How many Watts to heat 0.25 cuft of air 12*F above ambient....

[Lamont Cranston]

I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.

 

[Ricky]

On Friday, April 14, 2023 at 2:17:57 PM UTC-4, Lamont Cranston wrote:

I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft.. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.

If you are nesting two, same sized buckets, there will be very little air gap. Adding a few strips of wood can give you a quarter inch or so of air space. That will be fairly effective. You can even use some pencils.

You\'d still need to insulate the top.

I have no idea what temperature 30W would give you, but by seat of the pants, I\'m thinking it will get you to 80°F easily.

Are you planting an early crop? You could simply put the seedlings in the yard with a clear or translucent cover. At night, that traps the heat from the ground very well. I don\'t recall where you live, but many places in the US are already frost free for planting.

--

Rick C.

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[Lamont Cranston]

On Friday, April 14, 2023 at 1:41:34 PM UTC-5, Ricky wrote:

On Friday, April 14, 2023 at 2:17:57 PM UTC-4, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
If you are nesting two, same sized buckets, there will be very little air gap. Adding a few strips of wood can give you a quarter inch or so of air space. >That will be fairly effective. You can even use some pencils.

I just happened on two different brands of buckets, slightly different style, so a little more air gap then you would expect, but still not a lot, plenty of room for the strip heaters and wire.

You\'d still need to insulate the top.

That\'s still up in the air, the growing kit instructions say don\'t cover, but I want to minimize airborn bacteria or fungus, so would like a cover that still allows
CO2/ oxygen exchange, that\'s left for experimentation.

I have no idea what temperature 30W would give you, but by seat of the pants, I\'m thinking it will get you to 80°F easily.
yep seems like plenty, I guess I\'ll hook it up and see.

Are you planting an early crop? You could simply put the seedlings in the yard with a clear or translucent cover. At night, that traps the heat from the ground very well. I don\'t recall where you live, but many places in the US are already frost free for planting.

No plants just sprouts! My wife cooks several meals with mung bean sprouts, she has grown them in a steamer with paper towel as the wetting medium,
but lately they got slimy, I suspect a good bleaching and using clean water would solve the problem, but I decided to play with an Arduino and control the environment better to see if I could make they grow better. There is an industry of growing them in bulk.
Mikek

--

Rick C.

- Get 1,000 miles of free Supercharging
- Tesla referral code - https://ts.la/richard11209

 

[Fred Bloggs]

On Friday, April 14, 2023 at 2:17:57 PM UTC-4, Lamont Cranston wrote:

I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft.. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.

Correct me if I\'m wrong, but the two 12V 12W in series is 6V per element, 6^2 power, so you total to 4 x 6^2= (2x6)^2=12^2 for just 12W total and not 30W.

 

[Fred Bloggs]

On Friday, April 14, 2023 at 2:17:57 PM UTC-4, Lamont Cranston wrote:

I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft.. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.

Okay, so 4 x(9.5)^2 or (19)^2 is (19/12)^2x 12W = 30 W. That\'s better than running them off 12V.

 

[John S]

On 4/14/2023 1:17 PM, Lamont Cranston wrote:

I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.

The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.

 

[Ricky]

On Friday, April 14, 2023 at 4:01:11 PM UTC-4, John S wrote:

On 4/14/2023 1:17 PM, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.

I\'m confused. Where did you get the figure 0.05796 BTU/hour? Above that you calculated BTU, but you don\'t indicate any time frame.

Then, I\'m further confused by the idea that you need to continue to heat the air after it reaches 80°F. If you are taking into account the thermal losses through the insulated buckets, you need to know the thermal insulation properties of the insulated bucket.

In other words, you didn\'t get any of this problem right.

This problem has little to do with the thermal mass of the air in the bucket. It will take ENERGY to heat that up (not a specific amount of power). But to maintain that temperature, the heat flow through the bucket would need to be known. This is an unknown, so the problem can not be solved, only, perhaps, estimated.

--

Rick C.

