Respuesta :
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, [tex]c_d[/tex], is given by the formula;
[tex]c_d = \dfrac{2 \cdot F_d}{\rho \cdot u^2 \cdot A}[/tex]
Where;
[tex]c_d[/tex] = 0.28
[tex]F_d[/tex] = 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
[tex]F_d = \dfrac{c_d \cdot \rho \cdot u^2 \cdot A }{2}[/tex]
∴ [tex]F_d[/tex] = (0.28 × 1.06168037 × (58.1152)² × 1.83948)/2 ≈ 923.4 N
We have;
[tex]F_d = \dfrac{M_e \cdot \varepsilon _0 \cdot \eta _d }{r}[/tex]
Where;
[tex]M_e[/tex] = The engine torque
ε₀ = The overall gear reduction ratio = 2.5
[tex]\eta _d[/tex] = The drivetrain efficiency = 0.88
r = The wheel radius = 12.6 inches = 0.32004 meters
[tex]\therefore M_e = \dfrac{F_d \cdot r }{ \varepsilon _0 \cdot \eta _d}[/tex]
[tex]\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[/tex]
The engine torque = [tex]M_e[/tex] ≈ 134.33 N·m
The engine torque ≈ 134.33 N·m
2) The speed of the engine, [tex]n_e[/tex], is obtained from the following formula;
[tex]v = \dfrac{2 \cdot \pi \cdot r \cdot n_e \cdot (1 - i)}{\varepsilon _0}[/tex]
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
[tex]\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[/tex]
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.
