For TCL 780KW and atransformer of 1000KVA
2007-09-02 17:16:49 · 4 answers · asked by Eng Doori 1 in Science & Mathematics ➔ Engineering
There is a HUGE difference between the Total Connected Load and the actual demand. In most real-world cases the actual demand is less than 50% of the connected load.
You can't size capacitors until you know the initial power factor (PF) and the desired PF. Your 780 kW load could be all resistance heat (PF=1.00) or all induction motors (PF=0.80).
You also do not size capacitors to match the transformer kVA. You size capacitors to improve the power factor.
The first link below performs all the math for you once you provide enough of the parameters. The second link is a power triangle diagram.
While this is completely incorrect, I suspect your teacher is expecting you to compute an initial PF using 780 ÷ 1000 = 0.78. They then want you to size a cap bank to improve the PF from 0.78 to 1.00 with the 780 kW load remaining constant.
2007-09-03 02:44:37 · answer #1 · answered by Thomas C 6 · 2⤊ 0⤋
I like that one
2007-09-03 00:21:43 · answer #2 · answered by Anonymous · 0⤊ 0⤋
I can't WAIT to see the answers for this one! LMAO
2007-09-03 00:21:03 · answer #3 · answered by Whynot 5 · 0⤊ 0⤋
Do your home work
Read the book.
2007-09-03 00:24:16 · answer #4 · answered by Questionable 3 · 0⤊ 0⤋