+ Get 1,000 miles of free Supercharging
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[Ricky]

On Friday, April 14, 2023 at 2:56:54 PM UTC-4, Lamont Cranston wrote:

On Friday, April 14, 2023 at 1:41:34 PM UTC-5, Ricky wrote:
On Friday, April 14, 2023 at 2:17:57 PM UTC-4, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
If you are nesting two, same sized buckets, there will be very little air gap. Adding a few strips of wood can give you a quarter inch or so of air space. >That will be fairly effective. You can even use some pencils.
I just happened on two different brands of buckets, slightly different style, so a little more air gap then you would expect, but still not a lot, plenty of room for the strip heaters and wire.

If you are heating the space between the buckets, then you have only the plastic wall of the outer bucket as insulation. You are heating the air gap, so no insulation there.

You\'d still need to insulate the top.
That\'s still up in the air, the growing kit instructions say don\'t cover, but I want to minimize airborn bacteria or fungus, so would like a cover that still allows
CO2/ oxygen exchange, that\'s left for experimentation.

You can use an open fabric for that. With fine enough openings, you can perhaps stop some airborne seeds of plants, but there\'s little chance of stopping bacteria or fungi. They are far too small to be stopped by anything that isn\'t similar to the masks people wear. I believe you are overthinking this.

I have no idea what temperature 30W would give you, but by seat of the pants, I\'m thinking it will get you to 80°F easily.
yep seems like plenty, I guess I\'ll hook it up and see.

Are you planting an early crop? You could simply put the seedlings in the yard with a clear or translucent cover. At night, that traps the heat from the ground very well. I don\'t recall where you live, but many places in the US are already frost free for planting.
No plants just sprouts! My wife cooks several meals with mung bean sprouts, she has grown them in a steamer with paper towel as the wetting medium,

Ah, yes. Sprouts are great!


> but lately they got slimy, I suspect a good bleaching and using clean water would solve the problem, but I decided to play with an Arduino and control the environment better to see if I could make they grow better. There is an industry of growing them in bulk.

Yeah. So you are looking to control the temperature. Ok, that should work somewhat faster than room temperature. We\'ll see.

Just keep in mind, that unless the conditions are very unhealthy, plants can take care of themselves. They seldom get sick unless the environment is pretty bad, or the pathogen is pretty good. Aphids don\'t need to prey on sick plants. They just take over. Other diseases can\'t get a foothold with a healthy plant.

--

Rick C.

-- Get 1,000 miles of free Supercharging
-- Tesla referral code - https://ts.la/richard11209

 

[Lamont Cranston]

On Friday, April 14, 2023 at 4:14:22 PM UTC-5, Ricky wrote:

On Friday, April 14, 2023 at 2:56:54 PM UTC-4, Lamont Cranston wrote:
On Friday, April 14, 2023 at 1:41:34 PM UTC-5, Ricky wrote:

Ah, yes. Sprouts are great!

One of my favorite ways to eat bean sprouts is in a peanut butter sandwich, sounds a little strange, but it adds a great crunch and some moisture to help with the sticky peanut butter.

but lately they got slimy, I suspect a good bleaching and using clean water would solve the problem, but I decided to play with an Arduino and control the environment better to see if I could make they grow better. There is an industry of growing them in bulk.

Yeah. So you are looking to control the temperature. Ok, that should work somewhat faster than room temperature. We\'ll see.

Just keep in mind, that unless the conditions are very unhealthy, plants can take care of themselves. They seldom get sick unless the environment is >pretty bad, or the pathogen is pretty good. Aphids don\'t need to prey on sick plants. They just take over. Other diseases can\'t get a foothold with a healthy >plant.

I\'m not sure what has changed and it may only be the time of year, but last few batches she complained about them being slimy, she doing the same
old methods she has done for 35 years. But, I got a little interested and read a little about ideal temps and watering times at the same time my son was mentioning he would like to get an Arduino to program. I ordered an Arduino kit and with his help we have a program that will control temperature and turn the water on for a settable amount of seconds every settable amount of hours. With a little time we should find optimum settings.
My wife is a gardener with a real green thumb, fruit trees, root vegetables, leafy greens, squash, okra, tomatoes, flowering plants, etc. Strangers stop by just to ask her about her garden.

Mikek

 

[John Larkin]

On Fri, 14 Apr 2023 15:00:58 -0500, John S <Sophi.2@invalid.org>
wrote:

On 4/14/2023 1:17 PM, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.


The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.

It takes energy, joules, to heat an insulated volume of air some
delta-t.

