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olddawgsrule
Advanced Member


USA
1434 Posts

Posted - December 16 2013 :  16:43:55  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
When determining BTU of a heater..

Can I use the high / low temp, even though it may be of days, as a hourly Btu?

Being that the heater is holding or reaching a given temp, with a particular Heat Loss, and it's holding..

Would that not mean that at this point in time, it's producing "X" Btu's per hour?

I'm wondering if the over-all time is more important than what it's doing right now?

It seems to me that even though it may take 2 days to get there, it holds at a temp, and if it holds at a temp, then that's what it really produces each hour.

Still curious!



Old in age, not in mind, so
'Teach me something new'!

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49er
Administrator



USA
4443 Posts

Posted - December 16 2013 :  17:28:48  Show Profile Send 49er a Private Message  Reply with Quote
Hi ODR

I think in out cases it is the candle along we need to take a reading with on a gallon of water or 1 cubic foot of water and the see how long to make it rise 1 F and I think that can be figured out to BTU per hour. So its the size of the candle that is important not what your distributing the heat with per say. After that you can see if it is better with 1/2/3 or 10 LOL

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
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49er
Administrator



USA
4443 Posts

Posted - December 16 2013 :  17:34:26  Show Profile Send 49er a Private Message  Reply with Quote
Hi ODR

Just found this and it should be easy to figure out something from it.

You must be logged in to see this link.

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.
SKYPE doug.bennett49er
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 17 2013 :  18:13:50  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
I've been through that system of figuring, yet it's air we're dealing with.

2 schools of thought I have found so far:
First is it takes .02 Btu's to rise 1 cubic foot of air 1 degreeF
Second is it takes .03..

Conciseness seems to be the .02 so far.

In reality, we're dealing with air (room temp), so I've been following this way of measuring.

Question is; Are you 'all' comfortable with this?





Old in age, not in mind, so
'Teach me something new'!


Edited by - olddawgsrule on December 17 2013 19:27:56
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 17 2013 :  18:58:26  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Guys
Personally i think you should use the heating of water idea of Doug's.
Obviously the container must be well insulated to prevent heat loss.

ron
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 17 2013 :  19:26:30  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
quote:
Originally posted by ron_o

Hi Guys
Personally i think you should use the heating of water idea of Doug's.
Obviously the container must be well insulated to prevent heat loss.

ron



The issue then is simply how??

The current configuration of my heater will not allow for a container to sit on it.
Greatest heat comes from the fins.
If not in contact with the heater, then I don't see a good reading..

That's the main reason I headed towards BTU per cubic foot of air..

Has to be a simple way of doing this I'm not thinking of...

If I'm dealing with a flame (basic design, not current), then I can not enclose the heater.
Design will not allow for contact..
??

Options??








Old in age, not in mind, so
'Teach me something new'!

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kcarring
Moderator



Canada
1057 Posts

Posted - December 17 2013 :  19:55:39  Show Profile Send kcarring a Private Message  Reply with Quote
I definitely agree on the use of water in a calorimetric sort of setup. I also agree with 49er, its the candle, and nothing more that will set the BTU's.

I think with the candle heaters, one consideration might be, that a candle alone in absolutely still air, will essentially heat a relatively small diameter column of air, vertically, so it's room heating capability will be greatly diminished. I think radiating that heat outward sideways will improve the perception of heated area.

I remember years ago I had a very very cold apartment in the north just below Alaska, and I took the 40 watt incandescent light bulb out of my ceiling and replaced it with a 100 watt bulb and made sort of a heat radiating lampshade of louvered metal. Just having that lamp on the ground instead of way up in the ceiling, and adding the radiator made a big difference. I had no control of my main heat source, at all, so I'd leave that running in the day and it made a noticeable difference indeed.

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!

Edited by - kcarring on December 17 2013 19:56:31
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 17 2013 :  20:01:42  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

You would need an insulated chamber ( or room ) and a method to circulate the air within that chamber. Chamber needs to contain sufficient air ( oxygen ) so that the burn is uniform
Measure start temp and end temp then temperature gained = end temp - start temp
To make things easy run the experiment for one hour ( or measure a one hour periods heat gain )

It's rather a crude method but it would give an indication

ron

Edited by - ron_o on December 17 2013 20:03:23
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 18 2013 :  16:57:42  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
To KC, I agree the the water test is probably the best, just don't see how right now..
So I've gone to HVAC standards for 1 cubic foot of air to Btu.
These range .02 to .034 depending on moisture content..

I've also never disagreed that the candle is what it is.. 50 Btu (give or take)
I've done a calorimetric value on my candle (Doug has this on the spreadsheet I sent) and it comes close to the 50 Btu.

