Calculate the cost of the energy lost to heat per hour per meter

• NasuSama
In summary: So the time is already 1 hour, and you only need to multiply power by the $/KWh ratio, which is given as 9.5 cents. That's where the 7.2 comes from. :)In summary, the conversation discusses the cost of energy lost to heat per hour per meter in a small city that requires 17 MW of power. The calculations involve using the resistance formula, unit conversions, and the cost of electricity. The final answer is$0.684(W/V)^2(\rho/d^2\pi) per hour per meter.
NasuSama

Homework Statement

A small city requires about ##17 \mbox{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 ##9.5\, \mbox{ cents}/\mbox{kWh} ##

Homework Equations

• ##R = \dfrac{\rho \ell}{A}##
• ##P = I^2R##
• ##W = IV##

The Attempt at a Solution

Given that a small city requires that amount of watts ##W##

##W = IV##
##I = \dfrac{W}{V}##

By the resistance formula, rewrite it in terms of ohms per meter, which is

##R = \dfrac{\rho}{A}##

But we also know that the "face" of the wire is circle, so

##R = \dfrac{\rho}{\pi r^2}##

which is equivalent to

##R = \dfrac{\rho}{\dfrac{d^2\pi}{4}}##

where ##d## is the diameter of the wire. For the power rate, we have double the power rate of one wire, so

##P = 2I^2R##
##P = 2\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

The units of that ##P## are now ##\mbox{W}/\mbox{s}##. By unit conversion, we obtain

##P = 7.2\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

Given that the cost of electricity is around 9.5 cents per kilowatts hour,

##\text{Cost of lost energy}=7.2\cdot 9.5\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

##\text{Cost of lost energy} = 68.4\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

So converting cents into dollars, we finally obtain

##\text{Cost of lost energy} = .684\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

By substitution, I got around \$8156585 per hour per meter, but Mastering Physics marked it incorrect. I don't see where I went wrong.

Last edited:
For starters

r2 <> d/4

I don't understand the bit that says...

The units of that P are in W/s.

Power is measured in Watts not Watts per second.

Where does the 7.2 come from exactly?

CWatters said:
Power is measured in Watts not Watts per second.

I mean ##P## is treated as Watts per second as a rate.

CWatters said:
Where does the 7.2 come from exactly?

I used the unit conversions to convert second into hours and Watts into kiloWatts

Last edited:
Never mind. I found the solution. Anyway, thanks for the mistake catch!

To all those checking out this thread to figure out the solution to this kind of question (like I was), the problem here was that NasuSama multiplied the power by 3600s (the time in an hour). This is unnecessary because the question was posed in KWh, where you're to find how much power is lost in 1 hour.

1. How do you calculate the cost of energy lost to heat?

The cost of energy lost to heat can be calculated by multiplying the amount of energy lost per hour by the cost per unit of energy. This will give you the cost of energy lost per hour. To calculate the cost per meter, you will also need to know the area of the space where energy is being lost. You can then divide the cost per hour by the area to get the cost per hour per meter.

2. What factors affect the cost of energy lost to heat?

The cost of energy lost to heat can be influenced by a variety of factors, including the type and efficiency of the heating system being used, the temperature difference between the inside and outside of a space, the insulation of the space, and the cost of energy in your area.

3. How can I reduce the cost of energy lost to heat?

To reduce the cost of energy lost to heat, you can take steps to improve the efficiency of your heating system, such as upgrading to a more efficient system or properly insulating your space. You can also make changes to your habits, such as turning down the thermostat when you are away from home or using natural sunlight to heat your space during the day.

4. How can I accurately measure the amount of energy lost to heat?

The most accurate way to measure the amount of energy lost to heat is by using specialized equipment, such as a thermographic camera, to detect and measure temperature differences in your space. However, you can also estimate the energy loss by taking into account factors such as the type of heating system, insulation, and temperature difference.

5. Why is it important to calculate the cost of energy lost to heat?

Calculating the cost of energy lost to heat can help you understand the efficiency of your heating system and identify areas where you can make improvements to save energy and reduce costs. It can also help you make more informed decisions when it comes to choosing a heating system or making upgrades to your space.

• Introductory Physics Homework Help
Replies
23
Views
470
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
622
• Introductory Physics Homework Help
Replies
4
Views
971
• Introductory Physics Homework Help
Replies
12
Views
545
• Introductory Physics Homework Help
Replies
17
Views
631
• Introductory Physics Homework Help
Replies
2
Views
838
• Introductory Physics Homework Help
Replies
12
Views
764
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
21
Views
1K