Question

A 2WD tractor's driving axle experiences total dynamic normal load of 28 kN. The driving wheels have 0.60 m rolling radius. The total driving axle torque measured is 10.7 kN.m. At 18% wheel slip, the coefficient of net traction is 0.35. Tractive efficiency of the driving wheels, in %, is nearest to

• A. 45
• B. 35
• C. 55
• D. 58

Hint:

Combine the rim-pull to axle-force ratio with the slip factor to get tractive efficiency.

Answer 1

The Correct Option is A

Step 1: Find the rim pull produced by the axle torque.
Torque and rolling radius give the driving force at the rim: \( F = \dfrac{T}{r} = \dfrac{10.7}{0.60} = 17.83\ \text{kN} \).

Step 2: Find the net drawbar pull from the traction coefficient.
The coefficient of net traction relates net pull to dynamic weight: \( P = C_T \times W = 0.35 \times 28 = 9.8\ \text{kN} \).

Step 3: Relate axle power and drawbar power through velocities.
Tractive efficiency is drawbar power over axle power, \( \eta_t = \dfrac{P \times V_a}{F \times V_t} \), and slip links the two speeds by \( V_a = V_t(1-s) \), so \( \eta_t = \dfrac{P}{F}(1-s) \).

Step 4: Substitute the numbers.
\( \eta_t = \dfrac{9.8}{17.83} \times (1 - 0.18) = 0.5497 \times 0.82 = 0.4507 \)

Final Answer:
Tractive efficiency comes out close to 45%. \[ \boxed{\eta_t \approx 45\%} \]