal area = 6.0 ft 2
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?
Assumptions:
Equations:
For force balance analysis:
F x = max
Fy
= ma y Fz
maz
∑ ∑ ∑ =
Lift and Drag:
Coefficient
Coefficient
Solution:
a)
of
drag = C
D
FD
≡
1
ρV
2
FL
of lift = CL
≡
1
ρV
2
1. Find the density of air from Table A.9.
2
A
2
A
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 .
y
x
T
L
W
D

Numbers:

Problem 3
Given: Old airplane with wires used to strengthen design, with: U = 70mph, Awing =
148 ft 2 , 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:
FD
Coefficient
of drag = CD
≡
1 2
ρV
A
2
FL
Coefficient
of lift = CL
≡
1 2
ρV
A
2
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:
FD
Coefficient
of drag = CD
≡
1 2
ρV
A
2
FL
Coefficient
of lift = CL
≡
1 2
ρV
A
2
Solution:
1. Look up the kinematic viscosity and density of air at 15 o C in Table A.10.
−5
2
3
ν = 1. 45x
10 m / s,
ρ = 1.
23kg
/ m
2. Calculate the Reynolds number of the wire braces.
km 1000m
1hr
1m
150
6mm
UD
=
hr 1km
3600s
1000mm
= 17,
241
5
ν
1.
45x10
m / s
Re = − 2
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.
FD
CD
≡
1 2
ρV
A
2
FD
1.
2 =
3
2
( 0.
5)(
1.
23kg
/ m )( 150km
/ hr * 1000m
/ km * 1hr
/ 3600s)
( 0.
006m)(
60m)
F
D
=
461.
25
N
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:
FD
Coefficient
of drag = CD
≡
1 2
ρV
A
2
FL
Coefficient
of lift = CL
≡
1 2
ρV
A
2
Solution:
1. Look up the kinematic viscosity and density of air at 15 o C in Table A.10.
−5
2
3
ν = 1. 45x
10 m / s,
ρ = 1.
23kg
/ m
2. Calculate the Reynolds number of the wire braces.
km 1000m
1hr
1m
150
6mm
UD
=
hr 1km
3600s
1000mm
= 17,
241
5
ν
1.
45x10
m / s
Re = − 2
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.
FD
CD
≡
1 2
ρV
A
2
FD
0.
06 =
3
2
( 0.
5)(
1.
23kg
/ m )( 150km
/ hr * 1000m
/ km * 1hr
/ 3600s)
( 0.
006m)(
60m)
F
D
= 23N
5. Now calculate power necessary, knowing P = FV = (23 N)(41.7m/s) = 960 W.
6. Comparing this to the power in problem 4:
P − Pstreamline
19,
200W
− 960W
x100
% =
x100%
= 95%
less power necessary!!!
P
19,
200W

Problem 6
Given: Spherical hailstones, D = 10mm, falling through air at STP
Find: Terminal velocity
Equations:
(1)
Coefficien t of
Solution:
drag = C
D
FD
≡
1
ρV
2
2
A
−5
2
1. Look up properties for air in Table A.10: ν = 1. 45x
10 m / s,
ρ = 1.
23kg
/ m
2. Look up the specific gravity of ice in Table A.1 and calculate the density of ice.
= 0. 917 → ρ = SG ρ 2
3
3
= ( 0.
917)(
1000kg
/ m ) = 917kg
/ m
SGice ice ice H 0
3. Since the sum of the forces in y is zero when terminal velocity is reached, sum
forces:
∑
Fy = 0 = mg − FD
4. Rearrange equation (1) for the drag force and plug it into the force balance
equation just found to find Vt.
mg = C
D
( 0.
5)
ρV
2
t
A
→
V
t
=
2mg
C ρA
3
3 π ( 0.
01m)
−4
5. Calculate the mass of the hail: m = ρ V = 917kg
/ m
= 4.
8x10
kg
6
6. Calculate the area:
7. Now calculate Vt(CD):
πD
A =
4
V
t
2
π 0.
01
=
4
D
2
= 7.
85x10
−5
m
−4
2
2(
4.
8x10
kg)(
9.
81m
/ s ) 9.
87m
/ s
=
3
5 2
C ( 1.
23kg
/ m )( 7.
85x10
m ) C
= −
8. Assume CD is in the flat region of Figure 9.11 so that CD ~ 0.47.
9.
87m
/ s
9. Now calculate: Vt = = 14.
4m
/ s
0.
47
Vt D ( 14.
4m
/ s)(
0.
01m)
10. For this velocity calculate: Re = =
= 9931
−5
2
ν 1.
45x10
m / s
11. Now go back to Figure 9.11 and look up CD for this Reynolds number: CD ~ 0.41
9.
87m
/ s
12. Recalculate: Vt = = 15.
4m
/ s
0.
41
* Although hailstones are not perfectly spherical this is probably a good estimate
2
D
D
3

