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#57319  10/10/05 04:48 PM
another class /instructor stumped


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



#57320  10/10/05 11:01 PM
Re: another class /instructor stumped


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.



#57321  10/10/05 11:20 PM
Re: another class /instructor stumped

Junior Member

Joined: Oct 2005
Posts: 9
seattle wa


i think it must be three phase power since all 3PH has 1.73 as efficiency correction.
p=ie



#57322  10/10/05 11:29 PM
Re: another class /instructor stumped


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(3000²2700²) 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 10112005).]
[This message has been edited by Bob (edited 10112005).]



#57323  10/14/05 02:38 PM
Re: another class /instructor stumped


jfwayer How about giving us the answer to this problem.



#57324  10/17/05 03:02 PM
Re: another class /instructor stumped


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



#57325  10/18/05 10:03 PM
Re: another class /instructor stumped

Member

Joined: Feb 2003
Posts: 231
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?



#57326  10/19/05 04:39 PM
Re: another class /instructor stumped


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(3000²2700²) 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 10192005).]




