A sprayer is operating at 6 kph, covering 30 hectares in 6 hours, 20% of the time was spent on headland turning and tank filling. a. What is the effective swat boom of the sprayer? b. If the area is increased to 40 hectares, what should be the operating speed to finish spraying in 6 hours with 20% time loss? c. If the boom is 10 meters wide and delivers 25 liters/hour, what is the application rate in li/ha in (a) & (b)?
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A sprayer is operating at 6 kph, covering 30 hectares in 6 hours, 20% of the time was spent on headland turning and tank filling. a. What is the effective swat boom of the sprayer? b. If the area is increased to 40 hectares, what should be the operating speed to finish spraying in 6 hours with 20% time loss? c. If the boom is 10 meters wide and delivers 25 liters/hour, what is the application rate in li/ha in (a) & (b)?
Explanation:
A, Total time taken = 6 hr time lost in filling = 20 % of 6 hrs = ( 20 / 100 ) X 6 = 1.2 hr Time available for spraying t = 6 - 1.2 = 4.8 hrs = 17280 s Area covered = 30…
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