Problem 7
Given: Bicycle with frontal A = 3.9 ft 2 , 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)
Coefficien t of
Solution:
drag = C
D
FD
≡
1
ρV
2
1. Look up density of air in Table A.10:
2
A
ρ =
0. 00238slug
/ ft
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:
2
3
2 2
−4
2
FD
= 0. 5C
D ρV
A = ( 0.
5)(
0.
88)(
0.
00238slug
/ ft )( 039 ft ) Vrel
= 4.
08x10
Vrel
5. Get equation for power necessary to peddle bike:
P = V F =
=
bike
D
−3
2
2
( 22 ft / s)(
4.
08x10
) Vrel
0.
0898Vrel
6. Calculate power for peddling in still air:
2
2
P = 0.
0898Vrel
= ( 0.
0898)(
22 ft / s)
= 43.
5 ft − lb / s
7. Calculate power for peddling in 20mph headwind:
2
Prel = 0.
0898Vrel
=
( 0.
0898)(
51.
3
ft / s)
2
= 236 ft − lb / s
8. Additional power necessary: Prel –P = (236-43.5)(ft-lb/s)(1hp/550ft/lb/s) = 0.35 hp
3

Problem 8
Given: Power to overcome aerodynamic drag: P ~ U n
Find: n
Equations:
(1) Power = UFDrag
(2)
Coefficient
Solution:
of
drag = C
D
FD
≡
1
ρV
2
1. Combining equations (1) and (2) for power:
2
A
= 0. 5UC
ρU
P D
2. For many vehicles the drag coefficient is essentially independent of Reynolds
3
number, thus CD is not a function of U so that =
0. 5C
ρU
A.
So n = 3.
P D
2
A

Problem 9
Given: Cork ball in water with: d = 0.3m, SG = 0.12, angle = 30 o
Find: Current speed, U
Equations:
(1) For force balance for stationary object:
F x = 0 Fy
= 0 Fz
0
(2)
∑ ∑ ∑ =
Coefficient
Solution:
of
drag = C
D
θ
FD
≡
1
ρV
2
2
A
U
FD
W = mg
Buoyancy force = FB
Tension = T
1. Do some geometry and use equations (1) to find in the x-direction: FD = Tcos30 o ,
and in the y-direction: FB = W + Tsin30 o .
2. Calculate the buoyancy force and weight of the ball: FB = ρgVol = (9.80
kN/m 3 )(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 tan30 o , Or D = 0.189 kN, where D = CD ½ U 2 A = CD U 2 (1/2)
(999 kg/m 3 ) (π(0.3m) 2 /4) = 35.3CDU 2 N, where U ~ m/s.
HENCE: (eq. A) 5.3 CDU 2 = 189 (in N) or CDU 2 = 5.35
(eq. B) Re = UD/ν = 0.3m U/ 1.12x10 -6 m 2 /s =2.68 x 10 5 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 10 5 , 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 10 6 , 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 10 6 , 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)
Coefficient
Solution:
of
drag = C
D
FD
≡
1
ρV
2
1. Look up density of air at STP in Table A.10.
2. Rearrange equation (1) to find:
2
A
= 0. 5C
ρV
FD D
3. Plug in the given conditions and calculate the drag:
F D
ρ = 1. 23kg
/ m
3
2
=
( 0.
5)(
1.
3)(
1.
23kg
/ m )( 20m
/ s)
( 87.
5m)(
154m)
= 4.
31x10
2
A
3
6
N

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 = Cy 0.4
(2)
(3)
Coefficient
of
drag = C
total drag = FD
= dFD
Solution:
∫
D
FD
≡
1
ρV
2
1. Look up density of air at STP in Table A.10.
2
A
ρ = 1. 23kg
/ m
2. Rearrange equation (1) to find C, knowing u(h/2): C = 20/(77 0.4 ) = 3.52
3. Rearrange equation (2) to find:
= 0. 5C
ρV
FD D
2
4. Plug this into equation (3) to find: FD = ∫∫ 0. 5C
D ρ u dxdy
5. Now plug in equation (1) and known values and integrate from the ground to the
top of the building.
F
F
D
D
2
= 0.
5C
ρw
∫ u dy = 0.
5C
ρwC
D
154
0
D
⎡ y
⎢
⎣ 1.
8
2
1.
8
3
2 154
= ( 0.
5)(
1.
3)(
1.
23kg
/ m )( 87.
5m)(
3.
52)
⎤
⎥
⎦
154
0
1.
8
* Drag is just less than with a uniform flow profile
2
1.
8
A
3
6
= 4.
17x10
N

Problem 12
Given: Nuclear submarine cruising submerged
V = 27 knots
y
x
L = 107m
δ
w = pi*D = 34.6m
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:
Re transition = 500,
000
displacement
*
thickness = δ =
momentum thickness = θ =
∞
∫
0
∞
∫
0
⎛ u
⎜1−
⎝ U
⎞
⎟dy
≈
⎠
u ⎛ u ⎞
⎜1−
⎟dy
≈
U ⎝ U ⎠
Flow over flat plate parallel to the flow:
friction drag = F = τ dA
∫
plate surface
Turbulent Boundary Layer:
5 x 10 5 < ReL < 10 7 :
0.
0742
C D =
Re
1
5
L
ReL < 10 9 :
D
w
δ
∫
0
δ
∫
0
⎛ u
⎜1−
⎝ U
⎞
⎟dy
⎠
u ⎛ u
⎜1−
U ⎝ U
⎞
⎟dy
⎠

C D =
0.
455
(log Re
L
Lift and Drag:
Coefficient
Coefficient
Solution:
a)
of
)
2.
58
drag = C
D
FD
≡
1
ρV
2
FL
of lift = CL
≡
1
ρV
2
1. Look up value of viscosity of seawater in Table A.2.
2. Calculate ReL.
2
A
3. Calculate the ratio Rext / ReL which will give us the ratio xt / L.
2
A
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 ρ air

Fox Solution:

Problem 14

Problem 15