Electricital question
We just bought a cabin. It has a 100 amp service breaker, and a service
panel with 5 20 amp breakers. I know on appliances, all I have to do to find out the amperage is RTFM, but for lights and such, how do I calculate just how much wattage I can put on any breaker circuit? The lighting is inadequate. I will need more. I have a licensed electrician friend who will come and wire everything, so it will be done right and safe. I just need to get an idea how many lights we are talking about so I can do some shopping. I don't want to max everything out and put up a ton of lights. I just want to balance them, and not put so many that I am approaching critical mass. TIA Steve |
SteveB wrote:
We just bought a cabin. It has a 100 amp service breaker, and a service panel with 5 20 amp breakers. I know on appliances, all I have to do to find out the amperage is RTFM, but for lights and such, how do I calculate just how much wattage I can put on any breaker circuit? The lighting is inadequate. I will need more. I have a licensed electrician friend who will come and wire everything, so it will be done right and safe. I just need to get an idea how many lights we are talking about so I can do some shopping. I don't want to max everything out and put up a ton of lights. I just want to balance them, and not put so many that I am approaching critical mass. TIA Steve Wattage devided by the voltage (120) is amps. Or just add up the wattage. 20 amps x 120v = 2400 watts. Lights are easy since the wattage is written on each bulb. |
You don't want to exceed 80% of the capacity of the circuit. If you are
using a 15 amp circuit don't exceed 1200 watts on the circuit "SteveB" wrote in message news:j1aKe.84965$4o.53276@fed1read06... We just bought a cabin. It has a 100 amp service breaker, and a service panel with 5 20 amp breakers. I know on appliances, all I have to do to find out the amperage is RTFM, but for lights and such, how do I calculate just how much wattage I can put on any breaker circuit? The lighting is inadequate. I will need more. I have a licensed electrician friend who will come and wire everything, so it will be done right and safe. I just need to get an idea how many lights we are talking about so I can do some shopping. I don't want to max everything out and put up a ton of lights. I just want to balance them, and not put so many that I am approaching critical mass. TIA Steve |
"SteveB" wrote in message news:j1aKe.84965$4o.53276@fed1read06... We just bought a cabin. It has a 100 amp service breaker, and a service panel with 5 20 amp breakers. I know on appliances, all I have to do to find out the amperage is RTFM, but for lights and such, how do I calculate just how much wattage I can put on any breaker circuit? The lighting is inadequate. I will need more. I have a licensed electrician friend who will come and wire everything, so it will be done right and safe. I just need to get an idea how many lights we are talking about so I can do some shopping. I don't want to max everything out and put up a ton of lights. I just want to balance them, and not put so many that I am approaching critical mass. TIA Steve this is Turtle. In normal lighting circuits it is hard to over load normal lighting wattages to just light up a cabin. Just take 1 -- 20 amp breaker and circuit for lights only and that gives you 20 -- 100 watt light bulbs to light up the cabin. One Circuit is what I thinik you need for all lighting to the cabin. Now really 19.6 -- 100 watt light bulbs to be exact. TURTLE |
"TURTLE" wrote in message ... "SteveB" wrote in message news:j1aKe.84965$4o.53276@fed1read06... We just bought a cabin. It has a 100 amp service breaker, and a service panel with 5 20 amp breakers. I know on appliances, all I have to do to find out the amperage is RTFM, but for lights and such, how do I calculate just how much wattage I can put on any breaker circuit? The lighting is inadequate. I will need more. I have a licensed electrician friend who will come and wire everything, so it will be done right and safe. I just need to get an idea how many lights we are talking about so I can do some shopping. I don't want to max everything out and put up a ton of lights. I just want to balance them, and not put so many that I am approaching critical mass. TIA Steve this is Turtle. In normal lighting circuits it is hard to over load normal lighting wattages to just light up a cabin. Just take 1 -- 20 amp breaker and circuit for lights only and that gives you 20 -- 100 watt light bulbs to light up the cabin. One Circuit is what I thinik you need for all lighting to the cabin. Now really 19.6 -- 100 watt light bulbs to be exact. TURTLE Not the best idea: Pop the breaker an dthe whole place goes dark. They should be mixed on at least two breakers, and instead of 20 bulbs, that would be 16 bulbs. Over 80% usage will allow normal variations in the grid, cabin, over time, breakers, etc. to begin to heat the breaker, thus degrading it over time and leaving no safety overhead. So with a min two lines you've got 32 bulbs now, lots more than you'll need. Not sure where the 80% figure comes from, nec, ul, mfg, whatever, but it's reality. HTH, Pop |
