# X Watts to run = x Watts of heat?

• TSN79

#### TSN79

If some appliance needs let's say 500 Watts to run, does that automatically mean that it also produces 500 Watts worth of heat to its surroundings? Or would that be an indication that it's hopelessly inefficient?

No, you need to look at the useful work (actually power) done on a case-by-case basis. For example:

1. All of the electricity entering a toaster is converted to heat. Since the purpose of a toaster is to produce heat, it is actually 100% efficient at its intended task.

2. Electric motors are about 90% efficient, that is, 90% of the electric power is converted to mechanical power. Your garbage disposal transforms most of the electric power into grinding carrot peels, the fan motor in your furnace moves air, and so on.

3. To analyze the efficiency of heat engines like refrigerators or air conditioners, you need to know some thermodynamics.

The first and second laws of thermo are what you are getting at. You could also look into availability as well. It does depend on the control volume and the results you wish to obtain.

If an appliance needs 500 W to run, that means that a percentage, say 80% will usually go to the intended purpose to produce work. The rest will go into the irreversibilities and heat transfers of that system.

So, in short, unless you are talking about a heat producing element like a resistive heat element, the answer is no.

If some appliance needs let's say 500 Watts to run, does that automatically mean that it also produces 500 Watts worth of heat to its surroundings?
Yes.

a toaster [..] is actually 100% efficient at its intended task.

2. Electric motors are about 90% efficient, that is, 90% of the electric power is converted to mechanical power.
The toaster oven is not 100% efficient (only the space heater can be), because some of the toaster's heat goes direct to the environment (rather than first increasing the bread's temperature, then into the environment later as the toast cools). Though it may be possible in principle to make a toaster produce electricity overall, considering that oxidising bread (like other fuels) releases energy itself.

If an appliance needs 500 W to run, that means that a percentage, say 80% will usually go to the intended purpose to produce work. The rest will go into the irreversibilities and heat transfers of that system.

So, in short, unless you are talking about a heat producing element like a resistive heat element, the answer is no.
Incorrect. Sure, your 80% efficient appliance will do 400W of work, and produce 100W of heat internally. But that work all gets dissipated (mostly straight through simple friction - you're certainly not spinning anything up to infinite rpm) producing another 400W of heat. So, if your appliance gets 500W of power, it ultimately produces 500W of heat overall. Only loopholes are that you can choose whether to sit close by the hottest end of the appliance (reverse-cycle airconditioning), or you can change the potential of some consumable (evaporative airconditioning).

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Hmm, I don't know about blanket statements. Here's another example of extracting mechanical work: The motor driving a well pump (we have a lot of them here in the West) converts most of the electric energy into work to raise water. Not heat.

Incorrect. Sure, your 80% efficient appliance will do 400W of work, and produce 100W of heat internally. But that work all gets dissipated (mostly straight through simple friction - you're certainly not spinning anything up to infinite rpm) producing another 400W of heat. So, if your appliance gets 500W of power, it ultimately produces 500W of heat overall. Only loopholes are that you can choose whether to sit close by the hottest end of the appliance (reverse-cycle airconditioning), or you can change the potential of some consumable (evaporative airconditioning).
All power is not dissipated as heat. It is only necessary to have all power dissipated as heat for purposes where it is needed. An example would be some sort of heat element. Take an ordinary appliance where all power gets converted to heat, we have an inefficient device that will not perform its intended function becuase of wasted power. And also, all work does not eventually end up as heat. I'm not sure why you have this idea that it is converted to heat by friction.

marcus, that's what i meant by changing the potential of a consumable.
ranger, would you give an example?

Is all power in a battery charger lost as heat? If so, what's left for normal operation? Notice I said normal, becuase if we dissipate the equivalent of the max power the device uses as heat, we run a risk of damage.

