Chapter 11 The rock can be idealized as a cube of size

11.19: PROBLEM DEFINITION

Situation:

A large rock is located on the bottom of a river.

The current is strong enough so the rock moves along the bottom of the river.

Find: The speed (m/s) of the current required for the rock to roll.

Assumptions:

•The rock tumbles along the bottom (rationale: rocks will tumble rather than

slide).

•The rock can be idealized as a cube of size L×L×L,whereL=0.3m.(rationale:

large rocks are block like; the selected size is typical).

Properties:Water(10 ◦C), Table A.5 (EFM 10e), ρ= 1000 kg/m3.

PLAN

1. Relate the drag force to speed of the current using the drag force equation.

2. To determine when the rock will tip, sum moments about point O.

Drag Force

Weight

Po i n t “ O”

3. Find the speed of the current by combining results from steps 1 and 2.

SOLUTION

1. Drag force:

2=CDL2ρV 2

2(1)

2. Moment equilibrium (about 0).

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3. Combine Eqs. (1) and (2):

REVIEW

•This speed is about 9 mph.

•The major variable is the size of the rock (L). Larger rocks require faster rivers.

Notice that the speed needed to transport a rock is proportional to the square

root of a typical dimension. V∼√L.

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11.20: PROBLEM DEFINITION

Situation:

Wind Vo=35m/sacts on a tall smokestack.

Height is h=75m.Diameter is D=2.5m.

Find: Overturning moment at the base.

Assumptions: Neglect end eﬀects—that is the coeﬃcient of drag from a cylinder of

inﬁnite length is applicable.

Properties:Air(20 ◦C) ,Table A.3 (EFM 10e): ρ=1.2×99/101.3=1.17 kg/m3,

ν=1.51 ×10−5m2/s.

SOLUTION

Reynolds number

Drag force. From Fig. 11.5 (EFM 10e) CD≈0.62 so

Overturning moment

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11.21: PROBLEM DEFINITION

V=37.5m/s.

H=20m,D=0.08 m

Find: Moment at bottom of ﬂag pole.

Properties:Air(20 ◦C), Table A.3 (EFM 10e), ρ=1.2kg/m3,ν =15.1×10−6m2/s.

SOLUTION

1. Reynolds number

2. From Fig. 11.5 (EFM10e), CD=0.8.

3. Drag Force

4. Moment

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11.22: PROBLEM DEFINITION

Situation:

A man in a boat is pulling up an anchor made of concrete.

V=1m/s

D=0.3m,L=0.3m

Find: Tension in rope (in newtons).

Properties:

Water (10 ◦C), Table A.5 (EFM 10e), γ=9.81 kN/m3.

Concrete, γ=15kN/m3.

PLAN

Apply equilibrium. The forces are tension, weight, drag force and buoyancy force.

SOLUTION

Equilibrium

Solve for T

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11.23: PROBLEM DEFINITION

m=2.6×103kg,D=.038 m.

T=18◦C=291.2K.

s/m2.

Assumptions: Assume the ping-pong ball is stationary (stable equilibrium).

PLAN

Find density of the air jet. Then, to ﬁnd V, balance the weight of the ball with the

drag force. Solve the resulting set of equations using a computer program.

SOLUTION

Ideal gas law

Equilibrium

Drag force

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Cliﬀand Gauvin correlation (drag on a sphere)

Solve Eqs. (1) to (4) simultaneously. The computer program TKSolver was used for

our solution.

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11.24: PROBLEM DEFINITION

Situation:

Wind (V=35m/s)is acting on a sign post 3m×2m.

Center of the sign is 4m above the ground.

Find: Moment at ground level.

Assumptions: Neglect the drag force on the sign post.

Properties:Air(10 ◦C,1.0atm), Table A.3 (EFM 10e), ρ=1.25 kg/m3.

SOLUTION

1. Coeﬃcient of drag (from Table 11.1 in EFM 10e).

2. Drag force.

3. Moment at the base of the sign.

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11.25: PROBLEM DEFINITION

Situation: Wind is blowing normal to a boxcar.

Find: Speed of wind required to blow boxcar over.

Assumptions:T=10

◦C;ρ=1.25 kg/m3.

SOLUTION

Take moments about one wheel for impending tipping.

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11.26: PROBLEM DEFINITION

Situation:

Air speed is being determined in a popcorn popper.

Additional information is provided in problem statement.

Find: Range of airspeeds for popcorn popper operation.

