Solar Heat

Are you really an engineer?

why the question? Is there anything wrong with my question?

Isn't the engineer the guy that wrecks train?

Are you one, or did you graduate from Wiki, Google and a garage?

I am sure you do not know the answer but to show your omniscience you make these disparaging comments. It shows only your ignorance and contempt to all other opinions and queries. I also notice that you are generally the first to come up with your puerile diatribes.

To learn one needs humility first.

I see you haven't met lyn then?

1. It would depend on the latitude, altitude, time of day, ambient temperature, angle of incidence, and how black "black" is....

2. It would further depend on how thick and what type the glass is, and if it is say a Fresnel lens.

1. It will be good of you if you could show the formula and method, use variable for sun irradiation power , ambient tempt,etc.

2. I meant normal glass,assume the most normal condition, eg 5~6 mm glass

I don't think there is any such thing as "the formula" for this. Maybe a bunch of formulas, each with limiting conditions that may be hard to decide which are important. As Mikerho suggests, this is probably better measured than calculated.

It is always better to have a first cut rough estimation/calculation first before you fab a real set up.

It is always better to have a first cut rough estimation/calculation first before you fab a real set up.

However, if you don't know where to start, it can be a real time-waster.

However, if you don't know where to start, it can be a real time-waster.

You don't where to start, that's why you ask.

Sometimes, it's better and quicker to do a quick test...

90 degrees C if you want a figure, but there are so many variables which will be cooling it, distance between the glass and the surface, wind etc.
Maybe if you tell us the real application we can point you in the right direction.
No calculations were harmed in the process of making these... solar hot water panels
Del

Hi Codey,

Your answer provides the closest I was hoping. GA for you. Thanks !

My experience of leaving a metal object under direct sun will be too hot to touch, so, your estimation of 91 Deg C is not too far.

Thanks bravo88.

I should have written 3 watt*m^-2*ΔT ^-1.25, but I'm sure you know what I meant.

91 C is just a few of your centigrade celsius Fancy degrees below boiling point of water, so I say your estimate is congruent with your experience- one can be scalded in a second at temps of approximately 68 C; see graph in primitive US Fahrenheit degrees here: http://www.accuratebuilding.com/services/legal/charts/hot_water_burn_scalding_graph.html

Yours and Dels were good answers, Though I would add to the original poster that the idea of steady state is a false notion as the insolation changes in cloudy conditions as well as over the diurnal cycle.

Finally, Insulation in the collector and conductivity of materials used for enclosure (metal vs Glass) play a major role in minimizing losses of heat from the system.

Here is an article I wrote back in 1977 about my first efforts at constructing a solar collector:

http://www.motherearthnews.com/Do-It-Yourself/1977-11-01/Recycle-a-Refrigerator-into-a-Solar-Water-Heater.aspx

No where near as sexy as Del's, but it was a "hands on them thats doin" project back in the 1970's.

Unless you put in a means to stop it, (check valve) or covers over plates, these will reverse and thermosiphon overnight, reradiating all that heat back into space.

Milo

You are going to have to make more assumptions to reasonably manage the math and model the problem.

Do you know that there are three primary mechanisms of heat transfer? What are they? What is the main mechanism in your scenario?

The easiest way by far, would to obtain the data empirically by setting up a series of experiments.

The answers to both 1. and 2. are: "It depends." _{(on a lot of things)}

So far, this thread is a self-fulfilling prophecy.

Just get a blackboard, a window pane, a thermometer, and a sunny day--and measure.

A global experiment! Have certain subscribed persons from all over the globe submit data to a dedicated server.

Data to be recorded:

• Latitude
• Longitude
• Time of day
• Ambient temperature
• Local atmospheric pressure
• Ambient relative humidity
• Current local wind speed.
• Degree of Atmospheric Clarity (OK, I made that one up, but I think it's important).
• Make and model of recording instruments.
• Name and ID number of the technician(s) in charge of instruments and data recording.

We can then analyze the data and get plots that look like target practice silhouettes having been shot with a 12-gauge shotgun using ellipsoid pure lead shot from 30 feet away.

You will need to supply variables in order to get a solution. I thought of at least 15 in the last 60 seconds. So, although I don't want to further disrespect you, your question does not seem to be very well engineered.

