Need for some help before building solar furnace (kiln)

[Power User]

Hi buddies

I’m goanna to build a solar furnace for aluminum melting. I planto build this to recycle aluminum and before I start to build it I want to makesure that this will be work.

I plan to build a parabolic mirror with 2 meter radius. The problemis that I cannot make sure whether my calculations are correct or not! It isnot supposed to be such good results!?

Here is what I done:

Aluminum Mass = 10 (Kg)
Aluminum Heat Capacity = 0.897 (J g-1 K-1)
Aluminum Melting Point = 933 (K)

So, amount of heat we need to melt 10(kg) Aluminum is:
 
(10 ×1000)gr     ×0.897 ×933=8369   (KJ)                              
 
So, we need 8369 Kilojoule to melt 10 (kg) of Aluminum.

And according to: http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/Mathematical_Thinking/sun12.htm
energy from sun can be calculated from equation bellow:

Energy in (mill joules)                     =>           mJ = (mW/cm2) × (Aria incm2) × (Time in sec)
Mirror Area = 125663.70 = 120000
Solar Constant = 137 (mW/cm2)
So, amount of energy will be received to mirror is:
137   ×120000 ×60 = 986400000 mJ=   986 KJ   per Min 
987 Kilojoules in a minute.
If mirror works with 50% efficiency we will get 493 Kilojoules.
And if we lose 50% of heat in melting process we have 246 Kilojoules. 

I guess something is wrong but I don’t know what …
Because when we do that with fuel powered kiln it takes up to 40 min and maybemore.
I tried to calculate as good as I could (as I known for mybad calculation and also a bad English writer).
I could not found any other way to perform this calculation.I ask for:

- Is there any equation to calculate amount ofsun power in the focal point of a parabolic mirror?
- What about my calculation?
- And a technical question: is it better to make it in Fresnel mirror insteadof parabolic shape mirror as it simple to both build and maintain? Which onewill operate better?

[Mike90250]

build a large enough mirror, and i can melt the world !

The issue will be your heat receiver chamber, and it containing all the heat, and not re-radiating it away.

The Solar Death Ray cooker does dogs in a couple seconds, so I believe the heat is there.

You may need a quartz window to allow the heat in, and keep the air from convecting it away. That will be expensive.

The supports for the heat collector chamber will shade the mirror, and reduce the flux somewhat.



1502_1_600.jpeg5d6f12fb-e648-4f92-b9a9-486ff45d41da.large-profile.jpg

An array of heliostats may be a better choice for capturing more power without the crucible support blocking your mirror.

http://en.wikipedia.org/wiki/Solar_power_tower &

http://en.wikipedia.org/wiki/Solar_furnace

See how others have done it, before trying a different method.

[J.P.M.]

Hi buddies

I’m goanna to build a solar furnace for aluminum melting. I planto build this to recycle aluminum and before I start to build it I want to makesure that this will be work.

I plan to build a parabolic mirror with 2 meter radius. The problemis that I cannot make sure whether my calculations are correct or not! It isnot supposed to be such good results!?

Here is what I done:

Aluminum Mass = 10 (Kg)
Aluminum Heat Capacity = 0.897 (J g-1 K-1)
Aluminum Melting Point = 933 (K)

So, amount of heat we need to melt 10(kg) Aluminum is:
 
(10 ×1000)gr     ×0.897 ×933=8369   (KJ)                              
 
So, we need 8369 Kilojoule to melt 10 (kg) of Aluminum.

And according to: http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/Mathematical_Thinking/sun12.htm
energy from sun can be calculated from equation bellow:

Energy in (mill joules)                     =>           mJ = (mW/cm2) × (Aria incm2) × (Time in sec)
Mirror Area = 125663.70 = 120000
Solar Constant = 137 (mW/cm2)
So, amount of energy will be received to mirror is:
137   ×120000 ×60 = 986400000 mJ=   986 KJ   per Min 
987 Kilojoules in a minute.
If mirror works with 50% efficiency we will get 493 Kilojoules.
And if we lose 50% of heat in melting process we have 246 Kilojoules. 

I guess something is wrong but I don’t know what …
Because when we do that with fuel powered kiln it takes up to 40 min and maybemore.
I tried to calculate as good as I could (as I known for mybad calculation and also a bad English writer).
I could not found any other way to perform this calculation.I ask for:

- Is there any equation to calculate amount ofsun power in the focal point of a parabolic mirror?
- What about my calculation?
- And a technical question: is it better to make it in Fresnel mirror insteadof parabolic shape mirror as it simple to both build and maintain? Which onewill operate better?

A quick look/thoughts :

Your guesses about something being wrong and your questions can be best addressed by reading a textbook about solar energy. That's the start of the solution path. Proper answers are longer than space here permits. You written English is fine.