What I really get a kick out of is your statement of using a electric bulb.

The unstated experiment (until we agree on how to determine) is using a IR light bulb.

To all: But back to the main question..

Even though it took days to get to temp, and it's mantaining temp, would that not mean Btu per hour?



Old in age, not in mind, so
'Teach me something new'!

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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 18 2013 :  17:20:34  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
I'm not sure how crude this may be since there are HVAC Standards to go by, yet I will still be cautious of any results till we agree on method.

What I've done is used my Bedroom (3,000 cubic feet plus a few).
I know the moderate temp of the room at a given degree outside.
I now know how many degrees it will lose in a hour.
So I know now the heat loss in my Bedroom.

I have recorded heat gain in the space accordingly.

The basic question still remains..

Can I use high temp, as it maintains, even though it took days to get there, as an hourly Btu.

Since it maintains temp, hourly, I believe so.

Just so you know, it's no longer a candle source I'm working with.
Candle is just too weak..

quote:
Originally posted by ron_o

Hi ODR

You would need an insulated chamber ( or room ) and a method to circulate the air within that chamber. Chamber needs to contain sufficient air ( oxygen ) so that the burn is uniform
Measure start temp and end temp then temperature gained = end temp - start temp
To make things easy run the experiment for one hour ( or measure a one hour periods heat gain )

It's rather a crude method but it would give an indication

ron




Old in age, not in mind, so
'Teach me something new'!

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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 18 2013 :  17:41:18  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
The next question becomes:

What will we determine to be "Dry Air"?

Will it be <4% by definition or 40% by Relativity



Old in age, not in mind, so
'Teach me something new'!


Edited by - olddawgsrule on December 18 2013 17:42:47
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 19 2013 :  18:19:15  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

Firstly the "DRY AIR" question ... if you are using any form of flame as a heat source the moisture content of the air will increase due to the water being produced as a by-product of the combustion process.

Rate of loss of heat to surroundings... as the temperature difference between a warm body and its surrounding increases so does the rate of heat loss increase ( this is not a linear relationship ) If you want to know more then google "Newtons Law of Cooling"

The heating of the bedroom..... If the heat source is maintaining the room at a constant temperature then you can assume that an equilibrium point has been reached ( heat input = rate of heat loss to surrounding )
If you are working in BTU/h then it is the rise in temperature achieved after one hour of "burn time" ( it cannot be added together as such )

ron
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 20 2013 :  16:09:04  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Thank you RonO and I will look up and read Newtons Law of Cooling.
That should be interesting!
And may answer a few more questions I have.

I did not realize heat loss increases, I did believe it constant.
Again, Newton's Law will be an interesting read.

My moderate temp I speak of is what you have now taught me as being the equilibrium point.

The Btu/h is not a matter of adding all gains together, yet using the highest, maintainable point and using that number alone.
Since I have found it takes many hours for the Terra cotta to reach it's point of (real) transference (and no longer be a heat sink), it seems to me that I can use the temp, even though it took a day to get there.
It also takes time, for a low source, to heat the objects within a given area (also heat sinks).

Example would be:
Took 2 days to reach a temp of 58F.
Holds at 58F, with same outside temp.
Then it's running each hour, as long as it holds, at that Btu/h...

Logic (probably where I'm going wrong) tells me that it takes time to heat 'everything' in room (heat sinks).
Even though it may take days to warm 'everything' up, that final temp is the gain.

Once again, thank you RonO for your input.
I will read as advised and report back later..






Old in age, not in mind, so
'Teach me something new'!

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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 20 2013 :  17:20:46  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

You are quite correct in your assumption regarding the fact that heat is absorbed by everything in the room ( all substances have what is called a "Specific Heat Capacity" this is basically how much energy in Joules is required to raise a mass of 1000 grams through one degree Centigrade .... if you recall i have mentioned that Joules per second = Watts , which makes it very easy to calculate energy inputs or outputs )

So in your experiment the heat is being absorbed by the items in your room at different rates ( air is acting as transference medium ) thus making it difficult to calculate a true BTU figure based on air alone.

If you want some practical examples of use of Specific Heat Capacities let me know.

ron
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 20 2013 :  17:53:03  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
I am a student and learning (my second greatest passion, after curiosity).

I am using Specific Heat Capacity, yet only with a few of the items involved.
Please do if you have time, shoot an example my way.
I'd like to think I have this at least close..

The simplest way seemed to be, measure heat loss in a given amount of time (hr)
Then see what I could raise it to and hold.

Still learning




Old in age, not in mind, so
'Teach me something new'!

