Vehicle performance assignment 2

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Vehicleperformance assignment 2

Arear wheel drive car of mass 2200kg has a rolling resistance of 150N, a frontal area of 1.9m2, a drag coefficient of 1.00, atransmission loss of 7% and a maximum engine power of 59kW. Assumingthat the traction limit for each tyre is equal to the weight that itcarries (i.e. the tyre traction coefficient =1) and that 60% of thevehicle weight is on the rear wheels during acceleration. Take thedensity of air to be 1.2 Kg/m3 and g = 9.8 m/s2Calculations

Mass=2200kg

RollingResistance=150N

Frontalarea=1.9 m2

DragCoefficient =1.00

Transmissionloss=7%

MaximumEngine power=59kW

Tyretraction coefficient =1

Weighton the rear wheel =60%

Densityof air 1.2kg/m3

g=9.8 m/s2

  1. What is the maximum power available at the back wheels?

Drivetrain efficiency Enginepower = (1-0.07)59 kW=54.87kW=54870 Watts

  1. What is its theoretical maximum speed in mph on a level road?

Rearwheels power= Aerodynamic power loss + Rolling resistance + powerinto hills

Rollingresistance power = Rolling resistance Force velocity = 150 N V

AerodynamicPower loss =Drag Velocity=

Poweron level ground = 0

54,870Nm/s = 150V + ()

54,870= 150 V + 1.14V3

Toquickly solve for V I will use iteration method:

35.1m/s…. 54,562.65 Nm/s

35.15m/s…. 54,781.12 Nm/s

35.16m/s …54,824 Nm/s

35.17m/s ….54,868.67 Nm/s

35.18m/s ….54,912.49 Nm/s

So78.67mph (35.17 m/s) is the fairly accurate top speed.

  1. What is its theoretical maximum speed in mph up a 1 in 4 hill?

54,870Nm/s= Rolling resistance + aerodynamic power loss + power into hills

Powerinto hills (1 in 8 hill) = mass g change in height per unit time

Powerinto hills (1 in 8 hill) =2200 9.8 (1/4) = 5,390

Substitutingin the above equation

54,870Nm/s = 150V + 1.14 + 5,390

Toquickly solve for V I will use iteration method:

33.6m/s …53,673.68 Nm/s

33.7m/s …54,075.93 Nm/s

33.8m/s …54,480.49 Nm/s

33.9m/s …54,887.37 Nm/s

So75.83 mph (33.9 m/s) is the approximate top speed up 1 in 8 hill.

  1. What is the maximum tractive force that each of the rear tyres could provide under acceleration before slipping occurs? (NB this is per rear wheel not per axle!!)

Tractiveforce = Coefficient of friction Downward force

Tractionforce = 1(0.6022009.8) = 12,936N

  1. What would be the maximum, useable, tractive force at each of the rear tyres if full power was applied at 20 mph? (Hint: is this limited by tyre grip or by engine power?)

Useabletractive force will be the minimum of (force due to engine power,tire traction force)answer = min(54870Nm/s / 8.94m/s , 12,936N)answer = min(6,137.58 N, 12,936 N) = 6,137.58N

  1. What would be the wind resistance at 20 mph?

Windresistance =

=91.11N

  1. What would be the acceleration (in mph/s) at 20 mph on the level?

Force(F) = Tractive force – Aerodynamic Drag- Rolling resistance force

AerodynamicDrag = 91.11N

Tractiveforce =6,137.58 N

Rollingresistance force =150 N

6,137.58N – 91.11 N – 150N = 5,896.47 N

=5.995MPH/s

Whatwould be the acceleration (in mph/s) at 20 mph up a 1 in 4 hill?

Force(F) = Tractive force – Aerodynamic Drag- Rolling resistance-Gravity

Where

AerodynamicDrag = 91.11N

Tractiveforce =6,137.58 N

Rollingresistance force =150 N

Gravitationpull due to the hill = mass g (1/4) = 22009.8(1/4)= 5390N

6,137.58N – 91.11 N – 150N -5,390 N= 506.47 N

=0.2302=0.515MPH/s

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