DiversityFactor

[MineGuy]

Hi Guys,
It is my first time here and I am here because people with experience can answer these two questions.

Q1 I am doing a load flow study with ETAP. The project is for a location in Africa. I have selected the base case as freq=50 Hzs and standard as IEC. Suppose I do it the other way round, Freq=50 and standard as ANSI. Could you kindly advise me, where the difference will be. In load flow, short ciruit or protection coordination.

Q2 The basic single line diagram I have just shows the main buses and transformers and the rest of the stuff is in notes which has descrition of loads connected to transfromers. Now the situation is, for example T/F TR1 is feeding four switchboards and then the breakers in switcboard are connected to motors , pumps, panels etc.
I dont want to model the system to so small detail and it is not possible also as it will mess it up. I want to lump the loads and find out the voltages at transformer. Please advise me what diversity factor can I apply for this lumping idea.
Any word from you guys would be appreciated.

Thanks

 

[MineGuy]

This is my first time response. No answer, I did not know my question is so dumb.Everybody says here no question is stupid, so some body might have a word to drop.
Thanks

 

[itsmoked]

Show a little patience man... You're posting on a US holiday.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[KJvR]

Hi MineGuy, I have not worked with ETAP for almost two years now and will not be able to answer Q1. However, for the loadflow package I am using now, the following differences are:
(a) If ANSI is selected in EDSA, all data must be inserted in ft, hp etc. With IEC the data is required to be in the metric standard.
(b) The method for fault level calc differ between the ANSI/IEEE and IEC. You will get slightly different answers.

Q2. If data is unknown, we use 0.8. A recent survey on a large industrial plant showed the average diversity factor (duty factor*diversity factor) was 0.7 for about 30% of the motors. The rest of the motors diversity factor had to be set to 0.6 to ensure the total load correspond to main MV feeder loads. The diversity factor for actual mining equipment (shovels, draglines, drills etc) might be totally different.

Motors can be lumped together but it will influence the accuracy of the fault contribution of the motors. Some guidelines I am using when lumping motors together pertaining to fault level calcs - for the IEC world:
(a) Lock Rotor Current = Sum of indiv motor LRC (remember the 0.93 reducing factor (to include effect of cables) for LV motor groups - IEC 60909 if the motor subtransient reactance is required in stead of the LRC);
(b) Full Load Current = Sum of indiv motors FLC;
(c) X/R =2.38 (LV motors) and 6.67 (HV motors)- IEC 60909
(d) Number of poles = function of rated power per pole pair (eq. can be derived from IEC equations) of each motor in group - this will influence the 3 cycle symmetrical break current (decay factor q of IEC 60909). Using only the average number of poles for a lump model will over estemate the motor contribution to 3 cycle symm breaking current by up to 30%
(e) Lump same voltage groups together and try to keep the LRC/FLC ratios per group as close as possible (only if you want accurate decay factor "u" for the lump model (refer to IEC 60909).

Note (a) and (b) is used to calculate the subtransient and have nothing to do with load flow studies.

 

[MineGuy]

Thanks KJvR,

Here is what I did. I am citing an example here of what I did in one of the cases. Transformer TR06 at zone 4700 Z3 entrance is feeding a switchboard which has 8 sections and out of those 5 are in place with 400A, 50 A, 160 A, 150 A and 100 A breakers.

1. The 400A breaker is feeding another switchboard with
400A, 100A, 100A and 160 A breakers. Out of these the second 100A breaker is feeding a fan started and 160A is feeding Jumbo box and the third 100 A breaker is feeding a diamond drill starter.

2. The 50 A breaker in main switchboard is feeding a starter.

3. The third 160 A is idle.No Load connected

4. The 150 A breaker is feed two fan starters.

5. The 100 A breaker is idle. No load connected
I dont know the ratings of these starters.

Now what I have done is to get a lumped load. I added the main five breakers that sums to 860 A.

So at 550 V the KVA is 818.2 KVA. The transformer feeding this load is 750 KVA. So now, I intend to multiply this 818.2 kVA by .8 and that will give me 654.56 KVA. I zeroed it to 700 KVA. Please advise me, if this is correct. As this will be the approach I will be following for rest of the system.


Best Regards,

 

[KJvR]

Hi MineGuy, You must NOT add feeder sizes, espesially in an mining environment. They tend to have standard size gully boxes and most of the time it is not in used or over rated. You should use the actual motor sizes connected to the feeders.