It takes power, watts, to keep it hot, assuming the container isn\'t
perfectly insulated.

 

[Lamont Cranston]

On Friday, April 14, 2023 at 3:01:11 PM UTC-5, John S wrote:

On 4/14/2023 1:17 PM, Lamont Cranston wrote:

The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.

Thank you doing the math on that. So I have plenty of heat, assuming it doesn\'t lose a huge amount to the environment.
The heater order actually has 5 of the 12V 12W heaters, if I series them all and run on 19V, I\'m down to 6 watts with a
more even heat distribution. I may experiment with 6 watts to see how it cycles. I have the option of putting the heaters on the
inside of the inner bucket and I thought about covering them with 2\" copper tape to keep them dry and spread the heat.
Also, copper has some anti bacterial properties and is toxic to slime molds.. But there is a limit to the amount of copper in solution before it is toxic.
I don\'t envision much contact between the copper covering the heaters and the water though. But I could make that happen.
First iteration, https://www.dropbox.com/s/aztik4n6atl80g6/Bean%20Sprouter%20parts.jpg?dl=0
This shows a lot of room for the heaters on the inside.
Mikek

 

[Ricky]

On Friday, April 14, 2023 at 6:37:40 PM UTC-4, Lamont Cranston wrote:

On Friday, April 14, 2023 at 3:01:11 PM UTC-5, John S wrote:
On 4/14/2023 1:17 PM, Lamont Cranston wrote:

The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.
Thank you doing the math on that. So I have plenty of heat, assuming it doesn\'t lose a huge amount to the environment.
The heater order actually has 5 of the 12V 12W heaters, if I series them all and run on 19V, I\'m down to 6 watts with a
more even heat distribution. I may experiment with 6 watts to see how it cycles. I have the option of putting the heaters on the
inside of the inner bucket and I thought about covering them with 2\" copper tape to keep them dry and spread the heat.
Also, copper has some anti bacterial properties and is toxic to slime molds. But there is a limit to the amount of copper in solution before it is toxic.
I don\'t envision much contact between the copper covering the heaters and the water though. But I could make that happen.
First iteration, https://www.dropbox.com/s/aztik4n6atl80g6/Bean%20Sprouter%20parts.jpg?dl=0
This shows a lot of room for the heaters on the inside.
Mikek

So you actually have two buckets, and a third container inside the smaller bucket. With that much air space on the inside, you will have no worries about \"evenly\" spreading the heat. The air is the conductor of the heat and it\'s not a great conductor. It will be surrounded by the heating, so will be essentially isothermal, unless the insulation at the top or bottom is poor.

Why the huge hole in the bottom? Don\'t you only need a small opening for a drain?

Other than possibly leaving a bit more space between the two buckets, I think you\'ve got this nailed.

--

Rick C.

-+ Get 1,000 miles of free Supercharging
-+ Tesla referral code - https://ts.la/richard11209

 

[John Larkin]

On Fri, 14 Apr 2023 15:37:35 -0700 (PDT), Lamont Cranston
<amdx62@gmail.com> wrote:

On Friday, April 14, 2023 at 3:01:11?PM UTC-5, John S wrote:
On 4/14/2023 1:17 PM, Lamont Cranston wrote:

The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.

Thank you doing the math on that.

Makes no sense.

 

[Lamont Cranston]

On Friday, April 14, 2023 at 7:02:59 PM UTC-5, Ricky wrote:

So you actually have two buckets, and a third container inside the smaller bucket. With that much air space on the inside, you will have no worries about \"evenly\" spreading the heat. The air is the conductor of the heat and it\'s not a great conductor. It will be surrounded by the heating, so will be essentially isothermal, unless the insulation at the top or bottom is poor.

Why the huge hole in the bottom? Don\'t you only need a small opening for a drain?

The smaller diameter containers, 4 of them are the actual growing trays, the seeds are spread in them.
The top one (fifth tray ) was actually the water catch tray for the for other the 4 growing trays, but I have made that into
the drip tray for watering the other 4 trays. I will let the pump fill it and it will drip watering the other trays.
The trays have drain holes only on the outer section of the floor.
The huge hole is perfectly sized to catch the offset in the tray to make it stand up right and it will drain it to the larger bucket, when the time is right the pump will come on and pump water into the top tray for the watering process.
I got my heaters in the mail today, so may start temperature experimenting tomorrow.
Mikek

Other than possibly leaving a bit more space between the two buckets, I think you\'ve got this nailed.

