Subtopic: Cost analysis of implements and tractors
Question Statement:
The application rate of an 18-nozzle hydraulic sprayer is 1120 L/ha. Nozzle spacing and forward speed are 400 mm and 3.4 km/h. Operating pressure is 2.1 MPa and pump efficiency is 60%. If 10% of the pump output power is used for agitating the liquid, the power needed to operate the sprayer in kW (rounded to 3 decimals) is ______.
Official Answer Key:
2.900 to 3.000
Step-by-Step Solution & Explanation
Spray swath width, W = nozzles × spacing = 18 × 0.4 = 7.2 m.
Area covered per hour = (W × speed) = 7.2 m × 3.4 km/h = 7.2×3400 m²/h = 24,480 m²/h = 2.448 ha/h.
Required liquid flow = application rate × area/h = 1120 L/ha × 2.448 ha/h = 2741.76 L/h = 0.7616 L/s = 7.616×10⁻⁴ m³/s.
Hydraulic (fluid) power = pressure × flow rate = 2.1×10⁶ Pa × 7.616×10⁻⁴ m³/s = 1599.4 W = 1.599 kW.
Pump input power (accounting for 60% pump efficiency) = 1.599/0.60 = 2.666 kW.
Total power including 10% extra for agitation = 2.666 × 1.10 = 2.932 kW. Answer: within 2.900–3.000.
Section 3 — Farm Power
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