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I would approach this analytically: -you know it can't be 0 (that is the grounded phase) -you know since you have some pos/neg Z it can't be 100 That leaves 114 and 173. If you understand the fault scenarios you know the phase V will rise 173 if no pos/neg Z If we assume no pos/neg: If =...

The pf drop from 0.85 to 0.8 has little effect. Basically sqrt(ratio): if pf goes down, Z goes up (Vdrop goes up), and the inverse. From table Z = 0.057 Vdrop = 400 x Sqrt3 x 250/1000 x 0.057 = 9.87, net 470.13 adjust Z for pf ~ 0.057 x sqrt(0.85/0.8) = 0.0588 Vdrop 10.18, net 469.82 470...

Vdc = 3 sqrt2 208/Pi =280.9 280.9 - 60 = 220.9/10 = 22.1 Ohm 22.1-2.5 = 19.6 Ohm btw 0.955 x (sqrt2 x 208) = 280.9 0.955 = 3/Pi   

I call phase-phase the phase V, and phase to n, the line voltage I was just pointing out you could work the problem on a 3 phase basis or a single line equivilent... Zbase = V3^2/S3 = VL^2/S = 15kv ^2 / 30 = (15kv/sqrt 3)^2 / 30/3 (3's cancel so = 15kv^2 / 30, the same) where S = S3/3 and VL...

or if you do use the per phase power 30/3 = 10, you better use the line (not phase-phase) voltage = V/sqrt3 so Zbase = (V/sqrt3)^2 / 10 = (V^2 / 3) / 10 = V^2 / 30 same as V^2 / 30 can't mix phase to phase with phase to line

what you want is Zf pu = Zf / Zb, Z under a bolted fault... we assume an infinite buss and Vb remains the same, no droop... Sf = Vb^2 / Zf or Zf = Vb^2 / Sf and Sb = Vb^2 / Zb or Zb = Vb^2 / Sb so Zf / Zb = (Vb^2 / Sf) / (Vb^2 / Sb) = Sb / Sf as is illustrated in the problem... and it makes...

I come up with pf ~ 0.9518 i ~ 2.1633 I calc'ed the power triangle: P ~ 0.71 x 230 x 2.9 ~ 473.6 W S ~ 2.9 x 230 ~ 667 VA W ~ j sin(arc cos 0.71) x S ~ j469.7 VAR calc'ed new C reactive power added ~ V^2/Z ~ 230^2 / (1/j(2 Pi 60 15.9 x10^-6)) ~ -j317.1 VAR new W total ~ j469.7 - j317.1...

wow, been 1 year already... just closed the books for last year, better than last: base 110k + 20k in OT bonues/profit 150k 280k total...unimaginable 5 years ago, honestly... full health, 401 with 1:1 matching to max allowable ~13k expenses ~1k/month, give or take plus picked up another...

what is a motor? COILS of wire what is the operating mechanism of an INDUCTion motor? it's an inductor(s) or wound coils, albiet moving in relation to each other it can ONLY be LAGGING... Z = jwL (ignoring resistance) V = Zi = jwL i the phase angle ( tan^-1 ) for any positive number (wL)...

as you get closer to the generating source the voltage increases...that goes without saying...Vd ~ impedence, ie, V=IR you're not giving us any information: what are the prim/sec voltages? what wire sizes? they start a 3000 HP motor across the line? not reduced voltage, delta-wye, etc.? 21%...

so the xfmr only serves a 3000 HP motor? what kind of load, what is the motor driving and how is it started? VFD or across the line? transient or steady state? does the voltage only dip when starting up, or does it stay low when running? is the drop at the motor leads or xfmr terminals...

at the motor or transformer? transient or steady state? motor HP, xfmr KVA?

I was out >20 years, studied the ASCE CD and FE manual for ~100 hrs and passed first attempt... I heard horror stories about being out of school too long, the math, etc., but honestly I didn't think it was too bad... curious to hear from some of the 'old dogs', lol

aluminum or steel? http://www.mcnichols.com/products/grating/...eckplate_lt.htm vendors are a great source for this kind of data

really? now THAT may be a violation....

|Z| ~ sqrt(0.029^2 + 0.048^2) ~ 0.0561 at 0.85 pf per 1000' adjust for pf = 0.8 (it's linear) 0.8/0.85 x 0.0561 ~ 0.0528 adjust for D ~ 250/1000 x 0.0528 ~ 0.0132 Vd ~ 400 x (0.0132) ~ 5.28 adjust for 3 ph ~ 5.28 x sqrt3 ~ 9.14 V ~ 470.85...give or take

first, the current doesn't 'cancel' in a 240 circuit...let's say the load is 10 ohm, the current is 24A in each leg... you MUST have a loop or circuit path to conduct current... EACH leg carries 24A, it's ALTERNATING current... so the loss would be (Iline)(Zleg1)+Vload+ (Iline)(Zleg2), we...

on a tangent: anybody follow F1 racing? the season's 1st race is this weekend talk about technology they implemented sweeping (the largest in the sports history) rule changes to slow the cars down smaller rear diffuser smaller front wing no winglets lower engine rpm by 1000 standard ecu...

how much power for how long? you need 0.012 A at 5VDC ~0.06 W but how long do you need to supply this? 5 sec? 1 minute? ??? possibly a capacitor charged to a V >> 5 and bled thru a regulator chip

no, it's not a N+N (2N) if you lose the UPS, you have no back-up...no redundency a single compnent failure results in loss of back-up power... you need another UPS or 6 smaller ones of 200KVA each (N+1 or in this case 5+1)