Power Line Losses: Trivial Problem, Confusing Answer

[rdjohns12]

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 9.0 cents per kWh.

Relevant Equations

$$P=IV$$
$$P=I^2R$$
$$R=\rho\ell/A$$

This problem is trivial, but I cannot make sense of my answer (I am not even going to bother with the cost calculation.

First, I used P=IV with P=15 MW and V=120 V to find I=125 kA

So far so good. Then I calculated the resistance per meter as R=\rho\ell/A=(1.68\times 10^8\Omega\cdot m)(2m)/(\pi(0.005m/2)^2) =1.68m\Omega

Now, I calculate the power per meter dissipated in the transmission line as P=I^2R=(125 kA)^2(1.68m\Omega)=27 MW I have checked the solutions manual and so I am certain this value is correct.

This does not make sense to me. In my understanding, we started with the assumption that the transmission line losses were going to be small enough that we could use P=IV where P was both the power from the plant and the approximate power delivered to the city, but now we are finding line losses that are nearly double that power.

I could, obviously, go ahead and calculate the voltage drop across 1 m with V=IR=(125kA)(1.68m\Omega)=200V which is larger than 120V.

Is this problem poorly thought out, or do I fundamentally not understand something?

 

[haruspex]

we started with the assumption that the transmission line losses were going to be small enough that we could use P=IV where P was both the power from the plant and the approximate power delivered to the city,

That's not how I see it. You assumed P as the power the city needs, and that this needs to be delivered at 120V. The power the plant needs to generate is P + transmission losses. This will increase the voltage as generated at the plant. Nowhere did you assume the losses were small.

 

[Steve4Physics]

Then I calculated the resistance per meter as R=\rho\ell/A=(1.68\times 10^8\Omega\cdot m)(2m)/(\pi(0.005m/2)^2) =1.68m\Omega

Your resistivity for copper is a factor of $10^{16}$ too big because you've missed out the minus sign on the exponent!

The question is not very well written because in real life, voltage is transformed up and down at each end of the main power transmission line. This means the power is transmitted at a *much* higher voltage than 120V. This reduces the current and hence reduces the losses in the line. However, from the information supplied, it looks like you are only being asked about losses in the parts of the distribution system running at 120V (the wiring entering domestic residences). If so, your method is OK.

 

[haruspex]

Your resistivity for copper is a factor of 1016 too big because you've missed out the minus sign on the exponent!

Just a typo... the calculated resistance per length looks about right... except, not sure where the 2m length comes from. I guess it's the "2 wire line", so it doesn’t seem to be using Earth return.

The question is not very well written because in real life, voltage is transformed up and down at each end of the main power transmission line.

It says "Suppose that instead of using high-voltage lines to supply the power...". I.e. the point of the exercise is to demonstrate why high voltage is used in reality.

 

[Steve4Physics]

Just a typo... the calculated resistance per length looks about right... except, not sure where the 2m length comes from. I guess it's the "2 wire line", so it doesn’t seem to be using Earth return.

It says "Suppose that instead of using high-voltage lines to supply the power...". I.e. the point of the exercise is to demonstrate why high voltage is used in reality.

You're quite right. I should have read the question/checked the arithmetic properly.

I guess the '2m length' comes from the 'two-wire line' with one wire the live and the other wire the neutral (not ground).

 

[haruspex]

one wire the live and the other wire the neutral

The two would be opposite phases of a two phase supply. Both are live. Neutral only exists in the domestic domain and results from customers on different live phases sharing a neutral, which thereby carries little current.

 

[Delta2]

The two would be opposite phases of a two phase supply. Both are live. Neutral only exists in the domestic domain and results from customers on different live phases sharing a neutral, which thereby carries little current.

When you say both are live you mean they are + and - instead of + and neutral? Btw does this problem assumes DC? And how do you say that the neutral carries little current, to my best knowledge it carries all the return current...

 

[haruspex]

When you say both are live you mean they are + and - instead of + and neutral? Btw does this problem assumes DC? And how do you say that the neutral carries little current, to my best knowledge it carries all the return current...

It doesn't say DC, but I agree that is probably the intent.

In a conventional 2- or 3-phase AC distribution (substation to house, say), each house gets the neutral and one of the phases. The phases delivered along a street are interlaced, so that the current on the neutral tends to blend out. Back at the substation, the neutral goes to ground, so only live (or more correctly "line") wires connect back to the power station (which likewise has a ground connection).

 

[Tom.G]

If you want to be picky, the answer is Zero.

The fusing current of that 0.5cm wire (the current that causes it to melt) is around 1100Amps. That's going to be very fast-blow fuse.
(Oops!)

 

[Steve4Physics]

The two would be opposite phases of a two phase supply. Both are live.

Possibly. But the original question is unrealistic and oversimplified. In the same spirit, I interpreted the question as referring to a single-phase supply. One wire would be the live and the other wire (grounded at both ends) would be the neutral.

Neutral only exists in the domestic domain and results from customers on different live phases sharing a neutral, which thereby carries little current.

Agreed, providing you are referring to shared (common) neutral. Inside my single phase domain (i.e. my UK home), live and (unshared) neutral carry exactly equal currents.

[EDITed - minor change.]