Improving Power factor to compensate for resistance?

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aprice44

Member
Location
New Jersey
I have a question related to power factor. I tried it and got lost on the way. If someone could help I would really appreciate it.

A feeder from a distribution transformer is 500 feet long. It consists of 3#4/0 AWG wires having a resistance of 0.052 ohms per 1000 feet and current capacity of 225 Amps. The 3 phase load is balanced and draws 120 KW at 440 Volts with 60% PF lagging. Determine the KVA Rating of capacitors capable of correcting the power factor enough to allow rated current to flow.

Thanks

Andy
 

Besoeker

Senior Member
Location
UK
I have a question related to power factor. I tried it and got lost on the way. If someone could help I would really appreciate it.

A feeder from a distribution transformer is 500 feet long. It consists of 3#4/0 AWG wires having a resistance of 0.052 ohms per 1000 feet and current capacity of 225 Amps. The 3 phase load is balanced and draws 120 KW at 440 Volts with 60% PF lagging. Determine the KVA Rating of capacitors capable of correcting the power factor enough to allow rated current to flow.

Thanks

Andy
Exam question?
The resistance is probably thrown in as a red herring.
Think about what current a 120kW 0.6pf load will draw at 440V.
You might then see what's being asked.
 

Smart $

Esteemed Member
Location
Ohio
Are you certain you quoted the problem correctly? That is, are you certain its 120kW and not 120kVA?

I now see Bes' is thinking along the same line... ;)
 

Smart $

Esteemed Member
Location
Ohio
And what problem is that in terms of the question posed by the OP?
Perhaps meaning differs across the pond???

?prob?lem

Pronunciation: 'pr?-bl?m, -b?m, -?blem Function: noun
Etymology: Middle English probleme, from Latin problema, from Greek probl?ma, literally, obstacle, from proballein to throw forward, from pro- forward + ballein to throw ? more at PRO-, DEVIL
Date: 14th century


1 a : a question raised for inquiry, consideration, or solution b : a proposition in mathematics or physics stating something to be done

2 a : an intricate unsettled question b : a source of perplexity, distress, or vexation c : difficulty in understanding or accepting <I have a problem with your saying that>

synonyms see MYSTERY
 

Smart $

Esteemed Member
Location
Ohio
I rechecked and its certainly 120kw.
My apologies... I asked the question before I realized it was 3?.

PS: FWIW...

120,000W ? 440V ? sqrt(3) ? 0.6pf = ???A

PPS: Welcome to the forum... :thumbsup:

It is a generally accept policy here to only assist with test or coursework problem solving rather than provide the answer.
 
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Besoeker

Senior Member
Location
UK
Perhaps meaning differs across the pond???
Don't think so.
You may recall that my missus is from GA.
Some words do have different meanings, biscuit for example, and quite a few words have different spellings.

Anyway, back to the problem as stated.
Determine the KVA Rating of capacitors capable of correcting the power factor enough to allow rated current to flow.
If the 120kW was resistive it wouldn't take conductor rated current. And no correction would be required.
I don't see how you can construe that as a problem to be solved.
 

aprice44

Member
Location
New Jersey
My apologies... I asked the question before I realized it was 3?.

PS: FWIW...

120,000W ? 440V ? sqrt(3) ? 0.6pf = ???A

PPS: Welcome to the forum... :thumbsup:

It is a generally accept policy here to only assist with test or coursework problem solving rather than provide the answer.

This is the way I had done it and later lost my way:
I ignored the cable resistance thing.

PF =KW ? KVA

KVA = 120 = 200 KVA 0.6
KVAR = (sqrt)KVA?-KW? = (sqrt)200?-120? = 160 KVAR

KVA = 225 x 440 x (sqrt)3 = 171.473 KVA

KVAR = (sqrt)171.4?-120? = 122.5 KVAR

KVAR(capacitor) = 160-122.5 = 37.5kvar for capacitor.


120,000W ? 440V ? sqrt(3) ? 0.6pf = is going to give us 262A which is well above the cable rating of 225 Amps.
The question is probably asking capacitors to reduce current flow to 225 Amps.
 
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Besoeker

Senior Member
Location
UK
This is the way I had done it and later lost my way:
I ignored the cable resistance thing.