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There are many other losses than just heat. For instance, in a motor we can have vibrational loss, heat loss, wind resistance losses on the spinning armature, more losses from spinning the fan. Also, having weak magnets in a permanent magnet DC motor would mean more current would be required to drive the same load. I did an analysis on a series wound and shunt wound DC motor.

For the series wound motor, I managed to get a maximum of 53% efficiency with 147 watts lost at full load (9lb-in). The shunt wound motor had similar losses.

The efficiency that triden mentions is another point to prove that not all power is wasted. Take an amplifier (class A and D). A class A amplifier has the worst power efficiency. With only about 10%-25% of the total power going to output to do work. Compare this with a class D which is far more efficient. More that 90% of the power is used to do work. Whatever the varying degrees of power efficiency, not all is wasted as heat.

Incorrect. Sure, your 80% efficient appliance will do 400W of work, and produce 100W of heat internally. But that work all gets dissipated (mostly straight through simple friction - you're certainly not spinning anything up to infinite rpm) producing another 400W of heat. So, if your appliance gets 500W of power, it ultimately produces 500W of heat overall. Only loopholes are that you can choose whether to sit close by the hottest end of the appliance (reverse-cycle airconditioning), or you can change the potential of some consumable (evaporative airconditioning).
Ummmm...no. You need to rethink your thermo. Work in and of itself is an end. Think about a turbine that extracts work (power). The end result is a lower temperature and the only heat produced is due to inefficiencies. There is also sound production as well.

Is all power in a battery charger lost as heat? If so, what's left for normal operation? Notice I said normal, becuase if we dissipate the equivalent of the max power the device uses as heat, we run a risk of damage.
A battery charger charging a battery isn't an end-use, it is an intermediate storage. Do you have any examples of end-uses that don't eventually end up as heat? Your car, radio, lights, dishwasher, hvac, electric shaver, etc. - the energy used to power all of these devices eventually ends up as heat.

Is all power in a battery charger lost as heat?
This doesn't add more to marcus' example. (What do you do with a charged battery?)

There are many other losses than just heat. [eg. vibration]
There is a simple experiment where a propeller/stirrer is placed in an insulated cup of water, a string is wound around the propeller's axle, the end of the string is attached to a weight, and the weight is dropped to the floor. You calculate the difference in gravitational potential, then measure the heat change.

Sure, we could point a laser at the sky, and perhaps it's a matter of semantics whether you accept, ultimately, that you've just raised the temperature very slightly over whichever astronomical length of space it took to completely attenuate the signal (versus insisting that not all of the energy used in the laser pointer was converted to heat in the backyard). But in the case of ordinary home appliance use, there seems less to dispute.

[..] to prove that not all power is wasted. Take an amplifier [..]

Say your efficient amplifier outputs a joule of energy every second. This energy won't just keep piling up in the speaker. Say 100% is converted to sound in some room. As more and more energy is pumped in, will the sound inside the room get louder each hour? Or is the sound constantly disipating (at the same rate of a joule per seceond) into other forms of energy?

[..]The end result is a lower temperature[..]
I have no idea what you are thinking about.

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cesiumfrog, just think about a motor with wheels. When you use up the battery, some goes into heat, but some goes into the kinetic energy of the car.

You are saying everything eventually goes into heat in some form or another. Ok, fine...but that is a useless definition becuase then it makes no sense to have efficiency (thermodynamic or otherwise).

Take the space shuttle. When it gets into orbit it has used up all its chemical energy in the boosters and converted it into kinetic energy and potential energy. Is that lost to heat? No. (obviously, unless it decides to come back down)

Another way, I can put energy into form a new chemical. Is that lost as heat? Not really. Its STORED as bond energy.

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This doesn't add more to marcus' example. (What do you do with a charged battery?)
OK, so you and Russ are considering what will eventually be done with the battery, fine. When I gave my example about the battery charger, I did not care about the battery. I was simply proving that not all power will be lost as heat for the battery charger system. But if you insist on including the battery, is heat only option? What about kinetic energy or bringing about chemical reactions and so on?

