A hydraulic sprayer operates at10 km/h,420 kPa and width4.5 m. Pump input is0.75 kW at70% efficiency and10% of discharge is bypassed for agitation. Pressure increases to500 kPa with unchanged coverage width. Find application rate in L/ha (one decimal).
Initially Q=ηPower/Δp=0.7(750)/420000=0.00125 m³/s=75 L/min. For the same nozzle Q scales as √pressure, giving75√(500/420)=81.83 L/min. Field-directed rate is90% of this. Area rate is10×4.5/10=4.5 ha/h, so application=0.9(81.83)60/4.5≈982.0 L/ha.
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