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Percent and per unit are same. 5% means 5/100=0.05 per unit. Now their relation with actual quantity in case of Z (Impedance) is Zpu=Zactual/Zbase where Z base is KV**2/MVA so Zpu=Za*MVA/KV**2 or Za=Zpu*KV**2/MVA. This per unit or percent was not making any difference in the answer here because...

I have explained in red. Hope it will make sense. I have not fully solved it for you to make you understand the concept and use of formula. If it does not make sense, we can discuss further. Think about using the basic formula of equalizing I1Z1 and I2Z2 where Z are actual Zs not the per unit...

Both methods give same results of 2500KVA. Can you share your calculations to locate the error. 

If I have understood you correctly and if the question is pointing to my posts regarding the question in question, you are absolutely right. In this particular question the answer should be 6MVA (3*2 or 3+3). 

There is a word of caution here. Generally synchronous Gen do not share power like this because in that case other conditions of Frequency and Excitation Voltage etc comes into play. This simple method of power sharing ican be applied to Xmers, Batteries etc. but is not to be applied when other...

It is my pleasure.

To supply power while operation in parallel by any two sources, the terminal Voltage has to be equal because they are connected to same terminals. My advise will be instead of using the formula, understand the problem, go from basics of sharing load by two resistances in parallel.The condition...

Hope is has made sense. Thanks

yup. I do not know , but yes, there is lot of confusion around here. 

Can you post the questions please

I am sorry did not get time last evening.Thanks Omer.  @Omer has got it right.(Not verified the calculations). Now coming to better explanation, let me try- Any synchronous machine ( motor or Genr) can be represented by its equivalent ckt. which has an internal generated Voltage-Eo ( which is...

A small correction. It is not at rated current while working as capacitor; it is working at same reactive power that means at reactive component of 125A. The answer may not be 9.5A.

I will do it myself in the evening, but meanwhile the process can be excitation voltage at rated values can be found. This value of Eo is proportional to 6A. (assuming no saturation). Now find Eo with 90 leading rated current. The excitation current for this Eo can be calculated by unitary...

I do not have both but I understand the issue. Both the equivalent circuits are correct. The second one in which the stator impedance has been taken towards rotor side is more approximation. For larger Machines (Wildi says above 2hp) the approximate model is also okay. For short circuit...

Thanks @joy21 and @ptatohed for the inputs. This has fine tuned the stuff!!!

Thanks a ton @fireguy_PE for the elaborate reply. This certainly helps me in planning my things. I am at Main Place Mall on Main street. By the accounts you have given  I should start at 630 AM from here, that makes 7 o clock at center, searching, normalizing for 30 minutes and then thing starts.