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:

57.1?

Step-by-step explanation:

Answer 2

The input torque of the gearbox is approximately 23,278.46 Nm.

To find,

Input torque of gearbox.

Given,

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

Solution,

The problem states that the automobile engine is developing 70 kW of power at 4000 rpm. In order to calculate the input torque of the gearbox, we need to know the speed of rotation of the engine in radians per second. We can convert the engine speed from rpm (revolutions per minute) to radians per second by multiplying it by 2π/60 since there are 2π radians in a full revolution and 60 seconds in a minute.

So, the angular velocity of the engine is:

Angular velocity = (4000 rpm) × (2π/60) ≈ 418.88 rad/s

Next, we can use the formula for power:

Power = Torque × Angular velocity

To rearrange this formula to solve for torque, we can divide both sides by angular velocity:

Torque = Power ÷ Angular velocity

Substituting the values we have, we get:

Torque = 70,000 ÷ 418.88 ≈ 166.82 Nm

However, this is the torque output of the engine. We need to find the input torque of the gearbox, which takes into account the gear ratio between the engine and the gearbox. The gear ratio in first gear is given as 3.9:1, which means that the output shaft of the engine rotates 3.9 times for every 1 rotation of the input shaft of the gearbox.

To calculate the input torque of the gearbox, we can use the formula:

Input torque = (Power ÷ Angular velocity) × Gear ratio

Substituting the values we have, we get:

Input torque = (70,000 ÷ 418.88) × 3.9 ≈ 23,278.46 Nm

Therefore, the input torque of the gearbox is approximately 23,278.46 Nm.

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