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 seed drill has 11 furrow openers spaced $18\text{ cm}$ apart. The ground drive wheel has an effective diameter of $70\text{ cm}$. In a stationary calibration test of 100 revolutions of the drive wheel, $1.5\text{ kg}$ of wheat seed is collected from all tubes. If the expected field wheel slip is $8\%$, calculate the actual seed rate in $\text{kg/ha}$.
Answer Key:
37.45
Step-by-Step Solution & Explanation
Working width of drill:
$W = 11 \times 0.18 = 1.98\text{ m}$
Nominal distance without slip:
$L_0 = 100 \times \pi \times 0.70 = 70 \pi \approx 219.91\text{ m}$
Actual distance with $8\%$ slip ($s = 0.08$):
$L = L_0 (1 - 0.08) = 219.91 \times 0.92 = 202.32\text{ m}$
Actual field area covered:
$A = W \times L = 1.98 \times 202.32 = 400.59\text{ m}^2 = 0.040059\text{ ha}$
Actual seed rate:
$\text{Rate} = \frac{1.5\text{ kg}}{0.040059\text{ ha}} \approx 37.445\text{ kg/ha}$