Sowing and Planting Equipment Questions and Answers

Answer: B

The ground wheel turns at about 24.6 rpm.

The forward speed is 2.5 km/h = 41.67 m/min. With 10% skid the wheel slides, so each revolution covers 10% more ground than its circumference:

\[ \begin{aligned} \text{travel/rev} &= \pi(0.49)(1.1) = 1.693\ \text{m} \\ N &= \frac{41.67}{1.693} = 24.6\ \text{rpm} \end{aligned} \]

Without skid the wheel would turn at 27.1 rpm, which is not offered. 28.3 and 32.6 rpm would need a smaller wheel or a higher speed, and 20.3 rpm overstates the effect of skid.

Answer: A

The speed ratio is 0.125 : 1.

Allowing for 84% germination, the seeds to be sown are 44800/0.84 = 53333 per ha. With rows 0.75 m apart, one metre of row serves 0.75 m², so:

\[ \begin{aligned} \text{seeds/m} &= 53333 \times 0.75/10000 = 4.0 \\ \text{seeds/wheel rev} &= 4.0 \times 1.693 = 6.77 \\ \text{ratio} &= 6.77/54 = 0.125 \end{aligned} \]

The 1.693 m per wheel revolution includes the 10% skid from Q52, which is why the pair is linked. Ignoring germination or the skid lowers the ratio below 0.125, and the larger ratios would sow far too many seeds per metre.

Answer: B

The variation due to skid is 9.23%.

The drill has 5 furrow openers at 20 cm spacing, a width of 1.0 m. In the laboratory, 20 turns of the 0.5 m wheel cover

\[ \begin{aligned} L &= 20 \times 3.14 \times 0.5 = 31.4\ \text{m} \\ \text{area} &= 31.4 \times 1.0 = 31.4\ \text{m}^{2} \end{aligned} \]

so the calibrated rate, and the seed it should drop on 2 ha, are

\[ \begin{aligned} \text{rate} &= \frac{0.320 \times 10000}{31.4} = 101.9\ \text{kg/ha} \\ 2\ \text{ha} &: 203.8\ \text{kg} \end{aligned} \]

The field actually received 185 kg, because the drive wheel skidded and turned the metering mechanism less than the distance travelled. The variation is

\[ \frac{203.8 - 185}{203.8} \times 100 = 9.23\% \]

10.17% is the same difference divided by the field quantity (185 kg) instead of the expected quantity. The other values come from slips in the width or circumference.

Answer: A

The seed rate is about 96.6 kg/ha.

A 9×20 drill has 9 furrow openers 20 cm apart, so it covers 1.8 m. The fluted shaft turns once for every two turns of the ground wheel:

\[ \begin{aligned} \text{distance per shaft turn} &= 2 \times 3.14 \times 0.35 = 2.198\ \text{m} \\ \text{area} &= 1.8 \times 2.198 = 3.956\ \text{m}^{2} \\ \text{seed} &= 9 \times 4.25 = 38.25\ \text{g} \end{aligned} \]

\[ \begin{aligned} \text{rate} &= \frac{38.25}{3.956} = 9.67\ \text{g/m}^{2} \\ &\approx 96.7\ \text{kg/ha} \end{aligned} \]

The paper prints this as 96.62, a small rounding difference. Forgetting the 2:1 drive ratio doubles the rate to about 193 kg/ha, which no option matches; 141.55, 187.35 and 386.42 are all well above the rate the drill actually sows.

Answer: C

The flow rate is \(1.563 \times 10^{-3}\) m³.h^-1.

Four rows at 25 cm spacing give a working width of 1 m. At 2.5 km/h the drill covers

\[ \begin{aligned} 1\ \text{m} \times 2500\ \text{m/h} &= 2500\ \text{m}^{2}/\text{h} = 0.25\ \text{ha/h} \\ \dot{m} &= 5.0 \times 0.25 = 1.25\ \text{kg/h} \\ Q &= \frac{1.25}{800} = 1.563 \times 10^{-3}\ \text{m}^{3}/\text{h} \end{aligned} \]

\(1.172 \times 10^{-3}\) is what three rows (0.75 m width) would give, \(0.391 \times 10^{-3}\) is the flow from a single row, and \(1.953 \times 10^{-3}\) would need a width of 1.25 m, that is five rows.