Maximum Demand KVA measured at transformer primary and secondary 1

[edison123]

I have a sanctioned demand of 100 KVA and my transformer is 250 KVA.

The utility energy & MD meter is on the transformer primary side (11 KV).

I want to connect a MD meter on the transformer secondary side (415 V) for cross verification.

I understand the KWHR units will differ from primary to secondary by transformer losses.

But what about maximum demand (KVA) measured at 11 KV and 415 V sides? Will they be different because of transformer demand? If yes, by how much the MD measured at transformer primary will differ from the MD measured at transformer secondary?


Muthu

http://www.edison.co.in

 

[waross]

At no load the primary will show the magnetizing current demand and the secondary will show zero demand.
At the condition of interest the magnetizing current will be swamped by the transformer losses.
When loaded the losses will depend on the current.
You are concerned with KVA demand and the difference is in kWatts demand.
At unity power factor KVA demand will equal kWatts demand.
The lower the PF the more difference there will be between KVA and kWatts.
If you are looking for an exact correlation try measuring kWatts as well as KVA.
If you resolve the kWatts and KVA into kWatts and KVAR and then add in the kWatts you should be able to find a correlation between primary and secondary KVA.
Remember that transformer losses increase as the square of the current. Because of this it is unrealistic to look for a correlation between KWHrs and KVAHrs with a varying load.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[edison123]

Hi Bill

I am looking only for the difference in KVA maximum demand (KVA MD) when metered before and after transformer.

Muthu

http://www.edison.co.in

 

[waross]

I'll try a different way.
Consider the power triangle: kW-KVAR-KVA. kW and KVAR are at 90 degrees.
You can measure instantaneous kW and KVA.
Transformer losses are related to the square of the current.
Transformer losses add to the kW side of the power triangle.
To see the effect of the kW losses on the KVA you must use the Pythagorean theorem to determine the KVARs and then recalculate the KVA.
For loads at other than unity power factor, greater the load, the greater the difference between the primary measured KVA and the secondary measured KVA.
The lower the PF, the greater the difference between the primary measured KVA and the secondary measured KVA.
If you want to monitor your demand to avoid exceeding your KVA limit, then secondary metering should be close enough. (Add a few KVA safety margin. (Please run the numbers to verify this general statement in your specific case)
If you wish to verify the accuracy of the primary metering, you may have to consider the transformer losses.
Generally the max KVA demand will coincide with the max kW demand, however this is ageneral statement and depending on your load profile the max kW demand may not coincide with the max KVA demand. (Rare but possible)


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[cuky2000]

Yes, will be somehow close but different.

The secondary peak
>>>>>>>>>>>>>>> .[highlight #FCE94F]kVApeak_415V = kVApeak_11kV - kVApeak_losses[/highlight]
The peak losses are non-linear. See a typical efficiency & losses curve below.

 

[prc]

250 kVA transformer details - 4.5 % impedance-Iron/Copper losses- 0.8/2.5 kW ,Excitation KVAR- 2 % of rated kVA
Assume your maximum demand 75 kW at UPF. Secondary maximum demand = 75 kVA
Excitation KVA= 0.8 +j5 -constant irrespective of load on transformer
Reactance KVA= 2.5x0.3 square +j250x0.045x0.3= 0.225+j3.375 -varies with load
Add all the three to get primary kVA-
76.03+j8.4 = 76.48 kVA
If your load is capacitive primary kVA may reach less than secondary kVA.

 

[edison123]

Thanks prc, those are the numbers I was looking for. My secondary side PF is automatically set between 0.97 to 0.99 lag with a PF controller at the transformer secondary.

When the utility does the monthly reset this month, I plan to run the transformer on open circuit for about 4 hours and note down the KVA, MD KVA, KW, KWHR, KVAHR every 15 minutes. Will that be sufficient?

Any idea how much a 11 KV, 100 KVA outdoor type CTPT metering unit will cost in India?

Muthu

http://www.edison.co.in

 

[prc]

CT-PT metering unit (0.2 class) + Maximum demand meter -one lakh rupees +18 % tax

 

[edison123]

Thanks again, prc.

Muthu

http://www.edison.co.in

 

[jghrist]

You can get revenue meters that compensate for transformer losses.

See

https://www.landisgyr.com/webfoo/wp-content/uploads/2012/09/TLCpaperLG1.pdf

for a discussion of loss compensation.

 

[waross]

I plan to run the transformer on open circuit for about 4 hours and note down the KVA, MD KVA, KW, KWHR, KVAHR every 15 minutes. Will that be sufficient?

That will indicate the exciting current demand, not the full load demand.
From prc's numbers, 76.48 KVA pri. / 75 KVA sec. = 1.0197%
You are looking at less than 2% difference.
Are you able to see the demand on the utility meter?
If so, at a time of relatively high loading, check your KVA with a clamp meter and a voltmeter.
This should give you a quick indication of any gross primary metering errors.
That will give you an indication of the accuracy of the primary metering within about 2%.
Be aware that most demand meters do not respond instantly to a block change in demand.
Typically it takes 15 minutes for the demand indication to stabilize after a block change in demand.
If you are able to see the Utility meter, you may be able to get more usable information.
On the older electro-mechanical meters, there is a kH number on the face of the meter.
This is the Watt-hours per disk revolution. (NOT kilo-Watt hours)
I would time ten consecutive revolutions.
Ten equal times would indicate a steady load.
From that I could calculate the instantaneous kW demand.
Old school, there would also be a KVARHr meter.
From that I could determine the instantaneous KVAR demand and then the KVA demand and the PF.
The display on many, but not all, electronic meters will cycle through those values.
You may still often use the kH to determine the instantaneous kW.
There is a series of dots that progress across the display. Time a complete cycle of dots.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[edison123]

Thanks Bill. Yes, I can see the utility meter readings. I want to cross check the accuracy of the meter readings and hence this exercise.