To suggest your solution, you need not have the fixed values for the variables, just use variables in your method or formulae, and also you can always make real world assumptions. What is important is the method , not the exact numerical answer.

try here www.seris.nus.edu.sg

Thanks, good link, right at my home ground ! And do you believe it, it is just about 1 km away from my previous office last year !

Solar collectors use a vacuum, a special absorbent but not reradiating coating on the heat pipe tube, which looks black. I have personally seen more than 270 F in a synthetic oil system used to drive an ammonia refrigerating system for an off the grid house.

If you can get 270 F in solar collector, that's is very impressive and useful. Does it use a concentrating mirror or focus lens or Fresnel lens ?

Do you have more info about the set up you have seen ?

We frequently achieve 350 degrees F in solar collectors, if we fail to provide coolant flow. What is the steady state temperature of the black surface of the collector? It is only steady state when there are no variables.

Hi PFR,

The solar collector you used, does it use parabolic mirror or concentrating lens, or is it just plain black surface.

By the way, in my earlier thread , by the word variable, I meant mathematical symbols.

Hi bioramani,

Double GA for you!

It uses heat pipe technology, with mechanical heat transfer from copper to coolant. We use evacuated tube collectors, not because they achieve higher efficiency at converting solar insolation to high temperatures, versus some other technologies, but because the usefulness of the heat energy is the primary design goal. As was very accurately pointed out in the magnifying glass example, energy density, in addition to energy utilization, are much more important factors in designing a useful solar collector. Here is a key. Removing energy from a collector impacts all of the functions inherent in it's usefulness as an energy plant.

I point this out, because the it appears that the answers to your question about the basic solar properties of an absorbent material are being interpreted as criteria for solar design. The replies have been very good, and I thank all of you who contributed.

Hi Bravo,

The calculations are all here in other posts, but as a theoretical exercise, if your object is completely thermally insulated (i.e. radiation can get in, no energy can get out) then exposed to solar radiation it can reach the surface temperature of the sun (about 5700K). This is your theoretical maximum.

Obviously it won't get there - you will radiate heat more and more quickly away from the plate as the plate itself heats up. The steady state is simply the point where what goes in balances what comes out - and this depends on all sorts of things - what the object is made of, whether it's in a vacuum, what conduction / convection channels are available to it etc.

Concentrating thermal technolocgy works by concentrating the sun's radiation onto a small area collector to minimise heat loss (i.e. there is not much area to shed the heat away) - so lots goes in compared to the amount that leaves. They don't get to the theoretical maximum, but a few thousand degrees is not unheard of.

Hi RobertOz,

Many thanks for your input.

My question 2, if I use a glass box to house the solar collector, how much more effective it is.

If a method can be found to collect more sun energy without using parabolic mirror or lens, it will be very useful.

GA. RabertzOz. But I feel glass box with air may be better compared to vacuum because glass box is acting like a green house, trapping & reflecting back the infrared radiation from plate which will be converted back to heat. Air inside the glass box will hold the heat and transfer to plate by conduction and convection. Please clarify, if I am wrong.

and GA to your post (above)!

Hmmm... it's going to come down to the calculations... but I tend to fall on the side of wanting to remove the convection currents by making a vacuum. I'd also tend to worry that heating the air heats the glass - then the glass begins losing large amounts of heat to the atmosphere on the other side (hence the double glazing argument).

It might be a case where there is a limiting temperature - below Tlim the radiation transfer is low and the convection is the main problem, but above Tlim and the greenhouse effect kicks in and limits the effects of radiation and you're better off with the air... I'm not sure - but nice logic!

By the way - these things are often non-intuitive - for instance as you add lagging (insulation) to a pipe you effectively increase the surface area. In certain (very specific) cases the increase in radiation and convection heat loss caused by adding the insulation can more than offset the thermal resistance of the insulation... so a pipe losing 100W per metre length uninsulated may lose 110W per metre length once insulated. Wouldn't that leave you scratching your head??

An interesting point. This may be theoretically possible, but I suspect only if the insulating material is a poor insulator. (I haven't tried any sample numbers yet.)

I remember running the numbers once... can't remember the exact results, but it turned out to only come into play for quite small diameters... so in microfluidics or chemistry-lab stuff, or heat loss along electric cables the effect was quite real - even for good insulators... if you plotted heat loss against insulation thickness it went up for a while before dropping back down after some critical thickness was reached. Very non-intuitive...

