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A field sprayer having a horizontal boom with 20 nozzles spaced 40 cm apart is to be designed for maximum application rate of 650 l/ha at nozzle pressure of kPa and forward speed of km/h. What will be the pump capacity in l/min?

(a) 44

(b) 46

(c) 48

(d) 50

I had been asked this question during an online exam.

My question is from Plant Protection Equipments in division Plant Protection Equipments of Farm Machinery

1 Answer

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Best answer
Right answer is (c) 48

To elaborate: Flow rate = application rate*area covered by single nozzle

Area covered by single nozzle = speed * width = \(\frac{40}{100}*\frac{5*1000}{60} \) = 0.0333 ha/min

Total flow rate = 20*650*0.00333 = 43.29 l/min

Pump capacity = \(\frac{21.6452}{\frac{90}{100}} \) = 48 l/min.

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