12 volt vs. 120 volt

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well, OK, but theres not voltage number to use on the low voltage size.

Lets assume 18 volts

P = I E

50 = (i) 18

than 50 / 18 = 2.77

For 120 V

50 = (i) 120
than 50 / 120 = .04167

Eff. is output / input : seems like a badly worded question ?
 
cadpoint said:
well, OK, but theres not voltage number to use on the low voltage size.

Lets assume 18 volts

P = I E

50 = (i) 18

than 50 / 18 = 2.77

For 120 V

50 = (i) 120
than 50 / 120 = .04167

Eff. is output / input : seems like a badly worded question ?
Click to expand...

Look at the title 12v vs 120 v
 
Many factors influence efficiency, for example the type of light source being used (LED, incandescent, HID, etc.), and even limited to a single technology, different design selections will influence efficiency. Considering only incandescent lamps, things such as gas fill, filament design, rated lifetime, etc. will all influence efficiency.

However, considering incandescent lamps, with all other factors held constant, you can show that for a given power output level there is an 'optimal' voltage to use for efficiency. A higher than optimal voltage filament must be long and thin, and is therefore weak. The lamp has to be run a lower temperature and thus lower efficiency in order to reach rated life. A lower than optimal voltage filament is very short and thick, and thus quite strong, but now lots of energy is lost by thermal conduction to the fat low temperature supply leads.

I do not know the details of this analysis, and thus could not tell you the optimal voltage for a 50W lamp....consider the above just so much hand-waving. *grin*

-Jon
 
what the low voltage 12, 24, 36, don't speak 60V cause it won't be low voltage any more ... U oh me 1 :), May Be IOU 1 !

OK 12 it is... I've been burned on the title before... I O U
 
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Sorry, it was 12 volts, A customer keeps asking me to put in low volt lights ( recessed) and I can't convince him that there is not much difference, in the way of going green.
Am I wrong?
 
I posted some numbers , and there right !

The thing you have to remember that we work on some level or applied application, something needs to feed that system... As Jon pointed out there are many variables. It's not that there 12V 's it what can you get out of the application for the required service of 12V...

Were all crazed to do the right thing, but each particular application will address itself, we just get to apply what we can to get it to go!
 
Well I will catch heat for this but 120 volt system is 19,000% more efficient if the load, distance, and wire size are the same.
 
dereckbc said:
Well I will catch heat for this but 120 volt system is 19,000% more efficient if the load, distance, and wire size are the same.
Click to expand...


Now come on, you can't just throw that out without an explanation. Well I guess you can because you just did. :smile:
 
Dennis Alwon said:
Now come on, you can't just throw that out without an explanation.
Click to expand...
Well OK fair enough. Draw yourself a simple circuit diagram of a source voltage, a length or wire, and a load. For example use say 100 feet of 12 AWG. That 12 AWG wire will have resistance with current passing through it which develops voltage drop and power loss. Work it for both voltages and see what you come up with.
 
dereckbc said:
Well I will catch heat for this but 120 volt system is 19,000% more efficient if the load, distance, and wire size are the same.
Click to expand...
Get your glove ready.
dereckbc said:
Well OK fair enough. Draw yourself a simple circuit diagram of a source voltage, a length or wire, and a load. For example use say 100 feet of 12 AWG. That 12 AWG wire will have resistance with current passing through it which develops voltage drop and power loss. Work it for both voltages and see what you come up with.
Click to expand...
No need for a circuit. The loss difference is a function of the square of the voltages: P = (V^2)/R

Don't see the 19,000%, more like 100x

And using the circuit:
100 ft, 50W, #12AWG, 104 deg F
120 volt: 0.148 volt drop, 0.17739 ohms => 0.12319 W
12 volt: 1.48 volt drop, 0.17739 ohms => 12.319 W

add: you said efficiency so:
50.123 W vs 62.319 W for 50 W
99.75% vs 80.23% = 19.52% more efficient
 
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mivey said:
Get your glove ready.
No need for a circuit. The loss difference is a function of the square of the voltages: P = (V^2)/R

Don't see the 19,000%, more like 100x

And using the circuit:
100 ft, 50W, #12AWG, 104 deg F
120 volt: 0.148 volt drop, 0.17739 ohms => 0.12319 W
12 volt: 1.48 volt drop, 0.17739 ohms => 12.319 W

add: you said efficiency so:
50.123 W vs 62.319 W for 50 W
99.75% vs 80.23% = 19.52% more efficient
Click to expand...

But your formula is comparing both voltages at 100'. It's likely that the 12v integral transformer will only be a few inches from the lamp. I don't seen any reference to remote transformers in the OP.
 
infinity said:
But your formula is comparing both voltages at 100'. It's likely that the 12v integral transformer will only be a few inches from the lamp. I don't seen any reference to remote transformers in the OP.
Click to expand...
Because that seemed to be the essence of post #16. I found nothing interesting to add concerning the rest of the thread that had not already been said.

As for anything else:

Winnie pretty much covered most it in #5 when he essentially said there are too many variables.

You can find LV lighting can and lamp combos that are more efficient but on the other side of the coin, why not just use a regular can and install a bulb that is more energy efficient than the incandescent?

As for lumen output, I think I read that incandescents are about 15% efficient (a lot of heat, very little light) but LEDs can be something like 85%. I think the CFLs put out something like 4-5 times the lumens per watt of an incandescent (8-10 times for LEDs).

As the OP said in #7, it can pretty much be a wash but, in my opinion, if the customer wants LV, give 'em LV.
 
mivey said:
Because that seemed to be the essence of post #16. I found nothing interesting to add concerning the rest of the thread that had not already been said.

As for anything else:

Winnie pretty much covered most it in #5 when he essentially said there are too many variables.

You can find LV lighting can and lamp combos that are more efficient but on the other side of the coin, why not just use a regular can and install a bulb that is more energy efficient than the incandescent?

As for lumen output, I think I read that incandescents are about 15% efficient (a lot of heat, very little light) but LEDs can be something like 85%. I think the CFLs put out something like 4-5 times the lumens per watt of an incandescent (8-10 times for LEDs).

As the OP said in #7, it can pretty much be a wash but, in my opinion, if the customer wants LV, give 'em LV.
Click to expand...

I only mentioned it because someone is likely to see 19.52% more efficient and think that 19.52% actually applies to the question in the OP.

IMO the more important issue is what type of lamp will be used.
 
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