"Pop" wrote in message ... "TURTLE" wrote in message ... "SteveB" wrote in message news:j1aKe.84965$4o.53276@fed1read06... We just bought a cabin. It has a 100 amp service breaker, and a service panel with 5 20 amp breakers. I know on appliances, all I have to do to find out the amperage is RTFM, but for lights and such, how do I calculate just how much wattage I can put on any breaker circuit? The lighting is inadequate. I will need more. I have a licensed electrician friend who will come and wire everything, so it will be done right and safe. I just need to get an idea how many lights we are talking about so I can do some shopping. I don't want to max everything out and put up a ton of lights. I just want to balance them, and not put so many that I am approaching critical mass. TIA Steve this is Turtle. In normal lighting circuits it is hard to over load normal lighting wattages to just light up a cabin. Just take 1 -- 20 amp breaker and circuit for lights only and that gives you 20 -- 100 watt light bulbs to light up the cabin. One Circuit is what I thinik you need for all lighting to the cabin. Now really 19.6 -- 100 watt light bulbs to be exact. TURTLE Not the best idea: Pop the breaker an dthe whole place goes dark. They should be mixed on at least two breakers, and instead of 20 bulbs, that would be 16 bulbs. Over 80% usage will allow normal variations in the grid, cabin, over time, breakers, etc. to begin to heat the breaker, thus degrading it over time and leaving no safety overhead. So with a min two lines you've got 32 bulbs now, lots more than you'll need. Not sure where the 80% figure comes from, nec, ul, mfg, whatever, but it's reality. HTH, Pop This is Turtle. He was asking what number of breakers would be need for lighting for the cabin. I said 19.6 for a 20 amp breaker will support the 19.6 light bulbs and will be the 80% of the amperate of the 20 amp breaker. He can split up the bulb in all area , but all lighting would not need more than 1 --- 20 amp circuit, no matter how he run it. With two circiuts of 20 amps he could put not 32 -- 100 watt light bulbs but 39.2 --- 100 light bulbs. This would still be compliant of the 80% rule of the two 20 amp breaker circuits. Now the degrading of the breaker to trip at lower amperages is just a effect you have to deal with 20 to 30 years from now and just wait 20 years or so and think about them. TURTLE |
wrote in message ... On Wed, 10 Aug 2005 21:28:49 -0500, "TURTLE" wrote: With two circiuts of 20 amps he could put not 32 -- 100 watt light bulbs but 39.2 --- 100 light bulbs. This would still be compliant of the 80% rule of the two 20 amp breaker circuits. What 80% rule? This urban legend has taken on a life of it's own. There are lots of 80% rules in the NEC (even some 50% rules) but none of them apply to lighting fixtures in a dwelling. This is Turtle. When Speaking of a Electrical breaker in a breaker box suppling electricity and the 80% rule does not matter if the elctricity is supplied to a whole home or to a one light bulb. If you have a 20 amp breaker in the box. You better not put more that a 80% of 20 amp load on it. So You can have a 16 amp load to be supplied to the 20 amp breaker. i know now that your not a electrician for you would know exactly what I was speaking about and you would be telling me about the 80% rules. So 80% Applies to everything that will pull amps from a breaker and if you want to change anything. Please be welcome to do as you please. TURTLE |
Dear Steve, the formula is
volts times amps = watts So, you can say that the 110 volt circuit, times 20 amps = 2200 watts. Hint: Try using flourescent, or compact fluorescent bulbs. They use a lot less power. -- Christopher A. Young Learn more about Jesus www.lds.org www.mormons.com "SteveB" wrote in message news:j1aKe.84965$4o.53276@fed1read06... We just bought a cabin. It has a 100 amp service breaker, and a service panel with 5 20 amp breakers. I know on appliances, all I have to do to find out the amperage is RTFM, but for lights and such, how do I calculate just how much wattage I can put on any breaker circuit? The lighting is inadequate. I will need more. I have a licensed electrician friend who will come and wire everything, so it will be done right and safe. I just need to get an idea how many lights we are talking about so I can do some shopping. I don't want to max everything out and put up a ton of lights. I just want to balance them, and not put so many that I am approaching critical mass. TIA Steve |
