Electrical Engineering - 3 Phase Delta Unbalance circuit currents

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3 Phase Delta Unbalance circuit currents

Lew

2006-03-15, 7:22 pm

A SCR pulses 480VAC to the above resistive heating element arrangement.
Notice, I have 2 seperate deltas 300ohm legs and 2000 ohm legs. I
also have 575 ohm heaters across L1-L3 and L2-L3. Using an amp clamp,
I have 39,39, and 9 amps on legs L1,L2, and L3. How do I calculate
what I should have in theory?

L1----------------.----------------------.------------------------------------------------------------|
| |
|
| |
|
L2-----------.---------------------.---------------------------------------------------|
| 575
| | | |
| 575 |
| | | |
| |
L3-----.---------------------.---------------------------------------------------------|--------------|-
| | | | | |
| | | | | |
| | | | | |
| /\ | | /\ |
300 | / \ 300 | / \ |
| / \ | 2000 / \ | 2000
|/_____\| |/______\|
300 2000

gnoge

2006-03-15, 11:21 pm

I have a similar question to post as well. I have a bunch of 3-phase AC
SCR power controllers (CCI) connected as follows:
480 delta supply through SCRs delta output to a 480delta/480wye
isolation transformer to a 480 delta quartz lamp load (slightly
unbalanced).

I have a single-phase VT and CT in each cabinet, which I can move to
any place in the circuit as necessary. I would like to determine how
much total (3-phase) power (kW) I am delivering at any point in time,
based on some measurements from these single phase devices. In an ideal
situation, I would think Volts * Amps * SQRT(3), but I am not so sure
about the phase-fired SCRs.

I have taken some V, A, kVA, and kW readings with a Fluke 43 True RMS
Power Quality Meter/Scope, but am perplexed by the results. For
instance, a typical reading at the output leads from the transformer to
the load (Volts T3-T2, Amps on T3)
Volts RMS: 323.5
Amps RMS: 84.5
kW: 19.9
kVA: 23.2
PF: 0.86

So based on the meter kW, my three phase power would be 19.9 * 1.732 =
34.5kW. However, based on the meter V, A and PF
323.5 * 84.5 * 0.86 * 1.732 / 1000 = 40.8kW which is about 18% off.
Where did I go wrong?

One note, if I read line voltage (i.e. 480V) and amps through one leg
of the SCR, I my calculated results using volts and amps are much more
consistent with the power readings on the Fluke. Since the SCRs follow
the line voltage (when they are on), should I use the line volts (480)
and RMS amps to calculate my power?

p.s. If you want, I have some interesting images of pre/post
transformer waveforms I could email out if anyone is interested.

Ben Miller

2006-03-16, 12:21 pm

"gnoge" <jon.roper@ge.com> wrote in message
news:1142476159.002543.265530@e56g2000cwe.googlegroups.com...
> So based on the meter kW, my three phase power would be 19.9 * 1.732 =
> 34.5kW. However, based on the meter V, A and PF
> 323.5 * 84.5 * 0.86 * 1.732 / 1000 = 40.8kW which is about 18% off.
> Where did I go wrong?
>

Check the "Instrument Setup" on the Fluke 43. You can set power factor to be
total (including harmonics) or displacement (fundamental only). You can also
set the power to be total or fundamental only. You need to understand the
differences in the information you will get. Also, be sure your waveform is
within the crest factor rating of the instrument.

Ben Miller

--
Benjamin D. Miller, PE
B. MILLER ENGINEERING
www.bmillerengineering.com

Ben Miller

2006-03-16, 12:21 pm

"Ben Miller" <benmiller@worldnet.att.net> wrote in message
news:G_adneIgh-3rFYTZnZ2dneKdnZydnZ2d@comcast.com...
> Check the "Instrument Setup" on the Fluke 43. You can set power factor to
> be total (including harmonics) or displacement (fundamental only). You can
> also set the power to be total or fundamental only. You need to understand
> the differences in the information you will get. Also, be sure your
> waveform is within the crest factor rating of the instrument.

One correction... You don't select a choice for the power factor. You get
both readings on the screen. Which one did you use?

Ben Miller

--
Benjamin D. Miller, PE
B. MILLER ENGINEERING
www.bmillerengineering.com

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