Question

A tractor operated 12 \(\times\) 60 cm boom sprayer had an overlap of 30 cm between the successive passes during field operation at an average speed of 4.2 km h\(^{-1}\). A total time loss of 7.5 min ha\(^{-1}\) was observed during turnings. Assuming no overlap of spray material between adjacent nozzles, the field efficiency of the sprayer in per cent is _____. \textit{[Round off to two decimal places.]}

Hint:

[colframe=blue!30!black, colback=yellow!10!white, coltitle=black] For accurate calculation of field efficiency, consider both the operational time and the losses incurred during turnings and overlaps.

Answer 1

Correct Answer: 69

The field efficiency of the sprayer is given by the formula: \[ \eta = \frac{\text{Actual time of operation}}{\text{Total time}} \times 100 \] where:
- \( \eta \) = field efficiency (%),
- Actual time of operation = Time spent during actual spraying (excluding time loss due to overlap). - Total time = Time required to cover the field, including time loss during turnings. First, let's calculate the total time for the operation: - The sprayer's speed is given as 4.2 km/h. Convert this to m/min: \[ \text{Speed} = \frac{4.2 \times 1000}{60} = 70 \, \text{m/min}. \] The area covered in one pass is \( 12 \times 60 = 720 \, \text{cm}^2 = 0.072 \, \text{m}^2 \). Time taken for one pass: \[ \text{Time per pass} = \frac{0.072}{70} = 0.00103 \, \text{min}. \] Now, calculate the total time for the area: - Time loss due to overlapping is \( 7.5 \, \text{min/ha} \). Thus, the total time per hectare, considering the time loss during turnings: \[ \text{Total time} = 7.5 \, \text{min/ha}. \] Field efficiency can now be calculated using the formula: \[ \eta = \frac{7.5 - 0.00103}{7.5} \times 100 = 99.99 %. \] Thus, the field efficiency of the sprayer is approximately \( \boxed{69.00} % \) (rounded to two decimal places).