Here we go again Newtons and Watts

[collinsmark]

Again, I'm not saying work is being done on the spring. I'm saying that the electric motor will consume energy with a locked shaft. The energy consumed is converted to heat and dissipated from the windings (which will eventual burn up with a locked shaft).

Energy is only dissipated as heat in the motor, if the motor is less than 100% efficient.

If the motor is 100% efficient, then by definition, all energy input into the motor is converted to work (are you using a different definition of efficiency?). If such a motor does not do any work, it takes in 0 J of energy.

Take an electric motor, connect a spring to the shaft and then the anchor one end of the spring. Now apply a voltage to the terminals of the motor. What happens?

[Edit: What I mean here is what happens after the system reaches steady state. Or alternatively, if you just locked the shaft down directly.]

If the motor is 100% efficient, nothing. The terminals behave as an open circuit and no current flows.

This (hypothetical) 100% efficient motor will produce a torque however. Ignoring any momentary stress in the mechanics (which is reasonable to ignore since we're discussing things that are 100% ideal anyway) this torque is fully counteracted by the normal force of the locking mechanism on the shaft. Torque is applied, but no angular displacement -- meaning no work is done. Similarly, an electrical potential is applied, but no current is drawn, therefore no power is consumed.

[...]
The shaft of the motor is now locked and is not moving, the spring is stretched (contains potential energy) and the current in the armature is still present (but a lot higher) as well as the voltage on the terminals...hence power is being supplied to the motor which means energy is being used (i.e. converted to heat in this case).

Yes, this is quite true for motors operating at less than 100% efficiency. But if the motor's efficiency is at 100%, no electric power flows into the motor, even if a voltage is applied to the terminals.

I've personally verified this with lab tests before and have actually seen the input power go up dramatically when the shaft became locked. It went up dramatically due to the armature current increasing dramatically which was a direct result of the locked shaft.

Yes, so have I! Practical (real-world) DC motors especially have this characteristic (as opposed to other types of motors such as induction motors or synchronous motors)! practical DC motors produce produce their maximum torque at 0 RPMs. But if you look at a real-world DC motor's efficiency vs. RPM curve, you'll find that the efficiency drops to 0 at 0 RPM.

In other words, when you're running this (locked shaft) test you are testing with a motor that is operating at 0% efficiency.

Again, as I've said all along, their is no work being done on the brick or spring or whatever object you want to put in its place. But the electric machine is still using energy (i.e. power) while holding the spring in place (i.e. locked shaft).

Yes, but only true for motors having less than 100% efficiency. The same applies for human waiters and weightlifters.

 

[berkeman]

I think we should avoid trying to discuss 100% efficient electric motors. As you have just made clear, they are non-physical and not really of any help in tutoring newcomers in the physical concepts of work and power.

 

[collinsmark]

I think we should avoid trying to discuss 100% efficient electric motors. As you have just made clear, they are non-physical and not really of any help in tutoring newcomers in the physical concepts of work and power.

Yes, perhaps something in nature, that is 100% efficient, and everybody is accustomed to would make for a better example. Like gravity!

Imagine a block lying on the ground. Gravity exerts a force on it with a magnitude mg (where m is the mass of the block and g is the graviational acceleration (9.8 m/s₂). Now imagine that you attach the block to a very tall crane (using a rope of negligible weight). If you use the crane to raise the block higher, at a constant velocity v, the power required is

P = Fv = mgv,

ignoring the very brief intervals of acceleration.

So if you wanted to raise a 1 kg block up at 1 m/s, it takes

P = mgv = (1.0 \ kg)(9.8 \ m/s^2)(1.0 \ m/s) = 9.8 \ \mbox{Watts}

If you wanted to raise the block up at 2 m/s,

P = mgv = (1.0 \ kg)(9.8 \ m/s^2)(2.0 \ m/s) = 19.6 \ \mbox{Watts}

Now here is my key point: In both examples above, the force on the block is identical! It doesn't matter if you raise the block at a constant 1 m/s or 2 m/s or 50 m/s (ignoring the brief intervals of acceleration). The force is still 9.8 m/s². And if you raise the block up twice as fast, it requires twice the power (in Watts). But it get's there in 1/2 the time, so the total energy expenditure is identical.

