Can a device use more wattage than it's rated for?

[member 407692]

We're talking about high-wattage things like boilers, radiators, etc., things that use something like 500-1500W an hour, but aren't as advanced as computer hardware (where you can tune everything in the BIOS). Can such equipment, even though it's rated for say 800W, consume more (810, 850, 900, 2000, etc. Watts) electricity? I don't know much about physics, but I think that they can't, because the resistors are protecting them from getting more voltage in, which is a part of the wattage equation.

P.S. Please keep the explanation more on the "words" side rather than complex formulas, as I said, I'm not a physicist, I'm just a guy who has a mild interest in electronics. :)

 

[K^2]

For starters, rating for some devices is approximate. Cheap submersible water heaters, for example, can easily be 10%-20% off. But generally, a device that operates properly should not consume more power than it is rated for. A malfunction, however, can easily cause a device to consume a lot more. Short circuit being the simplest example.

 

[member 407692]

I asked because I have been arguing with a friend of mine for good half an hour that it CANNOT use more because the resistors do guard from more voltage to getting through. It can use less though, due to other poor components...

And we were not talking about malfunctions either, his argument was simply "the instructions and manuals are bull****, it uses more than they say it does." He couldn't offer any better argument, that's why I kept to my, not too confident, rationalization about why and how it is.

So... Basically, it can use more electricity if there is a malfunction, it can use less due to poor parts, and it can use a little more due to voltage fluctuation. Is that correct?

Thanks!

 

[K^2]

The voltage is constant. Whatever your outlet supplies. (Yeah, fluctuations in that do happen, so that could be a factor.) Under normal operation, the power consumption is determined by the current being drawn, which can change during operation. A cold electric heater will draw more power than a hot one, because hot coil has higher resistance. A motor will draw less power once it spins up to maximum speed. And so on.

But yes, power rating is the design maximum. It's what device shouldn't be exceeding under normal operations. If nothing goes wrong. In general, it's a little more complicated than simply having fixed resistors. There is more stuff going on depending on what it is that the device does. But overall, your statement is closer to the truth. If nothing is wrong, device will not draw more. It can certainly draw less.

 

[member 407692]

The voltage is constant. Whatever your outlet supplies. (Yeah, fluctuations in that do happen, so that could be a factor.) Under normal operation, the power consumption is determined by the current being drawn, which can change during operation. A cold electric heater will draw more power than a hot one, because hot coil has higher resistance. A motor will draw less power once it spins up to maximum speed. And so on.

But yes, power rating is the design maximum. It's what device shouldn't be exceeding under normal operations. If nothing goes wrong. In general, it's a little more complicated than simply having fixed resistors. There is more stuff going on depending on what it is that the device does. But overall, your statement is closer to the truth. If nothing is wrong, device will not draw more. It can certainly draw less.

I would be interested in hearing what there is other than resistors - just curiosity.

So, as I understand, I am right about the power then. For example, a 1000W water heater is likely to draw 1000W until it heats up, and while it's heating up, gradually decreasing the wattage, while at *set* temperature, it uses just enough wattage to not get cold, maybe something like 800W.

Another example, a motor will eat up the maximum allowed wattage for maybe 5s, until it spins up and is fully spinning, which is when it'll use just enough to keep spinning - for example, a 200W motor will use full 200W for 5 seconds, then it will use maybe 120W to keep operating.

Assuming I understood the concept correctly, are my numbers in the examples way off, or it's possible?

 

[K^2]

A motor can give you a brief spike that's outside of rating. If you ever turned on an appliance with powerful motor and noticed lights dimming, that's that. How much power the motor will draw when running at full speed will depend on the amount of mechanical work the motor is doing. If it's spinning freely, it will only draw enough power to overcome friction and some to be wasted on resistance of the coil.

A heater is trickier. Like I said, the resistance will increase as the heating element heats up. That means power consumption will be higher when it's cold, but not by much. The way it will maintain temperature, however, is by turning heating element on and off.

In terms of what will actually limit the current, it's going to be very different depending on what the device does and whether we are talking about AC only or if we consider what happens in DC stages of devices that have them. Seriously, this isn't something that can be just covered in a few lines of text. If you are interested, start learning more about electrical circuits.

 

[Integral]

I have had laptops which ran at the same voltage but had different power requirements. 1 of the laptops shipped with a 60W power supply the other shipped with a 95W ps. You could indeed run the 90W laptop with the 60W ps BUT if you did for a extended time the 60W supply would first overheat then you let the smoke out. That is it will burn up.

This is why they label devices with the power draw. If you look at your collection of wall warts you will find some rated for 12V and 500ma, another will be 12V and 1500ma. You need to take care and match the voltage and power (current draw) with the device to be powered . You can use a device rated to draw 500ma with a power supply rated at 1500ma but not the other way round.

