help with trades.exambank.com questions

Thread Starter

peeka

Joined Apr 25, 2009
21
okay got another question :p sorry for all these question , but only way to learn.

2 single phase motors are connected paralell ,total circuit apparent is 3750 VA at a laggin pf of .663, if motor 1 draws 1650 VA at a pf of .483 laggin then the pf of motor 2 will b ?

now i dont want the answer :p coz im really close to it i just know it.
first off i did.

total true power = apprent X pf
3750 X .663 = 2486.25
and i got reactive power = 2807.32

for Motor 1
true power = aparent x pf
1650 x .483= 796.23
reactive power i got 1445

now i thort i would need to use reactive powers, and 2807(total) - 1445 (m1)
whih got me 1362. and the same deal with apprent powers which got me
2100
so M2 has apparent - 2100
and reactive of 1362.32
then 1362/2100 = .64 pf

now the answer spose to b .778
could u just tell me where and why i went rong :p not to point out the obvious and im an idiot :p but u know.

btw i appreciat all this time you put in to helping me.
 

t_n_k

Joined Mar 6, 2009
5,455
now i thort i would need to use reactive powers, and 2807(total) - 1445 (m1)
whih got me 1362. and the same deal with apprent powers which got me
2100
This is where you went off course.

BTW I haven't checked you calculated values - just the process.

You need M2 reactive (you did this part correctly) and real powers.

The M2 reactive power would be correct.

The M2 real power would be total Real Power - M1 Real Power.

Knowing both the real and reactive powers for M2 will lead you to M2 apparent power .... remember the power triangle.

You should then get the solution for the M2 pf from there.
 

Thread Starter

peeka

Joined Apr 25, 2009
21
hmmm maybe its just me or im really tired :p but dont think that really points out where i went rong coz i did the same thing u said to do :p but if i dont hear from u soon ill b trying to work it out anyway coz im stuborn like that at least i know im on the right track just gota play around with the numbers a bit.

thanx again :)
 

Thread Starter

peeka

Joined Apr 25, 2009
21
lmao, wen i was doing the power triangle i was using sin instead of cos :p stupid me all stressing lol oh well thats wat happens. thanx again dude :D u really helped me out again, although my answer was .785 wen its spose to b .77 :p thats not a huge deal i meen com on lol oh well at least im on the same page with all my working out and udnerstanding of it.

thanx
 

t_n_k

Joined Mar 6, 2009
5,455
lmao, wen i was doing the power triangle i was using sin instead of cos :p stupid me all stressing lol oh well thats wat happens. thanx again dude :D u really helped me out again, although my answer was .785 wen its spose to b .77 :p thats not a huge deal i meen com on lol oh well at least im on the same page with all my working out and udnerstanding of it.

thanx
Less than 1% error is OK.

Looks like you are getting there.
 

Thread Starter

peeka

Joined Apr 25, 2009
21
hi again :p

well im prety confused on the 3 phase :p i know all the theory just cant put it together if u know wat i meen.

"a 208 volt 60 hz 3 phase supply has 3 wye connected loads to it.
- line a to n has 35 ohm resistor
- line b to n has 55 ohm inductive load at pf .707
- line c to n has capacitor with capacitive reactance of 60 ohms
wats total tru power in watts.


now ive thrown all that in a triangle :p
reactive side i get either 20674 coz 55 inductive u gota change to XL. then minus 60.
then bottom side is 35.
which is pretty much where i get stuk coz the reactive side is just bs
.ive been on it for 2 days :( so really suks.do u know any good books to help for canada ? im already getin forget the name but its got neally everything in it.but cant hurt to hav more books always good to learn stuff.
 

t_n_k

Joined Mar 6, 2009
5,455
Hi peeka,

It might be worth your while starting a new thread for each new problem - members are more likely to open new posts, which increases the likelihood of getting feedback.

On this current problem.

All phase loads are connected line to neutral so you can treat them as three single phase cases in isolation and add then the individual real powers together to give the total load power. Only phases A & B are supplying real power - as you rightly point out phase C is purely reactive power and can be ignored in the power summation.
 

Thread Starter

peeka

Joined Apr 25, 2009
21
yea ithort bout starting new thread :p thanx for advice once again can always count on u.ill try that after work today :)
 
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