Hi Bravo,

Just re-reading your question 2... imagine you have a magnifying glass. You focus the sun onto a piece of paper and in one second it starts to burn (i.e. you have temperature in excess of Farenheit 451). Now imagine the magnifying glass has a 50x magnification - i.e. it concentrates some area of sunlight onto an area 50 times smaller. This smaller bit is the bit that burns - let's say that is one square millimetre.

If you leave the piece of wood in the sun, that same square millimetre will absorb exactly the same amount of energy in 50 seconds as it did in 1 second under the magnifier. The only reason it doesn't burn without the magnifying glass is because the energy has had longer to dissipate back out of the wood (or deeper into it).

So you're not gaining anything by concentrating except higher temperatures - you don't get more energy, you just put it into a smaller area. This is very useful if you want to boil the water to create steam, but if you just want to use the water to heat your domestic hot water, you a far better off with a good flat plate, because the lower temperatures allow you to put more of the available energy into your water rather than risk radiating it back into space, which you do as soon as you move to extreme temperatures.

So for your black plate example, if you want to collect the energy for domestic purposes, you actually want lower temperatures not higher.

If you want to boil water to make electricity, you really have to use a concentrator in order to reach the temperatures you need... but in doing so you will have almost certainly reduced your efficiency while increasing your temperature, by vastly increasing the outward radiation.

So for pure efficiency, go for a glass-covered matte black plate (or a thermo-siphon) and keep your temperatures near to the required end-user temperature (say 60C for domestic hot water).

a

ok - then my bet would be you want to find a surface that is well-designed for accepting solar radiation but well-designed to not radiate it out again, mount it on a very good insulator and stick it in a glass box, surrounded by a second glass box. Evacuate the space between the two glass boxes (i.e. put your plate in a double-glazed glass box). I reckon that is about the highest temperature you could get with low-tech apparatus... let us know when you get some measurements!

Yes, that is also my conclusion , much appreciate your input !

Problem with glass cover: About 4% of incident energy is reflected back before it gets to the solar array. Add another sheet of glass and there is a further loss of 4% of 96%.

good point, but it is worth it if it prevent more loss from conduction.

Exactly my point, though I didn't elaborate as you have done in more detail.

GA!

Hot enough to cook burgers!

http://mealwormfarm.com/hqsolar/hqsolargrillonline.html

In addition to the amount of incoming radiation from the sun (which varies) the steady state temperature will depend on the properties of the object as well as its surroundings. You have not provided sufficient information allow a definitive answer. However there have already been given answers that represent practical situations that may resemble yours.

In the branch of physics referred to as thermodynamics. In the basic university courses on that subject there is introduced the concept of the control volume, whose utility for calculation purposes derives from the fundamental idea of conservation of energy. This forms the foundation theory for solution of your problem. The idea is that your 3 dimensional object with its black flat surface comprises a body whose average temperature at the black surface exposed to solar energy constitutes a control volume.

I think we can presume your interest is in the maximum temperature that volume of material will reach under conditions of solar energy input. And we'll assume that condition is reached when the heat radiated out of the body is the same as the solar energy absorbed. And steady state means that the body has a constant temperature and therefore a zero change in heat content

So we are looking at a "control volume" holding a constant amount thermal energy at some temperature which permits the solar heat into the volume (total solar heat radiation exposure minus amount of radiation reflected back into space.) to equal the sum of all heat that leaves the volume by radiation, conduction and convection. This so called "heat transfer" is usually dealt with as rates of heat flow of which there are numerous calculation methodologies available from simple to complex.

You have methods of determining the solar energy flux. What you don't know yet is how much of that will be reflected back from the black surface. It may be more than you think.

Also what you don't know is how much heat is radiated from the total surface including the black part that faces the sun. In addition there will be some heat conducted to the body's mounting points and to any air in contact or heat transfer media. If there is air flowing over the body a third path of heat transfer will be by convection. At steady state the sum of these three will equal the net amount of solar flux absorbed by the black surface.