Say your efficient amplifier outputs a joule of energy every second. This energy won't just keep piling up in the speaker. Say 100% is converted to sound in some room. As more and more energy is pumped in, will the sound inside the room get louder each hour? Or is the sound constantly disipating (at the same rate of a joule per seceond) into other forms of energy?
The entire point of power efficiency for an amplifier is to tell how much power will be used on the output to do work. But why limit the application to sound?

I think that you will find this thread going beyond home appliances, becuase it sounds like you are saying that everything will end up as heat.

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OK, so you and Russ are considering what will eventually be done with the battery, fine. When I gave my example about the battery charger, I did not care about the battery. I was simply proving that not all power will be lost as heat for the battery charger system. But if you insist on including the battery, is heat only option? What about kinetic energy or bringing about chemical reactions and so on?
If you want to be that nitpicky about it, no battery holds its charge forever, so even chemical energy stored in a battery still ends up as heat, even if it takes a few years.

And kinetic energy? What is heat...?

What does it mean to talk about efficiency?

It is defined as the ratio of the amount of useful work done by the system divided by the amount of work provided into the system.

When you say it all gets converted to heat, is meaningless in how we define efficiency. The energy getting converted into heat is after the fact it did something useful first.

I am not saying your wrong that it all eventually goes into heat; what I am saying is that you are wrong to say that it is inefficient, because that is not what efficiency means in the first place.

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I have no idea what you are thinking about.
An adiabatic turbine (which is not a bad assumption in a lot of cases) will, as with any expansion process, decrease the temperature of the flow that the work is extracted from. There is some unsed, available energy going out the back end of the turbine due to inefficiencies, but the work extracted goes to the actual motion of the turbine and to the processes downstream that use that work, the sound produced, bonds in lubricants get broken...For that component, the entirety of the input is not converted to heat. I can concede that if you follow the chain of events far enough to all of the items downstream that interact with it, that the universe experiences an overall increase in temperature because of it.

What does it mean to talk about efficiency?

It is defined as the ratio of the amount of useful work done by the system divided by the amount of work provided into the system.

When you say it all gets converted to heat, is meaningless in how we define efficiency. The energy getting converted into heat is after the fact it did something useful first.

I am not saying your wrong that it all eventually goes into heat; what I am saying is that you are wrong to say that it is inefficient, because that is not what efficiency means in the first place.
The word "efficiency" does not appear anywhere in my posts and you are absolutely right that what I'm saying doesn't have anything to do with efficiency.

But it is still an important concept to understand, especially in the HVAC business, since you need to know how much heat is going into a building in order to know how much cooling it requires.

And it seems to me that understanding where energy eventually goes is what the OP was asking about. Ie, the OP is more or less correct (tough to tell exactly, because the wording is a bit sloppy) that essentially all work ends up as heat, but what he was missing is that efficiency is based on the useful work output, not the eventual resulting heat.

Take my telescope, for example. Electric motors are upwards of 90% efficient at converting electrical energy to mechanical. The efficiency of the motor is measured at its output shaft, dividing mechanical work out by electrial energy in. But where does that mechanical energy go next? All of it is dissipated as heat from friction in the bearings and gears of the telescope.

I'm doing some work at a cocoa factory right now. They use 160kW motors in large grinding machines to grind up cocoa beans (it is a fascinating job). These motors are probably around 95% efficient, which means about 8kW is dissipated as heat from the motor jacket and the rest goes into the grinding of the beans. That remaining 152kW all becomes heat, which must then be removed from the grinding machines, through two coolant water systems, one on the vessel of the grinder and one in the gearbox and oil reservoir of the machine.

Right, the only reason I said that is because the OP said:

If some appliance needs let's say 500 Watts to run, does that automatically mean that it also produces 500 Watts worth of heat to its surroundings? Or would that be an indication that it's hopelessly inefficient?

