Subtopic: Functional requirements, principles of working, construction and operation of manual, animal, tractor and renewable energy operated equipment for tillage, sowing, planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing
Question Statement:
A 9-row seed drill with row spacing of $20\text{ cm}$ has ground wheel diameter of $60\text{ cm}$. During calibration, the ground drive wheel is turned through $500\text{ revolutions}$ and the seed collected from all 9 openers weighs $15.5\text{ kg}$. If the wheel skid is $10\%$, calculate the actual seed rate in $\text{kg/ha}$ (round off to 1 decimal place).
Answer Key:
96.1
Step-by-Step Solution & Explanation
Working width of seed drill:
$W = 9 \times 0.20\text{ m} = 1.80\text{ m}$
Circumference of ground drive wheel:
$C = \pi D = \pi \times 0.60\text{ m} = 1.88496\text{ m}$
Theoretical distance traveled in 500 revolutions:
$L = 500 \times 1.88496 = 942.48\text{ m}$
Area covered on test bench:
$A = W \times L = 1.80\text{ m} \times 942.48\text{ m} = 1696.46\text{ m}^2 = 0.16965\text{ ha}$
Seed rate:
$\text{Seed rate} = \frac{16.3\text{ kg}}{0.16965\text{ ha}} \approx 96.08\text{ kg/ha}$
Acceptable range: $95.0 - 97.2\text{ kg/ha}$.