75kva 208 wye load calculation "Balancing"

[kwired]

It doesn't matter. My point remains that he can't take more than 208A from any line.

True but if you don't know how to add the phase to phase connected loads you will think you ran out of capacity sooner then you actually do run out of capacity, 10 amps plus 10 amps does not always equal 20 amps.

 

[Besoeker]

True but if you don't know how to add the phase to phase connected loads you will think you ran out of capacity sooner then you actually do run out of capacity, 10 amps plus 10 amps does not always equal 20 amps.

OK.
How about a clip on ammeter and measure the current on each line?
No calculation required.

 

[topgone]

OK.
How about a clip on ammeter and measure the current on each line?
No calculation required.

That simple! Haha.

 

[kwired]

OK.
How about a clip on ammeter and measure the current on each line?
No calculation required.

In some ways that is putting the cart before the horse though. One might want to know what things will be before purchasing materials or planning other parts of the project. OP of this thread sort of lucks out and finds there is more capacity per phase then he maybe thought there was but you don't want to be in a condition that is the other way around and your misunderstanding of how the currents add up left you with an undersized supply.

 

[Carultch]

In some ways that is putting the cart before the horse though. One might want to know what things will be before purchasing materials or planning other parts of the project. OP of this thread sort of lucks out and finds there is more capacity per phase then he maybe thought there was but you don't want to be in a condition that is the other way around and your misunderstanding of how the currents add up left you with an undersized supply.

Agreed. It is cart-before-the-horse to wait until things are built before finding out that they aren't the ideal materials for the job.

But in a way, that is how manufactures eventually develop the nameplate/datasheet values that we use in our calculations.

 

[Besoeker]

That simple! Haha.

Thank you for taking it in the vein it was intended...............

 

[dorianthird]

220.18 (B)

220.18 (B)

each light with ballast ~ 1.1 kva ~ 53 kva < 75 kva good to go
in essence you have 3 25 kva single ph xfmrs
48 lamps / 3 phases = 16 lamps per phase
put 16 lamps on each ph
each ph 16 x 1.1 ~ 18 kva < 25 kva good to go

220.18(B) states that "For circuits supplying lighting units that have ballasts, transformers, autotransformers, or LED drivers, the calculated load shall be based on the total ampere ratings of such units and not on the total watts of the lamps."

besides most 208V ballasts for a 1000W lamp would draw less than 1000W/208V=4.81Amps right?
and where does the 1.1kva come from?

 

[Smart $]

...besides most 208V ballasts for a 1000W lamp would draw less than 1000W/208V=4.81Amps right?
and where does the 1.1kva come from?

Nominallly, no. The unit as a whole will draw more. The 1000W rating is for the lamp only, nominally. That could go up or down depending on the pertinent parameters' variance. The ballast itself in operation is not 100% efficient, so it will use a few to several watts. So nominally, over 1000W.

~1.1kVA comes from the ballast having reactive components. As a result, energy is transferred to and from the ballast without being converted to heat or light. The proportion by which that occurs is the power factor parameter (pf), which is never greater than 1.

P = V × I × pf

The kVA value is just (V × I).

Welcome to the forum... :thumbsup:

 

[Ingenieur]

a 1000 W ballast would draw more than 1000/208'A
if a modern hi eff ballast with 90 pf (1000/0.9)/208 A

 

[Ingenieur]

220.18(B) states that "For circuits supplying lighting units that have ballasts, transformers, autotransformers, or LED drivers, the calculated load shall be based on the total ampere ratings of such units and not on the total watts of the lamps."

besides most 208V ballasts for a 1000W lamp would draw less than 1000W/208V=4.81Amps right?
and where does the 1.1kva come from?

what they are saying is load must be based on va not W
in other words pf must be considered

P Power = S total power x pf so typically S va > P W since pf < 1 and for a common V S amps > P amps