Promotion Test

Status
Not open for further replies.
T

thomasf

Member
To All:

I am new here and hope I posted this in the right place.
Recently I was offered a promotion. One of the requirements was to take a test a competency test.
The test consisted of one question in two parts.

Here it is.

A 25kva transformer supplies a load of 12 kw at a power factor of .6 lagging. find the percentage of full load the transformer is carrying.

If additional unity power factor loads are to be served with the same transformer, how many kw may be added before the transformer is at full load.

Does anyone else think this is an unfair question. I was not able to complete the test. I only got the first part of 80 %, any additional help would be appreciated.
Thanks,
Tom
 
Re: Promotion Test

If the remaining load is in unity, the answer would be 5kw

EDITED after thinking for a minute. :)


Roger

[ October 08, 2003, 09:17 PM: Message edited by: roger ]
 
Re: Promotion Test

IF the remaining load is at Unity PF, the total power Pt would be 19.2 KW.
The additional power added to the original 12kw would be 7.2 kw.
Since the additional loads are at unity, the reactive component will stay the same.
And VA = S = P + jQ and the phase angle will necessarily change with the addition of a unity load.
Do I think it an unfair question , NO.

WmColt
 
Re: Promotion Test

I'll go along with Roger on this one.

The 12kw load at 60% PF would load the transformer to 20kva. That leaves capacity for another 5kva of load, which is equal to 5kw at unity (100%) PF.

Ed
 
Re: Promotion Test

ED:
The question says its a 25kva transformer.
Part 1 wants to know what percent of full load is used.
Part 2. says if additional unity power factor loads are to be served with the same transformer, how many kw may be added before the trans. is at full load.


Since .6 lagging is argcos .6 = 53.1 degrees, the VARS is 20sin53.1 = 16kvar, since the additional load is at unity the reactive power will not change.
With the addition of the new load ie up to the 25kva then the new angle is argsin(16/25)= 39.8 degrees.
Pt = 25Cos(39.8) = 19.2kw.
The power triangle in this case would be:
P = 19.2Kw + j16kvar = 25kva.
another way to prove it is square root sum of the squares..

WOC
 
Re: Promotion Test

William, to simplify your method it would be.

12000 + 60% = 19200

12000 + 7200 = 19200 or 76.8 %

this would mean 5.8 KW

The problem is this is the wrong calculation.

In this case we have to devide the .6 into the number.

12000 / .6 =20000

Roger
 
Re: Promotion Test

For the first question on percent the load on the transformer is 20 VA (12 kw/.6). This is 80% of the xfmr capacity (20/25*100%).

This gives us a load of S=12+16j kVA. Adding loads with no power factor will not change the reactive load of 16 kVA. It will change the pf. Wocolt is right on the money.

So for S=25kVA = P + j16. Solving for P....

P=(25^2 - 16^2)^.5 = 19.2 kW

Therefore answering the question of how many more kW (pf=1) of load can be added is 19.2 - 12 = 7.2 kW. The new p.f. for this scenario (S=19.2+j16)is 19.2/25 = .77 , which is 39.8-degrees.

Hope this helps.
 
Re: Promotion Test

Roger:
I am not sure I follow your method. In a power triangle you have to consider the Reactive component, in this case is 16kvar. and the 12kw this would result in 20kva. at .6 lagging.
If 5.8kw is correct then that would make 17.8kw and 16kvar for the quadrature power, then the total kva would be 23.9kva@ 41.9 degrees.
As I said another way to simplify it would be sq.root sum of the squares. Or

(Sq.Root)(17.8(Squared) +(16(Squared))
or 316.8E6 + 256E6 = 572.84E6 taking the square root of 573.84E6 = 23.9kva and the angle would be
ArgTan (16/17.8) = 41.9 degrees.
At 23.9 kva you have not reached full load on the transformer.

Wm.
 
Re: Promotion Test

WOC is correct, the additional load is 7.2kw at 1.0PF. The new total transformer loading will be 25KVA at .77PF.

This is an example of the importance of knowing the PF of the loads when calculating actual loading. And also how diverse loads may mitigate power problems at "points of common coupling".
 
Re: Promotion Test

This doesn't change the answers (80% and 7.2kW), but is this a capacitive or inductive circuit. ie. Should S = 12 + j16 or S = 12 - j16? It says pf .6 lagging. In an inductive circuit voltage leads current. In a capacitive curcuit voltage lags current. When the question says "pf = .6 lagging" does that mean it is a capacitive load, and wocolt and I should have said S = 12 - j16? Thanks.
 
Re: Promotion Test

I'll confirm that the answers are 80% and 7.2 KW. I think I could give an easier explanation, if anyone wants one. But to answer that last question, the conventions are as follows:
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">(*) Lagging power factor applies to inductors (e.g., motors)
    (*) In the formula, S = P + jQ, the ?Q? is considered positive for lagging power factor.
    (*) Therefore, the final expression of the fully loaded transformer is correctly given as 25 KVA = 19.2 KW + j16 KVAR.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">

[ October 09, 2003, 11:32 AM: Message edited by: charlie b ]
 
Re: Promotion Test

Thanks for the help on lagging pf. Seems funny that a promotion would be hinged on the ability to answer one question.

I'd say it was a fair question if you should know this stuff for your job. If you design telephone systems I'd say it is an unfair question. Strange boss you got there!
 
Re: Promotion Test

Roger, I guess it's back to school for us. :)

For the mathmatically challenged among us, try this solution.

PF.gif


Ed
 
Re: Promotion Test

Ed,
I guess it's back to school for us.
Click to expand...
LOL Maybe we can car pool. :)


Roger

[ October 09, 2003, 02:10 PM: Message edited by: roger ]
 
Re: Promotion Test

My friend interviewed for a company that had a one question hiring test. It was:

"Do you lie?".

The answer led to the real interview or you were immediately shown to the door.
 
Re: Promotion Test

Well done, Ed! I could not have said it better (especially since I do not know how you do your fancy graphics).
 
Status
Not open for further replies.
Top