--

Rick C.

-+ Get 1,000 miles of free Supercharging
-+ Tesla referral code - https://ts.la/richard11209

 

[Anthony William Sloman]

On Saturday, April 15, 2023 at 6:01:11 AM UTC+10, John S wrote:

On 4/14/2023 1:17 PM, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.

This is idiotic.Watts are units of energy flow rate. The answer to the question has to be Joules, not Watts which are Joules per second.,

0.25 cubic feet of air is 7.1 litres. At room temperature one mole of air occupies 24 litres so that is about 0.3 mole air molecules are diatomic so they store both rotational and translational energy so their heat capacity is 5/2R where R is the universal gas constant 8.314 J⋅K−1⋅mol−1.

So the heat capacity of a mole of air is 29.8 joule per mole per degrees Celcius.

So 0.25 cubic feet air has a heat capacity of 3.46 joule per degree Fahrenheit and you\'d need 41.6 joules to heat it from 68F to 80F.

What actually matters is the heat you\'d have to supply to keep it at 80F, which depends on the amount of insulation you wrapped around the whole set-up which will have a much higher heat capacity than 0.25 cubic feet of ari

Bill Sloman

 

[Fred Bloggs]

On Saturday, April 15, 2023 at 1:09:21 AM UTC-4, Anthony William Sloman wrote:

On Saturday, April 15, 2023 at 6:01:11 AM UTC+10, John S wrote:
On 4/14/2023 1:17 PM, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.
This is idiotic.Watts are units of energy flow rate. The answer to the question has to be Joules, not Watts which are Joules per second.,

0.25 cubic feet of air is 7.1 litres. At room temperature one mole of air occupies 24 litres so that is about 0.3 mole air molecules are diatomic so they store both rotational and translational energy so their heat capacity is 5/2R where R is the universal gas constant 8.314 J⋅K−1⋅mol−1.

So the heat capacity of a mole of air is 29.8 joule per mole per degrees Celcius.

So 0.25 cubic feet air has a heat capacity of 3.46 joule per degree Fahrenheit and you\'d need 41.6 joules to heat it from 68F to 80F.

What actually matters is the heat you\'d have to supply to keep it at 80F, which depends on the amount of insulation you wrapped around the whole set-up which will have a much higher heat capacity than 0.25 cubic feet of ari

It doesn\'t need to be that complicated. The air is nothing, unless he\'s blowing fresh air through it, which doesn\'t sound like he is. The heater only has to make up for conduction through through the boundary of the bucket. Most of these little plastic buckets are HDPE, which from what I can find has a thermal conductance of k = 0.45-0.5 W/(m-C). So practically speaking, a mean diameter D , is circumference pi x D which when multiplied by H gives the are of sides As= pi x D x H. Then for the end(s) it is Ae = pi x (D/2)^2. So he maybe has total area A= As + 2 x Ae = pi x ( D x H + 2 x (D/2)^2) Then he needs to break out a caliper and measure thickness of the material, t. The total heat loss becomes (k/t) x A x (Th-Ta), h= heated volume, a= ambient. This will not require a whole of power for a 12oF= 7oC differential. All measurements metric. Simplest control is to PWM the heaters down to the few watts needed.

Bill Sloman

 

[John Larkin]

On Sat, 15 Apr 2023 07:10:18 -0700 (PDT), Fred Bloggs
<bloggs.fredbloggs.fred@gmail.com> wrote:

On Saturday, April 15, 2023 at 1:09:21?AM UTC-4, Anthony William Sloman wrote:
On Saturday, April 15, 2023 at 6:01:11?AM UTC+10, John S wrote:
On 4/14/2023 1:17 PM, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.
This is idiotic.Watts are units of energy flow rate. The answer to the question has to be Joules, not Watts which are Joules per second.,

0.25 cubic feet of air is 7.1 litres. At room temperature one mole of air occupies 24 litres so that is about 0.3 mole air molecules are diatomic so they store both rotational and translational energy so their heat capacity is 5/2R where R is the universal gas constant 8.314 J?K?1?mol?1.

