Understanding KVarh and KWh: What Do These Measurements Mean?

[brstilson]

Hi there,

I'm not an expert on electricity by a long shot but I do understand most of the basics (i.e. AC voltage follows a sine wave, etc. etc.). Anyway, I'm helping my company analyse the data we're gathering from our various utility meters and our electricity meters are giving us a reading in KVarH in addition to KWH, which we are recording but so far have have absolutely no idea exactly what it signifies. I did a little research on it today and so far what I've come up with is that the KVARH is basically the energy that is wasted starting an inductive load like a motor, and that the ratio between KVarH and KWh is what determines power factor.

What I'm not getting is that it doesn't seem to relate very well to the actual numbers we're getting. I'm just trying to understand how the two measurements relate to each other. For instance, on a given day we'll have 12,000 KVARH and 19,200 KWH. I guess I'd just like to understand a little better what that actually means. Is that good, bad, indifferent, etc.? Do I add the two numbers together in order to get my actual usage? Do I do something different?

 

[dlgoff]

Welcome to PF brstilson.

A http://en.wikipedia.org/wiki/Volt-amperes_reactive" (the ratio of the real power flowing to the load to the apparent power) is less than one. So what your are measuring over time (in hours) is the real energy Kw-hr and the reactive energy Kvar-hr.

From the numbers you are seeing, it is clear that your power-factor is not good. In order to improve the pf and decrease the Kvar-hr measurment you might need to add capacatance if you have a lot of motors runing (inductive load).

To maximize transmission efficiency, vars must be minimized by balancing capacitive and inductive loads, or by the addition of an appropriate (off-setting) capacitive or inductive reactance to the load.

 

[ravioli]

For instance, on a given day we'll have 12,000 KVARH and 19,200 KWH. I guess I'd just like to understand a little better what that actually means. Is that good, bad, indifferent, etc.? Do I add the two numbers together in order to get my actual usage? Do I do something different?

Depending on your region, this kind of power factor can get you penalized. The pF for the numbers that you gave us is about 0.848. If you are being charged a penalty, I would recommend a synchronous condenser if you have any extra synchronous machines. This is a very easy way to improve your power factor, and you don't have to worry about switching capacitors on and off. Most switches "flip" when the current is equal to zero to prevent arcing. Also the voltage of the capacitor will be at its greatest when the current is equal to zero. These two facts alone can show you that any switch you use on the capacitor will require at least twice the voltage rating of your line.

Lastly, if the poor power factor is costing your company a lot, hire an expert.

 

[brstilson]

Depending on your region, this kind of power factor can get you penalized. The pF for the numbers that you gave us is about 0.848. If you are being charged a penalty, I would recommend a synchronous condenser if you have any extra synchronous machines. This is a very easy way to improve your power factor, and you don't have to worry about switching capacitors on and off. Most switches "flip" when the current is equal to zero to prevent arcing. Also the voltage of the capacitor will be at its greatest when the current is equal to zero. These two facts alone can show you that any switch you use on the capacitor will require at least twice the voltage rating of your line.

Lastly, if the poor power factor is costing your company a lot, hire an expert.

Thanks for your answer. At my location, the electric company charges a penalty for a power factor under 0.80 and gives a credit on the bill if the power factor is 0.90 or above. I've been studying up on this a lot more and I think I understand how it works.

A inductive load causes the current wave to lag behind the voltage wave, which is called the phase angle. This delay can be counteracted by a capacitor which acts the opposite way, the voltage wave lags behind the current wave. If the capacitor and inductor are put in series and both phase angles are identical, they effectively cancel each other out.

So the problem with putting capacitors on an entire building is that while it may compensate for the phase angle of the inductive load, the compensation is static while the inductive load changes. If the inductive load disappears (like at the end of the day when most of the motors are turned off), the capacitors actually hurt the pF because they're pushing the phase angle the opposite way the inductive load was.

Please correct me if I'm wrong about any of this. I'm still learning. Thanks!

 

[ravioli]

If the inductive load disappears (like at the end of the day when most of the motors are turned off), the capacitors actually hurt the pF because they're pushing the phase angle the opposite way the inductive load was.

Please correct me if I'm wrong about any of this. I'm still learning. Thanks!

That's correct. To prevent that, you would have to put some kind of control on either the capacitor bank or the synchronous condenser if you expect to have a large variations in your loads KVAR.

 

[m.s.j]

What is the reactive power conception?
For a conceptual dicaussion about reactive power, you can refer to General Electrical Riddle No.35 in http://electrical-riddles.com

 

[Liam]

Can you explain how you calculated a power factor of 0.848 from those kVarh and kWh readings. Thanks !

 

[ravioli]

We know the apparent power S is calculated from S^2=P^2+Q^2. We also know that pF=\frac{P}{S}. From there, it's just plug and chug.

 

[Liam]

sorry i get 0.78 ...maybe I am doing it wrong. If you wouldn't mind demonstrating it I would appreciate...thanks once again.

 

[lbrett]

sorry i get 0.78 ...maybe I am doing it wrong. If you wouldn't mind demonstrating it I would appreciate...thanks once again.

