Subtopic: Drying and milling of cereals, pulses and oilseeds
Question Statement:
A modern rubber roll paddy sheller has two counter-rotating cylindrical rolls of equal diameter $D = 250\text{ mm}$. The fast roll rotates at $N_1 = 1200\text{ rpm}$ and the slow roll rotates at $N_2 = 960\text{ rpm}$ (speed ratio of $1.25$). The linear relative shearing (rubbing) velocity between the surfaces of the two rubber rolls at the nip is ________ $\text{m/s}$ (round off to two decimal places).
Answer Key:
3.14
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Peripheral Speeds Difference**
1. Linear peripheral speed of the fast roll ($v_1$):
$v_1 = \frac{\pi D N_1}{60} = \frac{\pi \times 0.250 \times 1200}{60} = 5.0 \pi \approx 15.708\text{ m/s}$
2. Linear peripheral speed of the slow roll ($v_2$):
$v_2 = \frac{\pi D N_2}{60} = \frac{\pi \times 0.250 \times 960}{60} = 4.0 \pi \approx 12.566\text{ m/s}$
3. Relative rubbing velocity ($\Delta v$):
$\Delta v = v_1 - v_2 = 5.0 \pi - 4.0 \pi = 1.0 \pi \approx 3.1416\text{ m/s} \approx 3.14\text{ m/s}$
**Method 2: Factoring Angular Difference**
$\Delta v = \frac{\pi D (N_1 - N_2)}{60} = \frac{\pi \times 0.250 \times (1200 - 960)}{60} = \frac{\pi \times 0.250 \times 240}{60} = \pi \approx 3.14\text{ m/s}$