Chapter 11 The coeﬃcient of drag is not inﬂuenced by the car

11.33: PROBLEM DEFINITION

Situation:

A cartop carrier is used on an automobile.

Vc=100km/h=27.8m/s,

Vw=25km/h=6.94 m/s.

Find: Additional power required due to the carrier.

Assumptions:

Density of air is ρ=1.2kg/m3.

The coeﬃcient of drag is not inﬂuenced by the car.

CDis best approximated as a rectangular plate.

PLAN

1. Find CD.

2. Find the force of drag.

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

SOLUTION

1. Coeﬃcient of drag. The aspect ratio is

From Table 11.1 (EFM 10e)

2. Drag force equation.

3. Power equation.

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11.34: PROBLEM DEFINITION

Situation: The problem statement describes motion of an automobile.

Find: Percentage savings in gas mileage when travelling at 55 mph instead of 65 mph.

SOLUTION

Work is force times distance. Thus, energy is

The energy, E,perunitdistanceissimplytheforceor

Substituting drag force

Then energy savings are

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11.35: PROBLEM DEFINITION

Situation:

Acar(W= 2000 lbf) coasting down a hill (Slope =6%)has reached steady speed.

μrolling =μ=0.01

CD=0.29 AP=18ft

2

ρair =ρ=0.002 slug/ft3

Find: Maximum coasting speed.

SOLUTION

Slope of a hill is rise over run, so the angle of the hill is

where FD=drag force, Fr=rolling friction and W=weight of car.

Insert expressions for drag force and rolling friction.

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11.36: PROBLEM DEFINITION

Situation: The problem statement describes a car being driven up a hill

Find: Power required.

SOLUTION

The power required is the product of the forces acting on the automobile in the

direction of travel and the speed. The drag force is

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11.37: PROBLEM DEFINITION

Situation: A bicyclist is coasting down a hill–additional details are provided in the

problem statement

Find: Speed of the bicycle.

SOLUTION

Equilibrium (direction parallel to motion of the bicyclist)

Then

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11.38: PROBLEM DEFINITION

Situation:

A bicyclist is traveling into a 3 m/s head wind.

P=275W,A

p=0.5m

3,C

D=0.3.

Find: Speed of the bicyclist.

Assumptions:

Neglect rolling resistance.

All power is used to overcome drag. (neglect power losses in the chain and gears)

Properties:Air.ρ=1.2kg/m3.

PLAN

1. Relate the cyclist’s speed Vcto the drag force.

SOLUTION

1. Drag equation (use velocity of air relative to the cyclist).

2. Power equation.

3. Solving the cubic equation (we used a computer program) for speed gives

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11.39: PROBLEM DEFINITION

Situation: The problem statement describes a 1932 Fiat Balillo that is “souped up”

by the addition of a 220-bhp engine.

Find: Maximum speed of a “souped up” Balillo.

SOLUTION

From Table 11.2 (EFM 10e), CD=0.60.

“Souped up” version:

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11.40: PROBLEM DEFINITION

Situation:

To reduce drag, vanes are added to a truck.

CD(no vanes) = 0.78.

Vanes reduce drag by 25%.

Ap=8.36 m2,V=100km/h=27.8m/s.

Find: Reduction in drag force due to the vanes.

Properties:Air(20 ◦C,1atm), Table A.3 (EFM 10e), ρ=1.2kg/m3.

SOLUTION

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11.41: PROBLEM DEFINITION

Situation:

To reduce drag, vanes are added to a truck.

CD(no vanes) = 0.78.

Vanes reduce drag by 25%.

Ap=8.36 m2,V=100km/h=27.8m/s.

Total resistance is given by R=FD+Cwhere C=350Naccounts for bearing

friction.

Find: Percentage savings in fuel.

Assumptions: Fuel savings are directly proportional to power savings

Properties:Air(20 ◦C,1atm), Table A.3 (EFM 10e), ρ=1.2kg/m3.

SOLUTION

1. Power (no vanes).

2. Power (vanes are installed)

3. Fuel savings.

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Problem 11.42

Situation: An engineer is designing an object to fall at a speed of 1 m/s in seawater.

Find: Identify the variables that have the most inﬂuence on terminal velocity.