As mentioned by ravioli above S2 = P2 + Q2 i.e. s = rt((19200)2 + (12000)2)
s = 22641.55 kVA

pF = p/s i.e. 19200/22641.55
pF = 0.85 (0.847)

 

[saadsms]

Hello
Please, i want to ask the following question and give me the answer as much as possible
What is the difference between the watt hour meter and Var hour meter in terms of internal connection and work theory .
thanks

 

[dlgoff]

What is the difference between the watt hour meter and Var hour meter in terms of internal connection and work theory .

Volt Amp Reactive (VAR) are measured by determining the phase angel between the current and voltage waveforms. This measured angel is used to calculate the Power Factor which determines the VAR value.

Transducers are used to meter such parameters in the power industry and can be bought with as "2 element", "2 1/2 elements", or "3 elements".

You might be interested in looking into these devices. They preform the same calculations of modern Watt/Var meters.

Here's an earlier post on the subject:

https://www.physicsforums.com/showpost.php?p=2619637&postcount=4"

Regards

 

[saadsms]

Thank you dear
I understand the triangle of power components.
But my question for the analog meter with a dynamic disk. Do the same meter measures kw and kvar or there is a difference in internal structure and theory of work between each meter.
ie how can i know the meter if it is watt hour meter or varh meter without specification lable fixied on it.
Thank you

 

[dlgoff]

Thank you dear
I understand the triangle of power components.
But my question for the analog meter with a dynamic disk. Do the same meter measures kw and kvar or there is a difference in internal structure and theory of work between each meter.

I took this quote from the paper http://www.google.com/search?q="var...=org.mozilla:en-US:official&client=firefox-a".

The reactive energy meters, for both single-phase and
three-phase applications, can be mainly divided into electromechanic
and static types.
The traditional electro-mechanic (inductive) types are
realized with an induction watt-hour meter and an
equipment for the displacement by 90° of the voltage; this
can be an autotransformer or a RC circuit; for three-phase
applications, the measurement of the reactive energy can be
obtained from the active one, by means of a suitable
connection of voltage terminals, based upon the relationship
between line-to-neutral and line-to-line voltages in
symmetrical systems. The reactive energy measured by the
inductive meters is close to the one related to the
fundamental components of voltages and currents, even
when harmonic components are present...

I have emphasized the answer to the relevant part of your question in Bold Type.

 

[saadsms]

I took this quote from the paper http://www.google.com/search?q="var...=org.mozilla:en-US:official&client=firefox-a".



I have emphasized the answer to the relevant part of your question in Bold Type.

Thank you my friend
Do you mean that interconnect such as Star or Delta connections are Determined the work of the meter
If it internalconnected as a star for voltage coils it records kWh and if internalconnected is delta it records kilo varh ?

 

[dlgoff]

Thank you my friend
Do you mean that interconnect such as Star or Delta connections are Determined the work of the meter
If it internalconnected as a star for voltage coils it records kWh and if internalconnected is delta it records kilo varh ?

No and No.

The electromechanical meter (for 3 phase power) feeds the power of each line (phase) through the meter and the disk spins for either Watt-Hr or Var-Hr measuring. The only difference is for Var-Hr meters, there's "equipment for the displacement by 90° of the voltage; this can be an autotransformer or a RC circuit".

 

[saadsms]

you mean now that the kvarh meter has adesplacment angle ( -90 ) for current coil lagging the voltage in voltage coil .
and the kwh meter has adesplacment angle ( o) ie in phase ?
And nothing to do with the internal link
And the settings are but by special equipment at the factory to adjust the above ?
please correct.
thank you
very much

 

[salma66]

I'm also interested in this question, and it would be nice if someone posted the corrected result, please!

 

[lexus31rus]

Hi there. I did thru same calculations with my electric bills.
I compare my numbers for october or last year where i have
79920 kWH and 36400 kVARH
S=√79920^2+36400^2
S=87818
ph=79920/87818=0.91, which is not bad, right?
Then we replaced Cooling Tower and my numbers droped to
62960 kWH and 11920kVARH... uh, huge drop for the second number!
But then when i do the same calculation i come up with pF=0.97...
So, how is such a big drop of kVARH give me only 6% of pF?
Do that number even look right?

 

[jim hardy]

draw your right triangles and check the trig.
at small angles the hypotenuse isn't much affected by opposite side.
SRT scale on sliderule demonstrates that, as does cosine graph it's pretty flat at top.

i use a slightly different power factor algorithm that let's me do a sanity check at each step, and might be less keystrokes...

ARCTAN of (VARS/WATTS) gives angle between volts and current ,
COS of that angle gives power factor
after division it's only two keystrokes.

And if a get a large angle I've mis-typed something.

try it and see if if is comfortable for you.

36400kVARhrs/79920kWH = 0.455455 etc repeating decimal
arctangent of that = 24.487 degrees, reasonable enough
cosine of that = 0.91005

awareness of the angle might help you - i don't know...

 

[FOIWATER]

a practical way to add and take out the power factor correction capacitors, or at least how we do it, is to use a KVAR transducer that accepts inputs from two sources, a current transformer and a potential transformer and calculates the phase angle between the two. It outputs a 0-10vdc signal based on the difference in phase (of supply)

the output is connected to scr firing modules, and the scr's AK junction to the capacitor bank. we adjust the output to fire either one or a multiple of our four available banks.

We do this for power factor penalty purposes