Stormin Mormon wrote:
volts times amps = watts So, you can say that the 110 volt circuit, times 20 amps = 2200 watts. Hint: Try using flourescent, or compact fluorescent bulbs. They use a lot less power. At a lower power factor... watts = 0.4 x volts x amps, for some undercounter fluorescents. Nick |
In article , "TURTLE" wrote:
When Speaking of a Electrical breaker in a breaker box suppling electricity and the 80% rule does not matter if the elctricity is supplied to a whole home or to a one light bulb. If you have a 20 amp breaker in the box. You better not put more that a 80% of 20 amp load on it. So You can have a 16 amp load to be supplied to the 20 amp breaker. Wrong. i know now that your not a electrician for you would know exactly what I was speaking about and you would be telling me about the 80% rules. So 80% Applies to everything that will pull amps from a breaker and if you want to change anything. Please be welcome to do as you please. What you say here is completely wrong. It's quite obvious that *you* are not an electrician, or you would know that the 80% rule refers to continuous loads, defined by the NEC as "a load where the maximum current is expected to continue for three hours or more." This does *not* apply to residential lighting circuits, or indeed to most other loads. -- Regards, Doug Miller (alphageek-at-milmac-dot-com) Get a copy of my NEW AND IMPROVED TrollFilter for NewsProxy/Nfilter by sending email to autoresponder at filterinfo-at-milmac-dot-com You must use your REAL email address to get a response. |
"Doug Miller" wrote in message m... In article , "TURTLE" wrote: When Speaking of a Electrical breaker in a breaker box suppling electricity and the 80% rule does not matter if the elctricity is supplied to a whole home or to a one light bulb. If you have a 20 amp breaker in the box. You better not put more that a 80% of 20 amp load on it. So You can have a 16 amp load to be supplied to the 20 amp breaker. Wrong. i know now that your not a electrician for you would know exactly what I was speaking about and you would be telling me about the 80% rules. So 80% Applies to everything that will pull amps from a breaker and if you want to change anything. Please be welcome to do as you please. What you say here is completely wrong. It's quite obvious that *you* are not an electrician, or you would know that the 80% rule refers to continuous loads, defined by the NEC as "a load where the maximum current is expected to continue for three hours or more." This does *not* apply to residential lighting circuits, or indeed to most other loads. -- Regards, Doug Miller (alphageek-at-milmac-dot-com) Get a copy of my NEW AND IMPROVED TrollFilter for NewsProxy/Nfilter by sending email to autoresponder at filterinfo-at-milmac-dot-com You must use your REAL email address to get a response. |
"Doug Miller" wrote in message m... In article , "TURTLE" wrote: When Speaking of a Electrical breaker in a breaker box suppling electricity and the 80% rule does not matter if the elctricity is supplied to a whole home or to a one light bulb. If you have a 20 amp breaker in the box. You better not put more that a 80% of 20 amp load on it. So You can have a 16 amp load to be supplied to the 20 amp breaker. Wrong. i know now that your not a electrician for you would know exactly what I was speaking about and you would be telling me about the 80% rules. So 80% Applies to everything that will pull amps from a breaker and if you want to change anything. Please be welcome to do as you please. What you say here is completely wrong. It's quite obvious that *you* are not an electrician, or you would know that the 80% rule refers to continuous loads, defined by the NEC as "a load where the maximum current is expected to continue for three hours or more." This does *not* apply to residential lighting circuits, or indeed to most other loads. -- Regards, Doug Miller (alphageek-at-milmac-dot-com) This is Turtle. first i don't do residentiual electric work and only commercial HVAC which makes me do the electric for them. Second if you was running power to lights and the lighting wattage was 2,400 watts / 24 -- 100 watt light bulbs and then would you say you would hook all them up to a 20 amp breaker which the load would be matching a 2,400 watt load. Then or would you only connect 19.6 light bulb on the 20 amp breaker to not over load the breaker and trip if all the lights were turned on at one time. Awwwww that would be the 80% rule. When every you pull over 80% of the breakers rating. Your asking it to trip and you have to come back to rebalance the loads. Tell me what you would do with this problem ! TURTLE |
