The following information is about the surge impedance loading or SIL (in MW). The Surge Impedance Loading (SIL) of a transmission line is the MW loading of a transmission line at which a natural reactive power balance occurs.
The concept of SIL is explained as follows:
1) Transmission lines...
ScottyUk,
Thanks for your comments. I think that the generator circuit breaker is preferable in power plants with large units. There is no argument that there are cases the GCB is not cost effective. Agree that GCB is not a requirement.
When I responded to the question, I had in mind the...
We may start by saying that the major objective of operating a power plant is to achieve the highest possible plant availability at the lowest possible cost.
The use of a generator circuit breaker (GCB) could achieve this goal because of the following advantages:
1) Installation of the GCB...
You may consider the minimum clearances given on the National Electrical Code (NEC) table 490-24. This table provides the phase-to-phase and phase-to-ground minimum clearances of live parts for indoor and outdoor applications. For example the clearance for 7.2 kV ph-grnd is 4.0 inches and for...
Edison 123,
You are right. Last communication was in 2004. Another thread that interested you was the clearance between the coil/bar end turns. But I follow your discussions and learn a lot. Zlatkodo attachment is super.
Best regards,
kh2
Thank you greatly for the correction and the formula.
This means to use one turn/coil (roebel bar) and 2 circuits will result in doubling the number of slots.
I thought that substituting two turns/coil with one turn/coil of a cross section equal to the two turns, will keep the number of...
Thanks again.
The existing machine slots/pole/phase =2.75 and the grouping is 3.3.3.2/3.3.3.2/3.3.3.2 repeated 6 times. Coil sides per slot is 2.
If we used C1 = 198 slots; TPC1 = 2 & TPC2 = 1; N1 =2 & [N2= 1 or 2 or 3]
Therefore C2 = [C1x(TPC1)x(N1)]/[(TCP2)x(N2)]
C2 (for one circuit and one...
Thanks for your response.
If we decided to rewind using one turn per coil (roebel bars) and 3 circuits/phase, then we will have: T2 =1, T1 =2, N1 =2, and N2 = 3.
Then we apply T2/T1 = N2/N1, the result will be 1/2 = 3/2. This is an incorrect equation, and where it shows that the number of the...
The project is to upgrade a hydro synchronous generator 50 MVA, 24 poles, 3-phase wye, 13.8 kV L-L, lap winding, 2-turn/coil, 2-circuits, 198 slots. The upgrade shall include the rewind and to replace the stator core. The new rating shall be 60 MVA. The alternatives for the rewind are one turn...
To solve this problem you have to know the utility (source) 3-phase fault current or MVA. Assume that the source 3-phase fault is 2500 MVA. Then use the transformer MVA as the MVA base.
1) Source (utility) p.u. impedance = MVA base/SC MVA source = 20/2500 = 0.008.
2) Transformer p.u. impedance =...
Also, you may consider a good reference book called "Electric Machinery and Transformers" by Irving L. Kosow. On Chapter 9 and 10, the author provided an easy to follow, supported by questions and answers, the principles, characteristics, and construction of polyphase and single phase induction...
Elep,
I came across your post late, but felt that you may look at IEEE C57.116-1989 since your original question was about standards/references. It is called IEEE guide for transformers directly connected to generators. The guide describes the selection of Unit Transformers and Unit Auxiliaries...
Agree with all you wrote saladhawks. My original question to zhz was related to the formula that he used to calculate the LV side fault. I was trying to understand how the formula was developed. I am familiar with a simple approach which gives the same result as follows:
1) Use MVA base = 1MVA...
Hi zkz,
Please note the following correction to my last calculation of 926 amps for the fault current on the secondary side of the 1000 kVA, 24.94/12.47 kV( L-L), 3-phase, 5% Z, auto transformer:
The 926 amps is 1.37 times the 674 amps that dimartin obtained, and is 1.17 the 791 amps that you...