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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 20 2013 :  18:25:17  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

I will start with the "classic" heating of water example

Firstly the specific heat capacity of a substance is not a constant but varies with temperature , so an average value is normally used.
Water ... specific heat capacity = 4200 joules per kg per degree centigrade

the standard equation

Joule ( input ) = mass of substance x specific heat capacity of substance x temperature difference

So how many joules will it require to raise a mass of 2 Kg of water through 50'C

Joules(input) = 2 x 4200 x 50 = 420000

And likewise if the water in the above example were to cool back down it would release 420000 joules

Lets now introduce the Joule per second or Watts into the picture

Since one joule per second = one watt per second we can now calculate how long it would take for example an electric heating element to heat the water

In the above example it required 420000 joules ( = 420000 watt seconds ) of energy , lets say we have a 700 watt heater , so how long will it take to heat the water

420000 watt seconds / 700 watts = 600 seconds ( 10 minutes )

I will get back with other examples

ron



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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 20 2013 :  19:01:33  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Joule ( input ) = mass of substance x specific heat capacity of substance x temperature difference

I'm with you so far!!

How I'm doing this..
Still hoping I'm close on the rest..

Old in age, not in mind, so
'Teach me something new'!

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49er
Administrator



USA
4443 Posts

Posted - December 21 2013 :  14:23:52  Show Profile Send 49er a Private Message  Reply with Quote
Hi ODR

In your calculation's in your room did you add you and your wife as part of your BTU's?? finally got home.

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.
SKYPE doug.bennett49er
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 21 2013 :  18:20:22  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

You have mentioned that your room has reached a balance ( equilibrium ) temperature where heat supplied = heat lost to surroundings

This heat loss is known as Thermal Transmittance . The coefficient of thermal transmittance is called the "U" value and has units of Watts per square metre per degree Centigrade

The "U" values for windows ( inc frame ) are :-

single glazed = 4.5 W/m^2 'C
double glazed = 3.3 W/m^2 'C
triple glazed = 1.8 W/m^2 'C

Equation

Heat loss ( in watts ) = "U" value of material x surface area in metres^2 x temperature difference in degrees Centigrade

Note the "temperature difference" in the case of a house would be difference between inside and outside ( example 18'C inside and minus 3'C outside ... difference = 21 'C )

Example calculation

Double glazed window 1.5 x 1 metre , temperature difference = 21'C

Heat loss ( watts ) = 3.3 x 1.5 x 1 x 21 = 103.95 watts

Similar calculations for the walls , roof , floor and door(s) would be made and added together to determine the total heat loss of a building ( or room )

ron
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 22 2013 :  10:14:51  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
I did a complete Heat Loss analysis of my room.
It came within 4% of my easy way..

I Followed the HVAC standards of values, found the R's of each component and compiled that information for Exterior Walls, Windows and Ceiling.
Converted this information to U (transference/loss).

Each item (wall, window & ceiling) result was added together showing the room's loss.

This was done against a known outside temperature and only to the value the house maintains the room at (50F).

From that point upward in temperature would now be what the heater is doing.

Using HVAC standard (now using .0247) value to raise 1 cubic foot of air 1 degreeF.

Basically:
1,008 Btu/h Loss
Plus the 75 Btu/h to raise the volume of air 1 degree
Resulting in 1083 Btu/h to raise the temperature of given room 1 degree beyond 50F.

Let me re-state that..

If the house is maintaining the room at the 1,008 Btu/h rate.
Then is the heater only adding to or compensating and adding?





Old in age, not in mind, so
'Teach me something new'!


Edited by - olddawgsrule on December 22 2013 13:40:14
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49er
Administrator



USA
4443 Posts

Posted - December 22 2013 :  14:24:04  Show Profile Send 49er a Private Message  Reply with Quote
Hi ODR

We have had a ICE STORM power lines down so we do like the 3 candle heaters

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.
SKYPE doug.bennett49er
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 22 2013 :  14:39:23  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
And I would as well my friend!

That storm is headed my way for Christmas!

I'm hoping it brings snow instead of ice..

Heh, Likin' it!
Ice storm, lines are down, you're on the internet!
Still have some stuff to learn from you!!

Stay safe my friend!

quote:
Originally posted by 49er

Hi ODR

We have had a ICE STORM power lines down so we do like the 3 candle heaters

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.
SKYPE doug.bennett49er




Old in age, not in mind, so
'Teach me something new'!

