Question

The tractor develops a torque of 35 kg-m at an engine speed of 1350 RPM. Calculate the BHP of the tractor.

• A. 45
• B. 66
• C. 90
• D. 77

Hint:

Use the torque-power-speed relation for a rotating shaft, keeping torque in kgf-m and speed in RPM, to directly get horsepower.

Answer 1

The Correct Option is B

Step 1: Use the standard relation for power developed by a rotating shaft in terms of torque in kgf-m and speed in RPM, giving power directly in metric horsepower:

\[\text{HP} = \frac{2\pi N T}{4500}\]

where \(N\) is the speed in RPM and \(T\) is the torque in kgf-m (4500 kgf-m/min is the metric horsepower constant).

Step 2: Substitute the given values, \(N = 1350\) RPM and \(T = 35\) kgf-m.

\[\text{HP} = \frac{2\pi \times 1350 \times 35}{4500}\]

Step 3: Work out the numerator first: \(2\pi \times 1350 = 8482.3\), and \(8482.3 \times 35 = 296880.5\).

Step 4: Divide by 4500:

\[\text{HP} = \frac{296880.5}{4500} \approx 65.97 \approx 66\]

Step 5: So the tractor develops about 66 BHP. Option 1 (45) and option 3 (90) do not correspond to this torque-speed combination through the correct formula, and option 4 (77) would only appear if the torque or speed were used incorrectly. The value consistent with the given torque and speed is 66.