2.7 A 3400-lb car is traveling in third gear (overall gear reduction ratio of 2.5 to 1) on a level road at its top speed of 130 mi/h. The air density is 0.00206 slugs/ft3 . The car has a frontal area of 19.8 ft2 , a drag coefficient of 0.28, a wheel radius of 12.6 inches, a drive axle slippage of 3%, and a drivetrain efficiency of 88%. At this vehicle speed, what torque is the engine producing and what is the engine speed (in revolutions per minute)?

1) The engine torque is approximately 134.33 N·m

2) The speed of the engine  is approximately 4,469.15 revolutions per minute

Step-by-step explanation:

1) The drag coefficient, , is given by the formula;

Where;

= 0.28

= The drag force

ρ = The fluid density = 0.00206 slugs/ft³ = 1.06168037 kg/m³

u = The object's flow speed = 130 mi/h = 58.1152 m/s

A = The frontal area = 19.8 ft² = 1.83948 m²

∴ = (0.28 × 1.06168037 × (58.1152)² × 1.83948)/2 ≈ 923.4 N

We have;

Where;

= The engine torque

ε₀ = The overall gear reduction ratio = 2.5

= The drivetrain efficiency = 0.88

r = The wheel radius = 12.6 inches = 0.32004 meters

The engine torque = ≈ 134.33 N·m

The engine torque ≈ 134.33 N·m

2) The speed of the engine, , is obtained from the following formula;

Where;

v = The vehicle's speed = 130 mi/h = 58.1152 m/s

r = The wheel radius = 12.6 inches = 0.32004 meters

i = The drive axle slippage = 3% = 3/100 = 0.03

ε₀ = The overall gear reduction ratio = 2.5

The speed of the engine in revolution per minute = 60 seconds/minute × 74.486 rev/second ≈ 4,469.15 revolutions per minute

The speed of the engine  ≈ 4,469.15 revolutions per minute.