Properties: Air, Table A.3 (EFM 10e), 150oC, ρ=0.83 kg/m3and ν=2.8×10−5

m2/s.

PLAN

This device should lift popped kernels while leaving unpopped kernels on the screen.

1. Find the maximum air speed by equating the drag force on an unpopped kernel

its weight.

2. Find the minimum air speed by equating the drag force on a popped kernel with

its weight.

SOLUTION

1. Before popcorn is popped, it should not be thrown out by the air. Thus, let

where FDis equal to the weight of an unpopped kernel.

Section area of a kernel

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Assume CDw0.4.Then

Check Reynolds number and CD:

2. For minimum velocity let popped corn be suspended by stream of air. Assume

only that diameter changes. So

where Dp=diameter of popped corn and Du=diameter of unpopped corn

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11.27: PROBLEM DEFINITION

Situation: Wind loads act on a ﬂag pole that is carrying an 6 ft high American Flag.

Find: Determine a diameter for the pole.

Assumptions: The failure mechanism is yielding due to static loading.

SOLUTION

An American ﬂag is 1.9 times as long as it is high. Thus

Assume

Compute drag force on ﬂag

Make the ﬂag pole of steel using one size for the top half and a larger size for the

bottom half. To start the determination of dfor the top half, assume that the pipe

diameter is 6 in. Then

With an Re of 4.7×105,C

Dmay be as low as 0.3 (Fig. 11.5 in EFM 10e); however,

for conservative design purposes, assume CD=1.0.Then

Assume that the allowable stress is 30,000 psi.

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Bottom half, Assume bottom pipe will be 12 in. in diameter.

REVIEW

Many other designs are possible.

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11.28: PROBLEM DEFINITION

Situation:

Aspherical(D=1.5m) submarineismovingthroughseawater.

V=10knots =11.5mph =5.14 m/s.

Find: Power (in kW) to propel the sub.

Assumptions: All power is being used to overcome drag.

Properties:Seawater(10 ◦C), Table A.4 (EFM 10e), ρ=1026kg/m3,ν=1.4×

10−6m2/s.

PLAN

1. Find Re, then ﬁnd CDusing the Clift & Gauvin correlation.

2. Find the drag force.

3. Find power using P=FDV.

SOLUTION

1. Reynolds Number

Clift and Gauvin correlation (drag on a sphere):

2. Drag Force (use projected area as reference area).

3. Power equation.

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11.29: PROBLEM DEFINITION

Situation:

The problem statement describes a dirigible.

V0=30ft/s

d=80ft

Find:

Power required for dirigible.

SOLUTION

Reynolds number

Drag force

From Fig. 11.9 in EFM 10e (extrapolated) CD=0.05

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11.30: PROBLEM DEFINITION

Situation:

•A runner is competing in a 10 km race.

Find: Estimate the energy in joules and kcal (food calories) that the runner needs to

supply to overcome aerodynamic drag.

Properties: Density of air is 1.22 kg/m3.

Assumptions: Assume that the air is still—that is, there is no wind.

PLAN

Energy is related to power (P)and time (t)by E=Pt. Find power using the

product of speed and drag force (P=VF

Drag).Find time by using distance (d)and

speed (d=Vt).

SOLUTION

Find the time to run 10 km.

Drag force

Power

Energy

REVIEW

1. The drag force (7.72 N) is small, about 1.7lbf.

2. The power to overcome drag is small (31.9W).Based on one of the author’s

(DFE)experienceinsports,aﬁt runner might supply 180 W to run at a 6:30

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11.31: PROBLEM DEFINITION

Situation: A cylindrical rod is rotated about its midpoint–additional details are pro-

vided in the problem statement.

Find:

a) Derive an equation for the power to rotate rod.

b) Calculate the power.

SOLUTION

For an inﬁnitesimal element, dr, of the rod

0

but r0=L/2so

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Then for the given conditions:

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11.32: PROBLEM DEFINITION

Situation:

A truck carries a rectangular sign.

V=30m/s.

Find: Additional power required to carry the sign (as compared to not carrying the

sign).

Assumptions:

Density of air is ρ=1.2kg/m3.

The coeﬃcient of drag is not inﬂuenced by the airﬂow over the truck.

PLAN

1. Find CD.

2. Find the force of drag.

3. Calculate power as the product of drag force and speed.

SOLUTION

1. Drag coeﬃcient. From Table 11.1 (EFM 10e) for a rectangular plate with an

2. Drag force equation.

3. Power equation.

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