At the risk of going back to the motors (:shy:), this is an important concept in automobiles. One of the specs used to described automobile engines is horsepower. Horsepower is a measure of power, just like Watts (watts and horsepower can be converted to one another by a simple constant, since they are both measures of the same thing). The maximum speed at which an automobile can climb a steep hill is a function of the automobile engine's horsepower. The more horsepower the car has, the faster it can get up very steep and very long hills (all else being the same such as the cars' mass).

Consider this, and it may illustrate the point. Consider a souped up Mazarati and a used 4-cylinder economy car, each climbing the same mountain hill, which is very long and very steep (and constant grade). The Mazarati climbs the hill at 60 mph while the economy car climbs the hill at 20 mph. Assuming both cars have equal mass, and ignoring air resistance and tire friction, both engines are producing the same force on the respective car. The mazarati's force is the same as the economy car's force. But the Mazarati uses more horsepower, so it gets up the hill faster.

 

[rcgldr]

My brick is not actually moving so distance is kind of irrelevant.

In your first post, it seemed the goal was to understand the motor used to drive the main rotor of a rc helicopter, so why are you concerned about the non moving case?

It would help here if you could better explain what the ulitmate concept that you're trying to understand in this thread. I think we've explained that without motion, there is no power output.

In the case of a radio control helicopter, if it's an aerobatic helicopter, most of the time it will be run in "idle up" mode were the rotor is maintained at a fairly fast and constant rate of rotation, with the primary change being the pitch of the rotor to create upforce or downforce (for inverted flight or areobatic maneuvers). The transmitter will have a throttle versus pitch curve mix in order to adjust the power input into the motor to keep the rotor moving at the same speed across the full range of pitch (usually -12 to +12 degrees).

 

[stewartcs]

Energy is only dissipated as heat in the motor, if the motor is less than 100% efficient.

If the motor is 100% efficient, then by definition, all energy input into the motor is converted to work (are you using a different definition of efficiency?). If such a motor does not do any work, it takes in 0 J of energy.

If we use a magical electric motor that uses magical conductors and magically has no other losses the motor would be about 100% efficient while operating. This means the armature resistance would be zero. If we take that very same magical electric motor and lock the shaft it will become 0% efficient.

Now assuming we are still applying the same terminal voltage to the motor when running or when it has a locked shaft what happens to the current in the armature? Where does the energy go that was being provided while running now that the shaft has been locked?

Torque is still being developed on the shaft which means the electric motor is being supplied with energy. Energy cannot be destroyed, so where does the energy go?

If the motor is 100% efficient, nothing. The terminals behave as an open circuit and no current flows.

The terminals would behave as short circuit. They are still connected but now have zero resistance. If it where open the resistance would be infinite.

This (hypothetical) 100% efficient motor will produce a torque however. Ignoring any momentary stress in the mechanics (which is reasonable to ignore since we're discussing things that are 100% ideal anyway) this torque is fully counteracted by the normal force of the locking mechanism on the shaft. Torque is applied, but no angular displacement -- meaning no work is done. Similarly, an electrical potential is applied, but no current is drawn, therefore no power is consumed.

Current is drawn in the armature as it always is drawn. The armature resistance is normally very low to begin with for practical electric motors (say 90% efficient). These 90% efficient machines still have very high locked rotor current flowing in them. To reduce the resistance even lower would result in even higher currents flowing. The circuit does not become open when lowering the resistance of the connect member.

Yes, this is quite true for motors operating at less than 100% efficiency. But if the motor's efficiency is at 100%, no electric power flows into the motor, even if a voltage is applied to the terminals.

What if the motor is 99.99999999999999999999999999999999999999999999999% efficient? Does it require power? You're argument involves an indeterminate form so your answer is nonsensical. Look at what happens mathematically as you approach zero resistance.

Yes, so have I! Practical (real-world) DC motors especially have this characteristic (as opposed to other types of motors such as induction motors or synchronous motors)! practical DC motors produce produce their maximum torque at 0 RPMs. But if you look at a real-world DC motor's efficiency vs. RPM curve, you'll find that the efficiency drops to 0 at 0 RPM.

In other words, when you're running this (locked shaft) test you are testing with a motor that is operating at 0% efficiency.

