Question

A mouldboard plough is operated by a 2WD tractor with 6.0 kN pull. The ratio of lateral component to the longitudinal component of soil forces is 0.30, and the ratio of the vertical component to the longitudinal component of soil forces is 0.50. If the diameter of the driving wheels is 1.2 m and the rear axle rotates at 10 rpm, the drawbar power produced at 20% wheel slip is _________ kW. (Rounded off to 2 decimal places)

Hint:

When calculating the drawbar power, ensure to adjust for the wheel slip by multiplying by \( (1 - {Slip}) \) to find the effective power.

Answer 1

Step 1: Calculate the effective radius of the wheel
The diameter of the driving wheels is given as 1.2 m, so the radius \(r\) is:

\[ r = \frac{1.2}{2} = 0.6 \, {m} \]

Step 2: Calculate the wheel speed at 10 rpm
The rear axle rotates at 10 rpm, so the wheel speed in meters per minute is:

\[ {Wheel speed} = 10 \times 2\pi \times r = 10 \times 2\pi \times 0.6 = 37.699 \, {m/min} \]

Step 3: Calculate the longitudinal speed of the tractor
At 20% wheel slip, the longitudinal speed is reduced by 20%. The effective speed of the tractor is:

\[ {Longitudinal speed} = 37.699 \times (1 - 0.20) = 37.699 \times 0.80 = 30.159 \, {m/min} \]

Step 4: Calculate the drawbar pull
The tractor produces a 6.0 kN pull. The longitudinal component of the soil forces is given by the total pull. Therefore, the drawbar power is:

\[ {Drawbar power} = {Pull} \times {Longitudinal speed} \] Substitute the values: \[ {Drawbar power} = 6.0 \, {kN} \times 30.159 \, {m/min} = 180.954 \, {kN.m/min} \] Convert from kN.m/min to kW by dividing by 60: \[ {Drawbar power} = \frac{180.954}{60} = 3.016 \, {kW} \]

Step 5: Adjust for wheel slip
Since the tractor operates at 20% wheel slip, the effective power produced by the tractor at the wheels is:

\[ {Adjusted drawbar power} = {Drawbar power} \times (1 - 0.20) = 3.016 \times 0.75 = 2.30 \, {kW} \]

Thus, the drawbar power produced at 20% wheel slip is 2.30 kW.