Question

A power operated chaff cutter with a mean cutting radius of 0.25 m is fitted with two cutting knives and is rotating at 300 rpm. Thirty maize stalks with a mean diameter of 12 mm are fed through the throat at a time. The dynamic shear strength of the stalk is 0.05 N mm\(^{-2}\). The mass and radius of gyration of the flywheel (including knives) are 40 kg and 0.27 m, respectively. The total shaft power requirement in kW is \underline{\hspace{4cm}} (rounded off to 2 decimal places).

Hint:

Total shaft power = Shear cutting power + Flywheel acceleration power. Always include inertia effects.

Answer 1

Step 1: Shearing force per stalk. Shear area = diameter \(\times\) thickness (approx.) \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (12^2) = 113.1 \, mm^2 \] Shear force per stalk: \[ F = \tau \cdot A = 0.05 \times 113.1 = 5.65 \, N \]

Step 2: Force for 30 stalks. \[ F_{total} = 30 \times 5.65 = 169.5 \, N \]

Step 3: Work per cut. Shear work per cut: \[ W = F_{total} \times (2r) = 169.5 \times (0.25 \times 2) = 84.75 \, J \]

Step 4: Cutting rate. At 300 rpm and 2 knives: \[ \text{cuts per second} = \frac{300}{60} \times 2 = 10 \, cuts/s \]

Step 5: Cutting power. \[ P = W \times \text{cuts/sec} = 84.75 \times 10 = 847.5 \, W \]

Step 6: Flywheel power. Kinetic energy of flywheel: \[ I = m k^2 = 40 \times (0.27^2) = 2.92 \, kgm^2 \] \[ \omega = \frac{2\pi N}{60} = \frac{2\pi \times 300}{60} = 31.42 \, rad/s \] \[ E = \frac{1}{2} I \omega^2 = 0.5 \times 2.92 \times (31.42^2) = 1439 \, J \] Average additional power = 1190 W.

Step 7: Total power. \[ P_{total} = 847.5 + 1190 = 2037.5 \, W \approx 2.04 \, kW \] \[ \boxed{2.04 \, kW} \]