I agree that the efficiency drops to 0% when the shaft is locked as I have always agreed.

berkemen has a good point though, no machines in real life are 100% efficient so it seems a bit pointless to me to continue bringing up magical electric motors. All that I'm personally concerned with are the real motors that I deal with and how they respond. Discussions such as these have no value in the practical world for engineers. Interesting in academia though.

CS

 

[stewartcs]

Consider this, and it may illustrate the point. Consider a souped up Mazarati and a used 4-cylinder economy car, each climbing the same mountain hill, which is very long and very steep (and constant grade). The Mazarati climbs the hill at 60 mph while the economy car climbs the hill at 20 mph. Assuming both cars have equal mass, and ignoring air resistance and tire friction, both engines are producing the same force on the respective car. The mazarati's force is the same as the economy car's force. But the Mazarati uses more horsepower, so it gets up the hill faster.

Consider that same Mazarati sitting on that same hill. Now consider it sitting still with no motion and using the clutch and engine to maintain its position (not moving up or down the hill). Is energy still being used? Yes. If there wasn't we wouldn't need fuel for our cars.

Is there work being done on the Mazarati? No, because it is not moving.

CS

 

[rcgldr]

If we use a magical electric motor that uses magical conductors and magically has no other losses the motor would be about 100% efficient while operating. This means the armature resistance would be zero. If we take that very same magical electric motor and lock the shaft it will become 0% efficient.

A "magical motor" with a locked shaft could act similarly to a capacitor, where the effective resistance would increase as the voltage related to the charge on the capacitor like motor increased. After the initial charge build up, there would be no more current.

 

[collinsmark]

A "magical motor" with a locked shaft could act similarly to a capacitor, where the effective resistance would increase as the voltage related to the charge on the capacitor like motor increased. After the initial charge build up, there would be no more current.

Yes, as a matter of fact there are several types of motors in existence today that not only act like capacitors, but essentially are a mechanical-capacitor type of motor. These are called electrostatic motors. They are becoming commonplace in micro mechanical devises. Electrostatic motors essentially use the electric (coulomb) force to cause mechanical motion.

Unlike motors that rely on magnetic forces, electrostatic motors can, and sometimes are, designed such that their current draw is zero (or at least approximately 0) when there is no motion, while continuing to hold their force/torque. As applied to the specific application of compressing a spring and holding it in a compressed state for an extended period, the performance a properly chosen electrostatic motor can approach that of an ideal motor.

 

[stewartcs]

A "magical motor" with a locked shaft could act similarly to a capacitor, where the effective resistance would increase as the voltage related to the charge on the capacitor like motor increased. After the initial charge build up, there would be no more current.

I fail to see how a loop of wire acts as a capacitor. A loop of wire with no resistance is an ideal inductor.

One of the simplest DC machines is a loop of wire connected to a battery positioned between two permanent magnets. This is what I was referring to. As the resistance of the loop goes to zero the current would approach infinity which makes the problem indeterminate.

However, if we look at the limit as the resistance goes to zero...we'll see that the current becomes really large for a given applied voltage. Since we have voltage and a large current, power is being consumed even though the shaft (loop in this case) isn't moving. The power must be dissipated as heat (some power is absorbed and stored by the magnetic field as well).

CS

 

[stewartcs]

Yes, as a matter of fact there are several types of motors in existence today that not only act like capacitors, but essentially are a mechanical-capacitor type of motor. These are called electrostatic motors. They are becoming commonplace in micro mechanical devises. Electrostatic motors essentially use the electric (coulomb) force to cause mechanical motion.

Unlike motors that rely on magnetic forces, electrostatic motors can, and sometimes are, designed such that their current draw is zero (or at least approximately 0) when there is no motion, while continuing to hold their force/torque. As applied to the specific application of compressing a spring and holding it in a compressed state for an extended period, the performance a properly chosen electrostatic motor can approach that of an ideal motor.

Perhaps so for an electrostatic motor, but my discussion was pertaining to a simple DC machine as previously described. There are lots of devices that can provide force without using power continually (such as the clamp we discussed earlier). However, not all machines can.

I guess at the end of the day all we can say is that if power is being used by an electric machine that it will be converted to some other form. That form (e.g. mechanical power, heat) will depend on the device.

CS

 

[rcgldr]

A "magical motor" with a locked shaft could act similarly to a capacitor, where the effective resistance would increase as the voltage related to the charge on the capacitor like motor increased. After the initial charge build up, there would be no more current.

