Removing heat from a room 3

[treddie]

Hi.

I have a solution to a problem here, that I would like to have checked by someone in the know, if anyone can spare the time.

I was wondering about a simple problem...Let's say I have a room of a given size. Also assume that the walls of the room are perfect insulators. The heat energy that is in the room cannot leak out through the walls, and no heat from outside the room can leak in. It is more a huge box than a room with windows and doors, except that a "perfect" air conditioner is included too. All heat transfer from the room will occur through the air conditioner system.

Now, let's say that the temperature in the room is 90degF, and I want to drop the temperature down to 77degF. I know that there is latent heat in the system, but I don't care about that because I just want to remove 13degF of "visible" heat from the room, which is immediately sensed as temperature. Therefore, I assume the following equations should give me the equivalent amount of energy for one molecule in the room:
T_hot = 90 degF (32.222 degC) = 305.372 degK
T_cold = 77 degF (25 degC) = 298.15 degK
deltaT = T_hot - T_cold = 7.222 degK

deltaT = E_m / k, where:
deltaT = differential temperature (degK)
E_m = energy (joules) of the molecule
k = the Boltzmann Constant (1.38x10^-23 joule/degK).

Therefore, E_m = k*T

Doing the math:
E_m = 1.38x10^-23 joule/degK * 280.372 degK = 3.87x10^-21 joule, of kinetic energy in one molecule in the room.

Now, according to the Ideal Gas Equation, which should be fine for conditions relatively close to STP:
P*V = n*R*T, where:
P = Gas pressure (pascals) = 1atm = 101325 pascal
V = Volume of the room (m^3) = 5m * 6m * 5m = 150m^3
n = Number of moles of gas in the room (mol)
R = Gas Constant (8.314 joule/(degK*mol))
T = T_hot, temperature of room before cooling has begun (degK) = 305.372 degK

Therefore, n = (P*V) / (R*T)

Doing the math:
n = (101325 pascal * 150m^3) / (8.314 joule/(degK*mol) * 305.372 degK) = 5986.44 mol of particles in the room.

Then, the total amount of energy we want to remove from the room is:
E_tot = E_m * n * Avogadro's Number, where:
Avogadro's Number = 6.022x10^23

Doing the math:
E_tot = 3.87x10^-21 joule * 5986.44 mol * 6.022x10^23 = 13951482.23 joules

Converting to KW-Hr (where 1 joule = 2.78x10^-7 KW-Hr):
E_tot = 3.86 KW-Hr

Now assuming an air conditioner rated at 48000 BTU/Hr of power, and disregarding any losses:
1 BTU/Hr = .00029 KW, so 48000 BTU/Hr = 13.92 KW
Finally,
TimeToCool = E_tot / 13.92 KW = .28 Hr = 16.8 min

Does this sound right to anyone? I know the numbers are in the correct range, but is that a coincidence, or is my math sound?

Thanks for any responses!

 

[IRstuff]

I think your answer is optimistic

http://www.energystar.gov/ia/home_improvement/home_sealing/RightSized_AirCondFS_2005.pdf

TTFN
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[willard3]

See forum policies which begin as follows:

.>>>>>>>>These fora should not be used to bypass your own in-depth research on the issues that affect you, nor is it intended to be a substitute for appropriate professional assistance within your field or geographical region.

Hire an Engineer familiar with this.

 

[317069]

Doing the math:
E_m = 1.38x10^-23 joule/degK * 280.372 degK = 3.87x10^-21 joule, of kinetic energy in one molecule in the room.

From wher did you get 280.372 [K]
In Celsius 32.222 - 25 =7.222 [C]
In kalven 305.372 – 298.15 = 7.222 [k] too.

 

[treddie]

.>>>>>>>>These fora should not be used to bypass your own in-depth research on the issues that affect you, nor is it intended to be a substitute for appropriate professional assistance within your field or geographical region.

Hire an Engineer familiar with this.

No need to...This was a problem inspired by my HVAC biting the dust, to understand the underlying fundamentals of thermal energy in air and how that energy can be moved. Also, what you see WAS in-depth research into a physics topic.

I think your answer is optimistic

Yes, but it was meant to be, to simplify the initial math. I was simplifying the entire problem down to an ideal solution, where there were no losses. I can take that and optimize the problem for an accurate real-world situation, once the basic math is in place (please see next paragraph).

