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Lowering Substation Voltages


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#1 kev21903

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Posted 30 November 2005 - 09:31 PM

I have recently lowered the voltage on a distribution substation by approx. 5% using an off load tap changer. After the reduction in voltage we have seen approx. 10% reduction in power consumption (and I think the power factor has changed - improved).;d;

Can anyone explain why this has happened?

I know there will be savings from resistive loads and reduction in some losses.

I am particularly interested in the effect of the reduced voltage on induction motors. What effect has this had and why. Surely P=V x I x p.f. so if I lower the voltage the current should increase. This isn't what I have observed.
Is this due to a lightly loaded motor or some other reason?

Regards
Kevin
;p;

#2 johnbose

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Posted 02 December 2005 - 11:58 AM

For induction motors which are connected to load, current should increase with reduction in voltage as the output requirement remain same due to the machine trying/ continuing to run at slip speed. Thus to make the Power output same only way would be to draw more current. There is no contribution of voltage in the speed formula N=120f/p.

Reduction in power on reduction of voltage appears to be contribution of resisitive loadsand as you have indicated- motors not running to thier full load requirement- may be they are at 70 to 80% .

#3 jraef

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Posted 14 December 2005 - 09:42 PM

There is no free lunch.

By lowering the line voltage, resistive loads are going to use less power. But what is a resistive load doing? Usually it is heating something up, or keeping it hot. By lowering power all you did was integrate the time factor out further. At a lower wattage, it takes longer to heat something up; look at your stove or an example. So if you have stretched out the process time, you may have pushed the total kWH out into the next billing cycle, but eventually it will catch up unless the expected process output is reduced accordingly. If you still have resistive lighting such as incandescent, reduced voltage will use less power, but of course you will get less light, just like a dimmer switch. In either case it means that you are using less power, essentially because you are doing less work.

In motor loads, the reduction in power would reduce your losses somewhat IF the motors were lightly loaded or unloaded and running for no good reason. But again, that is because you already were doing less work that the system was designed for. If they are or become loaded, that savings will be quickly eaten up by having a permanently reduced voltage and then may end up costing you more in the long run.

What you have demonstrated is that you have an efficiency problem(s) in your process that is worthy of being further investigated and acted upon. Just lowering the voltage will mask the symptom, not cure the ill.

PS:
Just read your older post so I'll address that here. Why CVR works for utilities as an energy saving technique is that it is usually done as a temporary measure during a power shortage. It affects the user's process by essentially forcing a lower productivity output, but is a necessary eveil compared to a blackout or brownout. Permanent reduction just passes the problem downstream to the user in lower productivity and/or increased costs over time.
"He's not dead, he's just pinin' for the fjords!"

#4 kev21903

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Posted 16 December 2005 - 07:05 PM

I think there is more to this than simply the resistive load (heater or lamp etc) doing less work. A lot of the systems on the site have feedback control loops so there should be no change in useful power consumed.
What about the resistance in cables? Small as the resistance in cables is, I2R shows us that even a small resistance will mean reduced losses. This can't explain represent the full savings at approx. 10% but these savings have been demonstrated over a full year - so there is a sustained saving to be made.
My question is still - where are we saving 10%?
Cables- yes
Motor efficiency/p.f. - yes
There must be more than these contributions to get a 10% saving.

#5 marke

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Posted 17 December 2005 - 01:11 AM

Hello kev21903
As stated by othere above, the reduction in voltage will reduce the power consumed by resistive loads such as light and heat loads.
Motor loads will however draw the same power, or slightly more.
As Jeff said, there is no free lunch. Heting circuits will take longer to rech temperature etc, lightig levels will not be as high, but at the end of the day, if you are getting the results you want at a lower cost, then that is fine.

Best regards,




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