Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A tractor drawn boom type sprayer fitted with 12 nozzles at a spacing of 50 cm is used to carry out spraying at a forward speed of 3.5 km·h⁻¹ and nozzle pressure of 250 kPa for an application rate of 1.5 m³·ha⁻¹. Out of the total discharge from the pump, 12 % is bypassed to the liquid tank and the total pressure drop in all pipe lines from the pump is 10 % of the pressure developed by the pump. Considering the pump efficiency as 60 %, if the nozzle pressure is increased by 10 %, the power required to operate the pump, in kW, is __________. (Rounded off to 3 decimal places)
Official Answer Key:
0.520 to 0.540
Step-by-Step Solution & Explanation
Step 1 — boom width and field capacity:
W = 12 × 0.50 = 6 m
Field capacity = (W × S)/10 = (6 × 3.5)/10 = 2.1 ha·h⁻¹
Step 2 — nozzle discharge at the original pressure:
Q = application rate × field capacity = 1.5 × 2.1 = 3.15 m³·h⁻¹
Step 3 — effect of raising the nozzle pressure by 10%. Nozzle discharge varies as the square root of nozzle pressure (Q ∝ √p):
Q′ = 3.15 × √1.10 = 3.15 × 1.04881 = 3.304 m³·h⁻¹
Step 4 — pump discharge. 12 % of the pump output is bypassed, so the nozzles receive only 88 %:
Q_pump = 3.304/0.88 = 3.754 m³·h⁻¹ = 1.043 × 10⁻³ m³·s⁻¹
Step 5 — pump pressure. The new nozzle pressure is 250 × 1.10 = 275 kPa. Pipe losses consume 10 % of the pump pressure, so:
p_pump − 0.10 p_pump = 275 ⇒ p_pump = 275/0.90 = 305.56 kPa
Step 6 — pump power:
P = (Q_pump × p_pump)/η = (1.043 × 10⁻³ × 305,556)/0.60
= 318.6/0.60 = 531.1 W = 0.531 kW
SECTION 3: FARM POWER
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