Watts as a measure of work

[MFrankly]

Hello - I've got a basic conceptual question. I have a hard time wrapping my head around using watts as a measure for work - moving a weight a given distance - and as a measure for dissipating heat and light. I can't use my light bulb to move my toy car, so how can I say it is rated at 100 watts? Or that a resistor can be rated to survive producing 0.25 watts of waste heat? I certainly understand the idea of wasted work, but how are waste heat and waste work the same? Do I have to just buy into it?

<Background: I'm an artist and art educator who used to work at an electrical supply store. I've got it into my head to learn some electronics to make some little machines. I'm a big fan of Arthur Ganson from MIT, who can be googled to great effect.>

 

[MrAl]

Hi,

This is usually referred to as the mechanical equivalent of heat. Watts causes heat, mechanical action can cause heat. The two are equated through the heat they both produce. You'll also note that there is a conversion for watts into horsepower and vice versa.

 

[MFrankly]

So an electrical motor might be able to do 745 watts worth of work, but since nothing is perfect, it might also create 15 watts of waste heat. Would you then account for 800 watts in your circuit?

Voltage absent resistance is 0. So an unconnected battery rated for, say, 9 volts, (which is the same amount of energy as 9 newtons of *push* over 1 meter). But there is no circuit, so Ohm's law doesn't apply. But is it still said to have 9 volts of potential energy because that's what happens when you plug it into a circuit, or is that integral to the battery?

Sorry about all this thinking out loud, but I have to chew through things so that I have a mental image to work with. Thank you for your help!

 

[Ratchit]

MFrankly,

I've got a basic conceptual question. I have a hard time wrapping my head around using watts as a measure for work - moving a weight a given distance - and as a measure for dissipating heat and light.

You got a basic definition problem. Watts is a measure of power, not work. It is the rate of work or energy.

I can't use my light bulb to move my toy car, so how can I say it is rated at 100 watts?

That is the amount of power it is dissipating. It is dissipating 100 joules/sec.

Or that a resistor can be rated to survive producing 0.25 watts of waste heat? I certainly understand the idea of wasted work, but how are waste heat and waste work the same? Do I have to just buy into it?

Nope, you just have to understand the difference between work or energy and power. The above resistor is dissipating 0.25 joule/sec. Heat and work are different energy forms, but both are energy.

So an electrical motor might be able to do 745 watts worth of work, but since nothing is perfect, it might also create 15 watts of waste heat. Would you then account for 800 watts in your circuit?

No, you would have to supply energy at the rate of 745+15=760 watts to the circuit.

So an unconnected battery rated for, say, 9 volts, (which is the same amount of energy as 9 newtons of *push* over 1 meter). But there is no circuit, so Ohm's law doesn't apply. But is it still said to have 9 volts of potential energy because that's what happens when you plug it into a circuit, or is that integral to the battery?

Voltage is not energy, it is the energy density of the charge, measured in MKS units of joules/coulomb. 9 volts means it takes 9 joules to move 1 coulomb of charge from one point to another in a particular electric field.

Ratch

 

[MFrankly]

Okay, thanks. A few more:

- Work is a form of energy? This is the part I'm having a problem with forming a mental workspace around. Watts are unit-of-work per second. Can I think of the Brownian motion as accounting for the things being worked on? Will that mental image get me into any trouble?

- Yes, yes, 745+15 is 760.

- So will motors rated for output or input?

- What do you mean by 'particular electric field'? That's not necessarily a circuit, I'm guessing, or you would have used the word 'circuit'. Is that why you can refer to a 9v battery even though there is no resistance or current?

Pedant away!

 

[Dr_Doggy]

KWH= kilowatt hour = 1000 joules
1joule = watts/time = work
1000watts = 1HP(horsepower)
power = energy(watts )
voltage = potential energy (force exerted)
current = flow RATE (amps or velocity)
resistance = opposing forces (resistors dont go to 0 only current does, except in superconductors)

joules = P/t
P = VI
V = IR

resistance is everywhere all the time, 9v battery(unplugged) = I(unknown current) * R(HIGH load in air)
9v = I(amps) * 10,000,000ohms

= very low current draining from battery(unplugged), that is why a battery goes dead if you leave it for a year

 

[Ratchit]

MFrankly,

- Work is a form of energy? This is the part I'm having a problem with forming a mental workspace around. Watts are unit-of-work per second. Can I think of the Brownian motion as accounting for the things being worked on? Will that mental image get me into any trouble?

