Traffic and Highway Engineering
Traffic and Highway Engineering
5th Edition
ISBN: 9781305156241
Author: Garber, Nicholas J.
Publisher: Cengage Learning
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Question
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Chapter 3, Problem 15P
To determine

(a)

The additional horse power required by a passenger vehicle on the curve.

Expert Solution
Check Mark

Answer to Problem 15P

  P=43.66hp

Explanation of Solution

Given information:

Speed =55mi/hr

Weight of car =2500lb

Cross sectional area of car =30ft2

Radius of the curve =850ft

Concept used:

The curve resistance is calculated when the road surface has a horizontal curve. Then the power is calculated using the curve resistance force.

  Rc=0.5(2.15u2WgR)

  P=1.47Rcu550

  RcCurve resistanceWWeight of the vehicleRRadius of the curvePPower of vehicle

Calculation:

  Rc=0.5( 2.15 u 2 W gR)=0.5( 2.15× 55 2 ×2500 32.2×850)=297.03lb

The power of the passenger vehicle is determined by

  P=1.47Rcu550=1.47×297.03×55550=43.66hp

Conclusion:

Therefore, the additional horsepower developed by a passenger vehicle on the curve is 43.66hp.

To determine

(b)

The total resistance force developed on the vehicle.

Expert Solution
Check Mark

Answer to Problem 15P

  R=430.43lb

Explanation of Solution

Given information:

Speed =55mi/hr

Weight of car =2500lb

Cross sectional area of car =30ft2

Radius of the curve =850ft

Concept used:

The total resistance is the sum of air resistance, rolling resistance and curve resistance of the passenger vehicle.

  Ra=0.5(2.15pCDAu2g)

  Rr=(Crs+2.15Crvu2)W

  Rc=0.5(2.15u2WgR)

  RaAir resistanceCDAerodynamic drag coefficientACross sectional areauSpeedpDensity of air=0.0766lb/ft3

  RrRolling resistanceCrsConstant for cars=0.012CrvConstant for cars=0.65×106sec2/ft2WWeight of vehicle

Calculation:

Take aerodynamic drag coefficient for passenger car as 0.4

  Ra=0.5( 2.15p C D A u 2 g)=0.5( 2.15×0.0766×0.4×30× 55 2 32.2)=92.83lb

Rolling resistance force is calculated as

  Rr=(C rs+2.15C rvu2)W=(0.012+2.15×0.65× 10 6× 552)(2500)=40.57lb

Curve resistance force is calculated as

  Rc=0.5( 2.15 u 2 W gR)=0.5( 2.15× 55 2 ×2500 32.2×850)=297.03lb

The total resistance force is calculated as

  R=Ra+Rr+Rc=92.83+40.57+297.03=430.43lb

Conclusion:

The total resistance force on the vehicleis 430.43lb.

To determine

(c)

The total horse power developed by the vehicle.

Expert Solution
Check Mark

Answer to Problem 15P

  Ptotal=63.27hp

Explanation of Solution

Given information:

Speed =55mi/hr

Weight of car =2500lb

Cross sectional area of car =30ft2

Radius of the curve =850ft

Concept used:

The total resistance is the sum of air resistance, rolling resistance and curve resistance of the passenger car. Using the total resistance the power of the passenger vehicle is calculated. The power is calculated individually for the straight road surface and for the curve surface. Finally the total power is determined by adding both the powers.

  Ra=0.5(2.15pCDAu2g)

  Rr=(Crs+2.15Crvu2)W

  Rc=0.5(2.15u2WgR)

  P=1.47Rcu550

  RaAir resistanceCDAerodynamic drag coefficientACross sectional areauSpeedpDensity of air=0.0766lb/ft3

  RrRolling resistanceCrsConstant for cars=0.012CrvConstant for cars=0.65×106sec2/ft2WWeight of vehiclePPower of vehicle

Calculation:

Take aerodynamic drag coefficient for passenger car as 0.4.

  Ra=0.5( 2.15p C D A u 2 g)=0.5( 2.15×0.0766×0.4×30× 55 2 32.2)=92.83lb

Rolling resistance force is calculated as

  Rr=(C rs+2.15C rvu2)W=(0.012+2.15×0.65× 10 6× 552)(2500)=40.57lb

Curve resistance force is calculated as

  Rc=0.5( 2.15 u 2 W gR)=0.5( 2.15× 55 2 ×2500 32.2×850)=297.03lb

The total horse power is calculated as

  Pstraight=1.47( R a + R r )u550=1.47×( 92.83+40.57)×55550=19.61hp

  Pcurve=1.47Rcu550=1.47×297.03×55550=43.66hp

  Ptotal=Pstraight+Pcurve=19.61+43.66=63.27hp

Conclusion:

The total horsepower is 63.27hp.

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Traffic and Highway Engineering
Civil Engineering
ISBN:9781305156241
Author:Garber, Nicholas J.
Publisher:Cengage Learning