Good vid, but we learned in class to use P=true power(KW), Ps=apparent power(KVA), Pq reactive power( KVAR).Also impedance is measured in OHMS(Ω) why use PU? For ease of calculations due to the fact that you can't measure an∞bus?
When you say 19:45 that 75 kVa transformer in 208/120 config is capable of 208A, is that per hot lead? so basically 600A of 120v power? or if you are just using two of the leads? 600A x 120v = 72 kVa Or, being 3 phase, is a 75 kVa transformer 75000/1.732/120 = 360 So 360 max on the total of the 3 120v lines? I've got a 75kVa 5.5% IMP 208v transformer that is regularly loaded with 100 Amps on each of the 3 lines. Been running that way for 4 years.
i noticed on the one line your utility power is Delta to Wye but yet each transformer after is listed as Delta to Wye is there a reason it is not Wye to Wye once power is transformed. Cause it appears that the input to all transformers is Wye after utility transformer
The wye secondary from the utility is just how the transformer secondary winding is connected, it doesn't mean everything from that point on has to be connected in wye. You can take three phases from a wye connected transformer's winding and connect to another transformer or motor in delta if you want. Think of it as just the way something is connected, it doesn't totally change the system like your question suggests.
Thank you for the informative presentation . Kindly assist , got a 90kw, 400vac , 50hz motor , motor starter is a star-delta , A 250kva generator is failing to start the motor , everytime the motor is started , the contactors on the panel locked in and at one point , they got burnt . Is the generator correctly sized or there are issues with need to be addressed. when running on Mains power , the system works fine
you switchgear is not designed correctly. the generator is going to deliver, and the motor is demanding 6 x startup current. so all your line equipment must be rated to handle the current..
In my opinion your generator is not able to supply the Starting current of the motor. lets calculate, Your Generator is 250 KVA so max current it can supply is 250/ sq. root 3*(0.4) = 360 Amps . Your Motor Full load running current is 90/ Sq.rt of 3 *power factor *0.4 = 162A ( Considering power factor to be 0.8) . now in star delta starting current is around 2.6 time of full load running current. so it needs 421.2 A currents for starting. however your generator is able to supply only 360 Amps.
@@jamesbond0004 No Motor demands 6 x starting current in DOL starting in star delta it only takes around 2 - 2.6 times . Also his system runs fine on the mains it is problematic only on generator. So line and switchgears are fine.
Utility transformer or Customer owned? If Utility, typically use infinite bus or call utility company. Customer transformer would be secondary of transformer ahead of it minus the losses calculated in the conductors to get to the transformer primary.
Thank you. I watched the video well. The fact that I can make such a useful video I'm jealous again.I hope you'll continue to make-up videos I'd appreciate it if you could help me when I ask you.
Good tutorial Eaton team....but @ 18:44 you have used 3 ph formula for calculating the secondary full load current of X'mr...could u please reconfirm it...isn't 1 ph formula has to be used here..?
If you consider only one conductor of all three conductors and its voltage i.e., 120Volts (Line to neutral voltage) we will use 1phase formula... He considered all the three phases so he used 3phase formula to calculate full load current....
