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

CWatters

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

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

physxGuy

This thread has gotten really interesting. :)

K^2

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

http://physicsforums.com/showthread....t=meter&page=2

cjl

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

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/vacuum...04t-motor.html

russ_watters

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.