# kW or KVA

Discussion in 'General Electronics Chat' started by ectmlk, Mar 5, 2012.

1. ### ectmlk Thread Starter Member

Feb 11, 2008
21
0
Hi
I've been studying power in AC circuits, primarily 50/60Hz circuits. I have read and watched various sources on what a residential utility meter measures but I am still a bit confused.
I partially accept that these meters for the domestic consumer reads kW/hr (active power) and hence the domestic user doesn't get charged for the kVA (apparent power) they use.
However when I read the description for how a utility meter operates i.e. it takes the instantneous Voltage and Instaneous Current and gets the product of these it seems quiet clear that it is measuring kVA not kW where there is a power factor less than 1.

Is it not easier to measure kVA than kW as you would need to take into account power factor to get the kW (bearing in mind the old utility meters which are mechanical in nature)?
How does the meter respond to reactive power kVAR?
How can it account for power factor?

Something I have been having difficulty getting is the internal operation of a utility meter, some theory a bit more detailed than above.

Any enlightenment appreciated!!

Regards,
M

2. ### crutschow Expert

Mar 14, 2008
21,372
6,124
It is quite clear. You just made an incorrect assumption. If you multiply the average current times the average voltage you get VA. If you multiply the instantaneous Voltage types the Instantaneous Current you get real power. If you look at the waveforms for voltage and current with a 90 degree phase angle between them (0 PF) you will clearly see that the instantaneous multiplication of the two will give an average value of zero for a complete cycle.

The meter thus responds to real power and does not respond to reactive power.

3. ### ectmlk Thread Starter Member

Feb 11, 2008
21
0
Thank you, (we know what they say about assumptions!!!!)
If i multiply instantaneous V by Instantaneous I, and take that to be real power than that means I have negative real power which is incorrect. I can see this from an LTSpice simulation of a capacitor.

From my understanding the "negative" power represented the reactive power.

Also from what I have seen in AC power circuits real power is only different from VA by a PF (Watts = V*I*pf), not a difference in averaging or not averaging (maybe this is where i have been wrong so far)?

Regards,
M

4. ### Expat New Member

Mar 6, 2012
1
0
Ectmlk,

Draw the waveforms for a purely inductive circuit. You will see that the voltage is out of phase with the current by 90 degrees. you will also know that this is theoretically 100% reactive and 0% real power.

Now, look at the point which would be 120 degrees of the voltage. This point will be approximately 0.866 (root 3 / 2), and positive. The Current waveform will be 0.5. This gives a positive value.

Look at the point which would be 60 degees of the voltage. The voltage waveform is still root 3 / 2, but the current waveform is -0.5. Multiply the two of these, and you will get a negative value.

Therefore, in a purely inductive (100% reactive, zero watts) circuit, you can get both positive and negative instantaneous values for the product of current and voltage.

Graphically, you can see that from 90 degrees to 270 degrees, you have only positive current in a purely inductive circuit. Therefore, you can use this window to determine the level of VAR in a circuit. The area under the curve from 90 to 270 will be proportional to the VAR percentage. Any "negative" area will be real power, and will be subtracted from the VAR.

We use this circuit in some older motor controls to calculate real and imaginary (reactive) power. We charge a capacitor with the current waveform, using the voltage waveforms to "Open" or "Close" the 90 to 270 degree windows. The voltage on the capacitor is roughly proportional to the desired unit (watts or vars). We calculate wattage in a similar manner, with different start and stop points of the "window" (0 and 180).