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49er
Administrator



USA
4443 Posts

Posted - December 22 2013 :  16:20:08  Show Profile Send 49er a Private Message  Reply with Quote
Hi ODR

Its All BATTERY POWER

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.
SKYPE doug.bennett49er
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 22 2013 :  17:01:12  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
It's the 220v for the well that still means a generator for me..
Get past that and I'm getting close..

Stay safe my friend.
Since you answered, know you are!

Old in age, not in mind, so
'Teach me something new'!

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49er
Administrator



USA
4443 Posts

Posted - December 22 2013 :  17:39:12  Show Profile Send 49er a Private Message  Reply with Quote
Hi ODR

Yes still need the gen for that 220v well also

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.
SKYPE doug.bennett49er
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 23 2013 :  17:29:10  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
I've now read Newtons Law of Cooling and the question (a few posts up) still remains..

I can apply the Law accordingly..

With the house maintaining a given temp, compensating for loss..
When "A" degree of temp is achieved..
Does the heater also compensate loss for that degree?
Only the difference of loss?

I'm headed towards only the difference of loss.

If I can get this right, then I have a way to evaluate a few of my different experiments for heat production.




Old in age, not in mind, so
'Teach me something new'!

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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 25 2013 :  16:53:48  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

After a quick check on the internet regarding output power of a candle and a conversion factor or two we have the following info:-

I,m going to assume that your candle produced 80 watts of power , this is equivalent to an heat input into the room of approx 273 BTU/hr

1 joule ( = 1 watt/second ) = 0.00094781712 BTU
so therefore 1 watt hour = 3.412141633BTU/hr

given this information you can now back calculate the heat input per hour ( irrespective of how long it takes ) into the room.

But ( and remember there is always a "but" ) if the temperature differential between inside and outside changes so does the rate of heat loss , so at best you will only be able to calculate an average BTU/hr value

hope that helps

ron
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 26 2013 :  17:59:41  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
That's a confirmation of my numbers so far.

In the other topic, I did a calorific formula to check an average I saw on the internet.
It came close enough (mine 44Btu/hr, Doug's 46Btu/h) to the stated 50Btu/h for the basic style candle he's using and the ones I tested with.

Far too small a source of heat in my humble opinion..
Far too much mass for such a low source.
Takes more time to come into play then the candle's life span..

I'm about to make (because I can not find one yet), a liquid candle and attempt a source calculation, before I entirely leave the candle (should say flame) behind.

The fun part of this is the current unit running in my bedroom.
IR bulb source.
8" pot
5 fins over

At 250w of power, should equate to 853Btu/h and I'm calculating it in at just under 1,050Btu/h

I am now in stronger belief, after reading Newton's Law, that any degree obtained, beyond that of moderate, combats outside temp (and loss).
I also believe I can use highest temp, as long as, it maintains and outside temp maintains at given temperature (given it maintains such temp for an hour).

Those opposed to these beliefs please state why..
I'm still learning here!








Old in age, not in mind, so
'Teach me something new'!

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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 27 2013 :  16:59:37  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

I don't want to "butt" in at the moment , i want to see exactly where your heading.

I maybe totally wrong here in my interpretation of the power rating of the IR bulb .....
reason is how the power is stated . IR is of course invisible to the naked / unassisted human eye so you should see no light from the bulb. If you can then the quoted figure of 250 watts IR is a part of the total power consumed figure of the bulb.

ron
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 27 2013 :  17:59:58  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
RonO, I always hope you "Butt" in, you always help me learn and understand!
Got the pun!!

If I put the IR bulb on the meter it comes close enough to it's draw to say it's 250w.
Consumption of power only.

Now what does it really produce?
That is the question I assume your headed towards.

If place it in an air space you can only expect up to what it states (well not really, cause it effects only what it shown on).

Now if you place it in an area that will adsorb, like a clay pot.. (an area it effects)
There is a greater production/transference.

The statement is:
Watts are Watts.
No matter how produced.

My belief is now "How They Are Used" becomes the difference.

Still willing to bring an apple to class to learn more!






Old in age, not in mind, so
'Teach me something new'!

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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 27 2013 :  18:12:49  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

Thank you for the update. Things are , shall we say becoming clearer.

Watts are indeed Watts , my friend. So now we have a new question ( or two ) .... what is the conversion efficiency ( specific heat capacity and energy transference , what colour absorbs IR best and a black body radiator )

I think you will find the answer most interesting !

ron

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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - December 28 2013 :  13:52:24  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Here I go RonO..

First Specific Heat for terracotta alludes me.
The closest I can come is red clay.
.92 kj/kg C (which I believe can also be stated .92j/g C)

I'm putting the weight of the pot at 907.18grams

So using:
Weight x Temp Differencial x Specific Heat
907.18g x 1C x 0.92

I have transferred 834.61 Joules

As far as color goes..
This I'm only 75% agreeing with..

