# NCEES Power #536



## thewalt33 (Aug 11, 2010)

Hi everyone. I'm new to the board and didn't see a topic on this question. The question refers to a 3-phase transmission line where you have to solve for the capacitive reactance of the line, and then find the charging current. I'm confused on one thing:

In the solution, they show the math for solving for capacitance using the line to ground capacitance equation given in the question. The equation is C = 0.0388 / log(Deq/Rc). With Deq = 12.57 and Rc = .429, the solution shows the following:

C = .0388 / log(12.57)/(.0429 X 12) = .0152 uF/mile

Can someone help with this calculation? I'm not sure where the 12 came from, and not sure why the log calculation is different from the given equation. This seems like such an easy problem, but my brain has apparently quit on me this morning.

Thanks for any assistance!


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## Flyer_PE (Aug 11, 2010)

They botched the equation in their solution. The value rc is the conductor radius in feet. The problem gives you the conductor diameter in inches. All they are doing is a unit conversion from inches to feet.

The correct equation for their solution should be:

C = .0388 / log(12.57)/(.0429 */* 12) = .0152 uF/mile


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## thewalt33 (Aug 11, 2010)

Hopefully I don't overlook units in October. Thanks for the help!


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## Ship Wreck PE (Aug 11, 2010)

I worked on this problem for days and thought maybe I am just stupid or something??? Glad to see the book was wrong arty-smiley-048:


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## cableguy (Aug 12, 2010)

There is an errata page posted on the NCEES web site. It does include this problem.


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