I need to build an inductive load simulator with a 500 watt Tungsten
lamp as the primary load. I need to achieve a power factor around .
75. The system voltage is 120V/60Hz.
I'd like help in calculating the correct Henry rating for an inductor
to install in series with the 500W Tungsten lamp. Once I have that, I
see there are several web sites that give me directions on how to
build air core inductors with solid wire.
Thanks in advance.

On 12/6/07 8:42 PM, in article
snipped-for-privacy@e23g2000prf.googlegroups.com,
" snipped-for-privacy@sternerc> I need to build an inductive load simulator with a 500 watt Tungsten
You really do not give enough information. Putting an inductor in series
with a lamp will result in reduced power to the lamp. In turn that means
that the lamp will run at a lower temperature. In turn this means that the
lamp resistance will be considerably lower than what it would be a design
temperature.
The resistance of the lamp will be a nonlinear function of the inductance in
series with the lamp. That makes calculation considerably more complicated
than it would be for a resistor that is not temperature sensitive. If you
cannot do it, do not expect it to be don for you for free.
Bill

I see it does get quite complex.
The inductance wouldn't have to be in series. I assumed we could get
to where we need to be with minimal voltage drop across the inductor
by over-sizing the wire used in the air-core inductor and that a
paralleled inductor would be much larger.
I need to use a 500 watt Quartz lamp (because of the inrush
characteristics) and must end up with a circuit power factor of .75
when connected to a 120V, 60Hz supply line.
I can't think of any other information to provide. What additional
information would you like?
Steve

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If you have a series inductor such as to get the desired power factor, it
will definitely impinge unfavorably on the lamp voltage even if the wire
used has negligable resistance. Oversizing the wire won't help with the
voltage drop. Assuming the lamp resistance was constant then the voltage
across the lamp for a 0.75 pf would be 90V- however as pointed out by Salmon
Egg, the lamp resistance would not be constant-so lower lamp power, cooler
filament, lower resistance and lower power factor, etc -you could end up
with a very dim lamp.
You are concerned with the inrush characteristics but a series inductor will
change these as well. A parallel inductor would be a better choice. In
either case, an air coil inductance will be quite large.
There is also the question of why you want a pf of 0.75. If you are trying
to simulate an inductive load then why use a lamp?
If it is absolutely necessary to have a real power of 500 Watts at a pf of
0.75 then use a resistor which doesn't have the temperature variation of a
lamp and consider that you want 90 V across this with a series inductor and
a 120V supply. This is 5.6A so the resistor is 16.2 ohms so the inductor
must be 14.3 ohms or 0.038H at 60 Hz. A bit large for an air core coil. If
you use a lamp rather than a relatively constant resistor- then you have a
bit of a messy problem. --
Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer
Steve

I am conducting an in-house test of a switching device prior to
submittal to a testing lab...trying to simulate the labs tests.
The paramaters of the test require the load be characteristic of
Tungsten, achieving steady state current within 4.1mS with an inrush
of at least 10x steady state current. Resistors could be used, but if
they are, we would still need to simulate the inrush by shunting part
of the resistance upon switch closure. I would rather just stick with
Tungsten lamps. The test parameters require 500W lamps be used.
It seems we need to calculate a parallel inductor. That's what I need
help with.

On 12/6/07 10:16 PM, in article
snipped-for-privacy@s8g2000prg.googlegroups.com,
" snipped-for-privacy@sternerc> I see it does get quite complex.
Let me suggest that you go electronic. Use a triac with phase control to
limit inrush current. As your lamp heats up and its resistance increases,
you use 100% conduction angle. You could also use a relay to short out the
triac. That way you get unity power factor except at the start. It is likely
to be cheaper because an inductor of the size you need is going to be
expensive and much larger.
Bill

Your stated specs, 500 watts resistive load, 0.75 power factor and
120 volt 60 Hz supply, are enough to caculate a parallel inductor. There
are several steps and I'm not sure which one you're missing, so here's the
whole process:
1. Calculate reactive load: Draw a right triangle. The horizontal leg is the
500 watt resistive load (real power), the vertical leg is the inductive load
(reactive power) to be determined and the hypotenuse is the apparent
power. Power factor, 0.75 in this case, equals real power divided by
apparent power. Solve for apparent power then reactive power.
2. From reactive power, solve for impedance at 120 volts.
3. From impedance, solve for inductance at 60 Hz. I get a number
under 100mH.
4. Work out the inductor construction from the formulas. Verify with
a meter and protect with a fast-acting fuse. Any wire large enough for
the current will have minimal DC resistance and quickly turn into a
demonstration of a short circuit or a fusible link.
Good luck,
Mike

----------------------------
------
If you want the load test to be characteristic of tungsten then model the
v-i -time characteristic of a tungsten lamp. This will depend heavily on
the thermal balance involved and conventional R-L or R-L -C circuit models
won't work. Forget the inductance because the inrush characteristic is
dependent on the thermal behaviour of the lamp- not upon some unknown
inductance You have, initially, a cold resistance which is cold. and a
resulting high inrush current (sure some filament inductance is present but
this is typically a pretty minor effect.
You are trying to model a thermal effect by a model which doesn't apply
(and a steady state "power factor" model is even less applicable.
I suggest that you look carefully at what is actually occuring and try to
model that. Conventional (linear) circuit theory doesn't apply. Ohm's Law is
up shit creek without a paddle.

Or...
Consider an inductor in parallel with the lamp. An inductive load of 440
VAR in parallel with 500 watts should give a total apparent power of about
667 VA and a power factor of 0.75 (if I got my sums right). That way the
lamp voltage stays pretty much constant (assuming negligible line losses)
For 120V 60Hz, I think that would be about 0.087 Henry
daestrom

That's the way I calculated it too, but I got more than 100 mH. With an
inductive load of about 440 VAR, you need an impedance of about 32.7 ohms.
At 60 Hz, that's more than 100 mH.
daestrom

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--------
I am a bit confused- in an earlier message you got 87mh and now indicate
over 100mh. The 87mh (0.087h) figure was correct.
I do agree that the parallel construction is the way to go but I have a
problem in that if the objective is to simulate a tungsten lamp-why use an
inductor-just go with a tungsten lamp? A linear resistor/inductor
combination, series or parallel will not give the desired inrush
characteristic or the tungsten lamp steady state behaviour.
Would steve-gatz clarify the specific parameters to be met? Inrush as for a
tungsten lamp- OK. 0.75 pf (steady state-meaningless otherwise)-why? Is the
lamp 0.75 pf- if so- why? It appears that the intent is to use an inductor
(series or parallel-doesn't really matter) and resistor (even switched) to
simulate the actual characteristic- it won't.
If you want a simulation using linear circuit elements, then use a parallel
capacitive circuit where you can jig resistance in series with the capacitor
to get a high inrush current with about the right time constant and jig the
total for the right pf except capacitive This too has problems as the duty
on the switch is different than it would be for an inductive load.
His primal instinct to actually use a tungsten lamp is right on target. What
do they do in the lab tests that he is trying to simulate and is the
simulation an actual test or a computer model test?

No, I'm afraid I'm the one that was confused. I read 100 mH, but somehow
was thinking 100 uH (microHenries). Guess too much 'holiday cheer'. Sorry
'bout that.
Yes, that is a question. I was focusing on wanting 500 watts with a pf of
0.75 and was just thinking of a way to 'simulate' that load. Although it's
hardly a 'simulation' since what I proposed was to connect the exact load
the OP spec'd.
daestrom

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