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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 press machine has a work output rate of 288 Watts. A torque of 1.75 Nm needs to be provided by its 1750 -rpm driving motor.

Step by step solution

01

Compute Work per stroke

Calculate the work done on each stroke by the press using the formula Work = force 脳 distance. Given that the force is 8000 N and distance is 18 mm (or 0.018 m in SI units), the work done per stroke is \(Work = 8000 N 脳 0.018 m = 144 Joules\).
02

Compute Power output by machine

Find the power output by the machine using the formula power = work/time. As the machine delivers 120 strokes per minute, the time for each stroke is 1/120 minutes (= 0.5 seconds in SI units). Hence, power output of the machine is \(Power = 144 Joules / 0.5 s = 288 Watts\).
03

Compute Required Input power

Now, find the required driving power using the formula efficiency = output power / input power. Given that efficiency is 90% or 0.90, input power needed is \(Input Power = Output Power / Efficiency = 288W / 0.9 = 320 Watts\).
04

Compute Needed Torque

Finally, find the torque needed on the motor using the power-torque relation \(Power = torque 脳 angular velocity\). First compute the angular velocity as \(Angular Velocity = 2蟺 脳 rpm / 60 = 2蟺 脳 1750 / 60 =183 rad/sec\). Hence, the required torque is \(Torque = Power / Angular Velocity = 320W / 183 rad/sec = 1.75 Nm\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done by Force
When we talk about work done by a force, we're referring to the energy transferred when a force moves an object over a distance. The formula to calculate this is:
  • Work = Force 脳 Distance
In the case of the press machine, each stroke provides a force of 8000 N over a distance of 18 mm, which is converted to meters as 0.018 m. So, the work done per stroke becomes:
  • \( Work = 8000 \, \text{N} \times 0.018 \, \text{m} = 144 \, \text{Joules} \)
Work done is measured in Joules, which is a standard unit of energy. Keep in mind that a greater force or longer distance results in more work done. In machines like this press, understanding work done helps us know how much energy is being used with each motion of the machine.
Press Machine Efficiency
Efficiency in machines is a measure of how well a machine converts input energy into useful output energy. It's calculated using the formula:
  • Efficiency = Output Power / Input Power
It鈥檚 usually expressed as a percentage. If a machine has high efficiency, it means less energy is wasted. In this scenario, the press machine's efficiency is 90%. This tells us that:
  • 90% of the energy going into the machine is turned into work, while 10% is wasted, possibly as heat or sound.
To find the input power required, we rearrange the formula to:
  • Input Power = Output Power / Efficiency
With an output power of 288 Watts, the input power becomes:
  • \( Input \, Power = 288 \, W / 0.9 = 320 \, W \)
Understanding efficiency is crucial for optimizing machines and saving energy.
Average Torque Calculation
Torque is a measure of the force that can cause an object to rotate about an axis. It鈥檚 an essential concept in machines with rotating parts. Torque is linked to power through the formula:
  • Power = Torque 脳 Angular Velocity
To solve for torque, rearrange the formula:
  • Torque = Power / Angular Velocity
First, calculate the angular velocity from the motor's speed:
  • \( Angular \, Velocity = \frac{2\pi \times \text{rpm}}{60} = \frac{2\pi \times 1750}{60} = 183 \, \text{rad/sec} \)
Using this angular velocity, compute the required torque:
  • \( Torque = \frac{320 \, W}{183 \, \text{rad/sec}} = 1.75 \, \text{Nm} \)
This torque value tells us how much rotational force the motor needs to provide to meet the system's work output, taking into account the press machine's efficiency.

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