Instead of secondary side metering, I might go for a primary side metering (a bit pricier) to avoid any confusion.

Muthu

http://www.edison.co.in

 

[waross]

Hi Muthu. My thought was that if a coupe of spot checks showed readings within a few percent of each other, you may avoid the expense, including the expense of installation, of check metering.
When I see a question such as this, I often find that it is in response to suspicion as to the utility metering ccuracy.
I have found that it can go either way.
I have found that the customer was actually using or demanding the metered quantity.
I have found instances where the utility meter was faulty and was fortunate enough to negotiate a prtial rebate on past billing as well as a replacement meter.
With respect for prc, he has provided the information to calculate the difference.
I am suggesting an economical way to do a quick check for gross errors and to decide whether or not you need to spend the time and money to proceed with your own meters.
Example: For a 10% difference between primary and secondary metering at 1/3 the transformer capacity, many utilities will themselves investigate the issue and make it right.
For a difference of two or three percent, the primary metering is most likely accurate. Don't forget the line loss between the transformer and the panel.
Your call.
Let us know how you proceed.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

A note on demand metering.
It takes 15 minutes to stabilize to a block change in demand.
That is five time constants.
One time constant is 15/5 = 3 minutes.
In one time constant the meter will rise to 63% of the terminal value.
If you are using load shedding to avoid demand penalties, your relay time constant or delay should be less than three minutes.
I would allow enough delay to allow large motors to start.
Subject, of course, to plant conditions, I would start with a response time for excess demand of between 15 and 60 seconds.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[edison123]

Thanks Bill. Our utility is government owned with laws skewed in their favor and so ....

I have had a doubt about their metering for the past six months since our MD has shot up by 10 KVA consecutively without any change in the load pattern and hence I had to payout hefty penalties, which would justify the cost (and peace of my mind) of putting up cross-check metering.

I will keep your tips in mind and update at the end of this month.

Muthu

http://www.edison.co.in

 

[waross]

Thank you for the explanation Muthu.
Has the voltage been drifting up?
Have your KWHrs been drifting up?
For 10% jumps a quick check with hand held meters should verify metering accuracy.
If the quick check suggests that the metering is accurate then look for un-noticed changes in your plant.
If the problem is in your plant there is no need for the expense of the check meters.
Has anyone added air conditioning to his office?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[edison123]

Update. Today, the utility did the monthly reset and we noted the utility meter readings (every 10 minutes initially and then every hour) as below with only transformer primary energized and secondary disconnected.

Time/KWHR/KVA MD/Present PF/Cumulative PF/Remarks
09.30 am /942.99 /63.0 /0.900 /0.910 /Before reset
10.20 am /943.00 /1.0 /0.488 /0.600 /After reset first cycle (reset at 10.00 am)
10.30 am /943.00 /1.2 /0.503 /0.533 /After reset second cycle
10.40 am /943.00 /1.8 /0.497 /0.500 /After reset third cycle
10.50 am /943.00 /1.8 /0.494 /0.500 /After reset fourth cycle
11.30 am /943.01 /1.8 /0.491 /0.500 /After 1 hour 30 minutes
12.30 pm /943.02 /2.0 /0.515 /0.494 /After 2 hour 30 minutes

Why would the transformer open circuit KVA demand go from 1.0 to 2.0 KVA?

Why the present PF is different from the cumulative PF?


Muthu

http://www.edison.co.in

 

[waross]

20 minutes; demand = 1.0 KVA,- Demand is stabilizing
30 minutes; demand = 1.2 KVA,- Demand is stabilizing.
40 minutes; demand = 1.8 KVA,- Demand has stabilized. Possibly a 30 minute time to stabilize.
50 minutes; demand = 1.8 KVA,- Demand has stabilized.
150 minutes; demand = 2.0 KVA,- After 2 hours the transformer has cooled down and the resistance has dropped, increasing the I²R and the I^X.
Or the primary voltage has changed.
With over demand charges on a sanctioned 100 KVA demand and a demand the last month of only 63 KVA, I think that your time may be better spent looking at your plant loads. Something has probably changed in the operating sequence.

Without knowing how the cumulative PF is calculated and the information on which the calculations are based, I hesitate to speculate.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[prc]

Under no-load conditions I had estimated kVA as 0.8+j5 assuming 2% as excitation current. But in modern transformers this can be as low as 0.7 % .Then no-load KVA can be 0.8+j1.75 =1.9 kVA, same as noted by Edison. I don't know why demand stabilization takes such a long time. But temperature cannot vary this kVA as I is negligibly small and hence contribution in o.8 kW is not significant. Power factor 0.8/1.9 =0.42,against 0.5 seen by Edison. Cumulative pf is the pF of load.

 

[waross]

Try doing a short circuit test to verify the listed %imp of a small transformer when the transformer is cold.
There is enough difference between the cold and hot % impedance to render the results unacceptable.

Cumulative PF.
I would guess that is the monthly average power factor as determined by dividing the KWHrs by the root of (KWHrs² + KVARHrs²)
With only magnetizing current flowing, I suspect that any CTs may be operating in a non-linear portion of their curve.
Working with 0.02 KWHrs I would not give much credibility to PF numbers and I wouldn't waste time worrying about the indicated PF.


Bill
--------------------
"Why not the best?"
Jimmy Carter