Recent content by magoo2

The attachment says the opposite. They change from HPS to LED lights. HPS is an arc-type of light. If the glass shatters, you have an open arc. I'm guessing that is why they want to go with LED lights, but that is just a guess.

kVA = kW/PF Once they give you the power factor, you can see what kVA you need.

200 A loadbreak elbows are available at 34.5 kV. The 600 A elbows are non loadbreak. We used Cooper/Eaton.

I see 230 kV on the left side of your drawing, but what is the voltage level supplied to the motor? 25 kV, 35 kV? Also, what is the rating of the substation transformer and its impedance? A series capacitor will reduce the reactance to the motor and thus reduce the starting voltage flicker.

What is the subtransmission voltage and what is the MCOV of the surge arresters? How many times has this happened before (arresters blowing)?

You should be able to do this with standard jacketed URD cable. I'd probably go with EPR insulation and stranded conductor for flexibility. You don't want to special-order something like this unless you have a very long length that you're dealing with.

I agree with Bill's result.

https://www.dailymail.co.uk/news/article-9282215/United-Airlines-flight-makes-emergency-landing-dropping-debris.html Flight was able to land in Denver. No injuries reported. It was a Boeing 777.

I believe Allen J. Wood covered this in his book "Power Generation, Operation and Control". I don't have a copy but I believe it was published around 1983. We adopted his technique for a secondary network which had a high R/X ratio and conventional load flow techniques didn't work too well.

Accuracy? Do you mean efficiency?

The last edition of the GE distribution data book was GET 1008M, published in 1980. I'll have to look for my pdf version but don't hold your breath.

Normally you want to work with ohms per phase and then get the capacitance. Given the 3-phase Mvar and line-to-line kV of the bank, not the nominal p-p voltage ohms = kV^2/Mvar You can also take the per phase Mvar and the phase-to-neutral kV rating of the bank and get the same result...

It's lightning not lightening.

One further comment. DC offset is mostly a function of closing angle, i.e., the point of wave on which the fault occurs. So, whether you have a capacitor bank on the circuit or not, you will get DC offset strictly based on closing angle.

R/1000 = 2 X/1000 = 0.054 I'll ignore X because R/X is almost 40(37) so X is not significant. Using just R R/1000 * 0.45 * 2 = 2 * 0.45 * 2 = 1.8 ohms Using just the #12, you get the available fault current of 120/1.8 or 66.7 A So it looks like your calculations are close enough to being...