As above if you use a lower power rated power supply with a device which draws a higher current, it may work for a long time or it may not. You are taking a chance of burning up the PS.

 

[member 407692]

K^2: Uhuh, so let's take the heater as an example. It will constantly use power to heat up (1000W for example when it's rated for it), and then, when the element is hot enough, it will turn off and basically use 0W, but it will turn on say every 10min to keep the element hot.

Integral: Ok, that is interesting and it does make a lot of sense. I actually had a theory about that. :)

 

[Studiot]

The rated wattage is the steady running wattage.

Many devices, (eg refrigerators, flourescent lights, tungsten lights,) draw more initial power when starting and then settle down to the rated power in normal running. Of course devices such as a refrigerator will have periods of off or reduced power.

Devices with an electric motors may draw considerably more than rated power if overloaded. eg an electric door opener with a sticky door or an electric drill with a blunt cutter or a vacuum cleaner with a blocked filter...

 

[jbriggs444]

Devices with an electric motors may draw considerably more than rated power if overloaded. eg an electric door opener with a sticky door or an electric drill with a blunt cutter or a vacuum cleaner with a blocked filter...

A vacuum cleaner with a blocked filter will be running with essentially no load. You aren't running air through the fan, so it free-wheels. That's why you hear the motor wind up to higher speed rather than lugging down to a lower speed.

I would expect the power draw to be lower in that condition, not higher.

 

[Studiot]

J Briggs
I would expect the power draw to be lower in that condition, not higher.

Interesting comment in theory, but there is nothing like practice so I tested my vacuum cleaner.

Measured power, normal running, 940 watts : 4.3 amps

Measured power, simulated blocked filter, 640 watts : 3.3 amps

I did not record power factors.

So you are quite correct.

Which is interesting since the instructions warn of motor burnout on blockage.

 

[Drakkith]

Which is interesting since the instructions warn of motor burnout on blockage.

Something to do with airflow over the motor perhaps? And I think I remember something saying that a stalled motor uses more current, producing more heat. However I don't think this applies to the vacuum motor since it isn't stalled.

 

[AlephZero]

Which is interesting since the instructions warn of motor burnout on blockage.

If the blockage slowed down the motor, (e.g. by something getting stuck in the fan) the power would go up, not down.

But I wouldn't recommend testing that on your own cleaner, if you want it to survive!

The "back EMF" that the motor produces by acting as a generator is roughly proportional to the RPM. As the speed increases, the effective voltage across the motor (= supply voltage - back EMF) is lower and the current is lower.

There are more complicated ways to understand that effect, but solving Maxwell's equations should get to the same result (eventually!)

 

[Studiot]

And I think I remember something saying that a stalled motor uses more current, producing more heat

Yes I was thinking of a (partially) stalled motor in my examples, but J Briggs put us right.
Pulling against a partial vacuum the vac motor speeds up, unlike the other examples.

Something is seriously wrong, however, if anything gets into the fan in a vac cleaner - this is more than blocked filters.
But for fans more generally this often happens - especially to cooling fans in a PC.

 

[Steve Harris]

If the blockage slowed down the motor, (e.g. by something getting stuck in the fan) the power would go up, not down.

But I wouldn't recommend testing that on your own cleaner, if you want it to survive!

The "back EMF" that the motor produces by acting as a generator is roughly proportional to the RPM. As the speed increases, the effective voltage across the motor (= supply voltage - back EMF) is lower and the current is lower.

There are more complicated ways to understand that effect, but solving Maxwell's equations should get to the same result (eventually!)

We should remember another factor, which is that clogged filters in vacuum cleaners possibly act like clogged filters in clothes dryers. They help burn out the motor not by increasing the load and electrical power in (for reasons noted, this doesn't happen) but instead by cutting down on air flow which cools the motor (along with whatever else the airflow does), so the motor simply overheats and burns out from THAT.

 

[russ_watters]

Several things:

First, welcome to PF.

We're talking about high-wattage things like boilers, radiators, etc., things that use something like 500-1500W an hour...

I just said in another thread, this is like fingernails on a chalkboard to me. Its just watts. Not watts per hour. Watts is already a rate: Joules per second.

...but aren't as advanced as computer hardware (where you can tune everything in the BIOS). Can such equipment, even though it's rated for say 800W, consume more (810, 850, 900, 2000, etc. Watts) electricity? I don't know much about physics, but I think that they can't, because the resistors are protecting them from getting more voltage in, which is a part of the wattage equation.

Resistors don't protect these circuits nor do they provide any control. But they are so dumb, they do always run at the same wattage until they start burning out: If you always apply the same voltage to a resistor, you will always get the same wattage out.

