Emissivity

[Joe Dragon]

*******************

 

[homebrewer_99]

Wouldn't they also burn the contents faster?

Just a question...I'm not a Thermographer, but I have taken pictures of the sun...(and I'm blind in one eye...see... )

 

[david_42]

Although I understand what your teacher is saying, I suspect convection dominates the heat transfer. If it is windy, I'll wrap a shield about the burners & pot. Makes a big difference. Whereas, if transmission/reflection were the major considerations, it wouldn't.

However, a scorched outside will make some difference.

 

[david_42]

Run a controlled experiment & report back.

 

[Lil' Sparky]

I think I'll try to burn some soot onto my keggle to see if it helps.

Any paint will burn right off. I don't know what the temp of the bottom and around the skirt is, but it's smokin' hot!

 

[Nexus555]

Well this is obvious enough, but make sure there is relativity the same amount of heat being applied. If you have a pressure gauge of some sort, that could help to find more of an accurate assessment.

 

[casebrew]

Motorcycle cylinders are usually painted black. So are car radiators.

Transmission/reflectivity of which wavelengths? We don't care about the visible spectrum. It's hot gasses giving off heat... IR ?

Who wants to make beer the tastes like burnt paint?

Time for Googling...

 

[brewman !]

Look up "black body radiator".

You should change the title of your post to brewpot efficiency or something like that.

Here is what the alpine people use to increase the efficiency of their pots when cooking over a flame.
http://www.mec.ca/Products/product_...older_id=2534374302696309&bmUID=1175550386688

I've also seen a lot of tinfoil used to keep the wind out.

 

[casebrew]

Heat moves via three methods. Conductivity, directly form one substance to the other. Convection, due to currents in the fluid. And Radiation, emission/emissability is how it is measured. Thermography being the measure of radiated heat, via emissibility.

Propane heats our kettles via conduction. The Wort picks up the heat from the kettle via conduction as well as convection. The kettle gives off heat via Emission. SO, a black kettle would give off heat faster. Like a radiator, or the cooling fins on a black motorcycle cylinder. As far as heating more efficiently due to black coatings, it would depend on how much of propane's heat is via radiation, and how much you would lose to the air via outward radiation. The .04 vs .765 is a ratio of 19 times. I suspect that improving the radative perfomance by 19 times won't mean squat if propane heat is transmitted via conduction at a rate 600 times (a number I picked out of my tu-tu) more than via radiation.

Of course, when you are in a high performance positon where weight and size have limitations, black motorcycle parts could be important. A couple percent could be important to a racer. But notice that they do have fins, to increase conductive area. Plus airflow, to increase convection. NOT a smooth cylinder painted black? I'd say the same reasons apply to heating a kettle- heat transfer via radiation is minor compared to conduction and convection.

Electric heat may be different, the red element radiates a real storm. But is also close enough to the pot to cause convection and conduction to the air, then conduction/convection air-to-pot.

Hmmm. google more.

A litle more googling points out that a "reflector" would be heated by conduction, but 'radiate' heat back at the kettle.

A little more googling tells me that 10-15% of a gas flame's heat is radiative. (Apparently, from the CO2 and H2O molecules being heated to incandesence.)
But since radiation goes off in all directions, only a small part of the 10-15% is aimed at the kettle. Let's say 1/3? So, different surfaces could collect the more heat at: 1/3 of 15% is 5%, times .o4,(if polished) give .2% of your propanes heat radiated to the usual pot. Or 1/3 of 15%= 5%, times .765(black anodised pot) = 3.8% of your fires heat. But if you add a reflector to your 'stove', you could double that, and gain 7% more efficiency . You could anodise a kettle, build a reflector, and save 10 cents per batch... ooops, I forgot to allow for the heat radiated away from your black kettle.... more surface area away from the flame than towards it... no gain? ooo, bigger reflector? noooo, wait, the wort won't be heated to incandesence, so not much radiation?.... ah, I've done enough larnin' and thinking for 10 cents/batch....

But, then, if you built a solar 100% radiation) brew pot....

 

[Jester4176]

Heat moves via three methods. Conductivity, directly form one substance to the other. Convection, due to currents in the fluid. And Radiation, emission/emissability is how it is measured. Thermography being the measure of radiated heat, via emissibility.

Propane heats our kettles via conduction. The Wort picks up the heat from the kettle via conduction as well as convection. The kettle gives off heat via Emission. SO, a black kettle would give off heat faster. Like a radiator, or the cooling fins on a black motorcycle cylinder. As far as heating more efficiently due to black coatings, it would depend on how much of propane's heat is via radiation, and how much you would lose to the air via outward radiation. The .04 vs .765 is a ratio of 19 times. I suspect that improving the radative perfomance by 19 times won't mean squat if propane heat is transmitted via conduction at a rate 600 times (a number I picked out of my tu-tu) more than via radiation.