PLAN

1. Derive the governing equation by applying force equilibrium.

2. Use the governing equation to identify which variables inﬂuence terminal velocity.

SOLUTION

1. Force equilibrium.

2. Variables that inﬂuence terminal velocity:

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11.43: PROBLEM DEFINITION

Situation:

Aparachute(D=0.35 m) is falling through the air.

m=0.02 kg,C

D=2.2,A

p=πD2/4.

Find: Terminal velocity (in m/s).

Assumptions: Neglect mass of the chute.

Properties:Air(20 ◦C,1atm), Table A.3 (EFM 10e), ρ=1.2kg/m3.

PLAN

Develop an equation for terminal velocity by applying force equilibrium.

SOLUTION

1. Force equilibrium.

2. Drag force equation.

3. Combine Eq. (1) and (2).

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11.44: PROBLEM DEFINITION

Situation: A small air bubble is rising in a very tall column of liquid—additional details

are provided in the problem statement.

Find:

(a)Acceleration of the bubble.

(b)Form of the drag (mostly skin-friction or form).

SOLUTION

Equating the drag force and the buoyancy force.

As the bubble rises it will expand because the pressure decreases with an increase

REVIEW

As a matter of interest, the surface tension associated with contaminated ﬂuids creates

a condition which acts like a solid surface.

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11.45: PROBLEM DEFINITION

Situation:

Aball(D=0.08 m) falls through water.

Weight of the ball in air is 15 N.

Find: Terminal velocity (in m/s) of the ball.

Properties:Water(10 ◦C), Table A.5 (EFM 10e), ρ=1000kg/m3,γ=9810N/m3,ν=

1.31 ×10−6m2/s.

PLAN

1. Determine if the ball is falling or rising by comparing the buoyant force and the

weight.

2. Relate terminal velocity to weight and buoyancy.

3. Develop equations for CDand Re .

4. Solve the resulting set of equations using a computer program.

SOLUTION

1.Comparebuoyantforceandweight

2. Equilibrium

Drag force equation.

3. Clift and Gauvin correlation (drag on a sphere)

4. Solve Eqs. (3), (4) and (5) simultaneously (we applied TK Solver).

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11.46: PROBLEM DEFINITION

Situation: A weighted cube falls through water (see the problem statement for all the

details).

Find: Terminal velocity in water.

Assumptions: Density of water: ρ=1000kg/m3.

SOLUTION

From Table 11.1 (EFM 10e), CD=0.81.The drag force is

Equilibrium

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11.47: PROBLEM DEFINITION

Situation:

A spherical rock fall in water.

W=30N.Weight in water is 5N.

Find: Terminal velocity (in m/s) of the rock.

Properties:Water(20 ◦C), Table A.5 (EFM 10e), ρ=998kg/m3,γ =9790N/m3,ν=

1.00 ×10−6m2/s.

PLAN

1. Find diameter by using known buoyant force.

3. Solve the resulting equation using an iterative approach.

SOLUTION

1. Buoyant force equation

Now solve for diameter

2. Drag force equation

Equilibrium

Combine Eqs. (1) and (2)

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3. Solve Eq. (3). Begin by rewriting the equation

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11.48: PROBLEM DEFINITION

Situation:

A helium–ﬁlled balloon is ascending in air.

Standard atmosphere.

WBalloon(empty) = WB=3N.

Find: Terminal velocity (in m/s) of balloon.

Properties:Air(15 ◦C), Table A.2 (EFM 10e), ρ=1.22 kg/m3,ν=1.46×10−5m2/s.

PLAN

Thegoal(velocity)appearsinthedragforceequationThus,ﬁnd the drag force using

equilibrium and then solve for the velocity. This requires an iterative or a numerical

approach. The solution below is an iterative approach. The steps are

1. Relate the drag force to other forces using equilibrium.

2. Develop an algebraic equation for terminal velocity (Vo)using the drag force

equation.

SOLUTION

1. Force equilibrium (vertical direction).

2. Drag force equation (to ﬁnd equation for terminal velocity).

4. Reynolds number

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Reynolds number

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11.49: PROBLEM DEFINITION

Situation: A sphere 2 cm in diameter rises in oil at a velocity of 1.5 cm/s.

Find: Speciﬁc weight of the sphere material.

SOLUTION

Equilibrium

Reynolds number

Then from Fig. 11.8

Substitute drag force, weight and buoyancy force into Eq. (1)

Eq. (2) becomes

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