Load

[MineGuy]

Hi,
Could you please advise me how to convert 3.7 MW of cooling
load in to Electrical load. Please advise me, if you have any questions.
Cheers !

 

[davidbeach]

What work have you done on this?

 

[itsmoked]

Hard to say.. Why don't you more fully describe what you are attempting. If it is cooling 150F air into 120F air it may only take a fan..

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[RonShap]

Watts is watts is watts.
You already have it converted.

 

[MineGuy]

Well Thanks for the comments. I eventually spoke to mechanical engg and he advised me that he has provided us with the electrical load. A 3.7 MW load for a refrigeration unit for cooling is huge but any ways. Now the situation is;

A 220 KV line feeding two 15 MVA transformer with % Z of 8.5 and stepping it down to 6.6 K V.
These two transformers are feeding a 4000 A , 6.6 KV switchgear with a tie- breaker. A 2000A breaker is feeding a 2 fans with 1 MW each, 2 refrigeration units with 3.7 MW each. Same switchgaer section is feeding a 2 compressors each rated 1 MW and one leaching plant 1.37 MW.

The other side of switchgear after tie-breaker is is feeding the same load.
2 fans with 1 MW each, 2 refrigeration units with 3.7 MW each and 2 compressors each rated 1 MW.

I assumed during a simulation all the equipments operating at power factor of .8. I accept this may not sound practical.

Fans and Refrigeration units are supplied through an overhead line that is FALCON: 954 mm2 with 2 cond/Phase.

Same on both sides of switchgear. Compressors are directly connected.

When I ran the simulation, there was a huge voltage drop.

What surprised me at 220 Kv, a voltage drop of 8 % at secondary. My question is this because of .8 PF only?

Thanks

 

[davidbeach]

Nope. The voltage drop is caused by the source impedance, sounds like you have a system that is too weak for that starting condition. Are you trying to start everything simultaneously, or are you starting each load sequentially? Sequential starting will provide for much less voltage drop than simultaneous start.

 

[wareagle]

According to your figures, the load you have on each 15 mva
2 fans with 1 MW each = 2 mw
2 ref with 3.7 MW each = 7.4 mw
2 compressors rated 1 MW = 2 mw
1 leaching plant 1.37 MW = 1.37 mw

Total 12.77 mw
MVA @0.80 = 16 mva on a 15 mva Tr.

8%VD is correct for a transformer with Z = 8.5
You are going to have almost that drop, 7%, at PF = 1.0.
If that a problem maybe you need a regulator.

 

[MineGuy]

Thanks for your comments.
David Beach:
This is a steady state voltage drop. So I think starting issue is not in the picture. What my concern is, how this problem can be sorted out. By capacitor bank compensation ?


Waross:
You are right, Transformer is over loaded. Could you advise, where I can find more info about commerically availably regulators. But I think the last solution might be replacing the transformer.

How would you cope with a situation when u have such a big load and 6.6 KV voltage.

Thanks

 

[jraef]

One suggestion would be to convert one or more of your loads (the most constantly operating) to synchronous motor(s) for use as a synchronous condenser(s) to improve your overall pf. That will at least get you down to 12.77MW on that transformer.

http:/

http://www.jraef.com

Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[wareagle]

"You are right, Transformer is over loaded. Could you advise, where I can find more info about commerically availably regulators. But I think the last solution might be replacing the transformer."

What do you mean by " the last solution might be replacing the transformer."
You may need to find a solution to the overload on the transformer although in this case is not much over the rating.

To find a regulator do a google for power regulators.
This is Seimans Site

https://www.energy-m/irj/portal/ptd...rvice=997%26language=en%26country=US%26cmid=1

 

[MineGuy]

Hi Wareeagle:
Thanks for your comments.
"You are going to have almost that drop, 7%, at PF = 1.0."

I am a Jr EE, so not much experience. So should I presume that we need to have a regulator. There must be a way out to tackle this situation. For example in case where there are huge transformers in service, It by default requires a regulator. Kindly advise me. Am I missing here some thing.?

Tap changers might help but that measns, the transformer will be always on the tap position. So in all the places, where we have a transformer with a higher impedance, we will not be able to satisfy the requirement of 3% voltage drop. As the min. voltage drop will be equal to % Z of transformer. So how it works than. I would appreciate a word.

Thanks

 

[wareagle]

" So should I presume that we need to have a regulator."
Yes if 8% drop is too much.

