Home Page link

Re: basic question about HID lamps and power issues

Home Repair - - If it ain't broken, don't fix it. Otherwise look here. 

Bookmark this page:  YahooMyWeb Yahoo!  Google Google  Windows Live Favorites Windows Live  del.icio.us del.icio.us  digg digg  Add to Netscape Netscape
Subject Author Date
Re: basic question about HID lamps and power issues nicksanspam 01-16-2007
If you were  Registered and logged in, you could reply and use other advanced thread options
Posted by on January 16, 2007, 6:54 am



>Capacitive reactance is lower to higher frequencies...

>> How does that work? I thought they could only reduce IV phase angles.
>>
>>>Power Factor correction capacitors can help absorb some of the harmonics.

Interesting. I wonder how to calculate the power factor improvement for,
say, a 4.7 uF cap in parallel with a 150 watt 277 V load with a 10% 3rd
harmonic distortion.

The first step might be to calculate the original power factor, with no cap.

This seems different and more promising:

https://www.galco.com/circuit/PFCC_har.htm

Harmonic currents can be significantly reduced in an electrical system
by using a harmonic filter.

In its basic form, a filter consists of a capacitor connected in series
with a reactor tuned to a specific harmonic frequency. In theory,
the impedance of the filter is zero at the tuning frequency; therefore,
the harmonic current is absorbed by the filter. This, together with
the natural resistance of the circuit, means that only a small level
of harmonic current will flow in the network.

A 600 Hz filter might help a power supply that clips 10% of an 60 Hz peak.

Nick


Posted by Solar Flare on January 16, 2007, 5:42 pm


Notch filters are always good selective units. Our problem is
filtering 10MW ++ loads full of harmonics. That is a lot of heat to
disapate. hmmmmmm... building heat in the winter and hot water? Why
bother when tons of heat is thrown away in the main transformers via
cooling fans and pumps. Trouble is the utility buldings run metered in
with the big bulk of the consumers and is not apparent to anybody in
the utility. **SIGH**

>
>>Capacitive reactance is lower to higher frequencies...
>
>>> How does that work? I thought they could only reduce IV phase
>>> angles.
>>>
>>>>Power Factor correction capacitors can help absorb some of the
>>>>harmonics.
>
> Interesting. I wonder how to calculate the power factor improvement
> for,
> say, a 4.7 uF cap in parallel with a 150 watt 277 V load with a 10%
> 3rd
> harmonic distortion.
>
> The first step might be to calculate the original power factor, with
> no cap.
>
> This seems different and more promising:
>
> https://www.galco.com/circuit/PFCC_har.htm
>
> Harmonic currents can be significantly reduced in an electrical
> system
> by using a harmonic filter.
>
> In its basic form, a filter consists of a capacitor connected in
> series
> with a reactor tuned to a specific harmonic frequency. In theory,
> the impedance of the filter is zero at the tuning frequency;
> therefore,
> the harmonic current is absorbed by the filter. This, together with
> the natural resistance of the circuit, means that only a small
> level
> of harmonic current will flow in the network.
>
> A 600 Hz filter might help a power supply that clips 10% of an 60 Hz
> peak.
>
> Nick
>



Posted by on January 17, 2007, 8:17 am



>Notch filters are always good selective units...

The R in these series RLC circuts broadens the notch and uses real power.
How much would a 4.7 uF cap in parallel with a 150 watt 277 V load
with a 10% 3rd harmonic distortion raise the power factor?

With 1/(377C) = 564 ohms at 60 Hz and 277 V and L = 1/((2Pi180)^2x4.7x10^-6)
= 0.166 henrys in series to resonate at 180 Hz, limiting 60 Hz ripple current
requires Z = sqr(R^2+(wL-1/(wC)^2) = 564, ie R^2+(377x0.166-564)^2 = 564^2,
ie R = sqr(564^2-251291) = 258 ohms. With how many real watts?

How much would that raise the power factor?

Nick


Posted by Solar Flare on January 17, 2007, 11:44 pm


Man, you're pushing my math skills here. I used to gobble this stuff
for breakfast nut now I'm not even sure I could do it with a Geritol
injection...LOL

>
>>Notch filters are always good selective units...
>
> The R in these series RLC circuts broadens the notch and uses real
> power.
> How much would a 4.7 uF cap in parallel with a 150 watt 277 V load
> with a 10% 3rd harmonic distortion raise the power factor?
>
> With 1/(377C) = 564 ohms at 60 Hz and 277 V and L =
> 1/((2Pi180)^2x4.7x10^-6)
> = 0.166 henrys in series to resonate at 180 Hz, limiting 60 Hz
> ripple current
> requires Z = sqr(R^2+(wL-1/(wC)^2) = 564, ie R^2+(377x0.166-564)^2 =
> 564^2,
> ie R = sqr(564^2-251291) = 258 ohms. With how many real watts?
>
> How much would that raise the power factor?
>
> Nick
>



Similar ThreadsPosted
Question About "Temporary Power Taps" (Power Strips) April 15, 2008, 12:40 pm
hot tub basic question March 16, 2006, 3:37 pm
suggest basic func/signal generator for basic scope test/setup ? October 5, 2007, 11:08 am
Basic DC electricity question February 14, 2007, 10:21 am
Basic Home Electrical Question October 22, 2005, 4:11 pm
Basic hot water heater question February 8, 2006, 3:12 pm
a basic question on patio building 101 March 1, 2006, 6:50 am
Very Basic Circuit Breaker Question July 9, 2006, 6:01 am
basic - I mean basic washing machine repair January 30, 2006, 8:42 am
dish tv question- timer/programmerfor basic tuner? January 19, 2008, 4:24 pm

Contact Us | Privacy Policy

XML SitemapXML Sitemap