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Chapter 1: Problem 61

A punch press with flywheel adequate to minimize speed fluctuations produces 120 punching strokes per minute, each providing an average force of \(2000 \mathrm{~N}\) over a stroke of \(50 \mathrm{~mm}\). The press is driven through a gear reducer by a shaft rotating \(300 \mathrm{rpm}\). Overall efficiency is \(80 \%\). (a) What power (W) is transmitted through the shaft? (b) What average torque is applied to the shaft?

Short Answer

Expert verified
a) The power transmitted through the shaft is 96 kW. b) The average torque applied to the shaft is 305.66 N*m.

Step by step solution

01

Calculate power.

We calculate the power by the formula: Power = Force x Distance/time. Here force is 2000 N, distance is 50 mm (or 0.05 m for unit consistency), and time is 1/120 min (or 1/(2 x 60) s). Therefore, Power = (2000 N x 0.05 m) / (1/(2 x 60) s) = 120000 W or 120 kW. However considering overall efficiency which is 80%, the actual power transmitted would be 0.8 x 120000 = 96000 W or 96 kW. Therefore the power transmitted through the shaft is 96 kW.
02

Calculate the torque.

Now we need to calculate the torque applied to the shaft. Torque (T) can be found using the formula T = Power/(2*pi*shaft rotation per minute/60). Therefore, the torque = 96000 W/(2*3.14*300/60) = 305.660 N*m. Therefore, the average torque applied to the shaft is 305.66 N*m.

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