What Is the Cost of Energy Loss in a Low Voltage Power Supply System?

[alwaysdazed]

Homework Statement

A small city requires about 19 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 8.5 cents/kWh.
Cost = $ per hour per meter

Homework Equations

P=IV
P=I^2*R
R=(rhoe*L)/A

The Attempt at a Solution

I used P=IV to find the current (converting 19 MW to 19*10^3 to keep it in kW for the answer) and found I=158 A .. then when I go to find the resistance I figure to use pi*r^2 as area than multiply the whole thing by two for the two wires but I do not understand what to use for L because they provide no length after i find R i intend to plug it into the equation P=I^2*R .. i am so lost!

 

[Redbelly98]

Since they want cost per meter, you really just need R/L and P/L.

p.s. welcome to PF.

 

[alwaysdazed]

thanks but i do not understand what you are saying sorry :(

 

[Redbelly98]

Okay, let's back up a little.

From your 1st post, you said you tried to calculate R, but need L in order to do that. What equation for R are you using, that has an L in it?

 

[alwaysdazed]

ohhh yeah that's what my problem is i am not given L .. i am trying to use R=(rho(resistivity)*L)/A

 

[alwaysdazed]

L being the length of the wire and A being the cross sectional area of the wire

 

[Redbelly98]

Okay.

Since you need to calculate power and cost per meter, use L=1m.

 

[alwaysdazed]

okay so then i use l=1 to find the resistance and i take the current that i found usuing p=iv and plug both into P=I^2R .. once i find that P how do i incorporate finding time?