Complex Imaginary practice exam questions

I hope someone can help me with the followings.
 
1) On volume 1 question 5, I have I=1500/(2.4+j3.4)=360.43 at angle -54.78 or 207.85-j294.46. But the answer shows 358A. Please advise. 
 
2) On volume 1 question 13, what is SCA? Could you explain to me how it works? My understanding is that the transformer secondary side can only supply the max current 130A. 
 
3) On volume 1 question 30, when we calculate the real power, how come we don't include the power factor 0.78 lagging? 
 
4) On volume 1 question 69, I think the answer should be c. 200A. According to NEC 2014 240.4 (B), the next higher standard overcurrent device rating (NEC 240.6) shall be permitted. 180A is not even the standard size per NEC 240.6.
 
Thanks!
 

The questions are from 2nd edition.

My material is somewhere in my attic but if I can find it I'll take a look tonight.

Thanks!

Limamike, thank your for your reply. Please see below for my response.






Look below Supra

1) On volume 1 question 5, I have I=1500/(2.4+j3.4)=360.43 at angle -54.78 or 207.85-j294.46. But the answer shows 358A. Please advise. 
 
When the switch is in position #1. 1500/437 = 3.43i  , then the switch goes to position #2.  You now have 1500/(2.4+3.43i) = 358a
 
Sorry, I should be more clear. When I entered "1500/(2.4+j3.4)" in my calculator, I got 360.43 at angle -54.78. I guess 360.43 is close to 358.
 
2) On volume 1 question 13, what is SCA? Could you explain to me how it works? My understanding is that the transformer secondary side can only supply the max current 130A. 
You can do this via the MVA method or PU . If you use PU,  SC is (Line voltage of where the fault is at /sqrt 3) divided by (impedance (z) contribution of the components in the line)  So in this case 3.2%+8.4%
 
I think I got it now. SCA = short circuit current. The solution uses the MVA method.
 
If I were to use the PU method, I should have the followings.
 
Set S-base = 9MVA, V-base = 40kV
 
I-base = S-base/(V-base * sqrt 3) = 130 A
 
Then, I need to calculate the I-pu.
 
V-pu = V/V-base = 40kV/40kV = 1
 
I-pu = V-pu/Z-pu = 1 / (0.021 + 0.084) = 8.62 pu
 
I  = (I-pu)*(I-base) = (8.62 pu)(130 A) = 1119.77 A = 1120A
 
3) On volume 1 question 30, when we calculate the real power, how come we don't include the power factor 0.78 lagging?  Max power is at 90 degrees and thats why u use 90 degrees instead of angle at Cos^-1(.78)
 
Ok.
 
4) On volume 1 question 69, I think the answer should be c. 200A. According to NEC 2014 240.4 (B), the next higher standard overcurrent device rating (NEC 240.6) shall be permitted. 180A is not even the standard size per NEC 240.6.
 
If you have a wire whose rating is 190, and you dont trip the breaker till 200...what happens to the wire?
 
From safety stand point, I agree to use to 180A circuit breaker to protect 195A conductor. Actually, any circuit breaker that is less than 195A rating is acceptable. Overheating conductor might start a fire.
 
I am still not sure why NEC 2014 section 240.4 (B) says the next higher standard overcurrent device rating (NEC 240.6) shall be permitted.
 
When we are taking the PE exam, should we follow the NEC 2014 code? Or should be select the circuit breaker or fuse one size down?
 
 
Feel free to PM me if you have additional questions and I will do my best to answer.
 
Thanks!

Its funny you ask that questions "should we use a circuit break one size down"  I wrote in my book in LARGE CAPS - SELECT ONE SIZE DOWN....