Power Factor Correction 3

[greenheck1]

Hello

I ran simulations using CYMCAP to figure out the ampacities for each of our cables. I ran all of my simulations using 1.0 power factor assuming the worst case scenario. But now, my boss wants me to find the ampacities at 0.9 load factor instead! is there a table of multipliers of something of the sort that I can use to get the new pfs at 0.9 from 1.0? or do I have to run all of the simulations again (There are about 60 of them)?

Thanks
RRA

 

[waross]

Well some of us consider a unity power factor to be a best case scenario rather than the worst case. But then I am not familiar with your soft ware and what you are trying to determine.
For the present issue, make sure that you are clear on the difference between LOAD FACTOR and POWER FACTOR before wasting any more time.

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

 

[Skogsgurra]

Bill is absolutely right! Lean back and try to understand what you are supposed to do instead of throwing yourself over the keyboard. I would even think about asking my boss. That is what bosses are for.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[desrod]

CYMCAP does not consider PF of the load in its simulations directly. You either get the hottest cable temperature for a given loading in Amp or get the maximum cable ampacity/loading of your cables without exceeding the temperature class of the cables (ex.: 90°C or 105°C)

If your study was of the second type (getting max amps for a given temperature) you only have to check the obtained ampacity against your revised loading at .90 PF.

If you used the first approach, using unity PF, the smaller current value you ran your initial simulations with, naturally gave you lower final temperatures. If you want to save time, try looking at your most stringent case(s), run them at 0.9 PF. This will give you an idea if you still have enough margin left. Then talk to your boss.

Also look at what loading factor your are using. Default value is 100% which might be on the pessimistic side.

You can also call the guys at CYME for help/support if you have a valid maintenance contract for the software.

Desrod

 

[greenheck1]

I am sorry!
I meant 0.9 Load factor, not power factor! I posted the question here toward the end of my shift.
I ran some simulations in CYMCAP and for a 500 CU cable the difference in ampacity going from a 1.0 to 0.9 LF is 3.33% increase ... then I ran the same simulations but for a 2/0 CU cable and the difference now is 2.7%
Obviously, it is not a linear relation.
Any suggestions on what should I do?

Thanks

 

[desrod]

Not a linear relation indeed: losses are a function of the square of the current to start with. LF of 1 means that current going thru your cables stays at the same value. Once your start your installation, cables will slowing heat up and will reach a steady-state temperature. WIth a LF of 1, peak amp will be the same than average amp.

Using a lower LF will factor the impact of cooling down the cables to estimate the peak current. A good example is solar power plants. We performed a number of studies for cable runs there and managed to reduce the number of cables by using the proper load factor (lot of cooling at night!).

You should first explain what you are trying to do with your study: check for load increase? optimize design? emergency configuration?
What you will do next depends on that.
If your boss suggested reducing the load factor, I suspect he is trying to save some cables.

Talk to the customer and find out how long last their peak load or even try to obtain a load profile. Use CYMCAP to run the profile. It will produce the best results and more room for optimization.

Hope this helps

 

[greenheck1]

The reason I am doing this is because we are about to implement a new system called DMS (distribution load management) to make our grid more 'smart gridy', and so I need to find the current carrying capacity (ampacity) for each type of cable given normal temp of 90C and emergency temp of 130C. So i need to find the max current at these temps. I hope this clears it.

I did all of my studies at 1.0 LF, but my boss said we are never at 100% load ... so he suggested a 0.9 LF because it is more reasonable at peak time.

 

[jghrist]

What does he mean "never at 100% load"? The load you are comparing to the cable ampacity is the peak load on the circuit. If you are never at the peak load, then it isn't the peak load. The load factor is the average load divided by the peak load.

 

[waross]

Ampacity is ampacity.
Changing the load factor does not change the ampacity. Using a load factor does change the maximum amount of loads that may be connected to a circuit or feeder on the assumption that not all the various loads will be connected at any given time.
Using a 24 hour base for average current and then using that for cable sizing is a concept that many of us have not considered.
I would have thought that with a steady state current that cables would reach equilibrium temperature in an hour or less.
A commonly used time constant for transformer heating is 3 minutes. Steady state values are often assumed to be reached after 5 time constants or 5 times 3 minutes gives 15 minutes, the time that we assume the temperature will need to stabilize after a step change in current. A more forgiving figure of 30 minutes is occasionally used for transformers. I have to wonder what the soil conditions are that enable you to use 24 hours or 360 minutes.


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

 

[electricpete]

fwiw: IEEE 242 (1991) Buff book has some discusion of transient behavior of cable, although I don't think NEC allows to take advantage of it:

9.5.2.4 Development of intermediate characteristics Cable, with the thermal inertia of its own and of its surroundings, takes from 1 h to 6 h to change from initial to final temperature as the result of a current change. Consequently, overloads substantially greater than its continuous rating may be placed on a cable for this range of times.

Figure 9-16 suggests time constants Tau (T=Tf-(Tf-Ti)*exp(-t/Tau)) ranging from 0.5 hours for "small cable in air" to 6 hours for "large cable, direct buried"

=====================================
(2B)+(2B)' ?

 

[waross]

Thanks Pete. I had no idea that time constants could run that long.
I can see where a constant of 6 hours applied to the output of a solar array may allow a reduction in cable size.
What type of soil conditions allow the longest time constants?
I remember some sizing cables for distribution conductors. There was a cutoff point on the ampacity tables and it was strongly suggested that loadings above the cutoff line should only be used for short term emergency situations due to the high I²R losses. The losses are something to consider when pushing the ampacity limits on long cables.

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

 

[dpc]

FWIW, the NEC defines a continuous load as one lasting three hours or more.



David Castor

http://www.cvoes.com

 

[desrod]

Greenheck1: a quick comment about LF and diversity and your boss' comment about never being at 100% load. Remember that due to diversity, the total of individual cable/feeder peak values will be greater that the peak seen from the supply transformer because they don't peak all at the same time.Sizing cables or estimating cable temperature in a given duct bank based on non diversified peak values might be too conservative thus the comment from your boss.

General comment: Not all usage will benefit from tighter assets management. In all cases, the key factor is temperature rise. 24 hours Industrial loads are a good counter-example where the margin does not exist.

When I was working for a Canadian utility, we used to plan for transmission transformers overload up to 30-40% based on equipment initial temperature, ambient temperature and winter peak duration (3-4 hours).

Some cables can even be loaded to 130°C for as long as 5000 hr over the life of the cable. We did not implement the practice since it was very hard to reliably keep a running total of the hours when the cable was loaded at such a high level over a period of 30-40 years.

In all case, when not meeting Std, you need a proper study by qualified people.

 

[LionelHutz]

I still wonder if you are confused about your use of the load factor. You just posted you want to know the current where the wire reaches 90C and 130C. That answer is simply at a certain current and after a certain time.

Once you know the above, then you can use that to match the conductors to the loads, and this is where the LF could be used.

You certainly don't want to get an answer such as, for example, the wire reaching 90C with 462A at a LF of 0.9. Because this really just means the wire reaches 90C at 416A.

 

[waross]

I wish that I had said that, Lionel. lps

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

 

[greenheck1]

Thank you all for your replies.

Lionel, that is exactly what I want for my project. We are taking our peak load from our peak day of the year (usually in late july or early august) and we compare average loads from different days to that peak load day. Not a great way to do it, but that's what our resources allows us to do.
Dividing average load to that peak load, we got 0.83. So my boss decided to go with the 0.9