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Re: 7.1Hz, how the heck did Tesla succeed?

Original poster: Jim Lux <jimlux@xxxxxxxxxxxxx>

At 04:57 PM 7/19/2005, you wrote: 
Original poster: Ed Phillips <evp@xxxxxxxxxxx>

" >      I'd have to look up his reference but he says that "only a few
 > horsepower" are needed to excite his antenna; I take that to mean
loss.

Ah, I remember that.  You're right, it would take WAY more than "a few
horsepower" to run Wardenclyffe if nobody was loading it.  But the Corum
paper said ~1 megawatt.  Was "few horsepower" about Wardenclyffe or
about
some other unit?"

        I can't remember but think it may have been more general than just
Wardenclyffe.

"But if those total corona/resistive losses at present are in the
megawatts, then a Tesla system with megawatt losses wouldn't be ridulous
in comparison.  The world system isn't like a power line, it's like a
worldwide power grid, so we need to compare it to at least the USA power
grid."

        As I recall Tesla wasn't talking about very many megawatts
(megahorsepower?).

Ed


Megawatts is decahorsepower (or maybe tenths of kilohorsepower). 750 W = 1 HP approximately.

As for actual numbers for corona losses
Khalifa gives P = 3.73K/(D/r)^2 * f * V^2 * 1E-5 kW/conductor/km

Typical values might be 0.3 to 1.7 kW/conductor/km for 500kV lines and 0.7 to 17 kW/conductor/km for 700 kV lines.

Taking something like the lines carrying power between northern and southern california, we're talking about 3 conductors, 500 km. So let's take 10 kW/conductor/km as the number. That's 10*3*500 = 15 MW

That line carries about 5500 MW, so the corona losses are less than 1%. I imagine that IR losses are comparable.

The HVDC Pacific Inertie runs at 400 kV (each side) and about 1100A. They're upgrading from the existing mercury arc valves to solid state, and they expect to bump that up to 500kV at 2000A. The line is about 1000 km long. I have no idea off hand what the resistance might be. 0.01 ohm per km, for 10 ohms total? That would make 20kV voltage drop, which seems like a lot.