Need more information. On a AC supply you can't calculate anything from just those two pieces of information.
Need current & power factor
Need current & power factor
Lumens - Effective Lumens - RAW Lumens LED rating?
- Thread starter ericmark
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funny enough i asked what the "power correction factor" was mentioned in the instructions and my 2 electrician mates whern't sure what it was 
so what are you actually saying ??
these meteres measure nothing??
they are fairly innacurate??
they serve no usefull purpose ???
i would be happy if they give enough information as long at it lets me know which is using more power will it not do this to any level off accuracy??
adrian uk thanks for taking the time to respond
so what are you actually saying ??
these meteres measure nothing??
they are fairly innacurate??
they serve no usefull purpose ???
i would be happy if they give enough information as long at it lets me know which is using more power will it not do this to any level off accuracy??
adrian uk thanks for taking the time to respond
I guess what I'm saying is that if you want to compare the power that an actual example of these lamps is consuming with what it says on the tin then simply taking a current measurement isn't going to give you a meaningful answer.
If the supply was DC then it would be fine to multiply the volts by the current & call the answer Power. That's how its defined for DC.
AC isn't quite so simple.
For a simple resistive load (say a heater or an old fashioned filament lamp) then the current it takes from the supply is in phase with the voltage (ie. If you looked at the voltage and current waveforms on an oscilloscope then you would find that they tracked each other). For an inductive load this is not true. The voltage and the current will not be in phase anymore. This means that the simple act of multiplying voltage and current will not work anymore.
There are three kinds of AC power: Reactive Power, Apparent Power & Real Power.
The only one you can measure with simple meters is apparent power. Apparent power is measured in VA and is the simple product of volts * amperes.
Real Power is the 'work done' by a system (heat generation, light generation etc) and is measured in Watts. You cant measure this directly.
Reactive Power is taken by inductive & capacitive loads eg. motors, transformers etc. Reactive Power is measured in VAr. A motor will draw both reactive & Real power - the reactive part creates the magnetic field, the real part creates the torque on the output shaft & the heat within the motor. If you where to measure the volts across & the current into the motor you would be measuring the apparent power.
These three are all linked by the power triangle, whereby real power is the horizontal base, reactive power is the vertical & apparent power is the hypotenuse. The power factor is the cosine of the angle between the apparent & real powers. It is also the cosine of the angle between the voltage & current waveforms.
The relationship can also be expressed as pf = real power/apparent power.
In order to correctly measure the real power (Watts) you need a wattmeter. This does more than simply multiply the volts by the amps, it also takes account of the phase difference between the two.
I hope that's not got too heavy on the AC theory
The quantity which you can measure and compare easily is the VA. That will be of some value but it isn't the same as the Watts. This is why you are struggling to get the same value as the manufacturer. Inside the LED lamp will be a small switch mode power supply unit. The current waveform which this takes from the mains will not be sinusoidal. It will have a reactive content (due to the capacitors) and will probably be confusing your meter. The manufacturer will have used a calibrated Wattmeter to make his claims and this will measure the voltage & current as well as the angle between them.
The closest you can get, without spending silly money, would be to obtain one of the better plug-in type power meters. You need one which measures the voltage as well as the current (the cheap ones just measure current & assume volts - these will be out for the same reason that your tests are). These will also show VA & VAr as well as Watts & power factor.
If the supply was DC then it would be fine to multiply the volts by the current & call the answer Power. That's how its defined for DC.
AC isn't quite so simple.
For a simple resistive load (say a heater or an old fashioned filament lamp) then the current it takes from the supply is in phase with the voltage (ie. If you looked at the voltage and current waveforms on an oscilloscope then you would find that they tracked each other). For an inductive load this is not true. The voltage and the current will not be in phase anymore. This means that the simple act of multiplying voltage and current will not work anymore.
There are three kinds of AC power: Reactive Power, Apparent Power & Real Power.
The only one you can measure with simple meters is apparent power. Apparent power is measured in VA and is the simple product of volts * amperes.
Real Power is the 'work done' by a system (heat generation, light generation etc) and is measured in Watts. You cant measure this directly.
Reactive Power is taken by inductive & capacitive loads eg. motors, transformers etc. Reactive Power is measured in VAr. A motor will draw both reactive & Real power - the reactive part creates the magnetic field, the real part creates the torque on the output shaft & the heat within the motor. If you where to measure the volts across & the current into the motor you would be measuring the apparent power.
These three are all linked by the power triangle, whereby real power is the horizontal base, reactive power is the vertical & apparent power is the hypotenuse. The power factor is the cosine of the angle between the apparent & real powers. It is also the cosine of the angle between the voltage & current waveforms.
The relationship can also be expressed as pf = real power/apparent power.
In order to correctly measure the real power (Watts) you need a wattmeter. This does more than simply multiply the volts by the amps, it also takes account of the phase difference between the two.
I hope that's not got too heavy on the AC theory
The quantity which you can measure and compare easily is the VA. That will be of some value but it isn't the same as the Watts. This is why you are struggling to get the same value as the manufacturer. Inside the LED lamp will be a small switch mode power supply unit. The current waveform which this takes from the mains will not be sinusoidal. It will have a reactive content (due to the capacitors) and will probably be confusing your meter. The manufacturer will have used a calibrated Wattmeter to make his claims and this will measure the voltage & current as well as the angle between them.
