Question

The automobile engine is developing 70 KW power at 4000 rpm. If the gear

ratio in Ist gear is 3.9: 1, determine input torque of gear box?​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The input torque of the gearbox is 167.22Nm.

Explanation:

Power in terms of torque and angular frequency is given as,

$P=\tau\times \omega$         (1)

Where,

P=power of the automobile engine

τ=torque produced by the automobile engine

ω=angular frequency with which the engine is rotating

From the question we have,

P=70kW=70×10³W

Frequency(f)=4000 rpm

$\omega=2\pi f$       (2)

So,

$\omega=2\pi \times \frac{4000}{60}=418.6rad/s$      (3)

By substituting the values in equation (1) we get;

$70\times 10^3=\tau\times 418.6$

$\tau=\frac{70\times 10^3}{418.6}$

$\tau=167.22Nm$

Hence, the input torque of the gearbox is 167.22Nm.