1.) Safety first : If you have not already done so, BEFORE YOU DO ANYTHING ELSE, BUY AND USE EYE PROTECTION. I'd get a welder's hood.

2.) Your calculations are using the solar constant of 1,376 W/m^2. That's above the earth's atmosphere. That's incorrect.

3.) What you want to use for input is what's called the " Direct Normal" component of the solar irradiance at the earth's surface. That varies throughout the day and under a clear sky at sea level is closer to something like 850 to 900 W/m^2 max at solar noon on a clear day, and less at other times.

4.) For a "tight" focal spot you will need an accurate parabolic mirror. Such surfaces cost money. Home or lab made foil surfaced plastic dishes usually give more scatter than desired or workable. If your melting aluminum, sounds like sort of an industrial app.

5.) You may guess 50% eff. and that may be close, or be off by an order of magnitude. Get with a textbook on heat transfer and look for correlations that estimate convection and radiation losses.

6.) Depending on the quality of the dish and its ability to produce a "tight" focal spot, a "cavity" type receiver may be a workable solution to the heat loss problem. See the literature. Its a box with a relatively small hole where the concentrated light goes in. It's a cheater method to imperfectly imitate perfect black body absorption, but, depending on particulars, effectively less than perfect black body emission (that's good). If the outside of the box is insulated (and the insulating material is VERY high temp. rated), and the inside is reflective, so much the better. See the literature. Lots/most of details are left out here.

7.) Don't forget wind loads. Solar concentrators usually have some of the same characteristics as sails.

8.) Do a literature search before building. You'll learn as much or more than not doing one. You'll find that what you want to do has been done by many others long before you. Read and learn from their mistakes. It's cheaper, faster and safer. Besides, you can't live long enough to make all the same mistakes yourself.

9.) After you learn from those who went before, start with a small version of what you want and scale it up. It's also cheaper, faster and safer.

10.) Check out a site called "builditsolar.com" Lots of useful, practical stuff in there. Maybe some about concentrators.

To repeat: THINK SAFETY. WHAT YOU ARE WORKING WITH CAN BE HARMFUL !

[Power User]

Thank you for replies

I do some searches before posting this topic, and surprisingly I only found that equation I used to calculation in post 1. I also read a book about this topic [solar radiation energy and industrial application] but there is not any way to calculate energy for such case.

I also watch any video on YouTube that related to this. The problem with what I found from what other people done (As Mike90250 said See how others have done it) is that most of them do this as hubby and therefor they do not provide any information that could be used for calculation.

I don’t want this for hobby or making fun. I want to use it to reduce costs and increase speed, so it is important for me to know what setup meets my demands and prevent from spending much more money. And I should say that is not just about money if I cannot control amount of heat it may damage kiln or crucible.

The issue will be your heat receiver chamber ...

yeah, this would be a major problem ! and i sould found some solution for that ...

…For a "tight" focal spot you will need an accurate parabolic mirror. Such surfaces cost money. Home or lab made foil surfaced plastic dishes usually give more scatter than desired or workable. If your melting aluminum, sounds like sort of an industrial app.

I think in this case I need good but not too tight focus because as I said it may damage kiln or crucible. (In some videos I watched, they used a lens to concentrate rays that received from mirror and melt steel) we cannot concentrate directly on Aluminum.

….. You may guess 50% eff. and that may be close, or be off by an order of magnitude. Get with a textbook on heat transfer and look for correlations that estimate convection and radiation losses.

Actually, I use that estimation to show even with such large amount of lost energy (heat) , solar powered kiln melt aluminum so fast and this is Suspicious … in practice we isolate kiln to avoid any leak.

… a "cavity" type receiver may be a workable solution to the heat loss problem.

This sounds good. I read some about that and I kept it in mind in redesigning kiln.

…. Don't forget wind loads.

I think about this. As I asked in my first post about choosing between parabolic and Fresnel mirror. Maybe I am wrong with the terms (Fresnel mirror) but I meant something like this one:

fm.jpg

I think this design would be able to handle wind load much better than parabolic shaped mirror due to the space between mirrors and also by theory, its performance will be same as the parabolic mirror if configured well.

After you learn from those who went before, start with a small version of what you want and scale it up. It's also cheaper, faster and safer.

Well done and thank you J.P.M.
Modelling, I forgot this.
I’ll make a model and experiment with that …

[FunGas]

A 2m diameter parabolic reflector will reflect a lot of energy.

I made a thing like this with 36 1" mirrors in London which would melt solder in a budgie bell (just) in summer.

Containing the heat you collect is very difficult as someone already pointed out. Also, the angle that the energy enters the crucible is not always suitable for molten metals.