"For example in case where there are huge transformers in service, It by default requires a regulator. Kindly advise me. Am I missing here some thing.?"
I do not know what you mean. Utility distrubition transformers have voltage regulators. I do not consider them
"HUGE".

"Tap changers might help but that measns, the transformer will be always on the tap position."
It will always be on a tap position. Why is that a problem?

"So in all the places, where we have a transformer with a higher impedance, we will not be able to satisfy the requirement of 3% voltage drop."

Thats true if the transformer is loaded at close to 100% as is the case here.

"As the min. voltage drop will be equal to % Z of transformer."
Yes at full load.

 

[wareagle]

Another problem for you to consider is what happens when
one of the transformers has a problem. Where are you going to put the load? The other transformer does not have spare capacity.

 

[KJvR]

It sounds as if you are on one of the gold/platinum mines in S.A. - most of the older mines used ONAN transformers. If so, you can then possibly uprate the transformer to ONAF and get a few extra MVA. Also, the main incoming transformers normally have on-line tap changers that will easilly handle the voltage drop.

 

[MineGuy]

Thanks WAREEAGLE,
I never noticed this interesting phenomena, that%Z is not a constant quantity rather it is a factor of load. That means, it refers to full load. So, the less the loading , less is the voltage drop. So, It can be infered that if
I have to step 220 KV to 6.6 KV and I need only 3 % voltage drop, then transformer shoud be lightly loaded.
I hope I am correct this time. But this sounds an
uneconomical way, light loading of transformer to get right voltage drop. I could not track your voltage regulator link, I would appreciate, if you could advise me again.
Thanks

 

[Skogsgurra]

Yes, MineGuy. Actually, the %Z is a way of specifying the impedance of the transformer.

The %Z is not happening by chance. It is designed into a transformer to reduce short circuit current. It is not a problem building a transformer with, say 2%, impedance. But that would mean a very high short circuit current and would necessitate stronger busbar (higher electrodynamic forces), higher breaking capacity of switchgear and such things. So transformers are built with a %Z to reduce short circuit current, not to regulate voltage. You need a tap changer or induction regulator for that.

Gunnar Englund

http://www.gke.org

 

[MineGuy]

So, Will it be right to say that if a transformer is on a tap position, the %z will change and so will the Short-circuit current. That means, if I buy a transformer with
%10 impedance and then I operate it at %5 Tap in primary.
My impedance into the system will change and so will my short-circuit current. Hence, while doing a Short Circuit calculation,I should consider the %z with tapped position.
Please advise me, if I making some sense.
Thanks

 

[davidbeach]

Minor change in impedance with change in tap. Change in tap is a change in the voltage ratio of the transformer. So, instead of a secondary voltage of 6.6kV, you might have a secondary voltage 6.93kV (a 5% increase). With that 6.93kV secondary, if you have a 5% voltage drop, you will be back to very near 6.6kV.

 

[Skogsgurra]

Yes. Do not confuse the tap changer % setting with the impedance %. They have nothing in common (except that the % sign is used for both).

It has been said before, but I do not think that you got it: The %Z is the change in output voltage when going from no load to full load. It is the same as the (internal impedance times rated current)/(rated voltage).

The tap changer % is how much the secondary voltage changes when you move the tap from 0 to that position (often +/-2.5% and +/-5%).



Gunnar Englund

http://www.gke.org

 

[MineGuy]

Hi,
I guess ur right. May be I am missing some thing bigtime.
I did a basic simulation in ETAP.
I used 12.77 mw, MVA @0.80 = 16 mva on a 15 mva TR. %z=8.5
vol on sec bus= 88%

2. Same Load- TR rating 22 MVA same%Z voltage in sec. bus
Vol on sec bus-92.26%


3.Same Load- TR rating 30 MVA same%Z voltage in sec. bus
Vol on sec bus-94.26%

4. Same Load- TR rating 50 MVA same%Z voltage in sec. bus
Vol on sec bus-96.71%

Now if %z is a factor of transformer loading, then the load is 15 MVA and TR. is 50 MVA, In this case we should have minimum drop.

When we change the tap position of a transformer. the no, of turns change and No,of turns represents impedance. That measn the current and that measns the voltage drop.

So,when can a transformer with a %Z of 8 or 5 be used without a tap position. I mean, Can we ever use transformers normally.

I might be driving some of you guys crazy here. I guess I need to find a power system analysis book and do some math.
But If what I wrote is correct then, all the transformers in the world should be named as Tapped transformers instead of transformers.

Thanks