Question

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)?

Answer 1

Answer:

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, $c_d$, is given by the formula;

$c_d = \dfrac{2 \cdot F_d}{\rho \cdot u^2 \cdot A}$

Where;

$c_d$ = 0.28

$F_d$ = 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

$F_d = \dfrac{c_d \cdot \rho \cdot u^2 \cdot A }{2}$

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

We have;

$F_d = \dfrac{M_e \cdot \varepsilon _0 \cdot \eta _d }{r}$

Where;

$M_e$ = The engine torque

ε₀ = The overall gear reduction ratio = 2.5

$\eta _d$ = The drivetrain efficiency = 0.88

r = The wheel radius = 12.6 inches = 0.32004 meters

$\therefore M_e = \dfrac{F_d \cdot r }{ \varepsilon _0 \cdot \eta _d}$

$\therefore M_e = \dfrac{F_d \cdot r }{ \varepsilon _0 \cdot \eta _d} \approx \dfrac{923.4 \times 0.32004 }{ 2.5 \times 0.88} \approx 134.33 \ N\cdot m$

The engine torque = $M_e$ ≈ 134.33 N·m

The engine torque ≈ 134.33 N·m

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

$v = \dfrac{2 \cdot \pi \cdot r \cdot n_e \cdot (1 - i)}{\varepsilon _0}$

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

$\therefore n_e = \dfrac{v \times \varepsilon _0 }{2 \times \pi \times r \times (1 - i)} = \dfrac{58.1152 \times2.5 }{2 \times \pi \times 0.32004 \times (1 - 0.03)} \approx 74.486 \ rev /second$

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.