PF =KW ? KVA

KVA = 120 = 200 KVA 0.6
KVAR = (sqrt)KVA?-KW? = ?200?-120? = 160 KVAR

KVA = 225 x 440 x ?3 = 171.473 KVA

KVAR = (sqrt)171.4?-120? = 122.5 KVAR

KVAR(capacitor) = 160-122.5 = 37.5kvar for capacitor.


120,000W ? 440V ? sqrt(3) ? 0.6pf = is going to give us 262A which is well above the cable rating of 225 Amps.
The question is probably asking capacitors to reduce current flow to 225 Amps.

Which you have now found.
:thumbsup:
 

Smart $

Esteemed Member
Location
Ohio
...

If the 120kW was resistive it wouldn't take conductor rated current. And no correction would be required.
I don't see how you can construe that as a problem to be solved.
For one, part of the discussion was before I realized it was 3?. Major blunder on my part... posted amid my late night snack after having awakened for bathroom run. I should have went back to bed. :angel:

All the same, just asking posed a problem, i.e. a question. Whether it has a preconceived solution is another problem in itself.
 
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Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
I have a question related to power factor. I tried it and got lost on the way. If someone could help I would really appreciate it.

A feeder from a distribution transformer is 500 feet long. It consists of 3#4/0 AWG wires having a resistance of 0.052 ohms per 1000 feet and current capacity of 225 Amps. The 3 phase load is balanced and draws 120 KW at 440 Volts with 60% PF lagging. Determine the KVA Rating of capacitors capable of correcting the power factor enough to allow rated current to flow.

Thanks

Andy
There is more than one "problem" here for sure. I think it's a trick question, although lately a number of exam questions posted in here and other fora i haunt have been horribly flawed, making me fear for our future...

What is the kVA rating of a capacitor?

"Rated current flow" of what? The conductors? There is no such thing as "rated current flow" per se, there is rated current capacity of a conductor, and there is current flow in a conductor or maybe a transformer, and there is rated current draw of a machine, but "rated current flow" without context is ambiguous. Nothing in the problem changes what the RATED current capacity of the conductors is. The only thing that could change the current FLOW through them would be a change in the load. Did they mean that? Or did they mean the "rated current" draw of the myterious load (althouh no FLC was given, only Running kW)? Did they mean the "rated current" capacity of the mysterious transformer (although no size was given)?

Most likely they meant (but poorly stated) the following:

"What amount of kVAR would need to be added to the given circuit at full current flow of the given load to decrease the current flow in the conductors in order to not exceed their capacity?"

It's the only way I can make sense of the possible intent of the question.
 
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Besoeker

Senior Member
Location
UK
Most likely they meant (but poorly stated) the following:

"What amount of kVAR would need to be added to the given circuit to decrease the current flow in the conductors in order to not exceed their capacity at full current flow of the given load?"
So why complicate it?
The OP got the answer without doing so.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
Meh... Should have read page two before posting... I do that a lot when on the iPad. Looks like that's where everyone ended up in spite of the ambiguities of the question.

Circuit resistance however is not irrelevant. Resistance and distance will create voltage drop at the load end and given that it is obviously an inductive load, that means the current draw at the machine will increase as a result. You need to calculate the voltage AT the machine terminals as a result of that VD to use in your current draw calcs.

Assuming, but again not stated, that the 440V was as measured at the transformer end of those conductors, not at the load end...
 

Besoeker

Senior Member
Location
UK
Meh... Should have read page two before posting... I do that a lot when on the iPad. Looks like that's where everyone ended up in spite of the ambiguities of the question.
Not a lot ambiguous about as far as I can tell.

Circuit resistance however is not irrelevant. Resistance and distance will create voltage drop at the load end and given that it is obviously an inductive load, that means the current draw at the machine will increase as a result. You need to calculate the voltage AT the machine terminals as a result of that VD to use in your current draw calcs.
"The 3 phase load is balanced and draws 120 KW at 440 Volts"
That reads to me like the load draws......at 440V. If the load was not at 440V wouldn't it draw something different to 440V?
And, as you know, the voltage drop also depends on the X as well as the R component and that would depend on cable type and method of installation.
I think we can reasonably leave that speculation out of the question as posed, don't you?

Assuming, but again not stated, that the 440V was as measured at the transformer end of those conductors, not at the load end...
An assumption not justified - see above. So don't make it.

So, we are back to the straightforward PFC problem as stated in post #1.
 
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