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#57319 - 10/10/05 01:48 PM another class /instructor stumped
jfwayer Offline
Member

Registered: 05/15/05
Posts: 30
Loc: Fairmont, WV, USA
In the Journeyman's class I'm taking we had the following test question:

A plant has an existing load of 4500KVA with a 0.6 lagging PF. Use an 800HP synchronous motor with 80% efficiency to correct the power factor to 0.9. Calculate the PF for the motor.

so given:
total load 4500 KVA
current 0.6 PF
motor 800 HP
efficiency 80%

Calculations:

Plant KW = KVA * PF = 4500 * .6 = 2700KW
Plant KVAR = KVA - KW = 4500 - 2700 = 1800KVAR
Ugly's .6 to .9 PF = .848
needed KVAR = KW * .848 = 2700 * .848 = 2289KVAR

HP to KW conversion

HP = (V*A*Eff*PF*1.73) / 746
HP*746 / (Eff*PF*1.73) = VA
800*746 / (.8*1.0*1.73) = 432.21KW (1.0 PF)
PF=KW/KVA
PF=432 / 2289.6 = .19PF

But Ideal Electric's web site suggests a Syncro can produce 1.5*KW KVARs which means we are expecting 2289 - (432 * 1.5) = 1642 more KVARs than we can get from the Synchro.


Does this make any sense? Do I have errors in my calc or application of formulas.

If we provide the same amount of KVARs leading as KVARs lagging then don't we correct the PF to unity? Here we have 1800KVARs lagging and supply 2289.6KVARs leading to correct to .9 PF?

The instructor only has a numeric answer but no description of the calculation.

Thanks
_________________________
JFW

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#57320 - 10/10/05 08:01 PM Re: another class /instructor stumped
WFO Offline
Member

Registered: 09/03/05
Posts: 206
Loc: Cat Spring, TX
Plant KVAR = KVA - KW = 4500 - 2700 = 1800KVAR


First of all, you don't subtract kw from kva to get kvar's. It is a trig function, not an algebraic one. Use the pythagorian theorem to get 3600 kvars, or since .6 is the cosine of the phase angle, find the sine for that same angle and multiply it times the kva.

"If we provide the same amount of KVARs leading as KVARs lagging then don't we correct the PF to unity?"

True. Quite frankly, I got lost in your math. I suspect you have included more formulas than you need. For one, as long as you are dealing in kw, kva, kvar, and Hp (as equaling 746 watts) I don't think you need to incorporate the square root of three (1.73) at all.

Beyond that, I am totally lost.

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#57321 - 10/10/05 08:20 PM Re: another class /instructor stumped
danickstr Offline
Junior Member

Registered: 10/01/05
Posts: 9
Loc: seattle wa
i think it must be three phase power since all 3PH has 1.73 as efficiency correction.
_________________________
p=ie

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#57322 - 10/10/05 08:29 PM Re: another class /instructor stumped
Bob Offline
Member

Registered: 02/05/02
Posts: 182
Loc: Mobile, AL, USA
If the plant has a 0.60pf and the load is 4500 kva then the kvar = 3600.
As WFO says you can't substract KVA - KW.
KW = pf x kva = .6 x kva = 2700 kw.
KVAR = sqrt(KVA² - KW²) = 3600 kvar. The KW is the same regardless of the PF. If you improve the PF to 0.90 then the
KVA = KW/pf = 2700 kw/0.90 = 3000 kva.
KVAR = sqrt(KVA² - KW²)= sqrt(3000²-2700²)
KVAR = 1306 kvar.
The sync motor must be adding 3600 kvar -
1306 kvar = 2294 kvar.
(800 hp x 746 w/hp)/0.8 = 746 kw
KVA = sqrt(KW² + KVAR²) sqrt(746² + 2294²)
= 2412 KVA
PF KW/KVA = 746/2412 = 0.31 leading
This sounds strange.
jfwayer
What you say?




[This message has been edited by Bob (edited 10-11-2005).]

[This message has been edited by Bob (edited 10-11-2005).]

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#57323 - 10/14/05 11:38 AM Re: another class /instructor stumped
Bob Offline
Member

Registered: 02/05/02
Posts: 182
Loc: Mobile, AL, USA
jfwayer
How about giving us the answer to this problem.

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#57324 - 10/17/05 12:02 PM Re: another class /instructor stumped
jfwayer Offline
Member

Registered: 05/15/05
Posts: 30
Loc: Fairmont, WV, USA
The answer book gives 3600KVARs.
I think it sounds strange as well, since the max KVARs from a synchronous motor is 1.5 times KW according to Ideal Electric. 746KW*1.5 is no where near 3600KVARs

Thanks for your help
_________________________
JFW

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#57325 - 10/18/05 07:03 PM Re: another class /instructor stumped
RobbieD Offline
Member

Registered: 02/23/03
Posts: 238
Loc: Canada
I think bob got the answer. But my question is- By putting a motor in to correct the Pf instead of capacitors you are introducing another load to the plant load. So wouldn't the plant load now increase 746 kW? So the new total plant load would have to be increased. Then all the values change. Am I right?

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#57326 - 10/19/05 01:39 PM Re: another class /instructor stumped
Bob Offline
Member

Registered: 02/05/02
Posts: 182
Loc: Mobile, AL, USA
J
I don't think 3600 kvar is the correct answer. I may be reading the question wrong.
Take the origional information
4500 kva at 0.60 PF. This equates to
2700 kw and 3600 kvar. If you improve the PF to 0.90 then the
KVA = KW/pf = 2700 kw/0.90 = 3000 kva.
KVAR = sqrt(KVA² - KW²)= sqrt(3000²-2700²)
KVAR = 1306 kvar. So the new kvar must be
reduced to 1306 kvar. That is a reduction of 2294 kvar.
The sync motor must be adding 3600 kvar -
1306 kvar = 2294 kvar.

Robbie
If you do not load the sync motor then it will not add the 746 kw. However if you
do load the motor then you would have to add that load to the new reduced load of 3000 kva and as you say thing would change.


jfwayer
What was the answer for the sync motor PF?


[This message has been edited by Bob (edited 10-19-2005).]

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