And that last sentence had to be addressed. You guys are not wrong, but no one clarified that last part on efficiency.

Hmm, I don't know about blanket statements. Here's another example of extracting mechanical work: The motor driving a well pump (we have a lot of them here in the West) converts most of the electric energy into work to raise water. Not heat.
Yeah, that's a pretty good example, doing work against potential energy.

Doing work against chemical energy would be another, but that's a tougher one since it is process dependent.

If you want to be that nitpicky about it, no battery holds its charge forever, so even chemical energy stored in a battery still ends up as heat, even if it takes a few years.

And kinetic energy? What is heat...?

OK. I see your point about kinetic energy and heat. I know that heat can be said to be vibrations at a microscopic level, but this is not what I meant. And I was not talking about chem energy as stored in the battery, but in end uses of the battery. Here is another example, how about changing the potential energy of an object. Not all is lost as heat, if it were, then how would you explain the gain in PE?

I'm doing some work at a cocoa factory right now. They use 160kW motors in large grinding machines to grind up cocoa beans (it is a fascinating job). These motors are probably around 95% efficient, which means about 8kW is dissipated as heat from the motor jacket and the rest goes into the grinding of the beans. That remaining 152kW all becomes heat, which must then be removed from the grinding machines, through two coolant water systems, one on the vessel of the grinder and one in the gearbox and oil reservoir of the machine.

Can you take pictures/photos of the plant operations? I would like to see that. I've been to the UTZ potato chip factory once. They had long drums with serrated grooves. The potato would come in one end, roll around inside these long horizontal drums, and come out pealed on the other end. They would bring in potatos literally by the dump truck load.

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Here is another example, how about changing the potential energy of an object. Not all is lost as heat, if it were, then how would you explain the gain in PE?
That was marcus's example and it was a good one. There are few devices that are like that, though.

All I'm trying to do here is make sure people understand that for a very high fraction (but no, not all) of devices that use energy, all of the energy being used eventually (generally, very quickly) ends up as heat.

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Can you take pictures/photos of the plant operations? I would like to see that. I've been to the UTZ potato chip factory once. They had long drums with serrated grooves. The potato would come in one end, roll around inside these long horizontal drums, and come out pealed on the other end. They would bring in potatos literally by the dump truck load.
This place has a hydaulic platform that picks up a truck and tilts it back to dump the beans into a receiver (several truckloads a day).

I'm not sure about the pictures - not all that much looks that interesting and it is pretty spread out, but I'll see what I have.

I have seen this question coming up many times in many forums. We are all right that total energy shouldn't be considered as heat load when calculating heat load of a controlled space.

For example, if we place a water cooled chiller along with motor in a controlled space, as per ASHRAE, total power of motor is to be considered as heat load to the controlled space. Suppose, the energy is added to the controlled space as heat load, then cooling tower capacity just equals the refrigeration effect. In actual case, this heatload is added to cooling tower water.

However, both Carrier and ASHRAE stick to this conservative method and so we all go with the wind.

Can you cite where ASRHAE says that? It is, of course, incorrect and would lead to problems if a building were designed that way.

I had a client with an MRI facility, designed by another engineer, that was so humid it had water dripping down the walls in the room with the MRI machine. The MRI had it's own water cooled chiller located in an adjacent room (both rooms conditioned). The MRI room itself had essentially no load. The HVAC unit for the MRI room was a too big 2-stage light commercial unit and as a result, supplied 65 degree air intermittently to a room that was kept at 67.

I am not saying that ASHRAE specifies this method for chiller room ventilation.

However, I mean they don't deal with the situations on a case to case basis. The heat load from the equipment is categorized into 3 parts.
1. Machine in and driver out,
2.Machine out and driver in and
3. Machine in and driver in.

Even the recent ASHRAE HOF 2005 deals the same way and there is no specific information. But, I feel, as long as these kind of discussions appear, there will be less room for mistakes.