Interesting argument between friends

[russ_watters]

The suspense is killing me,

Ima guess on the fan blades,

That's where the electrical energy is transferred to the air (via the motor and fan shaft, which are better than 90% efficient together), but what I meant is where is it converted to heat. There is no consensus in my office.

All agree that due to the inefficiency of the fan, a decent fraction immediately becomes heat. But what about the rest? My position is that it is converted everywhere you see a pressure loss, in proportion to the loss: across coils and dampers, due to friction in ducts, etc.

 

[Floid]

In an ideal case with a computer actively consuming 1000W and a heater actively consuming 1000W the heat produced would be fairly close. Most power consumed by a computer is converted to heat. The exceptions I can think of are a minor amount stored in capacitors, spinning of disc drives/fans, LEDs and speakers, power that leaves to go to external to the computer (Ethernet for example), and stray radiation. In all of those cases the end result will be some additional heat generation (friction from the disc drives for example), but some is converted to mechanical, stored, or radiated energy.

Where the thought experiment falls short in reality is a computer with a 1000W power supply will rarely if ever actually draws 1000W.

 

[yungman]

I am not a power engineer. Please explain from my drawing what is the power drawn from the battery. Assuming capacitor C is ideal and lossless, assuming the on channel resistance of the MOSFET is 0Ω so no ohmic loss.

When IN is low, Q1 turns on, voltage from the battery is connect to the C and charge it up as shown in Loop1. When IN switch to high, Q1 turn off and Q2 turn on, the C is discharged by Q2 through Loop2.

Q1, Q2 and C are inside an isolated enclosed place in the dashed box in . Battery and the IN driver is outside of the box.

1) What is the power input to the box when IN is pulsing to charge and discharge the cap continuously?

2) If there is real power input, how is heat generated inside the box as there is no resistance?

3) If there is no power input, do we expect the battery to last forever in this ideal case?

 

[rbj]

I am not a power engineer. Please explain from my drawing what is the power drawn from the battery. Assuming capacitor C is ideal and lossless, assuming the on channel resistance of the MOSFET is 0Ω so no ohmic loss.

why not assume it has negative resistance?

the issue is, yung, that when that cap discharges, the energy stored in the cap ends up somewhere. where do you think it goes?

better clip on some heat sinks to the MOSFETs.

 

[rbj]

a similar question is had when you get some big capacitor (say, 1000 uF and that is good for 50 volts) and you charge it up with a DC supply to, say 40 volts, disconnect the DC supply and then, using a screwdriver, short out the cap. what happens? where did the energy come from? and where did it go?

 

[yungman]

why not assume it has negative resistance?

the issue is, yung, that when that cap discharges, the energy stored in the cap ends up somewhere. where do you think it goes?

better clip on some heat sinks to the MOSFETs.

My name is Yungman or Alan

I am using an ideal case, there is no ohmic loss as a start to look at whether there is power involve in an ideal case.

For real case, you can use a MOSFET with very low output impedance to sink or source large current with very little heat. eg, if on channel resistance is 10mΩ, 1A only generate 10mV and power dissipation is very low even if charging a cap to 10V.

In computer, almost all circuits are MOSFETs, the biggest biggest problem is to charge and discharge the input and the line capacitance to achieve the logic level. Input capacitance and drive is THE single major issue of the speed limit. They get away with higher and higher speed inside the CPU only when they shrink to transistor down to lower the input capacitance. That's the reason they manage to run GHz inside the chip. But any external bus is still slow because of the trace and input capacitance need to be driven. The output has to be buffered over and over so the speed goes down. I want to see how driving a cap don't consume power.

 

[rbj]

Alan, no matter what you want to assume, when the transistors are ON, their resistance is at least as much as a wire of short length. when Q2 discharges the capacitor, if there is no other resistance in the loop, that energy will be dissipated in whatever resistance that current sees. in the wires, even those inside the capacitor, the soldered nodes, and in any other conductive element in the loop (i.e. Q2).

specifically, getting back to the original question, a computer consuming 1 kW is rather unlikely, but a computer left on that continually draws W watts will dissipate nearly the very same amount of heat and heat the room just as well as a space heater that also consumes the same W watts. only energy that somehow crosses the boundary between what is inside the room and what is outside (sound leaking out, light going out a glass window), only that energy will not heat the room.

a battery charger (drawing W watts) will heat the room less if energy is going into and stored in the battery. an electric motor (drawing W watts) with a shaft going through the wall to the outside will heat the room less than a space heater consuming the same W watts.

