Subtopic: Selection, operation, maintenance and repair of I.C. engines
Question Statement:
A power tiller rotavator has an effective working width of $60\text{ cm}$. It operates at a forward speed of $2.4\text{ km/h}$. If the rotary tiller shaft absorbs $6.0\text{ kW}$ from the engine, calculate the specific work per unit tilled field area in $\text{kJ/m}^2$.
Answer Key:
15
Step-by-Step Solution & Explanation
Forward velocity $v = 2.4\text{ km/h} = \frac{2.4}{3.6} = 0.6667\text{ m/s}$.
Area tilled per second:
$a_s = \text{Width} \times v = 0.60\text{ m} \times 0.6667\text{ m/s} = 0.40\text{ m}^2/\text{s}$
Power absorbed $P = 6.0\text{ kW} = 6.0\text{ kJ/s}$.
Specific work:
$w_s = \frac{P}{a_s} = \frac{6.0\text{ kJ/s}}{0.40\text{ m}^2/\text{s}} = 15.0\text{ kJ/m}^2$