wrote in message ... On Thu, 11 Aug 2005 11:04:06 -0500, "TURTLE" wrote: This is Turtle. first i don't do residentiual electric work and only commercial HVAC which makes me do the electric for them. Turtle, I specifically addressed residential lighting loads. Do you have every light in your house on at the same time? I bet Turtle doesn't, but if he has kids, probably. Like at my house. Steve |
In article , "TURTLE" wrote:
"Doug Miller" wrote in message om... In article , "TURTLE" wrote: When Speaking of a Electrical breaker in a breaker box suppling electricity and the 80% rule does not matter if the elctricity is supplied to a whole home or to a one light bulb. If you have a 20 amp breaker in the box. You better not put more that a 80% of 20 amp load on it. So You can have a 16 amp load to be supplied to the 20 amp breaker. Wrong. i know now that your not a electrician for you would know exactly what I was speaking about and you would be telling me about the 80% rules. So 80% Applies to everything that will pull amps from a breaker and if you want to change anything. Please be welcome to do as you please. What you say here is completely wrong. It's quite obvious that *you* are not an electrician, or you would know that the 80% rule refers to continuous loads, defined by the NEC as "a load where the maximum current is expected to continue for three hours or more." This does *not* apply to residential lighting circuits, or indeed to most other loads. This is Turtle. first i don't do residentiual electric work and only commercial HVAC which makes me do the electric for them. Then perhaps you shouldn't be trying to answer questions about residential electrical work... Second if you was running power to lights and the lighting wattage was 2,400 watts / 24 -- 100 watt light bulbs and then would you say you would hook all them up to a 20 amp breaker which the load would be matching a 2,400 watt load. Then or would you only connect 19.6 light bulb on the 20 amp breaker to not over load the breaker and trip if all the lights were turned on at one time. Awwwww that would be the 80% rule. When every you pull over 80% of the breakers rating. No, that's just plain wrong. The 80% rule DOES NOT APPLY in this situation because it is NOT a "continuous load" as defined in the NEC. Your asking it to trip and you have to come back to rebalance the loads. Nonsense. Twenty 100W light bulbs is NOT going to trip a 20A breaker. Tell me what you would do with this problem ! I wouldn't do anything about it -- because it is NOT a problem. -- Regards, Doug Miller (alphageek at milmac dot com) It's time to throw all their damned tea in the harbor again. |
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"Doug Miller" wrote in message m... In article , "TURTLE" wrote: "Doug Miller" wrote in message . com... In article , "TURTLE" wrote: When Speaking of a Electrical breaker in a breaker box suppling electricity and the 80% rule does not matter if the elctricity is supplied to a whole home or to a one light bulb. If you have a 20 amp breaker in the box. You better not put more that a 80% of 20 amp load on it. So You can have a 16 amp load to be supplied to the 20 amp breaker. Wrong. i know now that your not a electrician for you would know exactly what I was speaking about and you would be telling me about the 80% rules. So 80% Applies to everything that will pull amps from a breaker and if you want to change anything. Please be welcome to do as you please. What you say here is completely wrong. It's quite obvious that *you* are not an electrician, or you would know that the 80% rule refers to continuous loads, defined by the NEC as "a load where the maximum current is expected to continue for three hours or more." This does *not* apply to residential lighting circuits, or indeed to most other loads. This is Turtle. first i don't do residentiual electric work and only commercial HVAC which makes me do the electric for them. Then perhaps you shouldn't be trying to answer questions about residential electrical work... Second if you was running power to lights and the lighting wattage was 2,400 watts / 24 -- 100 watt light bulbs and then would you say you would hook all them up to a 20 amp breaker which the load would be matching a 2,400 watt load. Then or would you only connect 19.6 light bulb on the 20 amp breaker to not over load the breaker