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what you are saying is absolutely right. the only issue with this problem is that, the PF should be taken at the load not at the source. the current should be 34.9 lagging the terminal voltage which have to be calculated. if we assume the PF at the source then what you mention is 100% right.

Using trigonometry there is no such triangle with these values specified.  in my pervious solution I assume the reference is Vt@0 current I is [email protected] lagging Vt when I solve (512∠14)-(47∠-acos(.82))*(.6+1.3i)= 447∠11.58 VT in the answer is not at 0 degree, that mean angle between I and VT...

I mean, if your reference is Vt, the problem is not solvable for the given condition. it could be solvable in other conditions. Much higher currents or less Eg.

@rg1 after thinking for a while I think my answer is wrong.. the book answer could be right if they specify in the question the PF is at the Eg. that is to say current lags Eg by 34.9. However, general convention for PF is always at the load if nothing mentioned, that is current lags the...

I got around 449 V for terminal phase voltage. (512∠14)-(47∠-acos(.82))*(.6+1.3i)= 449

Usually in this kind of problem they say (transformer values are being used as base values). in this question they mention high voltage side just for confusion.

Yes, you are right. only the MVA base is change. the base voltage HV side of XFR is 35 KV and the base on the LV XFR side and generator is 15 KV. 

Thanks @rg1. That was right to the point. Interesting.

Bringing back this question. I have some confusion to be cleared. This is a 3 phase problem and to solve for the internal voltage usually we use the single phase circuit. 1 pu voltage is 13.8 kv (since it is the base).So for the single phase, terminal pu voltage is 1/(sqrt3).  Doing the...

the answer provided by the book is not clear at all. as suggested by @rg1 , you need to calculate the excitation voltage based on the rating values first: Eg= Vt-I*X =460/sqrt(3) - [email protected] * 2i= 461@-25 at the rated condition the VAR generated= sqrt(3)*460*125*sin (36.9)= 60 KVAR since...

Talking about the Load, your reference is always the voltage, then you look at the current angle, if it is less than the voltage angle then your load is inductive and we describe the system as lagging. If current angle is bigger than the voltage then your load is capacitive and system is...

@rg1 it looks complicated. for the single phase transformer: I primary= 15000/480=31 I sec= 15000/120=125 for the new configuration: I x max= 125+31=156@0 I a max= 31@0 +125@120= 112@106 Vxy= 120+480@-120= 432@-106 there are high chances I am wrong..

it is very difficult for me to visualize the new reverse connection :blink: . anyway, calculating based on simply reverse the polarity of secondary P, I got this: my reference is E (AB) E12=347@120 E23=600@-90 E31=347@60

I couldn't locate this in my copy. Can you share?

Sorry  @rg1, I think the calculation is wrong. It is 3.13:1 for the first one. It should be straight primary phase voltage/ secondary. i don't know why most of the problems showing on the board nowadays having mistakes.?

As I understand, those are different transformers with different turn ratio for each specific connection but all can handle the requirements. i think turn ratios are ok. @rg1

I think they got this one wrong also. there is a direct relation between P and sin delta.  As P increase, so does the voltage angle and its sin. in this case P increased by 10% so new sin delta2 is also 10% more than sin delta1. therefore,  sin delta2= (1.1/1) sin delta.

Yes, you are right. I think book answer is wrong including r2 in the calculation of Zeq. Wildi is the best.

This post is to discuss about the differential protection on 3 phases transformer. specifically, star delta connection. I think it is a rich topic in the concepts and might well appear on the PE How to connect CTs, 30 degrees phase shift, instantaneous currents, zero sequence currents, triple...

see the below post for the same problem. I think they got it wrong for this one.