So the heat capacity of a mole of air is 29.8 joule per mole per degrees Celcius.

So 0.25 cubic feet air has a heat capacity of 3.46 joule per degree Fahrenheit and you\'d need 41.6 joules to heat it from 68F to 80F.

What actually matters is the heat you\'d have to supply to keep it at 80F, which depends on the amount of insulation you wrapped around the whole set-up which will have a much higher heat capacity than 0.25 cubic feet of ari

It doesn\'t need to be that complicated. The air is nothing, unless he\'s blowing fresh air through it, which doesn\'t sound like he is. The heater only has to make up for conduction through through the boundary of the bucket. Most of these little plastic buckets are HDPE, which from what I can find has a thermal conductance of k = 0.45-0.5 W/(m-C).

Air is about 0.03. The bucket is not sitting in a pool of stirred
water.

> So practically speaking, a mean diameter D , is circumference pi x D which when multiplied by H gives the are of sides As= pi x D x H. Then for the end(s) it is Ae = pi x (D/2)^2. So he maybe has total area A= As + 2 x Ae = pi x ( D x H + 2 x (D/2)^2) Then he needs to break out a caliper and measure thickness of the material, t. The total heat loss becomes (k/t) x A x (Th-Ta), h= heated volume, a= ambient. This will not require a whole of power for a 12oF= 7oC differential. All measurements metric. Simplest control is to PWM the heaters down to the few watts needed.

Practically speaking, it wouldn\'t make much difference if the bucket
was made of silver instead of plastic. Well, it would cost more.

 

[Fred Bloggs]

On Saturday, April 15, 2023 at 10:26:24 AM UTC-4, John Larkin wrote:

On Sat, 15 Apr 2023 07:10:18 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:

On Saturday, April 15, 2023 at 1:09:21?AM UTC-4, Anthony William Sloman wrote:
On Saturday, April 15, 2023 at 6:01:11?AM UTC+10, John S wrote:
On 4/14/2023 1:17 PM, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0..25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.
This is idiotic.Watts are units of energy flow rate. The answer to the question has to be Joules, not Watts which are Joules per second.,

0.25 cubic feet of air is 7.1 litres. At room temperature one mole of air occupies 24 litres so that is about 0.3 mole air molecules are diatomic so they store both rotational and translational energy so their heat capacity is 5/2R where R is the universal gas constant 8.314 J?K?1?mol?1.

So the heat capacity of a mole of air is 29.8 joule per mole per degrees Celcius.

So 0.25 cubic feet air has a heat capacity of 3.46 joule per degree Fahrenheit and you\'d need 41.6 joules to heat it from 68F to 80F.

What actually matters is the heat you\'d have to supply to keep it at 80F, which depends on the amount of insulation you wrapped around the whole set-up which will have a much higher heat capacity than 0.25 cubic feet of ari

It doesn\'t need to be that complicated. The air is nothing, unless he\'s blowing fresh air through it, which doesn\'t sound like he is. The heater only has to make up for conduction through through the boundary of the bucket.. Most of these little plastic buckets are HDPE, which from what I can find has a thermal conductance of k = 0.45-0.5 W/(m-C).
Air is about 0.03. The bucket is not sitting in a pool of stirred
water.
So practically speaking, a mean diameter D , is circumference pi x D which when multiplied by H gives the are of sides As= pi x D x H. Then for the end(s) it is Ae = pi x (D/2)^2. So he maybe has total area A= As + 2 x Ae = pi x ( D x H + 2 x (D/2)^2) Then he needs to break out a caliper and measure thickness of the material, t. The total heat loss becomes (k/t) x A x (Th-Ta), h= heated volume, a= ambient. This will not require a whole of power for a 12oF= 7oC differential. All measurements metric. Simplest control is to PWM the heaters down to the few watts needed.
Practically speaking, it wouldn\'t make much difference if the bucket
was made of silver instead of plastic. Well, it would cost more.

The heated exterior of the bucket will create a draft, however slight, but the heat loss calculated through the conductance stays unchanged. The only effect of the air and other material inside the bucket is to contribute to thermal capacity of the whole, which may be considerable, but won\'t change anything once he brings it up to temperature. Most of these home projects with heaters and light bulbs end up overheating the plants.