and trip if all the lights were turned on at one time. Awwwww that would be the 80% rule. When every you pull over 80% of the breakers rating. No, that's just plain wrong. The 80% rule DOES NOT APPLY in this situation because it is NOT a "continuous load" as defined in the NEC. Your asking it to trip and you have to come back to rebalance the loads. Nonsense. Twenty 100W light bulbs is NOT going to trip a 20A breaker. Tell me what you would do with this problem ! I wouldn't do anything about it -- because it is NOT a problem. -- Regards, Doug Miller (alphageek at milmac dot com) It's time to throw all their damned tea in the harbor again. This is Turtle. You missed the question of what you would do in this case. I stated that I would not put but 20 -- 100 watt light bulbs on a 20 amp circuit / breaker. The 20 light bulbs would be the max i would put on the 20 amp breaker because of the 80% rule. what is the max number of 100 watt light bulbs that you can put on a 20 amp breaker ? So , What is the answer to this question ? TURTLE |
Now, I am really confused. I never thought this would be such a complex
issue. It seemed like a simple question, but now I am not so sure. I am thinking of cancelling the sale because of all of this. I mean, I lie awake nights now, thinking of amperage, wattage, voltage, and things I didn't even know existed last week. The current debate has caused me to think of a lot of related issues. Will the atmospheric pressure change affect these bulbs? What if I buy them here in Nevada and take them to Utah? Will the change in elevation or time zone affect their performance? Will the "long life" bulbs violate the religious beliefs of the area where there are strong afterlife philosophies? Would I be in violation of local traditions by bringing in these "long life" bulbs? What happens if I violate this 80% rule? Say, by 2%? Was this one of the commandments that was dropped by Mel Brooks playing Moses in "The History of the World, Part One"? If I use bulbs made outside of the US, can I, in good conscience, ask my union electrician buddy to work on this project? There are five 20 amp breakers. That will give me up to 20,000 watts in a 1200 square foot cabin. Or, a mere 16,000 if I listen to the 80% philosophy. My question is, IF the Space Shuttle were flying, will it be visible from space? That's a lot of light. Should I just use Halogens? Do you think I will need eye protection? Do you think it will cause sunburn or carpet discoloration? In the meantime, I believe I will just stick to flashlights, candles, and kerosene lights. All this other stuff is just so complicated and confusing. Right now, I'm going to take six valiums and try to wind down. Steve |
George E. Cawthon wrote:
...watts = 0.4 x volts x amps, for some undercounter fluorescents. Do fluorescent lights have a power factor? Sure... 0.4 in the case above. But, that said, if the undercounter light fixture say 15W, then that is how much is used regardless of power factor Yes. and w=va still applies. or am I wrong... You am wrong. W = PFxVxA. Nick |
In article , "TURTLE" wrote:
"Doug Miller" wrote in message om... I wouldn't do anything about it -- because it is NOT a problem. This is Turtle. You missed the question of what you would do in this case. No, you missed my answer - right above. I stated that I would not put but 20 -- 100 watt light bulbs on a 20 amp circuit / breaker. The 20 light bulbs would be the max i would put on the 20 amp breaker because of the 80% rule. Yes, I know that's what you stated. But you're wrong. You're having a very hard time grasping this simple concept: the 80% rule DOES NOT APPLY in this situation. what is the max number of 100 watt light bulbs that you can put on a 20 amp breaker ? So , What is the answer to this question ? 20A x 120V / 100W = 24 bulbs. -- Regards, Doug Miller (alphageek at milmac dot com) It's time to throw all their damned tea in the harbor again. |
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George E. Cawthon wrote:
wrote: ...watts = 0.4 x volts x amps, for some undercounter fluorescents. Do fluorescent lights have a power factor? Sure... 0.4 in the case above. But, that said, if the undercounter light fixture say 15W, then that is how much is used regardless of power factor Yes. and w=va still applies. or am I wrong... You am wrong. W = PFxVxA. Did you go to embassy doublespeak school? Ignorance and belligerance are an unfortunate combination. ...You answered the 2nd question yes and the third saying I am wrong. Those answers are inconsistent. No... In the case above, the real power is 15W = 0.4x120Vx0.3125A, but "w=va" = 120Vx0.3125A = 37.5 VA overestimates it by a factor of 2.5. If the fixtures says 15 watts and it has a power factor, the powerfactor is already applied. Yes. I think I'll just stick with totaling up the wattage of fixtures/appliances. That's a good idea, in most cases. Nick |
