Greenhouse Gases Do NOT Violate The Stefan-Boltzmann Law

tmiddles wrote:I am saying that radiance from the cooler walls is being absorbed by the warmer person.

tmiddles wrote:The most important fact to establish first is:
The Stefan-Boltzmann equation for radiance IS a law.
P(out)=σeA*To^4
Qt=σeAT^4
RAD = T^4 * emiss * bolt * area

tmiddles wrote:Now since an adult human can be fairly given values in ...

IBdaMann wrote:

tmiddles wrote:radiance from the cooler walls is being absorbed by the warmer person

ERGO you are asserting that radiance from the cooler walls increases the temperature ...

IBdaMann wrote:
Radiance = P(out)/Area

IBdaMann wrote:
Why not go with human Emissivity = 0.7 and wall Emissivity = 0.8 ?

tmiddles wrote:No I'm not I'm proving that it's net transfer.

tmiddles wrote: Absorption means a single molecule has it's energy increased, temperature is the internal energy of a whole system. It's the average. In fact the same molecule could emitt, absrob, emitt, absorb.

tmiddles wrote: How else would themodynamic equilibrium play out?

tmiddles wrote:

IBdaMann wrote:
Why not go with human Emissivity = 0.7 and wall Emissivity = 0.8 ?

Why does the emissivity of the wall matter?

tmiddles wrote: I did supply the source for human emissivity

IBdaMann wrote:I don't know what else to say.

tmiddles wrote:

IBdaMann wrote:I don't know what else to say.

Solve the problem the right way. Nothing is stopping you.

IBdaMann wrote:
Are you saying that you chose to waste time with an example you don't wish to solve?

tmiddles wrote:I already solved it:
Qt=(5.67×10−8J/s⋅m2⋅K4)(0.97)(1.50m2)(306K)^4=−723J/s=−723W

That's the emittence by radiance. I stand by that.

Are you claiming I calculated it wrong?

IBdaMann wrote:I don't know. How did you get the 306K figure

tmiddles wrote:...an adult human can be fairly given values in the "normal range" ...a temperature of 91F/33C/306K,

tmiddles wrote:

IBdaMann wrote:I don't know. How did you get the 306K figure

tmiddles wrote:...an adult human can be fairly given values in the "normal range" ...a temperature of 91F/33C/306K,

IBdaMann wrote:=
Have you abandoned your bogus equation?

tmiddles wrote:
BODY PHYSICS: MOTION TO METABOLISM Textbook

P(out)=σeA*To^4 (the radiance out To as in Temperature for out)

P(in)=σeA*Tenv^4 (the radiance being absorbed from the environment)

P(net)=σeA*(Tenv^4-To^4) (the NET gain or loss between the two)

tmiddles wrote:

IBdaMann wrote:=
Have you abandoned your bogus equation?

I want to know your answer for the thermodynamics for the person.

IBdaMann wrote:
As far as I am concerned, you are simply enamored with "net flow"
.

Aliaka wrote:
Hi guys, this is Ali
I was searching online for works that we can do in order to stop or at least slow down the global warming, and found this video which I hope will help all of you.
You can share it if you like, I don't mind

https://dausel.co/r4suvx
https://dausel.co/r4suvx

tmiddles wrote:

IBdaMann wrote:
As far as I am concerned, you are simply enamored with "net flow"
.

Bottom line: You can't solve the problem following your own rules.

If you could you would.

And I notice ITN is reeeally quiet.


But go ahead. Make more excuses.

Aliaka wrote:
Hi guys, this is Ali
I was searching online for works that we can do in order to stop or at least slow down the global warming, and found this video which I hope will help all of you.
You can share it if you like, I don't mind

https://dausel.co/r4suvx
https://dausel.co/r4suvx

Aliaka wrote: I was searching online for works that we can do in order to stop or at least slow down the global warming,

tmiddles wrote:

IBdaMann wrote:
As far as I am concerned, you are simply enamored with "net flow"
.

Bottom line: You can't solve the problem following your own rules.

If you could you would.

And I notice ITN is reeeally quiet.


But go ahead. Make more excuses.

Into the Night wrote:
The problem as you have stated it is unspecified. There is nothing to solve.

tmiddles wrote:
"EXAMPLE 1.13
Calculating the Net Heat Transfer of a Person
What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient temperature is 22.0°C? The person has a normal skin temperature of 33.0°C and a surface area of 1.50m2. The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place.

tmiddles mindlessly copy-pasted without bothering to notice the glaring semantic issues of:
"EXAMPLE 1.13
Calculating the Net Heat Transfer of a Person
What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient temperature is 22.0°C? The person has a normal skin temperature of 33.0°C and a surface area of 1.50m2. The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place.

IBdaMann wrote:
Hint: ....Hint: ....Hint: ....Hint:

tmiddles wrote:You cannot figure out the thermodynamics because you'd have to admit you've always been wrong.

tmiddles wrote:IBdaMann and ITN simply explain how a person in a room radiates as per Stefan-Boltzmann and does not freeze to death.

IBdaMann wrote:
....Asked and answered.
.

tmiddles wrote:

tmiddles wrote:

IBdaMann wrote:
As far as I am concerned, you are simply enamored with "net flow"
.

Bottom line: You can't solve the problem following your own rules.

If you could you would.

And I notice ITN is reeeally quiet.


But go ahead. Make more excuses.

Hey Aliaka I'll bet you're an account IBdaMann set up to try to cover over his shame in this topic.

Not gonna work sorry bub.