Just did a quick measurement on a 5-gal HDPE: D=12\", H=14\" t=1/16\"

A=pi x ( 12x14 + 2x(12/2)^2)= 754 sq in., so heat loss= 0.5 x (1/16) x 754 x 7oC/ 39.36= 4 W. Maybe only 2 W for the 2-1/2 gal bucket. The 39.36 factor is inches/meter. And that\'s just the bucket, there may be leakage.. If he strings the two 12V 12W in series and applies 19V, he gets 2 x (9.5/12)^2 x 12W=15 W , so maybe a 15% duty on the heater just for the bucket.. If the pulse is safely 100ms, then the frequency would be about 700ms, 1.4Hz. That\'s just a little op amp/ logic circuit.

 

[Fred Bloggs]

On Saturday, April 15, 2023 at 10:26:24 AM UTC-4, John Larkin wrote:

On Sat, 15 Apr 2023 07:10:18 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:

On Saturday, April 15, 2023 at 1:09:21?AM UTC-4, Anthony William Sloman wrote:
On Saturday, April 15, 2023 at 6:01:11?AM UTC+10, John S wrote:
On 4/14/2023 1:17 PM, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0..25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.
This is idiotic.Watts are units of energy flow rate. The answer to the question has to be Joules, not Watts which are Joules per second.,

0.25 cubic feet of air is 7.1 litres. At room temperature one mole of air occupies 24 litres so that is about 0.3 mole air molecules are diatomic so they store both rotational and translational energy so their heat capacity is 5/2R where R is the universal gas constant 8.314 J?K?1?mol?1.

So the heat capacity of a mole of air is 29.8 joule per mole per degrees Celcius.

So 0.25 cubic feet air has a heat capacity of 3.46 joule per degree Fahrenheit and you\'d need 41.6 joules to heat it from 68F to 80F.

What actually matters is the heat you\'d have to supply to keep it at 80F, which depends on the amount of insulation you wrapped around the whole set-up which will have a much higher heat capacity than 0.25 cubic feet of ari

It doesn\'t need to be that complicated. The air is nothing, unless he\'s blowing fresh air through it, which doesn\'t sound like he is. The heater only has to make up for conduction through through the boundary of the bucket.. Most of these little plastic buckets are HDPE, which from what I can find has a thermal conductance of k = 0.45-0.5 W/(m-C).
Air is about 0.03. The bucket is not sitting in a pool of stirred
water.
So practically speaking, a mean diameter D , is circumference pi x D which when multiplied by H gives the are of sides As= pi x D x H. Then for the end(s) it is Ae = pi x (D/2)^2. So he maybe has total area A= As + 2 x Ae = pi x ( D x H + 2 x (D/2)^2) Then he needs to break out a caliper and measure thickness of the material, t. The total heat loss becomes (k/t) x A x (Th-Ta), h= heated volume, a= ambient. This will not require a whole of power for a 12oF= 7oC differential. All measurements metric. Simplest control is to PWM the heaters down to the few watts needed.
Practically speaking, it wouldn\'t make much difference if the bucket
was made of silver instead of plastic. Well, it would cost more.

Slight transcription error- multiplied by 1/16 thickness when it should be a divide. Now it\'s over a 1000W which is a ridiculous result.

 

[John Larkin]

On Sat, 15 Apr 2023 08:51:39 -0700 (PDT), Fred Bloggs
<bloggs.fredbloggs.fred@gmail.com> wrote:

On Saturday, April 15, 2023 at 10:26:24?AM UTC-4, John Larkin wrote:
On Sat, 15 Apr 2023 07:10:18 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:

On Saturday, April 15, 2023 at 1:09:21?AM UTC-4, Anthony William Sloman wrote:
On Saturday, April 15, 2023 at 6:01:11?AM UTC+10, John S wrote:
On 4/14/2023 1:17 PM, Lamont Cranston wrote:
I have ordered four 12V 12W Flexible Polyimide Adhesive Hot Foil Heating Film. If I wire two in series and then put that in parallel with two more in series and run it on 19V, (solving my 19V problem) I will have about 30 Watts of heating. I\'m heating the air of a 2 gallon plastic bucket ~0.25cuft. I\'m thinking (guessing) this will be plenty of heat to keep the bucket at 80*F with a minimum ambient temperature of 68*F.
I do have a temperature controller to shut off the heater when the desired temperature has been reached. It does have a small insulating air gap between the two buckets.
Your thoughts?
Mikek

As added info, I\'m building a small bean sprout grower with adjustable watering schedule and temperature control. it is all set inside a two gallon bucket, that is slipped into a second bucket.
The temperature difference is 80F - 68F = 12F.