"Doug Miller" wrote in message . .. In article , "TURTLE" wrote: "Doug Miller" wrote in message . com... I wouldn't do anything about it -- because it is NOT a problem. This is Turtle. You missed the question of what you would do in this case. No, you missed my answer - right above. I stated that I would not put but 20 -- 100 watt light bulbs on a 20 amp circuit / breaker. The 20 light bulbs would be the max i would put on the 20 amp breaker because of the 80% rule. Yes, I know that's what you stated. But you're wrong. You're having a very hard time grasping this simple concept: the 80% rule DOES NOT APPLY in this situation. what is the max number of 100 watt light bulbs that you can put on a 20 amp breaker ? So , What is the answer to this question ? 20A x 120V / 100W = 24 bulbs. -- Regards, Doug Miller (alphageek at milmac dot com) It's time to throw all their damned tea in the harbor again. This is Turtle. I see you would load up a break to it's max. amps and if the people in the house happen to turn on all 24 light bulbs. It will blow the breaker in about 2 or 3 hours and then call you to ask why did my break throw. You maybe able to do this by NEC but I will not load anything up to the max. what so ever. Now what I did up above by just putting 20 light bulbs on a 20 amp breaker is Nothing wrong with doing it. You had said it was wrong well i tell you it is nothing wrong with it if i wanted to do it. Now you can explain any wrong as NEC goes please explain it to me. Let me explain to you what right and wrong is. Right is you can do it. Wrong is you Can't do it. Now take these two words and explain wrong as you say above here. TURTLE |
"SteveB" wrote in message news:saWKe.697$DW1.169@fed1read06... Now, I am really confused. I never thought this would be such a complex issue. It seemed like a simple question, but now I am not so sure. I am thinking of cancelling the sale because of all of this. I mean, I lie awake nights now, thinking of amperage, wattage, voltage, and things I didn't even know existed last week. The current debate has caused me to think of a lot of related issues. Will the atmospheric pressure change affect these bulbs? What if I buy them here in Nevada and take them to Utah? Will the change in elevation or time zone affect their performance? Will the "long life" bulbs violate the religious beliefs of the area where there are strong afterlife philosophies? Would I be in violation of local traditions by bringing in these "long life" bulbs? What happens if I violate this 80% rule? Say, by 2%? Was this one of the commandments that was dropped by Mel Brooks playing Moses in "The History of the World, Part One"? If I use bulbs made outside of the US, can I, in good conscience, ask my union electrician buddy to work on this project? There are five 20 amp breakers. That will give me up to 20,000 watts in a 1200 square foot cabin. Or, a mere 16,000 if I listen to the 80% philosophy. My question is, IF the Space Shuttle were flying, will it be visible from space? That's a lot of light. Should I just use Halogens? Do you think I will need eye protection? Do you think it will cause sunburn or carpet discoloration? In the meantime, I believe I will just stick to flashlights, candles, and kerosene lights. All this other stuff is just so complicated and confusing. Right now, I'm going to take six valiums and try to wind down. Steve This is Turtle. I would just go ahead and Punt on First 10 . TURTLE |
"TURTLE" wrote I would just go ahead and Punt on First 10 . TURTLE I would, but I am afraid that I might hit a light bulb. Steve |
In article , "TURTLE" wrote:
Now what I did up above by just putting 20 light bulbs on a 20 amp breaker is Nothing wrong with doing it. You had said it was wrong well i tell you it is nothing wrong with it if i wanted to do it. Please read more carefully, Turtle. I never said that putting only 20 100W bulbs on a 20A breaker was wrong -- I said you were wrong to claim that 19 bulbs was the maximum permitted because of the 80% rule. Now you can explain any wrong as NEC goes please explain it to me. I've explained it several times already, but you're not paying attention. Let's try again. The 80% rule applies to continuous loads. Residential lighting is not a "continuous load" as defined in the NEC. Therefore the 80% rule does not apply to residential lighting circuits. Let me explain to you what right and wrong is. Right is you can do it. Wrong is you Can't do it. Yep, and by that definition putting twenty-four 100W light bulbs on a 20A 120V circuit is right. So is twenty bulbs. Or five bulbs. Or one. Now take these two words and explain wrong as you say above here. There's nothing wrong with loading a circuit to less than its capacity. What's wrong is your understanding of the capacity of a 20A circuit when used for residential lighting, and your understanding of the 80% rule. -- Regards, Doug Miller (alphageek at milmac dot com) It's time to throw all their damned tea in the harbor again. |
