Does reactive/inductive load gets paid? And what's the equivalent wattage... - Page 1
Forum section: electronics / beginners

author: TNb
author_rank: Regular Contributor
timestamp: September 30, 2014, 07:55:02 am
content: The question is related to this video at around 11:30 -So in theory capacitors and inductors just pump power back and forth from the wall outlet. It makes sense.Now what I don't understand is if we do not pay for VA(inductive/reactive power), but pay only for Watts(resistive power), then can I connect an AC motor to outlet and expect it to run for month and get more or less standard bill? I do acknowledge that part of this power goes to moving the shaft, but what is this part? For example if motor draws 3A how do we calculate what is the watt dissipation in moving shaft?

author: Rerouter
author_rank: Super Contributor
timestamp: September 30, 2014, 08:10:40 am
content: The part sloshed back and forth offsets the phase of the current vs the voltage, (capacitors lag the voltage or lead the current, inductors lag the current or lead the voltage depending on what you are using as reference) if you want to find what your true watts consumed are then you will need to know the phase (or power factor) to be able to normalize it to a resistive load,Also be aware that an AC motors power consumption is dependent on the loading,An unsafe example used in tafes in aus to demonstrate power factor involves using a 2 channel oscilloscope behind an isolation transformer, one channel on the active voltage, the neutral broken by a current shunt, with the second channel measuring the device side of the current shunt, and the scope ground on the neutral side of the shunt,

author: jahonen
author_rank: Super Contributor
timestamp: September 30, 2014, 08:23:15 am
content: Power taken from the mains by the motor must be larger (by the losses in the motor itself) than angular velocity times the torque so for example 2.2 kW motor with nominal 1400 rpm (angular velocity 2*pi*1400/60 =~ 146.6 rad/sec) reaches nominal power output (and input current) at torque of 2200 W/(2*pi*1400 rpm/60) =~ 15 Nm.Regards,Janne

author: IconicPCB
author_rank: Super Contributor
timestamp: September 30, 2014, 08:53:07 am
content: Short answer is NO.Explanation: the motor has losses in the form of power lost due to windage ( picking up air and causing it to spin and or flow through the motor), heat lost in the wires of the motor, heat lost in the iron core, power loss in spinning the bearings and sounds the motor makes while running.These losses will cost You money.Further the reactive current the motor draws has to flow through real wires which need to be of a larger gauge due to size of reactive current so You will lose more power in the feed line ...or pay more for a larger amperage conductor.To calculate the no load power requirement of the motor... hang a watt meter on it and it will give You a measure of these losses.Small machines are typically 705 to 80% efficient and as size goes up, motors and transformers start nudging high nineties.

author: TNb
author_rank: Regular Contributor
timestamp: September 30, 2014, 09:06:10 am
content: Quote from: IconicPCB on September 30, 2014, 08:53:07 amWhat do you mean under efficiency of a motor? Is it like 100% means motor consumes no power(wattage power) and 0% means it consumes all the power(hence inefficient)? I'm asking because usually motor efficiency means ratio of electrical power to mechanical power, that is not of much interest in here.

author: mikerj
author_rank: Super Contributor
timestamp: September 30, 2014, 12:31:29 pm
content: Quote from: TNb on September 30, 2014, 09:06:10 amEfficiency is still of interest because the power factor of most loads is greater than zero i.e. even if the apparent power is higher than the real power, a low efficiency still means you will be consuming more real power.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: September 30, 2014, 01:40:57 pm
content: QuoteI guess you mean, can I run the motor for free?Yes, if you had some fictional ideal motor with no load attached.No otherwise, you will have to pay for the motor losses( friction + resistance + leakage + whatever else) and also for the power delivered to the motors load.

author: Dago
author_rank: Frequent Contributor
timestamp: September 30, 2014, 02:09:44 pm
content: I think the OP is thinking a motor as a purely inductive load which it is notYou can think of the "useful power" as consumed by a resistor in parallel with the motor inductance. This is the power that does actual work (turns the motor etc.).

author: TNb
author_rank: Regular Contributor
timestamp: September 30, 2014, 03:25:13 pm
content: Quote from: Dago on September 30, 2014, 02:09:44 pmI do understand that we have "resistive" load, I wrote it in the original post.Quote from: TNb on September 30, 2014, 07:55:02 amSo the question is how to calculate the (watt)power taken by motor. Though I tend to get simple watt-meter to save some headacheOr I could rephrase the question I guess - HOW(what part of it) inductive is the motor?

author: Dago
author_rank: Frequent Contributor
timestamp: September 30, 2014, 03:47:11 pm
content: Quote from: TNb on September 30, 2014, 03:25:13 pmI doubt you can easily calculate it. If the motor is totally unloaded the datasheet for the motor most likely will specify some kind of no-load current draw.The actual power consumed goes to the losses in the motor (bearings, resistive losses, eddy current losses etc.).

author: SeanB
author_rank: Super Contributor
timestamp: September 30, 2014, 04:32:35 pm
content: You can easily measure power factor, using pretty much quite a few industrial power meters that measure both current and voltage simultaneously, and which give power factor, true and apparent power in a series of readings. I got 2 of these at the scrapyard, will be looking at them when I get a round tuit.

author: IconicPCB
author_rank: Super Contributor
timestamp: September 30, 2014, 09:40:55 pm
content: Efficiency..shaft power out versus electrical power in ( in Watts not Vars ).Note some losses in the motor are load related ( motor current varies with load so resistive losses are load related also iron core hysteresis losses are flux that is ampere turn related ).

author: IanB
author_rank: Super Contributor
timestamp: September 30, 2014, 10:47:48 pm
content: Quote from: TNb on September 30, 2014, 09:06:10 amThe mechanical power is the power required to turn the motor. Even if you don't hang a load on the output shaft the motor itself is a load that requires power to turn (you have bearing friction, windage losses, cooling fan to consider).So the efficiency of an unloaded motor running at idle is going to be pretty bad--the motor consumes all the power and does no useful work.

author: ivaylo
author_rank: Frequent Contributor
timestamp: October 01, 2014, 06:41:27 am
content: If your household (assuming household) powermeter detects reactive power you pay for it, if not you don't. Depends on the country too. I come from a country where the home ones did not detect reactive power and the industrial ones did. And you always pay for the losses.

author: IconicPCB
author_rank: Super Contributor
timestamp: October 01, 2014, 07:07:45 am
content: ....and the industrial ones did....T^his goes to the main issue of low powerfactor loads... increased line currentand current is current .. something has to be in place to conduct it..no matter what the load quality is.So low powerfactor forces supply authorities to install thicker wires and they happily ask industrial users to contribute to cost by charging them for VARs not Watts

author: nuhamind2
author_rank: Regular Contributor
timestamp: October 01, 2014, 07:31:28 am
content: I think the OP want to rationalize how AC motor can consume power.Like somebody already said, if your motor is ideal motor and it is unloaded you will not consume any power. At that condition you are equivalent to connecting ideal inductor to power line. Current in inductor lag 90 deg to the voltage and zero real power.If you load your ideal motor,you'll consume electric power even if your motor has no mechanical loss (energy conservation law). At that condition (I might wrong about this), the powerline no longer see the inductor as purely inductive. Current and voltage relation might less than 90 deg and real power consumed.