Unbalanced 3 Phase Lighting Calculation

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KAZ 88

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I've been tasked with reducing lighting load at an industrial facility in an effort to reduce utility costs. Rather than hook up a power monitor on each circuit, I've been taking amperage readings using a clamp on ammeter. My theory is a little rusty and I'd like to run my method of calculating load on an unbalanced 3 phase circuit by you guys for accuracy.

Example:
480V 3 phase circuit breaker feeds 11 parking lot fixtures.
Fixtures are 480V single phase L-L (no grounded conductor)

Phase A: 7 amps
Phase B: 7 amps
Phase C: 5.2 amps

Ia + Ib + Ic /3 x 480 X 1.732 = total VA

7 + 7 + 5.2= 19.2 A

19.2 / 3 = 6.4 A

6.4 x 480 x 1.732 = 5.3 kva total

5300 / 11 = 482 VA per fixture

Am I way off base? I ran across a similar post on this forum a while back with a couple different ways of calculating unbalanced loads but can't find the thread using the search function. Any input or alternative calculations would be greatly appreciated.
 
Am I way off base? I ran across a similar post on this forum a while back with a couple different ways of calculating unbalanced loads but can't find the thread using the search function. Any input or alternative calculations would be greatly appreciated.
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That looks correct. What other load do you have that can be reduced? Are you billed a demand charge? Are you billed a power factor penalty?
 
KAZ 88 said:
I've been tasked with reducing lighting load at an industrial facility in an effort to reduce utility costs. Rather than hook up a power monitor on each circuit, I've been taking amperage readings using a clamp on ammeter. My theory is a little rusty and I'd like to run my method of calculating load on an unbalanced 3 phase circuit by you guys for accuracy.

Example:
480V 3 phase circuit breaker feeds 11 parking lot fixtures.
Fixtures are 480V single phase L-L (no grounded conductor)

Phase A: 7 amps
Phase B: 7 amps
Phase C: 5.2 amps

Ia + Ib + Ic /3 x 480 X 1.732 = total VA

7 + 7 + 5.2= 19.2 A

19.2 / 3 = 6.4 A

6.4 x 480 x 1.732 = 5.3 kva total

5300 / 11 = 482 VA per fixture

Am I way off base? I ran across a similar post on this forum a while back with a couple different ways of calculating unbalanced loads but can't find the thread using the search function. Any input or alternative calculations would be greatly appreciated.
Click to expand...

Dont you have to take the different phase angles of the L-L loads into consideration since, these are L-L loads and not really 3-phase loads?
 
Thanks for the replys. Yes, I work for this company, management would like a dollar figure on power saved by turning off unused lighting in parking areas and unused work spaces. You know how bean counters can get. Throughout this past year we've been reducing load where possible. I was just wondering if there was a fairly accurate way of calculating these types of loads without vectoring etc.

Thanks
 
Your actual kwh savings will be slightly less, since the utility measures kW x time for the energy charge, and you are calculating kVA. The actual kW metered would be the kVA x the power factor of the lighting, probably around 0.9
 
Lighting loads are fairly predictable. Why not add up the fixture watts instead of measuring the currents? Make sure to get the ballast watts included, not just the lamp watts.
 
The wattage ratings are already in the body of bulb. Just add'em together. Why do you take the complicated route to calculate them? However, the only way you can save energy is perhaps, changing the light bulbs to more y efficient ones (for the same voltage and lumens rating). There are no other cheaper ways you can reduce your energy bill.
 
The are few ways to save on electricity

The are few ways to save on electricity

1. Turn them off when possible - the easiest and most cost effective solution. Can be done with timers or presence sensors.
2. Relamping - replace the existing lamps with more efficient ones. If you calculate the labor cost, it is quite expensive. However, you can get 30%-50% reduction.
3. Voltage controller - change the voltage to the lamps, depending on the network voltage and light requirement. There are products that can do this for complete lighting line which makes it very simple and cost effective installation. Savings can be 20-35% with very fast ROI.

And of course, it is possible to combine all the above (particularly 1 and 3, as the controllers has built in timers and remote programming capabilities).
 
I suggest that everyone really read and understand how the POCO charges you for consumption of energy.

Understand what demand really means, what peak demand is and how it is applied, any penalities you may be paying for poor PF/capacity fees.

Some plants leave machines idling all weekend long, leave lights on, and cutting back on these won't save the time and costs it takes to start them up, even though their consumption would be reduced.
 
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