The mass of air in 0.25 cubic feet is 0.020175 lb.

The amount of heat required to raise the temperature of 0.020175 lb of
air by 12F is:

0.24 BTU/lb °F x 0.020175 lb x 12 F = 0.05796 BTU

Converting BTU to watts:

0.05796 BTU/hour x 0.293 watts/BTU = 0.017 watts (rounded to 3 decimal
places)

So, approximately 0.017 watts of power would be required to heat 0.25
cubic feet of air from 68F to 80F, assuming the air is contained in a
well-insulated container.
This is idiotic.Watts are units of energy flow rate. The answer to the question has to be Joules, not Watts which are Joules per second.,

0.25 cubic feet of air is 7.1 litres. At room temperature one mole of air occupies 24 litres so that is about 0.3 mole air molecules are diatomic so they store both rotational and translational energy so their heat capacity is 5/2R where R is the universal gas constant 8.314 J?K?1?mol?1.

So the heat capacity of a mole of air is 29.8 joule per mole per degrees Celcius.

So 0.25 cubic feet air has a heat capacity of 3.46 joule per degree Fahrenheit and you\'d need 41.6 joules to heat it from 68F to 80F.

What actually matters is the heat you\'d have to supply to keep it at 80F, which depends on the amount of insulation you wrapped around the whole set-up which will have a much higher heat capacity than 0.25 cubic feet of ari

It doesn\'t need to be that complicated. The air is nothing, unless he\'s blowing fresh air through it, which doesn\'t sound like he is. The heater only has to make up for conduction through through the boundary of the bucket. Most of these little plastic buckets are HDPE, which from what I can find has a thermal conductance of k = 0.45-0.5 W/(m-C).
Air is about 0.03. The bucket is not sitting in a pool of stirred
water.
So practically speaking, a mean diameter D , is circumference pi x D which when multiplied by H gives the are of sides As= pi x D x H. Then for the end(s) it is Ae = pi x (D/2)^2. So he maybe has total area A= As + 2 x Ae = pi x ( D x H + 2 x (D/2)^2) Then he needs to break out a caliper and measure thickness of the material, t. The total heat loss becomes (k/t) x A x (Th-Ta), h= heated volume, a= ambient. This will not require a whole of power for a 12oF= 7oC differential. All measurements metric. Simplest control is to PWM the heaters down to the few watts needed.
Practically speaking, it wouldn\'t make much difference if the bucket
was made of silver instead of plastic. Well, it would cost more.

The heated exterior of the bucket will create a draft, however slight, but the heat loss calculated through the conductance stays unchanged. The only effect of the air and other material inside the bucket is to contribute to thermal capacity of the whole, which may be considerable, but won\'t change anything once he brings it up to temperature. Most of these home projects with heaters and light bulbs end up overheating the plants.

Just did a quick measurement on a 5-gal HDPE: D=12\", H=14\" t=1/16\"

A=pi x ( 12x14 + 2x(12/2)^2)= 754 sq in., so heat loss= 0.5 x (1/16) x 754 x 7oC/ 39.36= 4 W. Maybe only 2 W for the 2-1/2 gal bucket. The 39.36 factor is inches/meter. And that\'s just the bucket, there may be leakage. If he strings the two 12V 12W in series and applies 19V, he gets 2 x (9.5/12)^2 x 12W=15 W , so maybe a 15% duty on the heater just for the bucket. If the pulse is safely 100ms, then the frequency would be about 700ms, 1.4Hz. That\'s just a little op amp/ logic circuit.

The thermal resistance from the interior of the bucket (assume it\'s
isothermal inside) to the world depends very little on the thermal
conductivity of the bucket material. The first inch of air outside of
the bucket has a huge thermal resistance. Convection or wind will
reduce the thermal resistance of the air but it\'s still big.

The solid bucket walls, plastic or silver, will have very low
temperature drop. Air wins.