Energy Harvesting: Calculating kWh from Motorway Traffic

[PeterC]

TL;DR

Energy Harvesting on the street

Hello there :)

if there were an energy converter for a motorway where 5000000 cars drive over it annually and that has an efficiency of 100 percent, how many kWh does the energy converter give me?

The energy converter would generate 4000 joules / s per pass. (P = 1400 * 9.81 * 0.02 / 0.07). 1400 kg = mass; 9.81 = m / s ^ 2; 0.02 (2 cm sink from the energy converter) 0.07 (time in which the energy converter converts the energy)

My approach: I know that 274 joules can be converted into kWh. However, the time in which that energy converter is in use is neglected?
Is it allowed to do the math? ==> 5000000 cars per year would be a result of 570 cars per hour.

If you generate 4000J / s of power per crossing, it should be 570 * 4000 for 570 cars. So 2283105 j / s ==> the whole divided by 1000 results in my 2283 kilojoules per second. ==> Because 1kw = 1kj / s = 3600 kj = 1kwh per h, I can now say that 2283 kwh are generated per hour or 2283 * 24 * 365 per year.

I would be very grateful if someone could help me.

 

[berkeman]

Summary:: Energy Harvesting on the street

Hello there :)

if there were an energy converter for a motorway where 5000000 cars drive over it annually and that has an efficiency of 100 percent, how many kWh does the energy converter give me?

The energy converter would generate 4000 joules / s per pass. (P = 1400 * 9.81 * 0.02 / 0.07). 1400 kg = mass; 9.81 = m / s ^ 2; 0.02 (2 cm sink from the energy converter) 0.07 (time in which the energy converter converts the energy)

My approach: I know that 274 joules can be converted into kWh. However, the time in which that energy converter is in use is neglected?
Is it allowed to do the math? ==> 5000000 cars per year would be a result of 570 cars per hour.

If you generate 4000J / s of power per crossing, it should be 570 * 4000 for 570 cars. So 2283105 j / s ==> the whole divided by 1000 results in my 2283 kilojoules per second. ==> Because 1kw = 1kj / s = 3600 kj = 1kwh per h, I can now say that 2283 kwh are generated per hour or 2283 * 24 * 365 per year.

I would be very grateful if someone could help me.

Welcome to PhysicsForums.

Before running through any numbers, you do understand that such an energy source is not free, right? You are forcing drivers to pay for extra gasoline and electric charging time in order for you to make them less efficient so that you can harvest some of their energy. Driving on a deformable surface that you can use for energy harvesting makes them use more energy (that the drivers have to pay for).

And since such energy harvesting is far from 100% efficient, it is wasteful. It is more efficient to directly use some of the gasoline/electricity that the drivers would have been paying for.

Does that make sense?

 

[jartsa]

Energy is a simple thing to calculate:

$E=mgh$

energy harvested in a year = mass of 5000000 cars * the g * the depth

 

[russ_watters]

The energy converter would generate 4000 joules / s per pass. (P = 1400 * 9.81 * 0.02 / 0.07). 1400 kg = mass; 9.81 = m / s ^ 2; 0.02 (2 cm sink from the energy converter) 0.07 (time in which the energy converter converts the energy)

My approach: I know that 274 joules can be converted into kWh. However, the time in which that energy converter is in use is neglected?
Is it allowed to do the math? ==> 5000000 cars per year would be a result of 570 cars per hour.

If you generate 4000J / s of power per crossing, it should be 570 * 4000 for 570 cars. So 2283105 j / s ==> the whole divided by 1000 results in my 2283 kilojoules per second. ==> Because 1kw = 1kj / s = 3600 kj = 1kwh per h, I can now say that 2283 kwh are generated per hour or 2283 * 24 * 365 per year.

I would be very grateful if someone could help me.

Is this homework? If it is, I can't tell what's given and what's assumed. There's at least one obvious problem, though: you have Joules and Joules per hour, not Joules per second. The time it takes a car to roll over the generator is irrelevant/a red herring. What matters is the force and distance only, for each pass.

 

[russ_watters]

Before running through any numbers, you do understand that such an energy source is not free, right? You are forcing drivers to pay for extra gasoline and electric charging time in order for you to make them less efficient so that you can harvest some of their energy.

We've had some people misunderstand that, so it is worth pointing it out, however there is a potential legitimate use for it, in powering road sensors or signals:
https://www.sciencedirect.com/science/article/pii/S1877050916301971

It's better to just use solar and batteries, though.