Energy is defined as the ability to do work. Brownian motion does not enter into consideration from the macroscopic viewpoint of electric circuits. Electromagnetic fields are what you want to study if you delve into the whys and wherefores of charge behavior.

- So will motors rated for output or input?

Eh???

- What do you mean by 'particular electric field'? That's not necessarily a circuit, I'm guessing, or you would have used the word 'circuit'. Is that why you can refer to a 9v battery even though there is no resistance or current?

I should have said "any" electric field. What does a 9 volt battery have to do with the resistance of a circuit?

Ratch

 

[Ratchit]

doggy,

KWH= kilowatt hour = 1000 joules

1 kwh = 3,600,000 joules

1000watts = 1HP(horsepower)

1 hp = 745.70 watts

power = energy(watts )

power = energy/time

voltage = potential energy (force exerted)

voltage = energy/charge

current = flow RATE (amps or velocity)

current = charge flow/time

resistance = opposing forces (resistors dont go to 0 only current does, except in superconductors)

resistance = opposition to charge flow

joules = P/t

joules = Pt

Back to school for you.

Ratch

 

[MFrankly]

Doggy - that completely cleared up the battery for me, thanks!

The question about motors (message typed one-handed as I rocked a baby to sleep) should be: If I look at a motor casing that reads 100 watts, will that indicate the wattage I should account for when I design the circuit, or will that be the amount of work that the motor can do, or is the difference negligible?

So: the heat is energy (the ability to do work) that is not being harnessed to do any work?

The thing (and this is not really a question, just something I'll have to ponder further - unless you have a great way to explain it, of course) that I keep coming back to is that the definition of work seems so specifically about moving a physical object, but in electrical circuits there are seldom moving parts. But there's a lot of watts floating around (work-per-second), so clearly a lot of work is, in fact, being done. I just need a way to think about it.

 

[Ratchit]

MFrankly,

The question about motors (message typed one-handed as I rocked a baby to sleep) should be: If I look at a motor casing that reads 100 watts, will that indicate the wattage I should account for when I design the circuit, or will that be the amount of work that the motor can do, or is the difference negligible?

The amount of work the motor is rated to do, unless you want to overload and abuse it. You should also design the power supply somewhat higher to account for starting and possible overloads.

So: the heat is energy (the ability to do work) that is not being harnessed to do any work?

The thing (and this is not really a question, just something I'll have to ponder further - unless you have a great way to explain it, of course) that I keep coming back to is that the definition of work seems so specifically about moving a physical object, but in electrical circuits there are seldom moving parts. But there's a lot of watts floating around (work-per-second), so clearly a lot of work is, in fact, being done. I just need a way to think about it.

A boiling water atomic reactor harvests the heat and converts it into electrical energy. You have kinetic energy, potential energy, electrical energy, rotational energy, force with motion energy, and countless others. They all have the ability to do work. Perhaps you should read a elementary text on thermodynamics, the science of energy.

Ratch

 

[ljcox]

If you put a rope over a pulley & attach a weight to it, then start pulling on the rope to raise the weight, you are doing work to raise it & thus converting chemical energy stored in your body into potential energy,

Question for you.

If you pull harder & therefore work harder, but raise the weight to the same height, have you done the same work or more work?

 

[misterT]

If I look at a motor casing that reads 100 watts, will that indicate the wattage I should account for when I design the circuit, or will that be the amount of work that the motor can do, or is the difference negligible?

The wattage rating of motor comes roghly from its nominal working point: Nominal voltage and nominal current.
- Nominal voltage follows from maximum no-load speed of the motor.
- Nominal current derives from the motor's thermally maximum permissible continuous current.
- It is ok to overload motors for short periods of time.

Electric motor drive converts incoming electric energy into outgoing mechanical energy. Losses occur during the transformation of the energy which in the simplest case are defined by the efficiencies of the corresponding components.. More electrical energy must be fed into the system than can ultimately be utilized.