Of course, when you are in a high performance positon where weight and size have limitations, black motorcycle parts could be important. A couple percent could be important to a racer. But notice that they do have fins, to increase conductive area. Plus airflow, to increase convection. NOT a smooth cylinder painted black? I'd say the same reasons apply to heating a kettle- heat transfer via radiation is minor compared to conduction and convection.

Electric heat may be different, the red element radiates a real storm. But is also close enough to the pot to cause convection and conduction to the air, then conduction/convection air-to-pot.

Hmmm. google more.

A litle more googling points out that a "reflector" would be heated by conduction, but 'radiate' heat back at the kettle.

A little more googling tells me that 10-15% of a gas flame's heat is radiative. (Apparently, from the CO2 and H2O molecules being heated to incandesence.)
But since radiation goes off in all directions, only a small part of the 10-15% is aimed at the kettle. Let's say 1/3? So, different surfaces could collect the more heat at: 1/3 of 15% is 5%, times .o4,(if polished) give .2% of your propanes heat radiated to the usual pot. Or 1/3 of 15%= 5%, times .765(black anodised pot) = 3.8% of your fires heat. But if you add a reflector to your 'stove', you could double that, and gain 7% more efficiency . You could anodise a kettle, build a reflector, and save 10 cents per batch... ooops, I forgot to allow for the heat radiated away from your black kettle.... more surface area away from the flame than towards it... no gain? ooo, bigger reflector? noooo, wait, the wort won't be heated to incandesence, so not much radiation?.... ah, I've done enough larnin' and thinking for 10 cents/batch....

But, then, if you built a solar 100% radiation) brew pot....

So you're saying that the absorbative net loss of the conductive reflection relative to the convection within the pot would be null?

I have another solution. Joe can report back with his findings, or I'll offer to do the esperiment. I have a double burner stand and 2 keggles. I can adjust the flame on each burner to have the exact same output. Then, I can put one keggle on each burner(after painting one on the bottom), and fill each with 5 gallons of water(premeasured). Then, set a timer on each in a semi closed garage(door cracked with fan blowing noxious CO vapors out) to negate the effect of wind on the burners, and see which boils first. What say the board?

 

[Chillbrook]

So you're saying that the absorbative net loss of the conductive reflection relative to the convection within the pot would be null?

I have another solution. Joe can report back with his findings, or I'll offer to do the esperiment. I have a double burner stand and 2 keggles. I can adjust the flame on each burner to have the exact same output. Then, I can put one keggle on each burner(after painting one on the bottom), and fill each with 5 gallons of water(premeasured). Then, set a timer on each in a semi closed garage(door cracked with fan blowing noxious CO vapors out) to negate the effect of wind on the burners, and see which boils first. What say the board?

Might sound stupid, but maybe taste the water after it's been boiled/cooled?

I'd love to see the results.

 

[Lil' Sparky]

Jester - I think it's a great experiment. If there's any "real" difference, then it would show up in this test. The hard part will be ensuring that the burners are really putting out equal heat. Honestly though, I doubt the time to boil will be substantially different.

 

[casebrew]

So you're saying that the absorbative net loss of the conductive reflection relative to the convection within the pot would be null?

I have another solution. Joe can report back with his findings, or I'll offer to do the esperiment. I have a double burner stand and 2 keggles. I can adjust the flame on each burner to have the exact same output. Then, I can put one keggle on each burner(after painting one on the bottom), and fill each with 5 gallons of water(premeasured). Then, set a timer on each in a semi closed garage(door cracked with fan blowing noxious CO vapors out) to negate the effect of wind on the burners, and see which boils first. What say the board?

Go for it. Can you swap the pots after a guesstimated half the time? That would moderate the differences between burners.

And, a good test would be blinded- have somebody else paint the pot, and msomebody who doesn't know the test theory actully do it- without looking at the bottoms. Not to be expected here, but try not to steer the results.

But, if the difference were really 19 times, don't you think all pots would be black from the factories? Especially today, when gas companies are buying light bulbs and paying electricians to change them, just to save a couple watts? Sempra isn't buying me new cookware, and I'm not buying a 19x improvement...

Lessee, 5 gallons, big burners, about 15-20 minutes? I'll place my money on one minute's difference, not enough to make up the time spent cleaning up flakes of paint.

 

[Jester4176]

Ok, I'll pick up some paint today, and try to get this done this week and report back. I'll even taste the water afterwards. Don't think that the flavor will be any different because the paint smell won't permeate through steel.

 

[casebrew]

Bump.

Anybody try this grand experiment this weekend, or did Easter resurect other duties?

 

[Jester4176]

Easter interviened on my part. Have the keggle painted, so going to try early this week.