Replacing light bulbs with LEDs - LED lightbulb information

Cliff

I am sorry Graham but you are close to trolling now.
I have given you clear examples. I have shown you AxV is not just for DC for a resistive load. It is valid for AC and PF of 1. Earlier in this thread we covered inductive loads compensating capacitive loads.

I also tried to explain that using a capacitive dropper introduces an undesirable PF.
I also stated that that the consumer pays for watts so doesn't lose out, but the power station, unless the PF can be brought back to 1, and compensated for, will suffer.

My last attempt to show the undesirability of the cheap capacitive solution.
Example.
LED consumes 12 watts DC (1 amp x 12volts)
Voltage dropped using a transformer or smps with a good PF near 1
Input  12w= 1 x 0.05 x 240v  (W = PF ×A × V)
Amps = 0.05 amps

The LED with a capacitive dropper may have a power factor of 0.5 (as per linked article)
So for a 12 watt LED the formula becomes 12 = 0.5 x A x 240v,
Amps = 0.1 Now it is 0.1 amps (twice the previous example) for the same 12 volt LED!

Of course, and pay attention at the back, the watts on your meter is still 12, but the grid sees a current of 0.1A and 240v i.e. 24VA



Anyway .. done to death now.

Cliff

Graham, ask yourself why we don't use capacitors for larger wattage LEDs? They don't get hot!

grahamlthompson

That is totally incorrect.  The turbo alternator unit is rated in megawatts, it is capable of that output at any power factor. The generator transformer that connects the unit to the local substation is rated in MVA. The rating is sufficiently high so as not to restrict the power output of the turbo alternator unit.  Transformers are typically up to 99% efficient.  Take a typical base load power station of the 70's like Ratcliffe On Soar, it has four 500MW turbo alternator units, and assuming there are no other issues like problems with boiler forced draft or induced fan problems it can deliver 2000MW 24 hours/day 365 days/year to the 400kV substation under any conditions.


Ratcliffe-on-Soar power station - Wikipedia, the free encyclopedia

Drax has 6 660MW turbo alternator units


Drax power station - Wikipedia, the free encyclopedia



Having actually sat in the generation despatch desk of a  CEGB control room pretty sure that I would know if the station had to reduce output as a result of having to generate more mvars.   I have already told you that the connected load is a tiny tiny proportion of the var loading on the grid system.   The excitation requirements of the thousands of connected transformers and hundreds of miles of 400/275Kv overhead lines is massively larger.  The only basic way to control the 400kV system voltage is by using generation excitation and the tap changer on it's associated generator transformer.

Maximum mvar output from generators is required at winter peak loadings, I rather think someone might notice if a generator can't actually reach the output it has, don't you.

National grid has the responsibility to warn of potential shortfalls of generation during the winter months. It has already said margins for the current winter are tight. It has nothing to do with your weird ideas, purely down to lack of generation due to closed and broken down plant.


Managing the network | National Grid


If you were really worried why don't you fit some shunt capacitors to your domestic system to correct your PF to 1 ?

You haven't yet explained how anyone buying LED bulbs can follow your recommendations.

You say the pf of the bulb you describe is 0.5 (is that -0.5 or  0.5).  It is really easy to find out, all you need is multimeter with the capability to read ac current.

grahamlthompson

I do not understand that question at all ?  How is the pf they operate at in way related to how hot they get ? They don't get hot because they are very efficient at converting the energy input into light.  Most of the energy input to a incandescent bulb produces heat not light.

You don't seem to get it.  Generating mvars does not require any additional power, the turbine and boiler are not in any way effected.  Any additional power required is related to increased losses in transmitting the slighty increased current across the network.  Copper losses on the grid compared to the actual connected generation are tiny. The extra losses associated with mvar transmission are imperceptible. Total losses due to transmission in England and Wales are just 1.65%.

Have you ever seen a vector diagram ?  It's a right angle triangle.  The in phase current is represented by the long side, the tiny quadrature current by the other side (up or down depending on whether leading or lagging).  The hypotenuse shows the phase and magnitude of the  resulting current. The actual current increase is tiny.

You are completely locked into your DC education.  You cannot add the two currents together.

See this diagram


electrical vector diagram - Google Search

The horizontal line OA represents the system load, the vertical line OC the quadrature current (which in our example will hardly be noticeable.  The length of OB shows the resulting current.  The example shows the relationship between reactance, admittance and impedance. It's just as valid for voltage or current.

Cliff

Ask yourself why we don't use capacitors for larger wattage LEDs?
I was talking about the capacitor.
My question was, why don't we use a capacitor every time we want to drop the voltage down?
The capacitor doesn't get hot, its cheap and it is simple to get mains down to LED voltages? Why bother with complicated smps or transformer?


Here is something by Caterpillar on generators and kw, kva and power factor.

http://www.cashmanequipment.com/UserFiles/Uploaded/cms/generator-power-factor.pdf

grahamlthompson

Because you would need an enormous capacitor to pass any sort of current at 50Hz.

That's what I have been telling you all along.  The engine (in this case the steam turbine) has enough horsepower to deliver say 500MW.  The alternator has sufficient current carrying capacity to deliver 500MW under all conditions.  The generator transformer is capable of transmitting the full power under any excitation current used from min to max of the operating envelope when used in combination with the generator transformer variable tap ratios.   The small generator you quote is not connected via a variable ratio transformer.

Cliff

So you don't think the problem is real?  
The article said...
As the world transitions from incandescent to solid state lighting (SSL) technology, utilities and government regulatory agencies worldwide are concerned that, as this large segment of the consumption base switches to SSL, it will increase infrastructure costs. This is due to the reactive nature of LED-based solid state lighting, which results in higher distribution currents that adversely affect power factor (PF) and, in turn create a larger demand on the power grid.

grahamlthompson

Only if you believe the same amount of energy will be used as is currently used.  The whole point of led is to reduce the lighting load from it's current levels.  If every incandescent bulb was replaced in the UK overnight the connected demand would fall dramatically even taking the extra current required in the network for the same final loading. Don't you think that incandescent lighting produces losses ?  A 5W led is capable of the same light output as a 60W bulb.  In any case the statement is flawed, it's the lower power factor that creates higher distribution currents not vice versa. Typical of the rubbish you see posted by journalists who don't have a clue what they are talking about.  Even by your pessimistic pf of 0.5 and interpretation that's a 6 times reduction in transferred current which is itself flawed.  Have a look at the vector diagram.

Note the expected fall in peak demands during the winter due to LED lighting.


ttp://www.carbonbrief.org/five-ways-the-uks-electricity-grid-is-changing

grahamlthompson

This is how you work it out. At pf of o.5 the VA vector is displaced 60 degrees anticlockwise for lagging PF and 60 degrees clockwise for leading pf.  If you have two devices with the same consumption and opposite but otherwise equal pf, the net effect is zero.


Electrical Power Factor | Calculation and Power Factor Improvement | Electrical4u

To correct using a shunt capacitor


Practical Power Factor Correction : Power Factor - Electronics Textbook

I guess the bulb you refer to has a circuit like the 3rd one down here.


How do LED light bulbs work?

Circuit described in detail here


Mains Powered White LED Lamp | EEWeb Community

Stuart Wright

I have a question for both of you. What % of people reading this thread will understand what you are arguing about?
Is it possible we could bring the thread back on topic? (It might still be, but I wouldn't know).
Or agree to disagree at least?