If you can't explain the heat loss of a person in a room your goose is cooked.

tmiddles wrote:

Into the Night wrote:
The problem as you have stated it is unspecified. There is nothing to solve.

Here you go again:

tmiddles wrote:
"EXAMPLE 1.13
Calculating the Net Heat Transfer of a Person
What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient temperature is 22.0°C? The person has a normal skin temperature of 33.0°C and a surface area of 1.50m2. The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place.

tmiddles wrote:

IBdaMann wrote:
Hint: ....Hint: ....Hint: ....Hint:

Hint: A person in a room is you and me. You cannot figure out the thermodynamics because you'd have to admit you've always been wrong.

tmiddles wrote:
IBdaMann and ITN simply explain how a person in a room radiates as per Stefan-Boltzmann and does not freeze to death.

tmiddles wrote:
That's your question.

tmiddles wrote:
You are in a hopeless pickle of your own making.

Into the Night wrote:
The person will primarily heat the room by conduction (assuming normal air pressure and humidity), not by radiance.

tmiddles wrote:

IBdaMann wrote:
....Asked and answered.
.

You get to say that when you've had an answer.

So come on Mr. Scientist.

If an average human radiates over 700 watts, which works out about 14,000 calories, or 6 large pizzas, of energy, how on Earth are we all not hypotheric?

I have given my answer (Well agreed with the college text book).
BODY PHYSICS: MOTION TO METABOLISM Textbook

You have put so much energy into dodging.

Paragraphs about my motives and trickery. Just answer it.

Do it the right way.

Into the Night wrote:
Because we are in warm rooms and we are not overloading our body's ability to generate thermal energy.

tmiddles wrote:
Never answered you know it Duck Dodgers.

tmiddles wrote:

Into the Night wrote:
The person will primarily heat the room by conduction (assuming normal air pressure and humidity), not by radiance.

Are you claiming that the person stops radiating?

tmiddles wrote:
You do realize Stefan-Boltzmann is quite clear that EVERY BODY RADIATES.

tmiddles wrote:
What radiance do you calculate?

tmiddles wrote:

Into the Night wrote:
Because we are in warm rooms and we are not overloading our body's ability to generate thermal energy.

How much energy is emitted through radiance?

I calculate over 700 watts.

How does a warm room help restore any of that 700+ watt loss?

Into the Night wrote:
Then he's dead. His body temperature is too low. His body temperature is approximately -123 deg F.

tmiddles wrote: You get to say that when you've had an answer.

Into the Night wrote:I already calculated it for you. Go back and look it up again.

IBdaMann wrote:it was in this very thread that I answered your question and I included a nifty picture.

tmiddles wrote:
So I would suggest we need to consider how the human body, even though it's hotter than it's surroundings, is able to absorb heat from those surrounding,

GasGuzzler wrote:

tmiddles wrote:
So I would suggest we need to consider how the human body, even though it's hotter than it's surroundings, is able to absorb heat from those surrounding,

Why do I sweat when my surroundings are cooler than my body?

Just "food" for thought.

GasGuzzler wrote:...
Yes, the only energy source is food, provided there is no light in the room.

tmiddles wrote:

IBdaMann wrote:it was in this very thread that I answered your question and I included a nifty picture.

You just love to hide the ball! "Oh hunt through my post to find the answer"

You never did. Of course not. The threads not that big.

So to keep this int he appropriate thread:

If a person has 2000 calories of food, which is 7/8ths of a pizza, per day

that works out to about 100watts of thermal energy.

24*60*60 = 86,400 seconds a day
1 Calorie = 1000 calories
1 calorie = 4.18 Watts
so 2000 Calories = 8,360,000 watts a day
so about 100 watts

Now to maintain body temperature and lose 700 watts a person would need 7 times the calories, 14000, or about 6 large pizzas a day:



Now I think I could do it!

But keep in mind this would be just to not go hypothermic in a room that is 20C, 68 F.

So I would suggest we need to consider how the human body, even though it's hotter than it's surroundings, is able to absorb heat from those surrounding, or how heat is reduced or however we want to tackle it.

Clearly there is a missing piece of your puzzle here.

GasGuzzler wrote:
As I stated earlier, these calculations are way over my head. With that said, I did a bit of searching and EVERY source I can find states that the human body emits only 100 watts. How did you get 700?

tmiddles wrote:
BODY PHYSICS: MOTION TO METABOLISM Textbook

P(out)=σeA*To^4 (the radiance out To as in Temperature for out)

P(in)=σeA*Tenv^4 (the radiance being absorbed from the environment)

P(net)=σeA*(Tenv^4-To^4) (the NET gain or loss between the two)

tmiddles wrote:
It gives an example:
"EXAMPLE 1.13
Calculating the Net Heat Transfer of a Person
What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient temperature is 22.0°C? The person has a normal skin temperature of 33.0°C and a surface area of 1.50m2. The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place.

Strategy
We can solve this by using the equation for the rate of radiative heat transfer.

Solution
Insert the temperature values T2=295K and T1=306K, so that

Qt=σeA(T2^4−T1^4)=(5.67×10−8J/s⋅m2⋅K4)(0.97)(1.50m2)[(295K)4−(306K)4]=−99J/s=−99W." It's actually: 98.5320

Now if you do that calculation without including T2, the surrounds, you get:
Qt=σeA(T2^4)=(5.67×10−8J/s⋅m2⋅K4)(0.97)(1.50m2)[(295K)4]=−723J/s=−723W , it's actually: 723.3221