In article ,
wrote: George E. Cawthon wrote: ...watts = 0.4 x volts x amps, for some undercounter fluorescents. Do fluorescent lights have a power factor? Sure... 0.4 in the case above. But, that said, if the undercounter light fixture say 15W, then that is how much is used regardless of power factor Yes. and w=va still applies. or am I wrong... You am wrong. W = PFxVxA. Nick QUESTION about this "power factor": (it has been *so long* since I understood any of that stuff, that I've forgotten all but a few words describing it.) With DC, the pf is 1.0? With AC, I'm not sure what it is. Of course there's the "rms" stuff, trying to get an average value of a sine-wave. The pf, I recall from *ages* (decades) ago, had something to do with the voltage and current "waves" getting out of sync with each other, due to a coil or a capacitor (one shifting in one direction, one in the other). I recall something about having to use trig to get the pf, maybe it was the sine or cosine of the degrees of lead or lag? If so, then since those functions range between plus and minus one, then maybe the pf I dimly recall is the reciprocal of yours? Anyway -- here's my question: How do you get a substantial pf for a fluorscent (sp?) light? (Huge electric motor, I understand how.) And a pf for an incandescent light, that would involve no phase shift at all? Obvously, I could use some mental fill-in! Thanks! David |
David Combs wrote:
With DC, the pf is 1.0? Yes. With AC, I'm not sure what it is. With AC, it's the sum of instantaneous products of current and voltage over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage. It's also the number you see when you push the PF button on a Kill-a-Watt meter. I did that with a "0.1A 120V" Little Giant fountain pump the size of a golfball and saw 0.42. The meter said the pump used 5 watts. Nick |
"With AC, it's the sum of instantaneous products of current and voltage
over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage. " David's memory is correct. For AC, its the cosine of the phase angle between the voltage and current. When there is zero phase shift, the power factor is 1.0 |
wrote:
"With AC, it's the sum of instantaneous products of current and voltage over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage. " David's memory is correct. For AC, its the cosine of the phase angle between the voltage and current. When there is zero phase shift, the power factor is 1.0 What's the phase shift of a switching power supply? :-) Nick |
On Sun, 11 Sep 2005 03:58:33 -0700, trader4 wrote:
"With AC, it's the sum of instantaneous products of current and voltage over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage. " David's memory is correct. For AC, its the cosine of the phase angle between the voltage and current. When there is zero phase shift, the power factor is 1.0 As Nick Pine eludes to, this isn't a good definition because it doesn't take into account the harmonic content of the waveforms. It works for purely sinusoidal voltage abd current though. Power factor is more appropriately the Power divided by RMS Volts * RMS Amps, or PF == P/VA. The power can always be found by averaging the instantaneous V*A over the cycle, as nicksanspam wrote above. -- Keith |
On Sun, 11 Sep 2005 02:08:43 +0000, David Combs wrote:
In article , wrote: George E. Cawthon wrote: ...watts = 0.4 x volts x amps, for some undercounter fluorescents. Do fluorescent lights have a power factor? Sure... 0.4 in the case above. But, that said, if the undercounter light fixture say 15W, then that is how much is used regardless of power factor Yes. and w=va still applies. or am I wrong... You am wrong. W = PFxVxA. Nick QUESTION about this "power factor": (it has been *so long* since I understood any of that stuff, that I've forgotten all but a few words describing it.) With DC, the pf is 1.0? PF is kinda meangless with DC, since P=VA. With AC, I'm not sure what it is. Of course there's the "rms" stuff, trying to get an average value of a sine-wave. Not average. When Volts is multiplied by amps (power) there is a squared term in there. RMS == Root Means Square (root of the mean square), so when you calculate