spin up exam -3 Q 17 & 18

[PE blues]

For calculating neutral currents, In=-(Ia+Ib+Ic). Can anyone explain why B is leading A by 120 and C is leading A by 240? I would assumed B lagging A by 120 and C lagging A by 240 for ABC sequence (assumed as phase sequence not given)

 

[spinup]

For calculating neutral currents, In=-(Ia+Ib+Ic). Can anyone explain why B is leading A by 120 and C is leading A by 240? I would assumed B lagging A by 120 and C lagging A by 240 for ABC sequence (assumed as phase sequence not given)

PE blues,

Looks like you are using the 1st edition, we are currently in the 2nd edition. There is an errata on the website www.spinupexams.com under the support tab.

Solution 3-17(Page 125) - Reported 3/5/12

Replace with IN =-( IA + IB + Ic)

=-( 54.8<(0 -36.9° )+ 0.25<(120°-5°) + 96<(240°-36.8°))

=-83.4<-122.3°

Change answer from "(A)" to "(E)"

Solution 3-18(Page 125) - Reported 3/5/12

Replace with IN =-( IB + Ic,)

=-( 0.25<(120°-5°) + 96<(240°-36.8°) )

=- 96<-15.9°

Change answer from "(D)" to "(E)"

For the second edition: there is no change. 3-17 answer is (A), 3-18 answer is (D).

Feedback can be provided to: [email protected]

Good Luck on your studies!

Joan

 

[PE blues]

my question is why 120 deg is used and not -120 deg?

 

[PSUEngineer]

my question is why 120 deg is used and not -120 deg?

Think of it more like, there's a 120 degree separation between the phases;

Va < 0

Vb < 120

Vc < 240

assuming positive sequencing.

When I run the math I get 83.4<57.7...but according to the answers provided that's wrong; so if you divide their answer by that answer I get 1<0 which leads me to say they're both right just a sign difference..

My question is how do you get their answer based off of 83.4<57.7??

 

[GreenNGold]

Can someone explain to me how this problem is different from PPI 1-67?

Here is the PPI problem: A distribution system is designed as a 3ph, 4 wire, neutral-grounded system. The phase-to-phase voltage is 12.8kV. The complex load power is expected to be 150+j75 kVA between phase A and neutral and between phase B and neutral. The expected neutral current due to the phase A and B, is most nearly:

A) 0A

B) 13A

C) 23A

D) 33A

Solution: -|In| = |Ia + Ib| = ((150+j75kVA)/(7.39kV<0)) + ((150+j75kVA)/(7.39kV<120)) = 22.70A<-33.....Answer is C

Why is there an additional phase shift in the spin ups problem? Is it from the voltage angle? And if so, shouldn't that have changed the answer for problems 3-11 and 3-16?

Thank you.

 

[Wheretostart]

my thought is, the solution is wrong, but the result is right.

for

Va < 0

Vb < 120

Vc < 240

it is negative sequencing.....