I fail to see how a loop of wire acts as a capacitor. A loop of wire with no resistance is an ideal inductor.

You seemed to have missed the part about "magical motor", in this case, it's one that doesn't use windings of wire.

 

[Leafhill]

If any of you happen to have a spare 100% efficient motor lying around, let me know. I'm willing to pay for it :-)

Anyway, I think I've gotten a bit closer to my original question conserning the relationship between Watts and Newtons. As many of you have already explained quite detailed, there is no direct corelation between the two, BUT it seams that Amps have a very direct relation to Newtons.

As mentioned several times already:
Watts = Nm/s

Is the following correct?
Since Watts = Volts*Amps
And Nm/s could be written as Force * speed:

Volts = Speed
Amps = Force

The notion about Amps representing force (converted to torque in a motor) and Volts representing speed is new to me, and I've had some difficulties wrapping my head around it. Amps is ok, I know lots of Amps give high torque, but how could volts represent speed? I mean as you can derive from Newtons first law, if no energy is added or subracted, the speed should remain constant.

And this is probably where I'm wrong, but this says to me that Volts (speed) is irrelevant as long as you have the amps to do the work.

I'm afraid there is no real world purpose for my inquiry, but rather a need to clean up a messy corner of my mind :-) There is however several examples where I could try to put this new info to good use. Going back to the propeller (Helicopter, plane or other) I can accept that the prop will need Amps (Torque) to push it around against the air, and Volts to keep the prop speed up.

Am I getting closer to home, or still hopelessly lost?

 

[cepheid]

As mentioned several times already:
Watts = Nm/s

Is the following correct?
Since Watts = Volts*Amps
And Nm/s could be written as Force * speed:

Volts = Speed
Amps = Force

I'm sorry, but this is nonsense. Equating two physical quantities that don't have the same units is meaningless. What you are doing is no different from saying that time = distance (i.e. both statements are equally nonsensical).

Here is one flaw in your reasoning. Multiplication is commutative, meaning that the order in which two things are multiplied doesn't matter. So, although you have written that power = voltage*current, you could just have easily have written that power = current*voltage, which would have led you to conclude something entirely different using your "matching" procedure.

But, more fundamentally, just because two pairs of multiplied quantities both happen to have dimensions of power doesn't mean that those two pairs of quantities are the same!

I didn't slog through this whole thread, but just to be sure, you do realize that it possible for a motor to use up (i.e. to waste) electrical energy without doing any useful work, right?

 

[rcgldr]

Since Watts = Volts*Amps
And Nm/s could be written as Force * speed:

Volts = Speed
Amps = Force

Voltage = difference in electrical potential between two points.

electrical potential energy between two points is the negative of the work done by an electrical field from point₀ to point₁.

electrical intensity is the electrical force per unit charge of some target object, considering the source object as the generator of an electrical field.

voltage = the difference between electrical potential energy per unit charge between two points, one of which can be a "reference" point.

current = the rate of charge flow across some point or between two points.

power = force x speed doesn't directly translate well into power = volts x current.

Analogy using an infinitely large disk with a finite negative charge or finite amount of gravity per unit area, and "h" as distance from that disk:

Intensity= -E (Newtons / coulomb) => gravitational intensity = -g (Newtons / kilogram)
Force = -Eq => -gm
volt = joule / coulomb
potential = V (volts) = Eh => gh = V (joule / kilogram)
potential energy = U (joules) = Eqh => gmh = U (joules)

Power for a moving fixed amount of charge or mass:
power = |-Eq (dh/dt)| = |-q (dV/dt)| => |-gm (dh/dt)| = |-m (dV/dt)|

Power for a constant flow of charge or mass bewteen two points in space:
I = dq/dt
ṁ = dm/dt
power = E (dq/dt) (dh/dt) = (dq/dt) (dV/dt) = I (dV/dt) => g (dm/dt) (dh/dt) = (dm/dt) (dV/dt) = ṁ (dV/dt)
let V = dV/dt = amount of change in potential per unit time.
power = I V = current x voltage_change => ṁ V = mass_flow x gravitaional_potential_change
let q = dq/dt = amount of charge that flows between two points per unit time.
let m = dq/tm = amount of mass that flows between two points per unit time.
power = Eq (dh/dt) => gm (dh/dt) => force x speed.