City requires about 15 MW of power problem

[skepticwulf]

Homework Statement

A small city requires about 15 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 12 cents per kWh.

Homework Equations

P=V^2 / R, R=r x l /A, P=I x V

The Attempt at a Solution

Can I have a hint at solving the problem? I know the formulas but I'm not sure how to approach. What does it mean by "energy lost to heat"?

 

[RaulTheUCSCSlug]

The reason why it gives you the diameter of the copper wire is to probably use the resistance of a copper wire to figure out how much is lost. Think of the copper being a very small small resistor. Unless it stated that this was a ideal wire, but I don't believe that's the case.

 

[skepticwulf]

R=r x l /A or R/l=0.0086 ohm/per meter.
I assume that means the wire will result that much resistance per meter. ( 2 times l and below 2 times A invalidates the multiple 2)

If I put this value to P= V^2 / R it gives me 16.83MW
Is this the power delivered per meter??

If city requires 15MW, the difference 1.83MW is what?? power lost per meter?? It can't be.
Where do I go from here?

 

[mooncrater]

Why it can't be? According to me its correct.
You are given resistance in the question. So you HAVE to put that in the formula. Potential difference too is given. So after putting values you get to know how much power is actually SUPPLIED. But since the village needs 15MW only then where will the rest power go? It will cause heating.

 

[skepticwulf]

1.83MW power loss per meter seemed way too much!

 

[RaulTheUCSCSlug]

Well also remember that a low resistance wire will heat up more then a higher resistance wire.

 

[insightful]

R/l=0.0086 ohm/per meter.

I think you mean 0.00086 ohm/m.

 

[RaulTheUCSCSlug]

I think you mean 0.00086 ohm/m.

I didn't check the numbers, did OP forget to convert cm to m?

 

[insightful]

No, I think it's a typo. Also, that's ohm/m for each wire, so double that for ohm/m of 2-wire line, assuming basis is one meter of electric line.

 

[insightful]

First calculate the amps needed to deliver 15MW at 120V.

 

[davenn]

What does it mean by "energy lost to heat"?

Pretty much all losses will be as heat

 

[skepticwulf]

It's a typo sorry it's 0.00086 ohm/m.

 

[skepticwulf]

First calculate the amps needed to deliver 15MW at 120V

Why? Why do I need to calculate the current? Why is it that P=V^2/R does not give the same power value as I^2 x R ?

I calculated the difference of two power values, and got 1.83MW as the power lost per meter, So multiplying that to 12 cents gave me 219.6$.
But it's not right according to solution manual. I'm lost.

 

[gneill]

Why? Why do I need to calculate the current? Why is it that P=V^2/R does not give the same power value as I^2 x R ?
.

Because the voltage at the end of the line is not the same as the voltage dropped across the line. Sketch the circuit.

Power is consumed in both the load and the line. You're interested in the power lost in the line. The straightforward approach then is to determine the current required by the load (which you should know from the load power and voltage), which must also be the current in the line since it's a series circuit.

 

[skepticwulf]

Thank you.