# Estimating cost due to power (application of electric current)

1. Jul 21, 2004

### Theelectricchild

I would like to understand this problem a bit better...

A small city requires 10 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 0.50 cm diameter copper wire, estimate the cost of the energy lost to heat per hour per meter. Assume the cost of electricity is about 10 cents per kWh.

Heres what im wondering: Should I find the area of the wire (using pi*diameter^2 all over 4) and and then use Ohms Law to solve for the amount--- knowing that V = 120?

2. Jul 21, 2004

### AKG

Do you have an equation that relates the heat released by a resistor and the voltage, power, or energy being sent through it?

3. Jul 21, 2004

### Theelectricchild

Well, I don't know what equation I should use to include heat.

4. Jul 21, 2004

### Staff: Mentor

energy consumed by resistance

First determine the current that must flow through those wires to deliver the stated power.

Then figure out the energy lost in the resistance of the wire as that current flows through it. (You'll need to figure the resistance of the wire.) The power consumed by the resistance is $P = V_{drop}I = I^2R$.

(Please post these kind of questions in the Homework Help forum!)

5. Jul 21, 2004

### Theelectricchild

Doc Al, thank you, and I apologize for posting in this part of the forum I wont do it again.

6. Jul 21, 2004

### Theelectricchild

I seem to get an answer of \$841.50! this is interesting--- would you say this is about right using the method you were suggested?

7. Jul 22, 2004

### AKG

According to this site, $\rho _{copper} = 16.8 n\Omega \cdot m$. I don't know why I missed how to do this question (of course, you don't need any formula to calculate heat from energy as I suggested, the heat is the energy!). Anyways:

$$R = \frac{\rho _{copper} \times 1m}{2\pi (0.0025m)^2}$$

$$P = V^2R$$

$$E = P \times 1\ hour$$

The cost, in dollars, is then E/10, as long as E is expressed in kWh. Your numbers look pretty different from mine. What do you have for the resistivity of copper?

8. Jul 22, 2004

### Theelectricchild

1.68e-8 ohms meters is correct, however i am wondering about your 2 times A in the denominator--- should it not be (rho*L)/A then that times 2? Then take that multiply by 120^2

Then take that answer divide by 1000 and then by 10 to get the answer?

9. Jul 22, 2004

### Theelectricchild

the answer is 1190--- btw its I^2 * R hehe, but thank you your explanation is perfect.

10. Jul 22, 2004

### AKG

You're right, I meant P = V²/R. Doing it this way, you don't even have to calculate current. And the 2A in the denominator seems right. Essentially, you're treating the two wires as one wire with double the cross-sectional area.

11. Jul 22, 2004

### Staff: Mentor

If you use $P = V^2/R$, V must be the voltage drop across the wire. This is not 120V.
Treating the two wires as a single wire with double cross-section won't work: Your value for resistance will be a factor of 4 too low.

12. Jul 22, 2004

### AKG

Oh, I see. My mistakes.

13. Apr 26, 2010