If your application is some kind of solar panel then you'll find plenty of info both here and on the rest of the internet as well as the many books on DIY solar power. If it is something new and creative you may have to get into some basic texts on heat transfer for the simple equations covering conduction and radiation and data on related properties of various materials. If you have to get into convection the math gets more complicated due to the complexity of fluid flow but there are lots of simple approximations as well as practical test methods once you get hardware together.

You can learn a lot from folks who have experimented with and built solar devices and a little bit of basic engineering will help you filter and adapt the info you receive to fit your own purposes.

Some here in this forum may want to suggest specific links or books to aid you. Not that heat transfer calculations though generally simple can be a real swamp when it comes to converting units, metric to English and orders of magnitude. You'll need to spend a lot of time rechecking your calculations to make sure you've got them right until you get real familiar with the drill. The nice thing about working in this area of engineering is that most of the time you are dealing with temperature measurement and the hardware for that is cheap and easy to get especially with the advent of DVM's with thermocouple readers and inexpensive noncontact IR thermometers.

Ed Weldon

When I am faced with an intractable problem like this, I find it helpful to just make a simple model and take measurments. All variables of construction and material and initial and ambient conditions will resolve themselves into the measurement I make of my model. Then, I change a material or add one etc. and measure again.
Sometimes a working model is better than a theory or a formula. It may even be faster and sometimes a whole lot more accurate because all of the variables are present, including the ones you couldn't know or account for. Also, the materials and conditions are easily changed and remeasured.
My own tests have shown that an aluminum plate with High-temp flat-black paint, in a box with Argon or Nitrogen or both will get pretty hot. But the same plate in a box with saline at saturation under glass will get even hotter, but take longer to heat up. Both tests assume insulation of the box and the plate is suspended within the space of the box.
A multi-layer saline pond will accumulate an enormous amount of heat, and is one of the better ideas for solar collecting.

Ref your test, do you mean the plate is immersed in saline solution contained in the box ? Or do you mean there is a sandwitched layer of saline solution within the glass?

The plate and the glass have saline in between. Got to coat the aluminum very well or anodize it to do this though...

I would set up an experiment using some different types of models ,one glass box one plastic box different shades of black flat surfaces etc. and at each one place a garden thermometer.and as far as getting the math right how about using Prism Math Maybe GM or some one of the big auto makers has some data on this hot box question.ds.

I was convinced collecting energy was about reflecting sun and black,the color Absorbs heat and makes a glass and metal enclosed auto hotter, but not sure heat is always energy.I wonder if you could use glass that is a partial mirror/ reflective and then using a black background behind it and maybe more mirror behind that like a prism or wall of mirrors . Some antique mirrors look as though they have a lead backing.
Maybe while youre gathering the data you could find out what panels send out cooling shade cover,wouldnt that would be great for this heat wave or solar storm or whatever it is.Nothing is impossible.

The saline does something like that with the mirrors. It has a very high refractive index that prevents the waveform from coming back out at the same angle it went in, so it becomes trapped! Heat builds up considerably, and from the right angle, the saline looks black to the eye because there is so little return/reflection from that direction. Layers of different saturated salines deposited one on top of the other create a very dark bottom...

Have a look at some of Gaiatech's threads, he covers much of this ground

I've run into a situation similar to the saline immersion design with double glazed insulated windows where the outer glass was impact glass with the lexan sandwiched/bonded to two layers of clear glass and the inside glass being Low-E clear glazing with the space being Argon filled.

The outer impact glass let the full spectrum in minus UV which the lexan layer filtered out and the Low-E inner glass reflected the heat but it became trapped between the layers and heated up the whole window to about double outside ambient temperature even though the HVAC inside the house was set at 70F.

Please stand by. I am trying to get my black cat flat enough to fit in between the storm windows.It was tough enough getting the thermometer to stay where I placed it.

ooohhh

Del's gonna be mad

Yes Dear,

You are lucky to get very good and accurate reply to your first question, specially by Code master GURU.

Your second question: If the object is put in a glass box, definitely the heat loses are reduced and steady temperature is ~120 degree cels. All the boundary conditions remains the same as for your first question.

Again if you put 2 glass covers on top and insulate the object on the sides and bottom, in this case steady temperature is ~140 degree cels. Further if you keep on increasing the no of glass covers on top, it will result in lowering the temp because the input radiation will be reduced.

LUV-nps