For household devices, there aren't many that have potential to exceed their nameplate wattage, but some devices with motors can. Taking what people said about the vacuum cleaner and applying it elsewhere:

Your air conditioner's evaporator fan, the fan that circulates air throughout the house, was designed for a certain airflow and pressure. If you remove the ductwork and run it, you will decrease the pressure and increase the airflow. The net change will be an increase in power that may exceed the nameplate rating on the motor. This is a bigger problem on commercial systems, where fan motors have been known to burn out if they are tested before the ductwork is installed.

The condensing unit's compressor power use depends on the temperature of the air on both coils. I suppose the motor must be constant speed and the torque dependent on that temperature. If it is too hot in your house when you turn it on or too hot outside, or the condenser fan fails, the compressor power use will go way up, potentially exceeding the nameplate wattage.

For non-household HVAC systems, this is a bigger problem. Fans are driven by belts and pulleys and the motor spins at constant RPM or dies trying, so you can easily exceed the nameplate wattage on the motor just by changing the pulley. You can burn out the motor -- or blow up the fan, for that matter by spinning it too fast.

And even if the fan pulleys are sized correctly, if the motor has a variable speed drive, most can go as high as 120 Hz. Since nominal is 60 Hz, you can easily overdrive the fan. VFDs have safeties on them, but the ones designed to protect the motor (not the circuit) are manually set.

 

[CWatters]

The blocked vacuum cleaner example is interesting...

Normally you would expect any blockage to increase the load on the motor causing it to draw more current. Indeed some makes have a pressure relief valve that opens if the pressure difference is too high (indicating a blockage) I think my Dyson has this.

However I suppose there might be another possibility... A fan is basically an aerodynamic device. If the airflow is blocked this might stall the fan blades causing the suction to reduce and consequently the load on the motor reduces automatically. I think some F1 racing cars stall their wings to reduce lift and drag (or was that only allowed in 2011?). I'm not sure this actually happens in a vacuum cleaner but I suppose it might.

 

[CWatters]

We're talking about high-wattage things like boilers, radiators, etc., things that use something like 500-1500W an hour, but aren't as advanced as computer hardware (where you can tune everything in the BIOS). Can such equipment, even though it's rated for say 800W, consume more (810, 850, 900, 2000, etc. Watts) electricity?

It depends what you mean by "rated"..

Normally the value specified on the power rating label is higher than actual consumption. In some cases the value is based on the fuse rating. eg It might be designed to consume 0.75A @ 230V = 172W but it might be fitted with a 1A fuse so the rating label might say 230W (eg 1A * 230V = 230W).

Something like a 50W light bulb may draw 50W when hot but it may draw more than that for a very brief period when switched on. That's because the resistance of the filament is temperature dependant, it's resistance is lower when cold.

In some/most countries regulations allow an electrician to apply a diversity rating. For example it's not unusual for an electrician to assume that a cook is unlikely to use all the rings on their electric hob/range/cooker at the same time. This may allow a big 6 ring hob to be installed on an existing circuit/wire that is apparently too small. Normally for safety reasons this circuit will be protected by an appropriate sized breaker. eg the breaker will be choosen to protect the wire feeding the hob not the rating of the hob itself. If the cook turns all the rings on at once the breaker pops. The alternative would be to rewire with a fatter wire if the supply to the house can take the increased load.

 

[jbriggs444]

However I suppose there might be another possibility... A fan is basically an aerodynamic device. If the airflow is blocked this might stall the fan blades causing the suction to reduce and consequently the load on the motor reduces automatically.

I have never inspected the impeller on a vacuum cleaner, but my assumption has been that it is basically a centrifugal pump. Air comes in in the middle and is flung to the outside. The power requirement comes from a continuous need to accelerate new air to the tangential speed of the blades at the rim. If you stop the airflow there's no new air to accelerate and the impeller can just spin in place doing little work.

 

[Studiot]

Since there is so much interest in the vacuum cleaner I have rerun the tests.

Every (mains) electrical device (sold) in the UK must bear a rating plate by law. This must state the voltage, number of phases if more than one and either the wattage or the current.

My vac states

Input 230V 1100W max
850 Watts input IEC
950 Watts total IEC input with electrical nozzle.

(Anyone know what an electrical nozzle is please?)

Results: The inlet tube was blocked by placing my hand over the end.
The pitch of the motor definitely increases on blocking, demonstating a speed rise.

XXXXXXXXX Offload Normal Blocked

Volts...241...241...241.6
Current...0...4.16...2.68
Power...0...940...617
Frequency...49.9...49.9...49.9
Powerfactor..1.0...0.93...0.95
VA...0...988...680

 

[member 407692]

This thread has gotten really interesting. :)

 

[K^2]

Input 230V 1100W max
850 Watts input IEC
950 Watts total IEC input with electrical nozzle.

