Question

Calculate the torque developed, if a tractor engine is operating at a speed of 3500 rpm and brake power recorded by the prony brake dynamometer is 44 kW.

• A. 120 N-m
• B. 220 N-m
• C. 360 N-m
• D. 440 N-m

Hint:

Remember the formula:
Power (W) = Torque (N-m) \(\times\) Angular velocity (rad/s).
Convert rpm to rad/s: \(\omega\) = \(2\pi N / 60\).
Always check the units.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find the torque developed by the engine.
We have the speed and brake power.

Step 2: Key Formula or Approach:
Power = Torque \(\times\) Angular velocity.
Angular velocity \(\omega\) = \(2\pi N / 60\).
So, Torque = Power \(\omega\).

Step 3: Detailed Explanation:
Given: Power = 44 kW = 44000 W.
N = 3500 rpm.
\(\omega\) = \(2\pi \times 3500 / 60\) = 366.52 rad/s.
Torque = 44000 366.52 = 120.04 N-m.

Step 4: Final Answer:
The torque developed is 120 N-m.
Hence, the correct option is (A).