Parking Lot Vd.

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steelersman said:
how can you use 3 phase for this application?
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Very common in large lots.

Fixture 1 on A&B

Fixture 2 on A&C

Fixture 3 on B&C

Fixture 4 on A&B

Fixture 5 on A&C

Fixture 6 on B&C

Fixture 7 on A&B

Supply it with a 3 pole breaker.
 
iwire said:
Very common in large lots.

Fixture 1 on A&B

Fixture 2 on A&C

Fixture 3 on B&C

Fixture 4 on A&B

Fixture 5 on A&C

Fixture 6 on B&C

Fixture 7 on A&B

Supply it with a 3 pole breaker.
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Ok thanks. I was kind of thinking that but just not sure. :)
 
The 400 watt ballast I linked to draws 2.2 amps at 208
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Yeah....like I said....just over 2 amps. ;)

The lights I work with would be 208V, H,H,G. No neutral

1000' is a long damn way. The last long run I did was pushing 1000'. The engineered plans called for #6's @ 240V. There were only about 6, 250W fixtures though. Add a few amps on the VD calculator and the numbers change a LOT.
 
Your load is progressive in that it's not all in one place. The formula you use for 1 phase load is: CM = (2 x K x I x D)/VD and 3 phase: CM = (1.732 x K x I x D)/VD. The CM is circular mills of your conductor Table 8, the 1.732 is the square root of 3, the K is resistance or 12.9 for cu is close, the D is distance, and the VD is maximum voltage drop like 3% of your supply voltage.
 
The 'center' of the load can be mathematically calculated.

Multiply each load by it's distance from the supply.

Add these products for all the loads, and divide the sum by the sum of in individual loads. The result is the 'distance' from the supply end of the circuit to the load 'center'. This figure is the 'length' of the circuit to use in your voltage drop calculation. See Fig. 33

If the loads are spread evenly along the circuit, then you can use the physical 'center' of the loads as a base figure. (Fig. 34)

voltdropmultiloads.jpg


Source: American Electricians' Handbook, Croft, 1953 edition.
 
480sparky said:
The 'center' of the load can be mathematically calculated.

Multiply each load by it's distance from the supply.

Add these products for all the loads, and divide the sum by the sum of in individual loads. The result is the 'distance' from the supply end of the circuit to the load 'center'. This figure is the 'length' of the circuit to use in your voltage drop calculation. See Fig. 33

If the loads are spread evenly along the circuit, then you can use the physical 'center' of the loads as a base figure. (Fig. 34)

voltdropmultiloads.jpg


Source: American Electricians' Handbook, Croft, 1953 edition.
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Sparky your on top of your game!:smile:
 
SEO said:
Also you will need to use the amperage off the ballasts .
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Correct... you can't use the lamp wattage as it is based on the secondary side of the ballast, which is nothing more than a transformer.
 
Wow, this Graphic has been in my mind all day.

Frankly it both cool and bothers me. I understand the diagram, and I understand the math, I'm just saying, "Here's why":

First its a diagram shown as a single pole service, but math wise state the distance of using three circuits this causes a lack of total valve of all the KCmil potentially obtainable if it where a two or three circuit service,I might prove why. In the same breath it does not include values for shared circuits that pass the first light in a two or three circuit MBWC. There?s no account of ?a? to ?d? and "b" to "e" of lights in series.

I don't care about the sum valves of the math, I'm just saying that all the math might not be in the graphic!
OK, I've never had just one circuit ususally its two to three in a parking lot.

One might also have a problems if the first leg is outright upsized by specifications/ required. Run #6 to first (light) device. Would one use a Mean average in this case as well would you? Use the "a" to "d" example again? I don't beleive so! And this might cause two calculations origin to "a" light then another to "d", all being with different KCmil's, etc.
 
cadpoint said:
Wow, this Graphic has been in my mind all day.

Frankly it both cool and bothers me. I understand the diagram, and I understand the math, I'm just saying, "Here's why":

First its a diagram shown as a single pole service, but math wise state the distance of using three circuits this causes a lack of total valve of all the KCmil potentially obtainable if it where a two or three circuit service,I might prove why. In the same breath it does not include values for shared circuits that pass the first light in a two or three circuit MBWC. There?s no account of ?a? to ?d? and "b" to "e" of lights in series.

I don't care about the sum valves of the math, I'm just saying that all the math might not be in the graphic!
OK, I've never had just one circuit ususally its two to three in a parking lot.

One might also have a problems if the first leg is outright upsized by specifications/ required. Run #6 to first (light) device. Would one use a Mean average in this case as well would you? Use the "a" to "d" example again? I don't beleive so! And this might cause two calculations origin to "a" light then another to "d", all being with different KCmil's, etc.
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Why would the distance between the loads make a difference? VD is function of source to load, not load to load on the same circuit. The calculation using the mathematical 'center' of the load would yield one size of conductor, not a diminishing size as the circuit progresses past the loads
 
480sparky said:
The 'center' of the load can be mathematically calculated.
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It seems that would be a quick way for estimating purposes but if time allows I would rather do point to point calculations and progressively reduce the conductor size.
 
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