Question:
A tractor-mounted boom sprayer carries a horizontal row of 12 identical flat-fan nozzles (spray angle \(110^{\circ}\)). To achieve uniform coverage, the required lateral overlap between adjacent nozzle footprints at the target plane is 30% of a single nozzle footprint. Boom height is 0.60 m above the target plane.
The tractor is set to travel at a theoretical forward speed of 8 km/h. Wheel slip is 8% and the field efficiency is 75%. All other losses are neglected.
The total time (in minutes) required to spray a 25 ha field is ________. (Rounded off to the nearest integer)
A tractor-mounted boom sprayer carries a horizontal row of 12 identical flat-fan nozzles (spray angle \(110^{\circ}\)). To achieve uniform coverage, the required lateral overlap between adjacent nozzle footprints at the target plane is 30% of a single nozzle footprint. Boom height is 0.60 m above the target plane.
The tractor is set to travel at a theoretical forward speed of 8 km/h. Wheel slip is 8% and the field efficiency is 75%. All other losses are neglected.
The total time (in minutes) required to spray a 25 ha field is ________. (Rounded off to the nearest integer)
The tractor is set to travel at a theoretical forward speed of 8 km/h. Wheel slip is 8% and the field efficiency is 75%. All other losses are neglected.
The total time (in minutes) required to spray a 25 ha field is ________. (Rounded off to the nearest integer)
Show Hint
Find the footprint width from the spray angle and boom height, reduce it by the overlap to get nozzle spacing, then compute effective field capacity.
Updated On: Jul 16, 2026
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Verified By Collegedunia
Correct Answer: 189
Solution and Explanation
Step 1: Find the ground footprint width of one nozzle.
Each nozzle sprays a cone of half-angle \(55^{\circ}\) from a height of 0.60 m, so the footprint width is
\[ w = 2h\tan(55^{\circ}) = 2(0.60)(1.4281) = 1.714\ m \]
Step 2: Find the spacing between adjacent nozzles.
A 30% overlap of the footprint means each nozzle only needs to be placed at 70% of the footprint width from its neighbour.
Spacing \(s = 0.70 \times 1.714 = 1.200\) m.
Step 3: Find the total effective spray swath.
With 12 nozzles spaced 1.200 m apart, the boom covers a swath of
\[ W = 12 \times 1.200 = 14.40\ m \]
Step 4: Find the actual forward speed.
Wheel slip of 8% means the tractor covers 8% less ground than the theoretical speed suggests.
\(V_a = 8(1 - 0.08) = 7.36\) km/h.
Step 5: Find the effective field capacity.
Theoretical capacity \(= \dfrac{W V_a}{10} = \dfrac{14.40 \times 7.36}{10} = 10.60\) ha/h.
Applying the 75% field efficiency, effective field capacity \(= 10.60 \times 0.75 = 7.95\) ha/h.
Step 6: Find the total spraying time.
\[ t = \frac{25}{7.95} = 3.146\ h = 3.146 \times 60 = 188.8\ min \]
Final Answer:
The field is sprayed in about \[ \boxed{189\ minutes} \]
Each nozzle sprays a cone of half-angle \(55^{\circ}\) from a height of 0.60 m, so the footprint width is
\[ w = 2h\tan(55^{\circ}) = 2(0.60)(1.4281) = 1.714\ m \]
Step 2: Find the spacing between adjacent nozzles.
A 30% overlap of the footprint means each nozzle only needs to be placed at 70% of the footprint width from its neighbour.
Spacing \(s = 0.70 \times 1.714 = 1.200\) m.
Step 3: Find the total effective spray swath.
With 12 nozzles spaced 1.200 m apart, the boom covers a swath of
\[ W = 12 \times 1.200 = 14.40\ m \]
Step 4: Find the actual forward speed.
Wheel slip of 8% means the tractor covers 8% less ground than the theoretical speed suggests.
\(V_a = 8(1 - 0.08) = 7.36\) km/h.
Step 5: Find the effective field capacity.
Theoretical capacity \(= \dfrac{W V_a}{10} = \dfrac{14.40 \times 7.36}{10} = 10.60\) ha/h.
Applying the 75% field efficiency, effective field capacity \(= 10.60 \times 0.75 = 7.95\) ha/h.
Step 6: Find the total spraying time.
\[ t = \frac{25}{7.95} = 3.146\ h = 3.146 \times 60 = 188.8\ min \]
Final Answer:
The field is sprayed in about \[ \boxed{189\ minutes} \]
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