power, the squared term is taken into account. Average voltage isn't all that useful; the average for sine wave is zero. ;-) The pf, I recall from *ages* (decades) ago, had something to do with the voltage and current "waves" getting out of sync with each other, due to a coil or a capacitor (one shifting in one direction, one in the other). Or a non-linear load adding harmonic content to the waveforms. I recall something about having to use trig to get the pf, maybe it was the sine or cosine of the degrees of lead or lag? Cosine of the angle between them, but that really only works for pure sine waves. If so, then since those functions range between plus and minus one, then maybe the pf I dimly recall is the reciprocal of yours? Anyway -- here's my question: How do you get a substantial pf for a fluorscent (sp?) light? The fluorescent is a non-linear load. It chops the current waveform, adding harmonics to it. The above formula isn't very useful in this case. (Huge electric motor, I understand how.) A "huge electric motor" should have a PF very close to one if it's heavily loaded. If it's creating no mechanical work, then it will have a worse PF. And a pf for an incandescent light, that would involve no phase shift at all? The phase shift will be small, but it will only conduct current during part of the cycle, chpping the current waveform. Since the current isn't being used during the entire waveform, the power uring this part of the cycle isn't useful. Obvously, I could use some mental fill-in! Thanks! -- Keith |
In article ,
keith wrote: On Sun, 11 Sep 2005 03:58:33 -0700, trader4 wrote: "With AC, it's the sum of instantaneous products of current and voltage over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage. " David's memory is correct. For AC, its the cosine of the phase angle between the voltage and current. When there is zero phase shift, the power factor is 1.0 As Nick Pine eludes to, this isn't a good definition because it doesn't take into account the harmonic content of the waveforms. It works for purely sinusoidal voltage abd current though. Power factor is more appropriately the Power divided by RMS Volts * RMS Amps, or PF == P/VA. The power can always be found by averaging the instantaneous V*A over the cycle, as nicksanspam wrote above. -- Keith Thanks, all! Now, where (on net) can I learn more, esp about that paragraph above? (Yeah, I can google, but I'm not quite sure for exactly what, and *maybe* one of you already have a good site in mind.) David |
In article ,
keith wrote: On Sun, 11 Sep 2005 02:08:43 +0000, David Combs wrote: .... Of course there's the "rms" stuff, trying to get an average value of a sine-wave. Not average. When Volts is multiplied by amps (power) there is a squared term in there. RMS == Root Means Square (root of the mean square), so when you calculate power, the squared term is taken into account. Yes, I now recall that. Average voltage isn't all that useful; the average for sine wave is zero. ;-) Indeed! The pf, I recall from *ages* (decades) ago, had something to do with the voltage and current "waves" getting out of sync with each other, due to a coil or a capacitor (one shifting in one direction, one in the other). Or a non-linear load adding harmonic content to the waveforms. .... .... How do you get a substantial pf for a fluorscent (sp?) light? The fluorescent is a non-linear load. It chops the current waveform, adding harmonics to it. The harmonics being like the terms in a (*very* dimly recalled) fourier series or transform that sum up to approximate the chopping-caused square or impulse or whatever waveform it is? As I asked in my other reply, do you know offhand any good sites that cover this stuff (with drawings, too)? The above formula isn't very useful in this case. So, what is? (Huge electric motor, I understand how.) A "huge electric motor" should have a PF very close to one if it's heavily loaded. If it's creating no mechanical work, then it will have a worse PF. Jeez, on the surface, that sounds like the *opposite*, maybe, of what one would expect??? Maybe you could say some more? (I knew way back then that going for a double-E would be way over my ability!) And a pf for an incandescent light, that would involve no phase shift at all? The phase shift will be small, but it will only conduct current during part of the cycle, chpping the current waveform. Since the current isn't being used during the entire waveform, the power uring this part of the cycle isn't useful. Wow... Maybe you could say a bit more about that! (This stuff is NOT simple!) Thanks to all for a fascinating look into the dark. David |
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