(Anyone know what an electrical nozzle is please?)

Probably an optional head, like one with electric brushes, or something.

I'm kind of more interested in why input power quoted here is significantly lower than what you are getting. The power factor is also a bit low. Have you checked the cap on that motor?

 

[Studiot]

Well you note the rating was for 850 watts at 230 volts and my mains voltage at the time was 241 volts.

So the expected power draw at 241 volts will be (241/230)² * 850 = 933 watts.

Since I could only read one parameter at a time and there was some degree of fluctuation I think that is pretty close, certainly close enough to the measured value of 940 watts.

940 watts is consistent with the measured voltage, current, VA and power factor.

It should be noted that 230 (-10, +15) volts is now the nominal standard voltage in the EU.

 

[K^2]

Am I reading this correctly? You got the power factor independent from power measurement? I suppose, you could measure phases independently and get your power factor that way, but that sounds like unnecessary extra work.

 

[Studiot]

No the instrument I used has one readout and you press buttons to display the required parameter for which it then provides continuous readout.

I have more accurate laboratory grade equipment but I didn't think it worth the setup in this case. The figures I got matched to better than 10 in 1000 or 1 in 100 or 1% which is pretty good for electrician grade measurement.

The meter is the one at the extreme left in the picture in post#5 of this thread

https://www.physicsforums.com/showthread.php?t=478363&highlight=meter&page=2

 

[cjl]

I have never inspected the impeller on a vacuum cleaner, but my assumption has been that it is basically a centrifugal pump. Air comes in in the middle and is flung to the outside. The power requirement comes from a continuous need to accelerate new air to the tangential speed of the blades at the rim. If you stop the airflow there's no new air to accelerate and the impeller can just spin in place doing little work.

This is exactly correct - a fan with a blocked air inlet will speed up, since the lack of available air decreases the load on the fan. You can try it with a hair dryer easily enough - block the inlet, and you can hear the fan speed increase. The danger is not due to increased load on the motor - the danger is because the vacuum cleaner (and most other devices which use a fan, including the aforementioned hairdryer) uses the airflow generated by the fan to cool the motor. If the inflow is blocked while the motor is running for an extended period, the motor can burn up due to a lack of cooling airflow, even though the load is lower than it normally would be. This also explains why Cwatters' Dyson has a relief valve - it isn't to lower the load on the motor, it's to provide cooling airflow to the motor.

 

[K^2]

No the instrument I used has one readout and you press buttons to display the required parameter for which it then provides continuous readout.

Ah, that makes sense. In that case, the power factor is better be in agreement with power measurement.

That brings me back to whether it's a bit low. Since you only have a 2% difference between clear flow and blocked, I would expect adjustment to within a percent to be rather simple. So .93 feels really low. Am I missing something? Or am I right to expect it to be higher?

I'm thinking about it as a theorist, though. I have very little engineering knowledge of this. There might be some obvious practical reason for this that I am missing.

 

[Studiot]

Here are some typical power factors in real life.

0.9something is well within this range.

http://www.controllix.com/display.cfm/p/50/pp/4

 

[K^2]

You think there is no capacitor at all to improve power factor? Seems kind of wasteful.

 

[Studiot]

I doubt there is any correction cap on the motor but it is more than 25 years since I last disassembled a vac motor. I don't know how this motor starts either.

I can only find suppressor capacitors listed, but this would appear to be the motor

http://www.espares.co.uk/part/vacuu.../1574/531/0/0/509299/tco-dl2-1104t-motor.html

 

[russ_watters]

You think there is no capacitor at all to improve power factor? Seems kind of wasteful.

It isn't common because a single device with bad power factor isn't wasteful in and of itself. Residentially, you aren't charged for bad power factor and commercially it is cumulative for the meter, so if the overall isn't below .9 it doesn't matter what the individual devices are.

 

[K^2]

It just seems like such an easy thing to fix for individual appliance that even the extra heating in the wiring between outlet and motor seems like a waste in comparison.

On the second point, though. The meter does charge based on RMS current, not power, doesn't it? Would it be at all practical to try and correct the power factor of the household? It'd have to be adaptive, of course, but the controller would be cheap. So long as capacitor banks required wouldn't be prohibitively expensive... I mean, we are talking 5-10% of the power bill here. That's not change.

 

[russ_watters]

Electric meters measure real power, not current. So a slight increase in amperage for something with very little resistance doesn't change the power enough to matter. It isn't like the manufacturer of the device is paying for the power anyway -- they'd rather pocket the extra $.50 per vacuum cleaner.

And adaptive power factor correctors -- commercial ones anyway -- are still pretty absurdly expensive.