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Solar Powered Hydrogen Engine


Joshcitylife

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Im trying to build an engine which runs off of water using compounded solar energy to heat/transfer energy. This will be used to create enough continuous energy to run an engine. This would mean that we can use the sun as energy to fuel a fossil-free engine . Yes we have used sun before for solar panels and etc…. However I believe I have found a method of compounding sunlight in order to generate a level of energy which is high enough to create an effective propulsion system. However, I need funds to generate this, Ive created diagrams which I have mailed to myself and saved. If anyone would like to team up to create this I believe we can!

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31 minutes ago, Joshcitylife said:

However I believe I have found a method of compounding sunlight in order to generate a level of energy which is high enough to create an effective propulsion system. However, I need funds to generate this

I don't think you're going to get a lot of takers with this vague description.

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9 hours ago, Joshcitylife said:

Im trying to build an engine which runs off of water using compounded solar energy to heat/transfer energy. This will be used to create enough continuous energy to run an engine. This would mean that we can use the sun as energy to fuel a fossil-free engine . Yes we have used sun before for solar panels and etc…. However I believe I have found a method of compounding sunlight in order to generate a level of energy which is high enough to create an effective propulsion system. However, I need funds to generate this, Ive created diagrams which I have mailed to myself and saved. If anyone would like to team up to create this I believe we can!

The problem I see with using solar power to move a vehicle is that the flux of incident radiation per unit area is fixed by the radiant power of the sun, so the available energy is determined by the surface area on your vehicle used to capture it. Unless you are talking of an energy storage system that accumulates energy over time before use, I suppose.

How much power are you expecting to need for propulsion of your vehicle?    

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1 hour ago, exchemist said:

The problem I see with using solar power to move a vehicle is that the flux of incident radiation per unit area is fixed by the radiant power of the sun, so the available energy is determined by the surface area on your vehicle used to capture it. Unless you are talking of an energy storage system that accumulates energy over time before use, I suppose.

How much power are you expecting to need for propulsion of your vehicle?    

That's a good point; I know there have been past discussions that show a voltaic panel solar-powered vehicle is impractical for operating with real-time propulsion, and some new method that doubled or tripled the efficiency still doesn't get you there

 

edit to add: power output of 100 kW here for an EV

https://www.renaultgroup.com/en/news-on-air/news/understanding-the-power-output-of-an-electric-motor/

 

250 W/m^2 insolation and 2 m^2 area of a car is just 500 Watts. Quite a gap.

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11 hours ago, Joshcitylife said:

However I believe I have found a method of compounding sunlight in order to generate a level of energy which is high enough to create an effective propulsion system.

That sounds like you may be running into an issue with the conservation of energy, if you mean by 'compounding' you can get more energy out than you put in.  If you are just talking about increasing efficiency, then the limitations that Swansont brought up will be an issue.

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I attached a very crude drawing that I made a few years ago. This would use “cells” which capture sunlight to compound within each other effectively multiplying. Ive been tweaking it using different materials for maximum energy. 
I guess to really simplify the idea is if u had two magnifying glasses capturing the sun and those focused into another magnifying glass it would in essence be compounded sunlight. Now imagine this doing it 100 times or 1000 times to generate enough energy for propulsion.

31D5F9A7-B0D6-43E4-847D-C2794678A1A8.jpeg

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My brother's Kia has a 64kw-hr battery.

He lives in London, where the average total daily solar energy available is 2.5 kw-hrs per square metre.

It follows that the total output of 64/ 2.5 ~ 26 square metres of collector for 1 day or the output of 1 square metre for 26 days would be required to recharge his battery, assuming all the sunlight is converted.

I reality, solar cells average no better than 10% efficiency so realistic figures of 260 square metres or 260 days would be required.

 

Our cousin in Alice Springs would fare better as the total insolation there is 3 times that of London.

 

These are the true figures of 'compounding'.

 

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2 hours ago, Joshcitylife said:

I attached a very crude drawing that I made a few years ago. This would use “cells” which capture sunlight to compound within each other effectively multiplying. Ive been tweaking it using different materials for maximum energy. 
I guess to really simplify the idea is if u had two magnifying glasses capturing the sun and those focused into another magnifying glass it would in essence be compounded sunlight. Now imagine this doing it 100 times or 1000 times to generate enough energy for propulsion.

31D5F9A7-B0D6-43E4-847D-C2794678A1A8.jpeg

OK this confirms what some of us suspected: you do not understand the principle of conservation of energy. A magnifying glass does not increase the amount of energy, it just intercepts energy from a wide are and focuses it (concentrates it) on a small area. That enables you to produce high temperatures, high enough to burn something, but only within a tiny area. The amount of energy flux in this small hot area is the same as the energy flux intercepted by the lens of the magnifying glass.  