 

[yungman]

I know your argument, I am just confused about the current need to charge and discharge the cap. I just take everything ideal and look at one part. Assuming there is no ohmic loss, is there power deliver into the box?

 

[Runei]

If there are no resistances anywhere in that circuit. The circuit will not consume any energy.

The charge stored in the capacitor will "return" to the flow when the capacitor is discharges and then charged again. We we were talking about positive charges moving, they would first move onto the positive part of the capacitor (charging it). Then they would move around loop 2 and end up on on the bottom (discharging the capacitor).

When the capacitor is charged again, these positive charges would flow from the ground node and through the voltage source.

So the circuit would be a perfect, forever circuit (if your voltage source was ideal also).

 

[yungman]

I don't see how, the current that charge the cap is from the battery with voltage, the current discharge the cap is by Q2, nothing to do with the battery. Far as the battery concern, it only see the charging cycle time after time.

I need to be convinced that there is no power transfer from the battery into the box using an idea case. Then worry about the real resistance that cause power loss and heat.

 

[Runei]

Actually it's not so easy at all. You would have infinite currents somewhere and its generally stupid to think about it this way.

There WOULD be resistances everywhere, and because of this, the energy is dissipated in these resistances.

As Russ has stated several times, almost all energy in a circuit will be dissipated as HEAT in that same circuit. A little bit of energy will perhaps be radiated as radiowaves, but again, these will turn into heat when they hit the walls of the house.

So for all practical purposes, the energy consumed by a computer will be dissipated in that computer as heat because of resistances in that circuit.

 

[Runei]

As soon as you add a resistance in the wires all these problems dissapear, and all power equations will be satisfied.

The power in supplied by the voltage source, will be dissipated by the resistances.

By wanting to have no resistances you present an impossible scenario with infinite current. Infinite current means the electrons move infinitely fast. They will bang into stuff when they reach the end of the wire and be dissipated as heat.

But as soon as you add just 1 Ohm of resistance in the wires everything becomes apparent = The power is dissipated in the resistances.

 

[russ_watters]

I know your argument, I am just confused about the current need to charge and discharge the cap. I just take everything ideal and look at one part. Assuming there is no ohmic loss, is there power deliver into the box?

Setting aside the issue of infinite current, if the capacitor is ideal and there is no resistance elsewhere, there us no power either used or dissipated.

But how does that help us?

 

[russ_watters]

I don't see how, the current that charge the cap is from the battery with voltage, the current discharge the cap is by Q2, nothing to do with the battery. Far as the battery concern, it only see the charging cycle time after time.

I need to be convinced that there is no power transfer from the battery into the box using an idea case. Then worry about the real resistance that cause power loss and heat.

If you write a conservation of energy statement to calculate that, it reduces to 1=0 or gives a divide by zero error, so that's why assuming no resistance doesn't clarify things:

I^2R=VI

Set R=0...

 

[cabraham]

With non-zero R value in the cap charge/discharge network, the power lost is the same regardless of value, as long as the RC time constant is short compared to the period of switching. An Rdson in the FET of 0.10 ohm, 0.010 ohm, or 0.001 ohm results in the same power dissipation.

If the Rdson is actually 0 ohm, the current is still limited, as it cannot be infinite due to the inherent inductance present. Since the energy stored in the cap is not returned to the battery it is either dissipated or radiated. In the R=0 case it is radiated. The parasitic inductance value of the FET, herein "L", forms an LC network with the capacitor. When the cap is fully charged w/ the FET off there is energy W = C*V^2/2. Then the FET turns on. The cap discharges into L, the FET inductance, which results in L storing energy. When the cap energy is depleted, V=0, but I continues through L charging C negatively. This is a resonant LC network.