The closest you can get, without spending silly money, would be to obtain one of the better plug-in type power meters. You need one which measures the voltage as well as the current (the cheap ones just measure current & assume volts - these will be out for the same reason that your tests are). These will also show VA & VAr as well as Watts & power factor.
thank you very much for taking the time to give a very full answer
i do understand very basic electrics
now i dont want to get things intentionally wrong
is there a moderately costing device /devices that can give me an answer within say 10% on ac currents[that is probably the wrong term lol]
i do understand very basic electrics
now i dont want to get things intentionally wrong
is there a moderately costing device /devices that can give me an answer within say 10% on ac currents[that is probably the wrong term lol]
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I have used a plug in power consumption meter which shows power factor, Amps, Volts, and Watts with CFL and the power factor was close to unity if not unity but the minimum reading is 1W so with LED lamps one has no idea is 2W or 2.9W so not really any help with 3W lamps.
The other problem with a lot of equipment in the wave form clipping so oscilloscope is the only way to really see what power is being used.
However the point is not so much what power is being used but how to select lamps where both watts and lumen have so little value in working out how much light a unit will emit. Both as new and at end of it's useful life.
With a tungsten bulb it either works or it fails but with CFL and LED the light output becomes lower and lower before it completely fails and the only way I can measure the array of lights output is with my camera I don't have a meter to measure lumens.
My film pictures do not contain meta data so memory is all I have but I know with a f2 28mm lens and 600 ASA film I as able to take indoor pictures without flash so I would guess around 1/30 second exposure time.
To day with a modern camera with f3.5 zoom lens equivalent to 28mm with 35mm film to get 1/30 second shutter speed I need to ramp the ISO (was called ASA) to 1600 to take the picture without flash so in spite of my room looking brighter in real terms it is dimmer than before. By time you remember lens has smaller aperture by about 1EV.
However I can't find any conversion between EV and Lumen with is not surprising as EV is the light in the room and Lumen is the light output of the bulbs.
But most people do have a digital camera and most will embed meta data on the image giving lens size, ISO rating and shutter speed which gives a very good indication of light levels in the room.
In theory we should be able to use the digital camera as a very good light meter specially if a picture of a standard item was taken. In the old days of film on the packet it told you standard setting for outdoors bright, cloudy, and in doors with special tungsten film so throughout my life time there has been a standard lighting level for in doors.
However I have never tried to quantise that light level as a photographer maybe I should?
The other problem with a lot of equipment in the wave form clipping so oscilloscope is the only way to really see what power is being used.
However the point is not so much what power is being used but how to select lamps where both watts and lumen have so little value in working out how much light a unit will emit. Both as new and at end of it's useful life.
With a tungsten bulb it either works or it fails but with CFL and LED the light output becomes lower and lower before it completely fails and the only way I can measure the array of lights output is with my camera I don't have a meter to measure lumens.
My film pictures do not contain meta data so memory is all I have but I know with a f2 28mm lens and 600 ASA film I as able to take indoor pictures without flash so I would guess around 1/30 second exposure time.
To day with a modern camera with f3.5 zoom lens equivalent to 28mm with 35mm film to get 1/30 second shutter speed I need to ramp the ISO (was called ASA) to 1600 to take the picture without flash so in spite of my room looking brighter in real terms it is dimmer than before. By time you remember lens has smaller aperture by about 1EV.
However I can't find any conversion between EV and Lumen with is not surprising as EV is the light in the room and Lumen is the light output of the bulbs.
But most people do have a digital camera and most will embed meta data on the image giving lens size, ISO rating and shutter speed which gives a very good indication of light levels in the room.
In theory we should be able to use the digital camera as a very good light meter specially if a picture of a standard item was taken. In the old days of film on the packet it told you standard setting for outdoors bright, cloudy, and in doors with special tungsten film so throughout my life time there has been a standard lighting level for in doors.
However I have never tried to quantise that light level as a photographer maybe I should?
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Sorry made an error. Left the polarising filter on the lens. Found original lens f2 at 28mm fixed with skylight filter on it and at ISO 800 shutter speed was 1/45 second in all rooms down stairs so rooms are approx the same light level as 30 years ago.
Living room moved from 200W to 27.2W now LED.
Dinning room moved from 120W to 48W now CFL.
Kitchen unchanged 116W florescent. (2 x 58W tubes)
Would be interesting to hear how other people set their light levels?
Living room moved from 200W to 27.2W now LED.
Dinning room moved from 120W to 48W now CFL.
Kitchen unchanged 116W florescent. (2 x 58W tubes)
Would be interesting to hear how other people set their light levels?
Living room, (2 rooms knocked into 1) 2 x 9.5w LED pendants + a 9.5w LED standard lamp.
Kitchen 1 x 58w fluorescent for the larger part, 2x 11w CFL for the smaller part.
Small workshop 2x 58w fluorescent (easily brightest room in house).
Landing 3x 4w LED.
Smallish 3-bed terrace.
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