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Chapter 1: Problem 59

What is the rate of work output of a press that delivers 120 strokes per minute, each stroke providing a force of \(8000 \mathrm{~N}\) throughout a distance of \(18 \mathrm{~mm}\) ? If the press efficiency is \(90 \%\), what average torque must be provided by a 1750 -rpm driving motor?

Short Answer

Expert verified
The rate of work output of the press is 288W, and the required average torque by a 1750rpm driving motor is 1.74Nm.

Step by step solution

01

Calculate the Work Output

Let's start by calculating the work output of the press.Work is calculated using the formula:\[ \text{Work} = \text{Force} \times \text{Distance} \]Given in the problem, the Force is \(8000 \, \text{N}\) and Distance is \(18 \, \text{mm}\) or \(0.018 \, \text{m}\). Substituting these values into the formula, the work done by the press in each stroke is:\[ \text{Work} = 8000 \, \text{N} \times 0.018 \, \text{m} = 144 \, \text{J/stroke}\]
02

Calculate the Power Output

The power output of the press can be calculated using the formula:\[ \text{Power} = \frac{\text{Work}}{\text{Time}} \]However, the problem provides the number of strokes per minute (120). Since the work calculated is for each stroke, we can calculate the total work done in a minute. As we know there are 60 seconds in a minute, we can set Time = 60s. Substituting the already calculated work per stroke and provided data into the formula, the power output then becomes:\[ \text{Power} = \frac{144 \, \text{J/stroke} \times 120 \, \text{strokes/min}}{60 \, \text{s}} = 288 \, \text{W}\]
03

Calculate the Required Torque

The second part of the problem requires calculating the torque provided by the motor. The formula for Power that relates it with Torque is:\[ \text{Power} = \text{Torque} \times \text{Angular Speed} \]Where, Angular Speed = 2蟺 * RPM / 60.It's given in the problem that the press efficiency is 90%, which indicates that the motor's power is greater than the actual work output. So the power output computed in the earlier step needs to be divided by the efficiency to get actual power provided by the motor.\[ \text{Motor Power} = \frac{\text{Power}}{\text{Efficiency}} = \frac{288 \, \text{W}}{90\%} = 320 \, \text{W}\]Angular Speed of the motor given as 1750 rpm can be converted to rad/s as follows:\[ \text{Angular Speed} = 2蟺 * \frac{1750}{60} = 183.26 \, \text{rad/s} \]Now substituting these values into the formula for Power and solving for Torque gives:\[ 320 \, \text{W} = \text{Torque} \times 183.26 \, \text{rad/s} \]\[ \text{Torque} = \frac{320 \, \text{W}}{183.26 \, \text{rad/s}} = 1.74 \, \text{Nm}\].

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Most popular questions from this chapter

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