Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
A power operated hydraulic sprayer operating at a pressure of 1.4 MPa has a boom with 11 nozzles spaced at 0.3 m interval. The input power of the pump is 1.5 kW and its mechanical efficiency is 75%. If the nozzle diameter and the discharge coefficient are 2.4 mm and 0.5, respectively, the average jet velocity of the nozzles in m s⁻¹ is _____.
Official Answer Key:
32.2 to 32.4
Step-by-Step Solution & Explanation
Hydraulic (output) power delivered to the spray liquid = input power × mechanical efficiency = 1.5 × 0.75 = 1.125 kW.
Since hydraulic power = pressure × total discharge, total discharge Q = P/Δp = 1125 W / (1.4×10⁶ Pa) = 8.036×10⁻⁴ m³/s.
Total nozzle flow area (11 nozzles, d=2.4 mm): A = 11×(π/4)(0.0024)² = 4.976×10⁻⁵ m².
Using orifice discharge relation Q = Cd×A×V: V = Q/(Cd×A) = 8.036×10⁻⁴/(0.5×4.976×10⁻⁵) = 32.3 m/s. This lies within the accepted range 32.2–32.4.
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