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 tractor-mounted seed-cum-fertilizer drill with a row-to-row spacing of $20\text{ cm}$ has drive wheels of $60\text{ cm}$ effective diameter. In a stationary calibration test, the drive wheel is rotated through 50 revolutions, and $720\text{ g}$ of seed is collected from all the tubes together. If the drive wheel slip during actual field operation is expected to be $10\%$, the actual seed rate in $\text{kg/ha}$ is:
Answer Key:
42.44
Step-by-Step Solution & Explanation
Given:
- Number of rows $n = 9$
- Row spacing $s = 20\text{ cm} = 0.2\text{ m}$
- Width of drill $W = 9 \times 0.2 = 1.8\text{ m}$
- Wheel diameter $D = 0.6\text{ m}$
- Wheel revolutions $N = 50$
Theoretical distance covered without slip:
$L = N \times \pi \times D = 50 \times \pi \times 0.6 = 30\pi = 94.248\text{ m}$
Effective distance when wheel has $10\%$ slip ($s_l = 0.10$):
In the field, when the wheel rotates 50 times, the actual distance travelled is:
$L_{\text{actual}} = L \times (1 - s_l) = 94.248 \times (1 - 0.10) = 84.823\text{ m}$
Actual area covered for 50 wheel rotations:
$A = W \times L_{\text{actual}} = 1.8 \times 84.823 = 152.68\text{ m}^2 = \frac{152.68}{10000}\text{ ha} = 0.015268\text{ ha}$
Seed collected $M = 720\text{ g} = 0.720\text{ kg}$
Actual seed rate:
$\text{Seed rate} = \frac{0.720\text{ kg}}{0.015268\text{ ha}} \approx 42.44\text{ kg/ha}$