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So, I picked up this book and am working through it by myself. I am a bit stuck on problem 1.6:

enter image description here

In the bottom you can see how I'm approaching this. I'm assuming the transmission cable goes across NY, and New Yorkers draw power from different points along the line. Since they each draw power at 115V, then the total voltage at the station, V@Station minimum, is 115 x 107V.

The power dissipated through one foot of cable is

(V@Station ^ 2) / Resistance per foot of cable

This worked out to be 2.645 x 1025W. And 1010W would be dissipated in 3.78 x 10-16 feet.

I haven't calculated the temperature because I don't know how to.

The last part of the question comments on these strange results and asks for a resolution to this problem. I'm guessing that, since power rises non linearly with voltage, we reduce the V@Station by the use of transformers in order to reduce power dissipation through the transmission cable.

Is this approach and the answer correct? Please add your own thoughts and point me in the right direction if I'm wrong.

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  • \$\begingroup\$ Have you considered the idea of power substations and the use of very high voltages between them? (And no, you don't multiply voltages that way -- if I gather you correctly.) \$\endgroup\$
    – jonk
    Nov 30 '16 at 9:05
  • \$\begingroup\$ Ok thanks. Could you please be a bit more specific about power substations. \$\endgroup\$
    – kchak
    Nov 30 '16 at 11:17
  • \$\begingroup\$ The voltage between a generating source and a substation isn't 110VAC. \$\endgroup\$
    – jonk
    Nov 30 '16 at 16:07
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As hinted by others in comments, that question is formulated so that you might reflect on the benefits of high-voltage power transmission lines.

Power is not transmitted over long distances at such a low voltage as 115Vac (230-240Vac in EU). This latter, relatively low, voltage is used in households for safety and practical reasons, but it is wasteful for long range transmission. You may find some mathematical result in this section of the article linked above.

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(a)

If we want to supply \$10^{10}\$ watts of power at \$115\$ Volts then the current would have to be $$ I=\frac{10^{10}}{115} = 86.956 \times 10^6 Amperes$$ .

Power lost per foot = \$I^2\$ (Resistance per foot)

$$\frac{P}{foot} = I^2\frac{R}{foot} = 7.561\times 10^{!5} \times 0.05 \times 10^{-6} $$

$$= 378.071 \times 10^6 \text{ Watt per foot}$$

(b)

So all the power \$10^{10}\$ is lost in

$$\frac{10^{10}}{78.071 \times 10^6} = 26.45 \text{ feet} \approx 8 \text{ metres}$$

(c)

The value of the Stefan–Boltzmann constant is given in SI units by

$$σ = 5.670367×10^{−8} \frac{W}{m^{2} K^{4}}$$

Emissivity of copper \$ε = 0.023\$. Note that the diameter of the wire is \$0.3048\$ metres.

Surface area of cylindrical wire $$A = 2\pi rh = 0.3048 \times \pi \times 8.06196 = 7.71 m^2$$

Power radiated from heated body $$= ε σ A T^4 = 1.0068 \times 10^{-8}T^4 = I^2R = 10^{10}$$

$$T^4 = 9.932 \times 10^{17}$$

$$T=31,569.22 \text{ Kelvin}$$

To put that into perspective, copper melts at \$1,358\$ Kelvin while the surface of the sun is \$5,778\$ Kelvin hot.

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First off, you have a misconception about the voltage at the power station. Assuming the power station is directly feeding the customers (which is what the question is implying), then the voltage at the power station would have to be 115V. Customers are connected to the grid in parallel, not series.

(A) Calculate the power lost per foot from the I2R losses

We know the total power (10^10 watts) and we know the voltage (115V), so using the handy dandy P=VI formula, we can figure out how much current has to be flowing through the cable. A little algebra later and we come up with I=P/V. 10^10/115 = 86,956,521.74 amps (side note: that is a LOT of amps).

But continuing on, we know the resistance of our cable is 0.05uOhms per foot. P=(I^2) * R, so plug & chug and we come up with 378,071,833.6 watts per foot (that's a lotta watts).

(B) Calculate the length of the cable over which you will lose all 10^10 watts.

That's easy. We know we lose 37,8071833.6 watts per foot (as calculated in A). So what's 10^10 / 37,8071833.6? 26.45 feet. Seems pretty preposterous, huh?

(C) How hot will the cable get?

I'll simplify it: Hot enough to turn into plasma. One does not simply dissipate 10 billion watts in 26 feet. It just doesn't end well for everyone involved.

What is the solution to this puzzle?

I will leave you to figure that out, but here's a hint: do all of the calculations again (excel can help greatly here), but try different voltages, like 220V, 1kV, 100kV, etc, and see what happens to the current needed.

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  • \$\begingroup\$ @RoyC I'm off by a factor of 10? I originally did all these calculations in an Excel spreadsheet. I double checked everything and redid all of the calculations by hand and came up with the same number. 10^10 is 10 billion, unless I've forgotten my 5th grade math (which is entirely possible) \$\endgroup\$
    – CHendrix
    Jan 9 '17 at 12:25
  • \$\begingroup\$ My bad confusing ^ and exp. Deleting my comment. \$\endgroup\$
    – RoyC
    Jan 9 '17 at 14:03
  • \$\begingroup\$ @RoyC Not a problem. It's always good to have someone double-checking math. \$\endgroup\$
    – CHendrix
    Jan 9 '17 at 18:22

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