Oscillation takes place & energy is radiated. The small value of L results in a high resonant frequency. If the energy is not dissipated as heat, as in the nonzero R case, then it is radiated due to LC resonance. It does not recharge the battery because there looks to be no path for that to happen.

A real world FET will have nonzero Rdson, so I don't think we should dwell on radiated power. The energy in the cap is dissipated in Rdson on a per cycle basis. Anything unclear can be explained if needed.

Claude

 

[jim hardy]

i agree with russ

thought experiment:

place that computer (tv, stereo, whatever) inside a black box in center of the room.

Whatever energy goes in (1kw of electrical) must come out into room

light will impinge on walls of black box and warm them
sound will echo around inside till it is converted to heat warming the air(recall Newton's error in calculating speed of sound)
RF will be absorbed by the computer's internal shielding and warm it

the heater FEELS warmer because it radiates heat at higher temperature than the others

but recall temperature is not heat.

my 2cents

old jim

 

[JustNobody]

This seems to be the apex of the argument. While I agree that the "black box" concept is true, recall that our goal is to HEAT the room. To do this we must add thermal energy to the room faster than the surroundings can take it away. While at the end of the day all the energy will go to thermal energy (as everything does in the universe), a heater turns "nearly" all of it into thermal energy, while with something like a computer or a television the energy gets "sloshed around" via other means and does not contribute to heating the room when the heat is needed

 

[Floid]

But again that goes back to the difference in heat and temperature. I don't think anyone would argue that a heater is more efficient at changing the temperature in a room than a computer, but that doesn't mean that the two aren't producing the same amount of heat.

The biggest difference there is a heater is designed to transfer thermal energy to the room very rapidly.

In a computer the primary source of heat transfer is heat sinks. Heat sinks are designed to have a higher thermal mass, meaning they transfer thermal energy slowly (a high thermal mass can store a lot of thermal energy). This means they aren't going to quickly change the temperature of a room, even though the same amount of heat is involved in the two scenarios because the heat sink stores the thermal energy and releases it over a longer period of time.

 

[russ_watters]

This seems to be the apex of the argument. While I agree that the "black box" concept is true, recall that our goal is to HEAT the room. To do this we must add thermal energy to the room faster than the surroundings can take it away. While at the end of the day all the energy will go to thermal energy (as everything does in the universe), a heater turns "nearly" all of it into thermal energy, while with something like a computer or a television the energy gets "sloshed around" via other means and does not contribute to heating the room when the heat is needed

While strictly speaking you are correct, in practical terms the difference is negligible since the only portion that 'sloshes around' for more than a few seconds, is the heat absorbed by the mass of the computer. That's a small amount and it only causes a few minutes of lag for that small amount of heat it holds.

 

[sophiecentaur]

I have just laboured my way through this thread, reading as much as I could bring myself to and scanning through the rest of it. It amazes me that otherwise sensible people seem to be arguing about the principle of Conservation Of Energy in the specific case of someone's front room! How can it not apply here, same as in the nucleus and everywhere else?
Unless the energy leaves the room / data centre in the form of sound (I don't think so) or EM radiation (a few Watts at the most) it must end up by heating up the air, walls or people in there.
Why are you wasting time trying to show that energy in a circuit magically disappears because Mosfets and Capacitors are involved (are the REALLY that ideal?). It's crazy; isn't this supposed to be a Physics and not a Magic forum?

Where heating is concerned, there are a lot of details involved in where the heat ends up in the room and how our body actually becomes aware of it. Cold feet in an otherwise warm room is not comfortable, for instance. But that is an issue that heating engineers and serious environmental engineers know all about. A watt hour is a watt hour and it's not how you produce it that counts - it's how you use it.
End of rant.

 

[LawRooney]

If we are talking about purely heat, then yes, I guess they would give off the same amount of heat.

But if some guy is trying to argue that using a 1000W computer will keep you as warm as a 1000W heater will, I'd say that guy definitely needs to switch majors.

I guarantee a heater gives off more heat per second than a computer.

 

[Averagesupernova]

If we are talking about purely heat, then yes, I guess they would give off the same amount of heat.

But if some guy is trying to argue that using a 1000W computer will keep you as warm as a 1000W heater will, I'd say that guy definitely needs to switch majors.

