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I must be missing part of the problem statement because without efficiency data, I would assume 1kVA = 1hp (see IEEE std 399-1997, the "Brown Book", Table 7-3) for the 1,200hp induction motor. Obviously kW indicates input power (at least in the US) for the synchronous motor. Thus the complex...

Flyer brings up an excellent point! So, in general mathematical terms, how does the load connected to this delta configured transformer differ from the 500kVA load placed between phases B and C on the 3-phase, 4-wire system given in NCEES #110? DK_PE's approach is very elegant, however isn't...

Here's what I got assuming N1/N2=10/1 and ANSI standard wye-delta connection (high voltage phasors lead low voltage phasors by 30 degrees) Primary side: IA=17.3<(-arccos(0.94))=17.3<(-20) VAN=1732/sqrt(3)<(-30)=1000<(-30) Secondary side...

I'm thinking that's a mistake. The problem states that the generator voltage is 45kV (line-to-line). So I used 1 per unit for the generator voltage and got S=(14.2+j2.59)MVA for the complex power delivered by the generator. Does that sound reasonable? I don't agree with the solution to #10...

Understood, however it is true that zero-sequence currents will circulate in the delta windings of a transformer if a grounded-wye set of windings exist in the transformer.

True or False? a. Zero-sequence currents do not exist in the delta windings of a three-phase transformer when the other windings are connected in wye. I believe the correct answer to this is another question or it depends. Are the wye windings grounded, either solidly or through an impedance...

Unfortunately, it's not free. However, I think the power sample from Kaplan was worth the cost for the extra problems. I agree with Benbo regarding some errors, although overall valuable study material.

For Kaplan power sample #3, the total reactive power is given in the solutions as sqrt(3)x[P(WB) - P(WA)]. After researching many different sources, including some old circuits II lab material, and the excellent handout recommended by Flyer, I'm still not sure why the wattmeter B power is...

For #135, the assumed terminal voltage is given as 1.0 pu. It then follows that because this is a transient stability study at rated MVA, rated voltage, and unity power factor, the current through the terminal leads is assumed to be (1.0 + j0) pu. Please advise if this assumption is valid.

So regardless of the system connection, the impedance is always in reference to a phase value, correct? This takes us back to the basics of per unit analysis and logically follows that the base impedance can be calculated as either 1) Z(base) = (Base kV L-N) / (Base kA) 2) Z(base) = (Base kV...

I realize that each of these problems is extremely fundamental (as are all of the sample problems when one recognizes which one or two basic concepts is in play) and I apologize for not recognizing the difference here. NCEES sample problem #132 is identical to EESE (Camera) power #69. NCEES #513...

That's why shorting blocks are so important when dealing with CTs! So for problem #139 one could actually solve the problem using the relationship Is/Ip=Np/Ns and Is=Ip(1/80)=8000/80=100A, correct?

CT's are expressed as a ratio of the rated primary current to the rated secondary current where the primary is actually one (1) turn connected in series between the source and load. Therefore, using the turns ratio may not work as neatly as it does for a VT which is connected in parallel with...

Finally for problem 540, my solution seemed extremely fundamental: T1Zpu,new=0.15(834/933)=0.134 Ssc,pu=VpuxIpu=1x(1/Zpu)=1x(1/(0.23+0.134))=2.75pu Ssc,actual=Ssc,puxSsc,base=2.75x834=2291MVA Do you agree with my methodology? It seems as if all of the sample problems involve fundamental...

Switching to transmission line problems. What exactly is the difference between NCEES sample problems 132 (L-N voltage, 60kV/sqrt(3)) and 513 (L-L voltage, 12kV). Is it the transformer in 513, in contrast to #132 where there is no voltage transformation between stations?

I figured it out; much easier than I thought. In fact this is a very basic problem.

Not sure where to begin. The polarity dots in the solution look arbitrary: VA=2300(21.74+416.7)=1008kVA Please advise if you have a strategy for solving this one and more importantly for rationalizing these types of problems (e.g., autotransformer, transformer polarity schemes).

Here's the figure

Question: A single-phase transformer rated 50kVA, 2,300/120V is connected as an autotransformer as shown in the figure to boost the voltage of a 2,300V bus. What is the kVA rating of the autotransformer? Any suggestions?

IEEE/ANSI C57.12.90