A two-wheel drive tractor having a total weight of 20 kN is pulling an implement at an actual speed of 3 km·h⁻¹. The ratio of axle power (kW) to dynamic weight (kN) on the rear axle is expressed as 1.79/V, where V is the theoretical forward speed in km·h⁻¹. During pulling the implement, the weight distribution on the front and rear axles of the tractor is 40 % and 60 % of the total weight, respectively, with wheel slip 12 % and coefficient of net traction 0.42. The tractive efficiency, in %, is __________. (Rounded off to 2 decimal places)
Official Answer Key:
66.00 to 67.00
Step-by-Step Solution & Explanation
Step 1 — theoretical forward speed. Actual speed = theoretical × (1 − slip):
V = 3/(1 − 0.12) = 3/0.88 = 3.409 km·h⁻¹
Step 2 — dynamic weight on the rear (driving) axle:
W_r = 0.60 × 20 = 12 kN
Step 3 — axle (input) power from the given relation:
Axle power = W_r × (1.79/V) = 12 × (1.79/3.409) = 12 × 0.5251 = 6.301 kW
Step 4 — drawbar (output) power. Net traction force = coefficient of net traction × dynamic rear-axle weight:
F_net = 0.42 × 12 = 5.04 kN
Actual speed = 3/3.6 = 0.8333 m·s⁻¹
Drawbar power = 5.04 × 0.8333 = 4.20 kW
Step 5 — tractive efficiency:
TE = drawbar power/axle power = 4.20/6.301 = 0.6666 = 66.66 %
SECTION 4: SOIL AND WATER CONSERVATION ENGINEERING
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