I guarantee a heater gives off more heat per second than a computer.

A large portion of this thread has had discussion that disagrees with you.

 

[uart]

I have just laboured my way through this thread, reading as much as I could bring myself to and scanning through the rest of it. It amazes me that otherwise sensible people seem to be arguing about the principle of Conservation Of Energy

I totally agree. I've given up on this thread already. Russ probably needs to post the facts one final time and lock this.

Maybe I could suggest that many of the responders here could then head on over to the general math forum and start a new topic on 0.\dot{9} \neq 1. Hurkyl would just love that.

 

[milesyoung]

I guarantee a heater gives off more heat per second than a computer.

Of course it won't. A heater might be more effective at circulating the air in the room but that doesn't change the fact that all but a few watts of the power the computer draws is heat transfer to its surroundings. At least argue why you think otherwise.

 

[sophiecentaur]

If we are talking about purely heat, then yes, I guess they would give off the same amount of heat.

But if some guy is trying to argue that using a 1000W computer will keep you as warm as a 1000W heater will, I'd say that guy definitely needs to switch majors.

I guarantee a heater gives off more heat per second than a computer.

That is rubbish, I'm afraid. How is one KiloWatt different from another Kilowatt? The only possible difference in the energy supplied to the room would be in the light power that escapes through the window from the monitor (perhaps a Watt).

We have already been into the fact that supplying 1kW to a room is not the whole story when it comes to making people feel warm but that is not the issue.

Hmm - fools rush in, you know.

 

[Windadct]

All I can say is that a 1000W computer - is it's rated value, full power. So just because the rating is 1000W, does not mean you are using all of that 1000W - If you max out the power supply - then ALL of the 1000W it consumes is left as residual heat.

I would also like to add that Sound and EM are not the only ways to consume the energy - think of all of the ways you can do work ( affect a change in a system) and not give off heat. For example all of the ways to convert to Potential Energy - for example a 1000W Battery charger does NOT give off 1000W of heat.

 

[russ_watters]

If we are talking about purely heat, then yes, I guess they would give off the same amount of heat.

But if some guy is trying to argue that using a 1000W computer will keep you as warm as a 1000W heater will, I'd say that guy definitely needs to switch majors.

I guarantee a heater gives off more heat per second than a computer.

That is self-contradictory: either the numbers are equal or they aren't.

 

[sophiecentaur]

All I can say is that a 1000W computer - is it's rated value, full power. So just because the rating is 1000W, does not mean you are using all of that 1000W - If you max out the power supply - then ALL of the 1000W it consumes is left as residual heat.

I would also like to add that Sound and EM are not the only ways to consume the energy - think of all of the ways you can do work ( affect a change in a system) and not give off heat. For example all of the ways to convert to Potential Energy - for example a 1000W Battery charger does NOT give off 1000W of heat.

Obviously 'rating' is not what counts and a battery charger is a valid exception but that's only finding loopholes in an argument which was dealing with the notion that dissipated energy is somehow different, according to what's written on the front panel label of the equipment.

And, before questioning just how much heat a computer can generate, just consider the vast amount of trouble they have in getting rid of all the heat in a big Server Room. It's a serious problem.

 

[rbj]

If we are talking about purely heat, then yes, I guess they would give off the same amount of heat.

But if some guy is trying to argue that using a 1000W computer will keep you as warm as a 1000W heater will, I'd say that guy definitely needs to switch majors.

I guarantee a heater gives off more heat per second than a computer.

if the room is windowless and if the room is completely sound insulated (note neither of these conditions are about either the heater or the computer) and if both the heater and the computer have watt-meters attached and both are drawing the same power of W watts, then I guarantee that you're incorrect, Rooney. they will both heat the room exactly the same.

 

[LawRooney]

if the room is windowless and if the room is completely sound insulated (note neither of these conditions are about either the heater or the computer) and if both the heater and the computer have watt-meters attached and both are drawing the same power of W watts, then I guarantee that you're incorrect, Rooney. they will both heat the room exactly the same.

Yeah, I don't see how I would be incorrect. I made no statement whatsoever about an "ideal room" and simply stated that a heater would obviously warm you up better, even if the total amount of heat given off ended up being the same.