Voltage Multiples of 11?

Because the transformation ratios are so decided in the transformer as per international standards.

It's all in the history:

First we humans have this obsession with the decimal system, perhaps its because we have 10 fingers (or is it 8 fingers and two thumbs???)...

Then came batteries that sourced DC voltage (before AC - Alternating Current)...

Then came the need for standards / reference voltages such as 10Volts DC (although Mother Earth did not like to comply to provide physical cells that would source 10.000 Volts)...

Then came AC... and how does AC and DC relate to each other???... RMS voltage of AC needs to be used to transfer the same power as DC voltage... i.e. RMS voltage is 1.11 times the average DC voltage... (and so instead of using 100 volt DC to source power, it was required to use 111 volts AC to provide equivalence to battery sources...

Then came the transformers which are normally integer relationships (you know we humans like to standardise!!!): 11kV:110V has 100:1 ratio, or 132kV:33kV has 4:1 ratio, or 3.3kV:380V has 8.684:1 ratio... Oh dear the integer rule was broken, don't know how that one slipped through???

Any way this is my story of electrical history, and I'm sticking to it!!! (heee...heee...heee...)

NeilJ

Your storey sounds good but where do you get the peak voltage is 1.11 times the RMS voltage. RMS stands for Root Mean Square and translates to the square root of the mean of the voltage squared. The reason for the square root of the squared value is to get rid of the negative values.

Now for a sin wave the RMS value is the peak value divided by the square root of two or

V_RMS = V_Peak ÷ Sqrt(2) = V_Peak ÷ 1.414

So by your logic you should be using 140VAC instead of 110VAC

I too have no idea where the 1.11 factor came from.

As to how AC and DC relate to each other: well its a little more complicated than that. Masu is correct on the meaning of the RMS value of AC voltage ....

However, that ONLY relates an AC voltage to an equivalent DC voltage for a purely resistive circuit! It does not equate them for circuits or loads containing inductive or capacitive elements. So in the case of a motor for instance, you can not determine the power consumed by measuring the amperage and RMS voltage because they will be out of phase with each other. In that case you could apply the correct power factor for that motor to those values to get the power ..and that's only for single phase ..The best way is a power meter which is designed for that purpose. A 3 phase motor requires special meters to determine the power consumed.

Masu:

I didn't use the term "PEAK", I used the term "AVERAGE". i.e. the AVERAGE of a half cycle of sine wave is about 0.636 of the PEAK value; and the RMS value of a half cycle of sine wave is 0.707 of the PEAK. Now if you express them as a ratio, 0.707/0.636, you get the 1.11 value I talked about.

The average of say 100 volts DC is 100 volts; the RMS of 100 volts DC is also 100 volts, so the RMS to AVERAGE ratio is unity.

The average of say 100 volts AC RMS is however less than 100 volts; by the factor 1.11 times.

So AC sinusoids, and DC have different ratios of RMS to AVERAGE.

NeilJ

You used the word "AVERAGE" for DC, NOT AC as shown below:

"Then came AC... and how does AC and DC relate to each other???... RMS voltage of AC needs to be used to transfer the same power as DC voltage... i.e. RMS voltage is 1.11 times the average DC voltage... (and so instead of using 100 volt DC to source power, it was required to use 111 volts AC to provide equivalence to battery sources..."

That is simply WRONG! RMS AC voltage EQUALS the DC voltage!

The only part that is corrrect is "RMS voltage of AC needs to be used to transfer the same power as DC voltage"

The AVERAGE you talk about has absolutely no meaning here.

A True RMS meter reads the AC and gives a voltage value would produce the same power as a DC voltage of the same value for a purely resistive load, and does so irrespective of the waveform.

An "Averaging" AC meter is calibrated to yield the same reading as a true RMS meter, BUT ONLY FOR A PERFECTLY SINUSOIDAL WAVEFORM ... which is not commonly found in real world AC measurements (there is almost always some distortion)

A perfectly sinusoidal waveform AC that will produce the same power (through a resistive load) as 100 VDC will read 100 VAC on both RMS and averaging meters, and the peak will be 141.42 ... volts period!

Greg,

Of course you are correct, I was having a senior moment; DC should have read AC where you underlined.

Thanks for pointing out my error

sir,

according to me you are not giving correct answer

the ratio of rms value to average value is 1.11

That was a good approximation. But add in the effects of standard rectifiers of the time, and it works out even closer. (No, it's not going to be exact).

It might be to do with the old 100 volts DC days as measured by a voltmeter.

Then came the alternator and when run to give 100 volts measured with the same meter it was later found to be 110 volt when measured with an dedicated AC meter.

