flood lighting problem

[marty12345]

I HAVE A PROBLEM I WANT TO WORK OUT LOAD ON A FOOTBALL PITCH LIGHTING
Philips ‘OptiVision’ MVP507 2KW Floodlight. there are 72 no off lights they are 400v 2 phase
the load is split over the three phase i though it would be 72 * 2ooow = 142000 so142000/400v=355 355/3= 118.33 amps but it is taking around 200 amps per phase
any ideas

 

[Paul.M]

Welcome to the forum Marty. You've put your question in the wrong section but I'll ask a couple of the lads to have a look at this instead of me making an half arsed attempt. (i get it at 208A but I'm not 100% sure so I won't put up my findings).

 

[marty12345]

thanks for speedy reply how did you work out the calculation

 

[Animal]

think paul's got ot . 72 x 2000 =144000 divide by 400 = 120A then there's the ubiquitous root3 gives us 207.84A

 

[davelerave]

your total lamp wattage is 144KW

as already said
that's 207 amp line current straight away

without any pf or losses added

 

[accordfire]

think paul's got ot . 72 x 2000 =144000 divide by 400 = 120A then there's the ubiquitous root3 gives us 207.84A

THat's about right - the calc is P = U * I * sqrt(3) * cos(phi) - where sqrt3 is angle of rotation and Cos(Phi) is inductive load factor - from memory.

For resistive loads like a floodlight, Phi will be zero. For inductive loads, such as flourescents, is can be around 0.95 or so.

Square root of a nominal 415V is 1.732., so:

144,000/(415*1.732) = 200.57A nominally. These are RMS values, by the way.