How much power is lost in heating the wires of a power line

In summary, a generating station is producing 1.4e6 W of power to be sent to a town 7.0 km away through two wires with a resistance per kilometer of length of 5.0e-2 ohm/km. Using a 1100 V transmission, the power lost in heating the wires is 5.7e5 W. With a 70:1 step-up transformer, the voltage is raised and the power lost in heating the wires becomes 1.6e4 W.
  • #1
gamesandmore
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A generating station is producing 1.4e6 W of power that is to be sent to a small town located 7.0 km away. Each of the two wires that comprise the transmission line has a resistance per kilometer of length of 5.0e-2 [tex]\Omega[/tex]/km.

(a) Find the power lost in heating the wires if the power is transmitted at 1100 V.

(b) A 70:1 step-up transformer is used to raise the voltage before the power is transmitted. How much power is now lost in heating the wires?


Heres what I did:
a) P = 1.4e6 W
L = 7.0 km
R/L = 5.0e-2 ohm/km
Vp = 1100 V

R = (5.0e-2 ohm/km)*(7.0km)
R = .35 ohm
I = P/V = 1.4e6W/1100V = 1272.73A
P = I^2 * R = (1272.73)^2 * (.35ohm)

P = 5.7e5 W

But it is wrong...
 
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  • #2
Please note that there are two power lines, not one. Each has a resistance of 0.05 ohms/km. Do they represent parallel resistances, or simply divide the power?
 
  • #3
Aren't they in series, so I could just then do:
5.0e-2 ohm/km * .35 for each one and then add them,
wouldn't that still come out to .35 ohm though?
 
  • #4
No they are in parallel - assume the electricity goes down both of them to the town.
 
  • #5
Alright, so:
1/Rt = 1/.35 + 1/.35
Rt = .175 ohm,
using this resistance still gives me the incorrect answer.
 
  • #6
The wires are in parallel, but you're using the wrong approach. How does the current that you have calculated divide between the 2 wires? Then find the heat loss in each, and the total of the losses.
 
  • #7


The wires are in series, you just multiply the resistance/km (.05) times the distance (7km) to get .35ohm, but then multiply that product by two to account for the other wire, so instead of .35ohm use .7ohm and you should get it. (At least that's what I did and I got the right answer according to webassign)
 

1. What causes power loss in heating the wires of a power line?

The power loss in heating the wires of a power line is mainly caused by the resistance of the wires. As electricity flows through a wire, some of the energy is converted into heat due to the inherent resistance of the material. This heat energy is then dissipated into the surrounding environment, resulting in power loss.

2. How is power loss in heating the wires of a power line measured?

The power loss in heating the wires of a power line is measured using a unit called "line loss". This unit represents the difference between the power transmitted at the source and the power delivered to the end user. Line loss is typically expressed as a percentage of the total power transmitted.

3. What factors can affect power loss in heating the wires of a power line?

Several factors can affect the power loss in heating the wires of a power line, including the length and thickness of the wire, the type of material used, and the temperature of the environment. Higher temperatures can increase the resistance of the wire, leading to higher power loss.

4. How can power loss in heating the wires of a power line be reduced?

There are a few ways to reduce power loss in heating the wires of a power line. One method is to use thicker wires with lower resistance, as this will decrease the amount of heat generated. Another option is to use materials with lower resistance, such as copper instead of aluminum. Additionally, keeping the wires at a lower temperature can also help reduce power loss.

5. Is it possible to completely eliminate power loss in heating the wires of a power line?

No, it is not possible to completely eliminate power loss in heating the wires of a power line. This is because some amount of resistance and heat generation will always occur when electricity flows through a wire. However, through proper design and maintenance, power loss can be minimized to ensure efficient power transmission.

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