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KWH to KVA conversion

W

Wes Faul

Jan 1, 1970
0
I am working on a transformer load balancing program, and need to find some
info that I have not been able to find on the internet. Meters read in kWh,
but the transformers are rated in KVA. For Single-Phase meters, I have been
told that you multiply the kWh times .0035 (customer wasn't sure where this
number came from - if anybody has any ideas, please shed some light on this
subject). For example, 4,760 kWh is equal to 16.66 KVA (4760 * .0035 =
16.66). Is this correct?
Assuming that part is correct, how do I calculate the KVA from the kWh for
3-phase power? The customer said that the .0035 conversion factor doesn't
work for 3-phase meters. For example, if I have a 3 phase meter that used
46,951 kWh, does it overload a transformer with a capacity of 150 KVA? How
do I calculate that?
Thanks,
Wes
 
D

daestrom

Jan 1, 1970
0
Wes Faul said:
I am working on a transformer load balancing program, and need to find some
info that I have not been able to find on the internet. Meters read in kWh,
but the transformers are rated in KVA. For Single-Phase meters, I have been
told that you multiply the kWh times .0035 (customer wasn't sure where this
number came from - if anybody has any ideas, please shed some light on this
subject). For example, 4,760 kWh is equal to 16.66 KVA (4760 * .0035 =
16.66). Is this correct?

NO.

The *easiest* way to get kVA for a single phase load is to measure the
current it draws and the voltage supplied. kVA then is just current times
voltage, divided by 1000.

For three phase, use line to line voltage, and the highest phase's current
reading and calculate....
kVA = Vline * Iline * sqrt(3) / 1000

To use the kWh, first, you must find the kW of the load, not the kWh. To do
this you need to know over what time period it took to accumulate those
4,760 kWh. And you need to know if that was running continuously at a fixed
loading, or if it varied. If fixed, just divide kWh by the number of hours
and you have kW. If varying or cycling on/off, then you're much better off
getting a kW/kVA meter. Many such meters exist and can display either kW or
kVA (some even will measure kWh, but not all).

If you don't get a kW/kVA meter, but deduce the kW rating some other way,
then to convert kW to kVA, simply divide by the power factor. Problem there
is that to know the power factor, you need to have a meter that measures
that. Again, a good kW/kVA meter often has the option for reading the power
factor, but if you have kVA meter, you don't need to proceed any further.
Assuming that part is correct, how do I calculate the KVA from the kWh for
3-phase power? The customer said that the .0035 conversion factor doesn't
work for 3-phase meters. For example, if I have a 3 phase meter that used
46,951 kWh, does it overload a transformer with a capacity of 150 KVA? How
do I calculate that?

Using 46,951 kWh in a steady, continuous load for 30 day billing cycle of
the utility (720 hours), would mean your load draws 46951 kWh / 720 h =>
65.2 kW. If it runs at a 0.9 power factor, that would be 65.2 kW/ 0.9 =>
72.5 kVA.

But if it uses those 46,951 kWh by running only 1/3 of the time, then 1/3 of
720 hours is only 240 hours. *That* would be a load of 46951 kWh / 240 h =>
195.6 kW. If it runs at a power factor of only 0.7, that 195.6 kW load
would be 195.6 / 0.7 => 279.5 kVA.

You really need a proper kW/kVA meter to do this. Guessing at the power
factor or the number of hours the load runs to use 46,951 kWh is just that,
guessing.

Frankly, that 0.0035 'conversion factor' sounds like BS. It could *only* be
right if the customer knew how many hours and exactly what power factor it
takes to use that 4760 kWh. Chances are, he/she doesn't. If they did, they
wouldn't be asking you to figure the kVA rating.

daestrom
 
W

Wes Faul

Jan 1, 1970
0
Time is usually 30 days (720 hours). Where would I get documentation on the
power factor at? Should it be in the documentation for the meters?

Does this sound correct?
4760 KWH / 720 H = 6.61 KW.
Power Factor = 6.61 KW / 16.66 KVA
Assuming that the KVA they gave me was correct, that means the power factor
is roughly .4? Everwhere I've seen defaults to .85 for the power factor.
Is anybody out there familiar with electricity meters that power companies,
pud's, etc... use?
Thanks,
Wes


| I am working on a transformer load balancing program, and need to find some
| info that I have not been able to find on the internet. Meters read in kWh,
| but the transformers are rated in KVA. For Single-Phase meters, I have been
| told that you multiply the kWh times .0035 (customer wasn't sure where this
| number came from - if anybody has any ideas, please shed some light on this
| subject). For example, 4,760 kWh is equal to 16.66 KVA (4760 * .0035 =
| 16.66). Is this correct?

There are two unknowns you need to convert with. Time and power factor.
http://ka9wgn.ham.org/ |
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W

Wes Faul

Jan 1, 1970
0
I have the customer checking with his engineer to see if this sounds about
right. Is the calculation from KWH to KVA the same for 3 phase as it is for
1 phase? Or do I need to divide it by 1.732 (square root of 3)?
Thanks,
Wes

| Time is usually 30 days (720 hours). Where would I get documentation on the
| power factor at? Should it be in the documentation for the meters?

Power factor is the relationship between volt-amps (power drawn) and watts
(power used). It affects the phase lag (or lead) of the current relative
to the voltage. A power factor of 1.0 indicates all drawn power is used,
and the current is in phase with the voltage. Resistive loads have this
characteristic. Reactive loads will drawn more power then give some back
at other parts of the voltage wave cycle.


| Does this sound correct?
| 4760 KWH / 720 H = 6.61 KW.

6.61111111111111 :)


| Power Factor = 6.61 KW / 16.66 KVA
| Assuming that the KVA they gave me was correct, that means the power factor
| is roughly .4? Everwhere I've seen defaults to .85 for the power factor.
| Is anybody out there familiar with electricity meters that power companies,
| pud's, etc... use?

That's quite a low power factor, but motors could do that. Perhaps default
values listed are just typical sums resulting from a common mix of resistive
and reactive loads.

Power drawn and returned is still costing the power company something,
such as I2R line losses in transmission and distribution.
http://ka9wgn.ham.org/ |
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W

Wes Faul

Jan 1, 1970
0
Paul Hovnanian P.E. said:
kWh is a measure of energy consumed. 4760 kWh over what period. One
month?
Lets assume that the formula is based on the monthly consumption. The
0.0035 factor appears to be a rule of thumb relationship between kWh and
peak consumption and would only be valid for a certain class of loads,
in this case probably residential loads. The 0.0035 factor is based
(among other things) on a peak to average load ratio of 2.5.

The period is 1 month (720 hours - 30 days/month * 24 hrs/day). This works
out to be roughly 6.61 kW. Assuming the 16.66 kVA is correct, this works
out to roughly .4 for the power factor. Does this sound right?
Or for different load types either. Even single phase commercial loads
will have different load characteristics and power factors. So don't use
that number if you don't know what its for.


Is 46,951 kWh a monthly consumption? If so, divide by 720 (hours per
month) to get the average kW demand. Then multiply by the proper peak to
average load factor for the particular load type (good luck finding
these numbers). Finally, divide by the load power factor (which also
depends on the load type). Using your example, if the load factor is
greater than 2.3, or the power factor worse than 1.0, your transformer
is overloaded.

Thanks for the info. I just scheduled a conference call with this
customer's lead engineer to try to hammer some more of this stuff out.
Wes
 
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