Something to do with average sine waves maybe.

Does that make sense.

That sounds more than plausible - measuring a sine wave using an 'ideal' rectifying meter results in the AC measurement being 2*sqrt(2)/pi of the RMS value. That corresponds to the actual level being multiplied by about 1.111 to achieve the original meter reading.

The voltages you cited are not generating voltages. Power is typically generated at 12 kV and immediately stepped up at the generating substation to a transmission voltage; anything from 115 - 500 kV. AEP's 765 kV system voltage (which is the highest AC transmission voltage in North America) is obtained by stepping up from 345 kV. From transmission, voltage is stepped down to sub-transmission (69 kV and below) and finally, distribution.

If the reason was in fact due to meter errors then I would suspect that the problem was due to the rectifier that need in a moving coil meter so it can read AC voltages.

Theoretically a moving coil meter will show the RMS voltage as it uses the applied voltage to do work by moving the needle. There are however other factors involved that distort the reading of a moving coil meter but for the frequencies we use for power distribution the reading will be close to the RMS value. However a If you push AC through a moving coil meter it will show zero so as it cant cope with the reverse part of the cycle so you need to introduce a rectifier into the circuit. Depending in whether you used a full or half wave rectifier you in theory should see the half or the full RMS value less the loss in the rectifier.

I am not familiar with the type of rectifiers that were available when they were developing AC as a means for power distribution but if they were 90% efficient then it would explain the 11, 22, 33 etc voltages for example 100 V + 10% is 110V.

Hi Masu

No need for rectifier imperfections. A DC meter with an ideal rectifier will give a
factor of 1.111 [=pi/2/sqrt(2)]. See for example contributions 5 and 6.

fyz

friends.

i don't think the discussions above doesnot give perfect answer to the question.please some one helpout to this question

thanx

I honestly don't know the real answer, but I doubt it it has anything to do with highly complex reasoning.

Perhaps it is to do with volts drop. 110 volts in the old days might have been chosen to deliver 100 volts at point of use. That allowed for 10% volts drop - which became the norm - from which everything else evolved since.

It is like the railway track gauge of 4ft 8½ ins. Nobody chose it as such - it sort of happened. The original gauge was 5ft - nice round number - but it was to the outside edge of the wheels that ran inside a channel made of 2 inch angle iron.

Then they found it better to put the flange on the wheels instead - which had to be on the inside edge in order to use the existing tracks - and because the angle iron was 1/4 inch thick the inside edges of the flat section were found to be 4ft 8½.

The upright flange on the outside edge of the track became redundant to leave a flat top rail that we have to this day.

"It is like the railway track gauge of 4ft 8½ ins. Nobody chose it as such - it sort of happened. The original gauge was 5ft - nice round number - but it was to the outside edge of the wheels that ran inside a channel made of 2 inch angle iron."

I had heard that the standard gauge railway lines came from the fact that when the original piece of test track was laid it was laid down a section of unused road and the ruts from the cart wheels were 4' 8½" apart. The reason for the ruts being that distance apart was that the cart wheels were that far apart. The cart wheels were that far apart because they ran in the groves made by the Roman chariots and if you used a different width you would shake your cart to pieces. The reason that the Roman chariot wheels were 4' 8½" was because that was the width of two horses.

Taken together, the answers seem to amount to the following:
a) The available meters were designed to give sensitivities that were suitable for 50V, 100-V ... DC supplies (probably peak reading about 20% more than nominal).
b) They were easily converted for joint DC/AC applications using rectifiers
c) The AC supply levels could be set up so that the same range settings could be used for both existing DC and newer AC supplies. This would minimise the number of meter ranges; more importantly, it minimised the chances of errors as people moved between installations or reset their meters (large errors being relatively obvious). In order to use the same visible meter readings for correct and over-Voltage supplies, the RMS values of the AC supplies would have to be 1.111 times the DC Voltages
d) As there was no good reason to do otherwise, the AC supply levels were set up for these levels
f) When the need for generation workers to have dual-purpose meters vanished, the existing infrastructure made it impractical to change the AC transmission Voltages

I'm not saying it is what actually happened (I'm not a power-generation historian), but I do think it makes a coherent and sensible story.

So - if this you know this is wrong, please say so. If you just feel it is inadequate, please say what are your reservations.

Fyz

Dear Friend,

This was answered about 6 weeks back.

Pl. use the search tool here and find the answers.

In Electrical Engg. there is a FACTOR CALLED "FORM FACTOR" which is 1.1 and hence multiples of 11 used. Pl. refer standard Electrical Engineering Text BOOK.

Thanks,

RAJESWARI.