Darker the better, as we all know, absorbs more heat from light.
The 25% that bothers me is the difference in how White light heats vs IR light. I just makes sense that IR is more efficient..

I have not tested darker interiors.
My running IR unit has not seen a candle yet.
Could be an interesting test to run one and soot it up a bit.


Have a feeling I'm bringing 2 apples next visit..


Old in age, not in mind, so
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - December 31 2013 :  17:12:21  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

After much internet miss information ! Here we go...

Firstly 1 BTU will raise 55 cubic feet of air through 1 'F ( the type of air , initial density or specific heat capacity do not seem to be stated however ! )

1 cubic metre of dry air at 0'C has a density of 1.275 and a specific heat capacity of 1.006 kj / 'C kg
So to raise the 1 cubic metre of air ( as stated above ) will require 1.006 x 1.275 x 1
which gives 1.28265 Kjoules

Also stated 1 BTU is approx 1055 joules

so 1.28265 x 1000 / 1055 joules = 1.215782 BTU

A temperature differential of 1'C is equivalent to a temperature differential of 1.8'F

Thus to raise 1 cubic metre of air by 1'F would require 1.215782 / 1.8 = 0.6754 BTU

1 cubic metre = 35.314 cubic feet ( approx )

so 1 BTU = 1 x 35.314 / 0.6754 = 52.28 cubic feet ( through 1'F )based on the original stated physical constants

Just out of intrest if the initial air temperature was 50 'C the result would yield 61.715 cubic feet of air ( through 1'F )

This is why it is so difficult to calculate the true effect of your heater !

ron
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - January 01 2014 :  10:08:47  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Almost all the numbers here matchup to what I'm using.
Greatest difference is the:
1 BTU will raise 55 cubic feet of air through 1 'F

This would equate to; .018Btu per cubic foot.
I'm currently using a blend of the best HVAC Standards I've found which puts me at: .0247Btu

I also found a chart that works density & specific weight at different temperatures.
It's claim is 'Dry Air', thus became a question of specifying Dry Air.

HVAC uses <40%

My room is within the mark of Dry Air by HVAC Standard.

By changing the value of Degree per cubic ft, I have a 120Btu difference.

All other values Matchup.

The difficulty for me separating the Heat source of the house from the Heat source of the heater.

I've lived here long enough to know what the room will maintain at certain outside temp ranges (singles, teens and twenties)'F.

Given the room size 3,072 cubic feet
Knowing I'm transmitting 1008 Btu/h through my walls
At a 36'F temp difference (inside to outside)

3072 x .0182 = 55.9 Btu to raise this amount of air 1'F
1'F also raises my heat loss by 28 Btu's
Making my total Heat Loss at 1,036 Btu's

So either the heater is producing:
60 Btu plus the difference loss (28 Btu's) = 88 Btu/h
Or the entire loss 1,036 Btu's plus the gain 60 Btu = 1,096 Btu/h

Since the Human body gives off 250 Btu/h or greater
I'm finding it difficult not to use the 1,096 Btu/h result..
Being in the room, I do not raise the temp..








Old in age, not in mind, so
'Teach me something new'!

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49er
Administrator



USA
4443 Posts

Posted - January 01 2014 :  10:26:43  Show Profile Send 49er a Private Message  Reply with Quote
Hi ODR

Its a new year take a break for the day.. LOL

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
You must be logged in to see this link.
SKYPE doug.bennett49er
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - January 01 2014 :  16:47:49  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Even with the 'kiddo's' showin' up
Especially with the Kiddo's Showing up
And asking questions...
Kinda hard to 'Take a Day off'

To you I'm known as ODR
To them I'm known as Uncle Bob
And he's always doing something interesting
Auntie's food, my strange mind..
Hey, they keep coming back!

There is no day off for Uncle Bob
And I like it this way!

They keep me active!

Enjoy your day your way
We'll get back to this

Me and the 'Kiddo's' are playin' with somethin'

Only showed up because I wanted to show them a 'Cool Site' to visit and ask questions
They were taken back from your post...



Old in age, not in mind, so
'Teach me something new'!

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ron_o
Moderator



United Kingdom
1052 Posts

Posted - January 01 2014 :  17:33:53  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

Glad to hear that we are in the same ballpark ... the value of 55 cubic ft by 1'F per BTU seemed to crop up on most sites and is often quoted as a standard ( but the physical properties are not stated )

I hope you have enjoyed being Uncle Bob today but i trust ODR will be back tomorrow !

All the best

ron
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