From wher did you get 280.372 [K]
In Celsius 32.222 - 25 =7.222 [C]
In kalven 305.372 – 298.15 = 7.222 [k] too.

DoH! You are correct...Dumb high-school error there. I had caught the error earlier, but hadn't fixed my written "scratch" sheet. So that results in values that no longer seem in the ballpark. The energy to remove out of each molecule is now only 9.97x10^ -23 joule, which results in total room energy to remove = 348121 joules (.097 KW-Hr). This would only take about 25 sec to cool. That is way off. So I have to assume my math is just plain wrong, or it IS correct, but that real-world losses will bring the answer back into the expected range. If the latter is correct, than I would tend to think that insulation losses would be the main culprit. Inefficiency of the compressor (not accounted for here) shouldn't prevent my numbers from at least getting into the ballpark.

 

[Zesti]

So, what we are really doing here is cooling 150 m3 of air...

m.c.dT = Volume.density.c.dT = 150 x 1.2 x 1008 x 7.222 = 1,310,359 Joule

13.92 kW = 13,920 Joule/s

1,310,359 / 13,920 = 94 seconds to cool 150 m3 of air 7.222 degrees C

So, your value of 23 seconds is closer to the real answer than you think.
Now all you have to do is find where you went wrong by a factor of 4.

 

[chicopee]

The problem presented by Treddie seems to be a thermo assignment from his thermo. professor or teacher. I am saying that because the problem has too many assumptions.

 

[treddie]

"So, what we are really doing here is cooling 150 m3 of air..."

Hold on. I am working the problem from a different angle and making it more real-world. Am running into problems with differing ways of describing R-values as including BTU, or BTU-in! I HATE conversions!

"The problem presented by Treddie seems to be a thermo assignment from his thermo. professor or teacher. I am saying that because the problem has too many assumptions."

Sounds like it, but in this case, I'm my own professor and student. I can't say the professor part is contributing much! My A/C bit the dust last month, and it got me thinking about heat transfer in general. So I challenged myself to figure out some model that would eventually end up being a fairly accurate representation of outdoor heat fighting through walls with an air conditioner. What I have so far is something that numerically integrates heat transfer into a room, without air conditioning in the loop, yet. Currently, it comes up with rates that are 1.7-to 2 times too fast for how long it takes to get heat through the walls and raise the room temp. And that includes an assessment of building materials used, window spaces and air leaks. Right now, I am assuming 20% losses. I base my comparison on a stopwatch test of my living room temp increasing, with the A/C turned off, and a measurement of the outside temp. My program includes pressure at altitude, but assumes dry air. I have deferred the A/C cooling part until I get the heating model working properly.

 

[willard3]

This is a forum on the web. If you want to understand thermodynamics, go take a course in thermodynamics.

This forum is not a substitute for an Engineer because you don't want to pay for one nor is it an Engineering classroom because you don't want to pay for a course.

 

[IRstuff]

These forums are intended for work-related questions, hence the pushback. Consider RF'ing this post and posting in forum1495 which is intended for non-work-related questions.

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[IRstuff]

because it's a Pub link, it doesn't autoconvert the forum name:

http://www.eng-tips.com/threadminder.cfm?pid=1495

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[treddie]

IRstuff > Thank you for the clarification. I was a little taken-aback by a couple of negative comments. I really appreciate the help of those who were interested in engaging a dialog with me, and apologize for misplacing my OP in the wrong forum. What I will do is just start a new post over in forum1495, per your suggestion. Especially since my delving into this has led to an entirely different solution path since I started. I much more successful one, I might add.

 

[IRstuff]

Just be aware that the Pub is a wild and woolly place ;-)

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[treddie]

Oh...The Barbary Coast! I can take it! )

 

[IRstuff]

You might post this in forum1528 which does not require Pub access

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[treddie]

Thanks, IRstuff...I registered but I think I was going to have to wait till the end of the weekend to get in. So I put off posting my new thread. I think I'll go ahead and try 1528.

 

[treddie]

OK, I created the new thread, once I got my new program running. It's in

http://www.eng-tips.com/threadminder.cfm?pid=1528,

and is titled, "Heat Transfer in a Room and Air Conditioning".

 

[treddie]

Direct link:

thread1528-350467