| Confusing Units of Measure: Work, Power and Torque |
|
| Written by Joe Evans, Ph.D. | |
|
Page 2 of 2
So far so good, but there is another factor that can affect the torque generated by that simple wrench. Suppose we apply that force on the handle at a 45-deg angle, rather than at the 90-deg shown in Figure 2. In this case, some of the force would be directed downward, just as before, but some would also be directed toward the nut. How do we calculate the torque produced? Fortunately, some simple trig can expand our equation for torque and make it useable regardless of the angle of the force. Its new form becomes t = F(r sin θ), where θ is the angle. The sine of 90-deg is 1, so when the force is perpendicular, torque is simply Fr. The sine of 45-deg is 0.707, and at this particular angle torque becomes F(r x 0.707). A good example of the effect that the angular and radial components of force can have on torque is the piston engine. When the spark plug fires, the piston is at the top of the cylinder and the crankshaft is at 0-deg. As the piston moves down the cylinder the crankshaft rotates, torque is generated and reaches its maximum at 90-deg (halfway through its down stroke). During the second half of the down stroke, torque decreases and disappears at 180-deg. If you would like to see an excellent animation that illustrates this sequence, go to http://science.howstuffworks.com/fpte4.htm. In a linear system we saw that work = Fd. In a rotational system torque (t) - which consists of force and radius - replaces force (F), and distance (d) becomes the angle through which the torque acts. This means work = tθ, where θ is the angle. In the case of an electric motor, θ is 360-deg or one complete rotation. It should follow that the power or horsepower produced by a rotational system is simply work divided by time. Unfortunately it is not that simple, because time and rotational distance are related by angular speed. This adds a bit of mud to the equation and power becomes the product of torque and angular speed, and angular speed is measured in radians per unit of time! Fortunately we can convert angular speed to a more familiar unit: rpm. Now our equation becomes P = t x 2π x rpm and power is in ft-lb per minute. Why 2π? Because there are 2π radians in one complete rotation. If you divide the result by the work done by Watt's horse in one minute, you end up with horsepower. But there is an even easier way to calculate horsepower. Simply stated, hp = (t x rpm) / 5250. So where did that convenient conversion factor of 5250 come from? Well, we just divided the work done by Watt's horse (33,000) by 2π and got rid of both of them! If we know the horsepower, we can rearrange that equation to t = (hp x 5250) / rpm and calculate torque. As you can see from the equation in the previous paragraph, if we keep horsepower constant, torque will vary inversely with rotational speed. For example, a 100-hp electric motor rated at 3500-rpm produces approximately 150-lb-ft of torque. If it were rated at 1750-rpm, its torque would increase to about 300-lb-ft. This is exactly what we would expect since horsepower or power is the rate at which work is done. If a 1750-rpm motor is to accomplish the same amount of work in the same amount of time as its 3500-rpm cousin, it must do twice the work per rotation. This figure shows the torque and current curves of a typical Class B AC motor. The Y-axis shows the percent of full load torque and current while the X-axis shows the percent slip and the synchronous speed (Ns).
Synchronous speed is the speed at which the magnetic field rotates about the stator, and it is equal to (120 x Hz) / number of poles. At 60-Hz, a two-pole motor has a synchronous speed of 3600-rpm. For a four-pole motor, that speed is reduced to 1800-rpm. Slip or slip speed is the actual rotational speed of the rotor and it is typically 3 percent to 5 percent lower. The reason that an induction motor operates below synchronous speed is pretty simple. If the rotor were to move at the same speed as the rotating magnetic field in the stator, there would be no relative motion between them. Therefore, no flux lines would cut across the rotor bars and no magnetic field would be induced. The torque produced by an induction motor is proportional to the strength of the interacting magnetic fields in the stator and rotor, and varies substantially from zero to rated slip speed. The locked rotor torque (or starting torque) is the minimum torque that is developed when the rotor is at rest (rpm = 0). The current required to produce this starting torque ranges from five to seven times the actual full load current. Pull up torque is the minimum torque developed by the motor as it accelerates from rest to the speed at which maximum torque occurs. As you can see, current drops very little during this period of acceleration. As the rotor approaches about 75 percent of its rated slip speed, torque reaches its maximum (breakdown torque) and current begins to drop. As the rotor continues to accelerate, torque and current drop quickly and reach their full load values at the rated slip speed of the motor. The figure also shows us what happens when a motor is overloaded. Take, for example, that 100-hp / 1750-rpm motor we spoke about earlier. As the load is increased, torque increases and will reach its maximum when rotational speed decreases to about 25 percent slip. If you substitute the new values of torque and speed into the horsepower equation, you will find that horsepower increases from 100 to about 230! If the load continues to increase, torque will begin to decrease and rotation will come to a grinding halt. Chances are that you would not see this happen due to the large volume of smoke exiting the motor housing! Hopefully this has made work, torque, and power in a rotational environment a little more understandable. I have also had a little fun trashing the metric system, even though I believe we should have adopted it a long time ago. However, those who expected it to be an easy transition must have been wearing blinders. Can you imagine going to McDonald's for a 0.11339 kilogramer (maybe even with cheese)? Or, how about first down and 9.144 on the 45.72 (that would be midfield!). If we are going to make this system the standard, it will have to start in grade school. Joe Evans is the western regional manager for Hydromatic Engineered Waste Water Systems, a division of Pentair Water, 740 East 9th Street Ashland, OH 44805. He can be reached via his website at www.pumped101.com. If there are topics that you would like to see discussed in future columns, drop him an email. Comments (0)
 Write comment
|
The amount of rotational force that is applied to a nut by a wrench depends upon the amount of force (F) applied to the handle and where that force is applied (r). The torque that is generated is directly proportional to both. If 10-lbs is applied one foot from the nut, the result is 10-lb-ft of torque. If you move that force in by 0.5-ft, torque is reduced to 5-lb-ft. But, if you increase the force to 20-lbs, the torque at 0.5-ft returns to 10-lb-ft.
Subscribe to this comment's feed
