Maker Pro
Maker Pro

AC mains input impedance

S

Scott Ronald

Jan 1, 1970
0
legg said:
There's a measurement technique suggested by Kwasniok, Bui, Kozlowski
and Stuchly in the IEEE Transactions on EMC Vol35 #1 Feb'93.

RL


Suppose that someone wanted to generate 3kW on the power line the
equation to calculate this power would be

{[(VgeneratorPeak - VlinePeak)/sqrt(2)]^2}/LineImpedence = 3kW

Is this correct?

If the impedence is high then the generator peak voltage must also be
high. Unless you had a way to change your generator peak voltage you
cannot reliably generate 3kW, since you are at the mercy of the local
line impedence.

Scott
 
L

legg

Jan 1, 1970
0
Don't seem to be able to get a link posted through the mail server.

Google for an article by Montoya on Power Line Communications.
Theres a nice graph showing impedance ranges measured above 1KHz..

If you are talking about pumping power into the mains, there are a
growing number of standards that cover this kind of grid linked
interface. Google some terms.

Weak-Grid, micro-grid, nanogrid, distributed generators, grid
interconnections standard, grid-conected inverter.

The instances where client a would be expected to produce exactly
3.00KW into a distribution node are infinitessimal. Thats not how it
works. The 'generator' has a reference frequency and phase, with
acceptible distortion limits.

Running voltage sources in parallel requires compensation, if only
the crudest of sharing chokes. If the load on the two sources is
nonlinear, it's not the responsibility of just one of those sources to
compensate, it's a shared responsibility. Under those cicumstances,
your generator should probably be configured to act as a sinusoidal
current source, locked to the grid frequency, as there's no way your
own impedance will match the capability of the larger utility.

RL
 
T

The Phantom

Jan 1, 1970
0
legg said:
There's a measurement technique suggested by Kwasniok, Bui, Kozlowski
and Stuchly in the IEEE Transactions on EMC Vol35 #1 Feb'93.

RL


Suppose that someone wanted to generate 3kW on the power line the
equation to calculate this power would be

{[(VgeneratorPeak - VlinePeak)/sqrt(2)]^2}/LineImpedence = 3kW

Is this correct?

If the impedence is high then the generator peak voltage must also be
high. Unless you had a way to change your generator peak voltage you
cannot reliably generate 3kW, since you are at the mercy of the local
line impedence.

Scott

The impedance at our service entrance was measured as 1/10 ohm.

The way to avoid the problem you are anticipating is to sell power to the grid
from a current source rather than a voltage source.

Selling from a voltage source whose waveshape is a good sinusoid can also lead
to too high distortion of the current supplied to the grid because of the grid's
own distortion.
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
The said:
legg said:
Hi

Does anyone know what a ballpark value for the input impedance at 60Hz
for North American AC mains? Or does this vary wildly from region to
region.


There's a measurement technique suggested by Kwasniok, Bui, Kozlowski
and Stuchly in the IEEE Transactions on EMC Vol35 #1 Feb'93.

RL


Suppose that someone wanted to generate 3kW on the power line the
equation to calculate this power would be

{[(VgeneratorPeak - VlinePeak)/sqrt(2)]^2}/LineImpedence = 3kW

Is this correct?

If the impedence is high then the generator peak voltage must also be
high. Unless you had a way to change your generator peak voltage you
cannot reliably generate 3kW, since you are at the mercy of the local
line impedence.

Scott

The impedance at our service entrance was measured as 1/10 ohm.

That's a reasonable figure.

Assume a 15 kVA, 240V single phase transformer with a 2.5% impedance.
That comes to 96 milliOhms impedance. Throw in a little more for the
service drop and you get in the neighborhood of 0.1.

On the other end, residential service equipment is rated for a fault
duty of 10,000 A. That gives you an impedance of about 24 milliOhms.
 
Top