# Calculate the cost of the energy lost to heat per hour per meter

1. Sep 20, 2013

### NasuSama

1. The problem statement, all variables and given/known data

A small city requires about $17 \mbox{MW}$ of power. Suppose that instead of using high-voltage lines to supply the power, the power is delivered at $120 V$.

Assuming a two-wire line of 0.60cm -diameter copper wire, estimate the cost of the energy lost to heat per hour per meter. Assume the cost of electricity is about $9.5\, \mbox{ cents}/\mbox{kWh}$

2. Relevant equations

• $R = \dfrac{\rho \ell}{A}$
• $P = I^2R$
• $W = IV$

3. The attempt at a solution

Given that a small city requires that amount of watts $W$

$W = IV$
$I = \dfrac{W}{V}$

By the resistance formula, rewrite it in terms of ohms per meter, which is

$R = \dfrac{\rho}{A}$

But we also know that the "face" of the wire is circle, so

$R = \dfrac{\rho}{\pi r^2}$

which is equivalent to

$R = \dfrac{\rho}{\dfrac{d^2\pi}{4}}$

where $d$ is the diameter of the wire. For the power rate, we have double the power rate of one wire, so

$P = 2I^2R$
$P = 2\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)$

The units of that $P$ are now $\mbox{W}/\mbox{s}$. By unit conversion, we obtain

$P = 7.2\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)$

Given that the cost of electricity is around 9.5 cents per kilowatts hour,

$\text{Cost of lost energy}=7.2\cdot 9.5\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)$

$\text{Cost of lost energy} = 68.4\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)$

So converting cents into dollars, we finally obtain

$\text{Cost of lost energy} = .684\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)$

By substitution, I got around \$8156585 per hour per meter, but Mastering Physics marked it incorrect. I don't see where I went wrong.

Last edited: Sep 20, 2013
2. Sep 20, 2013

### CWatters

For starters

r2 <> d/4

3. Sep 20, 2013

### CWatters

I don't understand the bit that says...

Power is measured in Watts not Watts per second.

Where does the 7.2 come from exactly?

4. Sep 20, 2013

### NasuSama

I mean $P$ is treated as Watts per second as a rate.

I used the unit conversions to convert second into hours and Watts into kiloWatts

Last edited: Sep 20, 2013
5. Sep 20, 2013

### NasuSama

Never mind. I found the solution. Anyway, thanks for the mistake catch!

6. Feb 16, 2017

### zexxa

To all those checking out this thread to figure out the solution to this kind of question (like I was), the problem here was that NasuSama multiplied the power by 3600s (the time in an hour). This is unnecessary because the question was posed in KWh, where you're to find how much power is lost in 1 hour.

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