AGRICULTURAL AND BIOSYSTEMS ENGINEERING
PROBLEM:
A 6-NOZZLE BOOM SPRAYER IS MOUNTED AT THE BACK OF A FOUR-WHEEL TRACTOR. THE NOZZLES ARE ARRANGED 150 CM APART. EACH NOZZLE HAS A 60-DEGREE SPRAY ANGLE. THE RICE PLANT CANOPY IS 50 CM TALL. THE TRACTOR TRAVELS AT 5 KPH WITH 95% FIELD EFFICIENCY
At what height must the boom be raised, relative to the ground level, so that the spray from each nozzle does not overlap upon reaching the top of the rice plant canopy?
a. 108 cm
b. 180 cm
c. 93.3 cm
d. 309.8 cm
e. 136.6 cm
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Step 1: Calculate height above canopy (h).
To avoid overlap, the spray width of one nozzle must equal the nozzle spacing (150 cm).
Formula: h = Spacing / (2 * tan(Angle / 2))
h = 150 / (2 * tan(30°)) = 150 / 1.1547 ≈ 130 cm
Step 2: Calculate height relative to ground.
Total Height = Height above canopy + Canopy height
Total Height = 130 cm + 50 cm = 180 cm
Result: b. 180 cm
What is the effective swath of the boom sprayer under the condition above?
a. 150 cm
b. 750 cm
c. 75 cm
d. 900 cm
e. 90 cm
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Calculation:
Effective Swath = Number of nozzles * Spacing
Swath = 6 * 150 cm = 900 cm
Result: d. 900 cm
If the tractor travels at 5 kph with 95% field efficiency, what will be the spraying capacity (in ha/h) under the condition above?
a. 4.25
b. 4.52
c. 4.28
d. 4.70
e. 4.75
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Formula: C = (W * S * Ef) / 10
W (Width) = 9 m (900 cm)
S (Speed) = 5 kph
Ef (Efficiency) = 0.95
Calculation:
C = (9 * 5 * 0.95) / 10 = 4.275 ha/h
Result: c. 4.28 (rounded)
If each nozzle delivers one liter per minute, how many will the boom sprayer deliver per hectare under the conditions above?
a. 84.11 li/ha
b. 84.71 li/h
c. 79.65 li/ha
d. 76.60 li/ha
e. 75.79 li/ha
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Step 1: Total flow rate.
Total flow = 6 nozzles * 1 li/min = 6 li/min
Total flow per hour = 6 * 60 = 360 li/h
Step 2: Application rate.
Rate = Total Flow per hour / Spraying Capacity
Rate = 360 / 4.275 ≈ 84.21 li/ha
Result: a. 84.11 li/ha (closest value)
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