The amount of the sun's energy you intercept is determined by the area of the capturing device. Nothing you do with it subsequently can increase the amount of energy you have. You are not "multiplying" anything.  

Edited by exchemist
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46 minutes ago, exchemist said:

The amount of energy flux in this small hot area is the same as the energy flux intercepted by the lens of the magnifying glass.  

The amount of energy flux in this small hot area less than the energy flux intercepted by the lens of the magnifying glass.

..because part of the photon energy is reflected and part is absorbed by the glass..

 

Quote

The amount of energy flux in this small hot area is the same as the energy flux intercepted by the lens of the magnifying glass.

This statement could be true if there were 100% efficiency..

 

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20 minutes ago, Sensei said:

part of the photon energy is reflected and part is absorbed by the glass..

A photon can't lose part of it's energy, usless it's redshifted in space. It would be more accurate to say that some of the photons are reflected, and some are absobed.

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@Sensei   and @exchemist

 

Whilst both of you have vailid points, I think you are being a little hard on the OP.

Concentrators (I prefer concentrators to compounders) are in use today in respect of solar energy.
However these are mostly reflective collectors rather than refractive concentrators as suggested by the OP.
Nevertheless whatever means of collection is employed it is necessary to collect solar energy over a wider area than where it is applied and a reasonable estimate of the area required is given in my calculation above.

I think this idea is indicated in the posted diagram, although the OP may not have fully realised thes size of the area required at the top of his 'funnel'.
So he would need a 260m2  magnifying glass to 'compound' (=focus ?) enough energy to recharge my brother's car in a day.
This would mean a 10 metre radius glass.

 

:)
 

 

Edited by studiot
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4 hours ago, Joshcitylife said:

I attached a very crude drawing that I made a few years ago. This would use “cells” which capture sunlight to compound within each other effectively multiplying. Ive been tweaking it using different materials for maximum energy. 
I guess to really simplify the idea is if u had two magnifying glasses capturing the sun and those focused into another magnifying glass it would in essence be compounded sunlight. Now imagine this doing it 100 times or 1000 times to generate enough energy for propulsion.

31D5F9A7-B0D6-43E4-847D-C2794678A1A8.jpeg

The total energy or power incident on the device does not increase. If the collector is 4m^2 and the intensity is 250 W/m^2, you have 1 kW. You can “concentrate” this to make a spot that’s 1 m^2, and the intensity jumps to 1 kW/m^2, but the power is still 1 kW

There might be applications where using this energy is more efficient at higher intensity, but you will never exceed the 1 kW value in whatever mechanical conversion you do. (You will always be lower than that, since y]that conversion can’t be 100% efficient)

1 hour ago, mistermack said:

A photon can't lose part of it's energy, 

True, but “photon energy” ≠ “a photon’s energy”

Sensei’s point is correct.

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2 hours ago, Sensei said:

The amount of energy flux in this small hot area less than the energy flux intercepted by the lens of the magnifying glass.

..because part of the photon energy is reflected and part is absorbed by the glass..

 

This statement could be true if there were 100% efficiency..

 

Fair point, in reality a number of photons will be reflected, and those in the appropriate frequency bands in the UV and IR will be partly absorbed by the glass.

What I was trying to focus (😁) our poster on is the concept of concentrating the radiation, as opposed to the incorrect notion of amplifying or "multiplying" it. 

 

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On 7/22/2022 at 5:24 AM, swansont said:
!

Moderator Note

Soliciting funding and people without actually delving into discussion of the science violates rule 2.11. “We are here to discuss science, in the open”

 

Sorry didnt know that rule. 

On 7/23/2022 at 10:24 AM, swansont said:

The total energy or power incident on the device does not increase. If the collector is 4m^2 and the intensity is 250 W/m^2, you have 1 kW. You can “concentrate” this to make a spot that’s 1 m^2, and the intensity jumps to 1 kW/m^2, but the power is still 1 kW

There might be applications where using this energy is more efficient at higher intensity, but you will never exceed the 1 kW value in whatever mechanical conversion you do. (You will always be lower than that, since y]that conversion can’t be 100% efficient)

True, but “photon energy” ≠ “a photon’s energy”

Sensei’s point is correct.

Thanks alot guys, im glad if nothing else that this sparked conversation. Im not trying to be famous or anything im just trying to SAVE OUR PLANET. 
Way to let me down gently 🤒

However I will still continue to redesign, find new materials for max efficiency to capture and use solar.

Edited by Joshcitylife
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