Considering only the power rating of a motor can be very misleading. For example with brushless DC motors, which are assigned a high power rating on account of the high maximum speeds. If this high speed of rotation is not needed or not utilized, it is often the case that a motor that is appropriate based on its power rating is unable to produce the required torque. The essential requirement is to meet the demand for speed and torque independently.

I'm in the middle of designing a drive system for a mobile robot. I have chosen a 60W brushless motor for the job, and my calculations show me that I need a max load current of 2.6 A and max motor voltage of 21V.. So, I need a motor controller that can meet those requirements. In addition to that the controller must be able to handle a starting current of 17A for a short period of time.

Short answer to your question: Don't use the wattage rating for anything when designing a motor controller circuit.

 

[MFrankly]

Len: My instinct says that is the same work since the same weight moved the same distance against the same gravity. However, a variable has changed: the muscles are putting more energy into the system, so something else has to give. I would say that time has to be what gives. Same work; faster rate. Watts are the rate of work (joules are the actual measure of work itself, right?) So we've increased the wattage.

Coming back to MrAl's comment: the thing that ties all of this together is heat; the mechanical equivalent of heat. That I'll have to investigate further; but for now I can accept that I can measure all of these with the same unit because they convert into heat and back.

Ratch: I'm already reading an introductory textbook (electronics) more or less for fun. I'll put thermodynamics next on my list, thank you, but for now there seem to be people here who are willing and able to explain some of my questions. If I stop and go to a more fundamental text every time I have a question, I'll wind up in pure mathematics, from which there is no escape. <grin>

 

[ljcox]

Len: My instinct says that is the same work since the same weight moved the same distance against the same gravity. However, a variable has changed: the muscles are putting more energy into the system, so something else has to give. I would say that time has to be what gives. Same work; faster rate. Watts are the rate of work (joules are the actual measure of work itself, right?) So we've increased the wattage.

Yes, the extra effort goes initially into Kinetic Energy as the weight is moving faster.

But then, the person pulling has to reduce his effort as the weight approaches the required height so the KE is transferred to PE.

Thus the same work is done.

Of course, you have to assume that the human body is 100% efficient in converting chemical energy into work & that there is no loss of energy in the pulley & rope.

 

[Roff]

The body doesn't work any harder to move the weight faster. It expends more peak power, but over a shorter period of time. Work is power x time (actually, work is the time integral of power).

 

[ljcox]

Roff,
You need to check your Physics book.

Mine states that W = F . d where F & d are the force & distance vectors.

 

[MFrankly]

Okay! This has been useful.

 

[Roff]

Roff,
You need to check your Physics book.

Mine states that W = F . d where F & d are the force & distance vectors.

Well, in the textbooks, the work required to lift an object is definitely W=F*d, but F is defined as mass*acceleration, where acceleration is that of gravity, ≈9.8m/sec².
The work required by the added acceleration to get the object up to a constant speed is apparently cancelled by the reduced work required when the object is decelerating to rest.

 

[ljcox]

Well, in the textbooks, the work required to lift an object is definitely W=F*d, but F is defined as mass*acceleration, where acceleration is that of gravity, ≈9.8m/sec².
The work required by the added acceleration to get the object up to a constant speed is apparently cancelled by the reduced work required when the object is decelerating to rest.

Roff,
You're confusing 2 issues.

F = m a indicates that if a force F is applied to a mass m, then the acceleration is a m/sec/sec.

If you want to express a force in Newton, then F = 9.8 * f Newton where f = the force in kg weight.

So a force of 1 kg wt = 9.8 N.

eg. If we accelerate a mass of 6 kg with a force of 18 N, then the acceleration is a = 18/6 = 3 metre/sec/sec

 

[Roff]

Roff,
You're confusing 2 issues.

F = m a indicates that if a force F is applied to a mass m, then the acceleration is a m/sec/sec.

If you want to express a force in Newton, then F = 9.8 * f Newton where f = the force in kg weight.

So a force of 1 kg wt = 9.8 N.

eg. If we accelerate a mass of 6 kg with a force of 18 N, then the acceleration is a = 18/6 = 3 metre/sec/sec

All I'm saying is, the work W required to lift (raise) a mass m (on Earth), at a constant rate, by a distance d, is W= m*d*G, where G is the acceleration of gravity on Earth.
The force required to do this is m*G, which is, of course, the weight of the object.