From the descriptions I assume you are working on a steep inclined mine. Some additional comments:
(a) For the exploration drill (also development and production drills), you should find more info e.g. what size of motors is used (normally 1 or 2 large hydraulic pump motors and a small water pump) - talked to one of the electricians about normal load current etc. They have good knowledge what the load will be under the different operating modes.
(b) A large number of motor starters will be dewatering pumps and will almost always have a standby pump. You should ignore standby loads.
(c) Minning equipment tend to move around. It depend on what you aim to accomplish with the model. If it is to determine the worst load conditions in a certain area of the mine, you have to determine what is the maximum number of machines they will put in production in the particular area - asked for a few shift reports or talked to the shift boss.
(d) Spend some time to get to know the operations - this will help you with modeling the load much better.
(e) It is always good practise to compare your results with some measurements on the HV distribution and do some fine tuning of the model.

What type of mine is this - Mining method and product?

 

[waross]

If you have not been trained on the software you may do better to forget it for a year or so. Add your motor currents.
Multiply the total by a safety factor. The minimum factor is +25% of the largest motor. You may wish to use a larger factor to allow for future expansion.
Alternately, for transformer sizing, sum your motor horsepowers. Multiply by a factor to get KVA for transformer sizing.
I did a layout once where the utility used a factor of .5 KVA per kilowatt. The utility had experience with the diversity factor for this type of industrial plant.
If the customer owned the transformer instead of the utility the Canadian Electrical Code required transformer sizing based on rated motor currents plus 25% of the largest motor.
The resulting conversion factor would be closer to 1 KVA per kilowatt. (1 hp. = 746 watts + power factor + efficiency)
respectfully

 

[MineGuy]

Thank KJ and Waross. This situation is arising because
My supervisor is not in and I was asked to do this.
We are expanding the distribution underground and at present I am simulating an existing system and after we figure out the voltage drops, we will plan the next stage.
The example I gave you earlier is existing and in service but the transformer connected to it is 750 KVA but if you add the feeders it comes out 818.2 KVA. From here I infer that some diversity factors have been added. Please advise me to make a wise decison as this situation is every where in the system in all levels of mine.

Kindly advise me a source or literature link so that I could get a feel for mining equipments used in the indurstry and how they work, so that I could figure out some thing on my own also.

Thanks again for sparing some moments from your precious time.
Regards,

 

[MineGuy]

One more, I forgot to mention. I have worked in medium voltage distribution(utility) stuff
and I have realized that dievrsity factor can be 60 % and that is how transformers are sized.

What is the actually thumb rule while sizing a transformer?
Should be fully or 75 % or what percentage loaded all the time. Please put few words.
Thanks

 

[KJvR]

MineGuy, I can not comment on the transformer sizing. What I did learn about the underground mines is that every thing is highly mobile and they have standard sizes of transformers. Nobody is actually calculating how much load they will be able to connect onto the transformer. It is more about what transformer is the closest and is a spare breaker available - else they simply install a new transformer closer to the load.

I include a link to a site for face drills (development and production). The sizes varies considerable - the mine I have modelled a few years ago used Boomers for development with a single 37kW hydraulic pump motor and a 1.1 kW waterpump. The production drills were Simba (also Atlas Copco) with two large motors (55 kW but I am not sure). The exploration / diamond drills were Demag L90 drills - with a 55 - 90 kW motor.

This is only examples and would vary depending on the size of the tunnels and minning technique. Best will be to contact the mine you are doing the study for and ask some details on it - also the number of each drill rig and how many are normally used in a specific area.

The study I have done was also to determine voltage drop on the underground distribution network. The network was normally fine untill some production issue. They will send about twice as many machines into that specific area. You need some load criterea and the mine is the only guys who will be able to help you. A mechanised mine is unfortunately not a utility network with constant load.


Note that I have assumed you are looking at a fully underground mechanised mine because you have mentioned a diamond drill.

http://www.boomer-rig.com/

Hope this will help

 

[tulum]

MineGuy,

It used to be that in an underground mining environment the transformers were only loaded to 50-60%. Now days most areas are buying 80degC transformers and running them at the extended kVA.

Good rule to remember is that at 150degC transformer can usually push it's kVA in HP @ 600vac (Canada). A misconception is that a 80degc rise transformer can push it's EXTENDED kVA rating in HP @ 600vac....for instance a 80degC 500/650 can not start 650HP...

P.S. We have 150, 200, 250, 350, 500, 650, 750, 800, 1000kvA transformers where I am.... not a real standard...


Regards,
TULUM