Question Bank • QB_FM_ADV_072 NAT • 2 Marks

Section 2: Farm Machinery

Farm Machinery

Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses

Question Statement:

A centrifugal spinning-disc fertilizer broadcaster has a horizontal disc of radius $R = 0.30\text{ m}$ rotating at $N = 600\text{ rpm}$. Fertilizer granules exit the disc vanes with a tangential peripheral velocity $v_t = \omega R$ and a relative radial velocity along the vane of $v_r = 12.0\text{ m/s}$. The magnitude of the absolute resultant exit velocity of the fertilizer particle in $\text{m/s}$ is ________ (round off to 2 decimal places).

Answer Key:

22.35

Step-by-Step Solution & Explanation

``` Spinning Disc Particle Velocity Vectors: Disc Center ---> Radial Vector v_r Disc Rim ---> Tangential Vector v_t v_res = sqrt(v_r^2 + v_t^2) ``` **Step-by-step Calculation:** 1. Angular velocity of the spinning disc: $\omega = \frac{2 \pi N}{60} = \frac{2 \times \pi \times 600}{60} = 20\pi = 62.8318\text{ rad/s}$ 2. Tangential velocity component at the disc rim ($R = 0.30\text{ m}$): $v_t = R \omega = 0.30\text{ m} \times 62.8318\text{ rad/s} = 18.8496\text{ m/s}$ 3. Radial velocity along the vane: $v_r = 12.0\text{ m/s}$ 4. Absolute resultant exit velocity ($v_{\text{res}}$): $v_{\text{res}} = \sqrt{v_t^2 + v_r^2} = \sqrt{(18.8496)^2 + (12.0)^2} = \sqrt{355.306 + 144} = \sqrt{499.306} = 22.345\text{ m/s} \approx 22.35\text{ m/s}$
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