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> Calculating Ambient Heat in a bedroom, Its too hot in there.
searcherrr
  Posted: Oct 24 2005, 03:46 PM


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Given a room 10W x 14L x 7H I need to calculate the ambient heat generated by the utilities running within it. Also bare in mind that this room has a window facing the sun from about 9AM till 3PM everyday and is yet covered with 3 layers (mini blinds, pull down blind, and curtain). Calculate heat generated and compare to something in English which describes the amount of heat generated by the objects just being plugged in and not necessarily in full operating state as well as in full operating state (2 answers). The room has the following properties which are always plugged in:

4 alarm clocks always on smile.gif
3 fans (2 small oscillating; 1 large stationary; all always running)
1 Coffee pot with timer/alarm
1 lamp usually off
1 lamp usually on at low power
1 ceiling light usually off
1 64 bit laptop usually always on and running at full operating state
1 36" Sony Trinitron picture tube TV (on 30% of time)
1 1.8 cubic ft refrigerator
1 microwave
1 dvd player
1 dvd recorder/vhs recorder combo
1 ps2
1 desktop hp printer
1 external enclosure hard drive

Other mass taking up space:
A sofa, computer desk, bunk bed, shelf, medium size trash can, large 32 qt rubbermaid container, large suitcase, full closet, computer monitor, broke *** desktop computer, and misc smaller items.

The room has always been hot even before all this "stuff" was in it and it was always thought it was due to the sunlight hitting the window and that side of the house all day, but given recent cold weather while the rest of the house is cool in every room this is the only room that remains warmer than comfortable (sweating heat). Maybe I'll get laughed off here, but I figured if I provided enough information that someone could tell me what I need to do to keep the room cool enough to be comfortable without removing stuff or without adding to noise by buying a portable a/c unit. The room has a central A/C vent which is about 12" x 6" and I feel it is not sufficient. I'm also suspect of not having enough insulation above and on the side of the room walls.
Anyway if someone can figure this out for me and let me know or tell me how to figure it out myself I'd greatly appreciate it. It just throws me that while most of the "powered" stuff is usually OFF that the room could be that much hotter anyway from those devices.
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searcherrr
Posted: Oct 24 2005, 10:27 PM


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Nobody aye? LOL - Well at least I gave it a shot. I figured someone who likes volume and temperature formulas would give this a go.
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Foxx
Posted: Oct 24 2005, 11:01 PM


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QUOTE (searcherrr @ Oct 24 2005, 10:27 PM)
Nobody aye? LOL - Well at least I gave it a shot. I figured someone who likes volume and temperature formulas would give this a go.

Hi Searcherr. Welcome to the forums. I'm sorry that I myself can't help with your question, but don't get discouraged. I'm sure someone with expertise in this area will be along to help you. If you look at the member list you will see that there are quite a few, and not everyone checks in every day. THE particular person who would like to and be able to assist probably just hasn't checked in yet. Give it some time.

Perhaps it is a paranormal phenomena ? biggrin.gif


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Hold ON,
Hold on to yourself...
for (the TRUTH) is gonna hurt like hell.


http://video.google.ca/videoplay?docid=-6714356054823827684&q=911
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adoucette
Posted: Oct 25 2005, 12:33 AM


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It doesn't appear that you have that much heat being generated in the room but unless you post the wattage of each device its hard to add it all up. The TV and the computer would seem to be the big offenders followed by the microwave (if you use it a lot) and the refrigerator.

You could consider a booster fan that fits in the air duct leading to the room, if the room is a long way from the Central Air unit, the flow might be a bit low.

As to the fans, if you put one of the fans so it will blow air OUT of the room (preferably near the top of the door), it will then draw in cooler air from the rest of the house, thereby making the room more liveable.

For the window, consider a reflective coating, applied to the inside pane, to reduce solar heating, you can find them at hardware stores.

Arthur


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"We cannot prove that those are in error who tell us that society has reached a turning point; that we have seen our best days. But so said all before us, and with just as much apparent reason. On what principle is it that, when we see nothing but improvement behind us, we are to expect nothing but deterioration before us?"

Thomas B. Macaulay
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faber
Posted: Nov 30 2005, 03:57 PM


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Ok, here is a similar case.
Much simplier though :)

The idea is to put a computer system in a small room, out of bedroom where it produces too much noise and heat.
I want the system to run 24/7, but room I have for it is really small, and I am worried about the ambient temperature raise during my absence at home...

I would love to get a formula to calculate what time does it take to raise an ambient temperature by 1 degree Centigrade if you put 1000BTU/hr heat source in a 1m/1.5m/2.5m room.

I suspect that a temperature raise it is not linear, so formula would be the best solution.

Perhaps Data Center specialists have some methods of calculating such things?

TIA
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Issachar
  Posted: Nov 30 2005, 08:15 PM


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In response to the first question.
If the room is hot at night during the winter, you know you have more to deal with than too much exposure to sun on that side of the house, though limiting the exterior heat transfer during daylight hours will aid in making it more comfortable.

As Arthur pointed out, positioning the fans to improve the circulation in and out of the room will also be of significant benefit.

You basically have a number of heat sources drawing a current on all the time in a small room the sum of which add up. The computer, refrigerator, TV when in use, and lamp as mentioned seem to be the biggies, but everything else adds up as well. Turning off when not in use or lower wattage bulbs will help, but you want to use this room.

Is there a utility room containing a hot water heater, etc. adjacent to a wall?
For all your heat sources given the size of the room, it seems that hvac system is not properly balanced to keep enough air flowing into the room. You need to increase air flow into the room to take away the higher temperature air.
The booster fan in the duct is a 'Fantastic' idea, but it might be as simple as damping some of vents in rooms with good air flow to increase the duct pressure allowing more air to flow to this room. Air flow through the room's small duct will need to be overcompensated by forcing more into this room. Do you keep the door open if there is not a return air vent in the room? Changing air filters on a regular basis, something I forget to do, also helps ensure the current system's efficiency is not compromised. Plan B might be upgrade.

Faber, I might be able to dig out a formula later, if someone else doesn't have one handy for you. If there is no proper ventilation in and out of your proposed small room, you might consider adding a vent to the door to prevent overheating.
Best of luck!
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adoucette
Posted: Nov 30 2005, 08:46 PM


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QUOTE (faber @ Nov 30 2005, 03:57 PM)
Ok, here is a similar case.
Much simplier though smile.gif

The idea is to put a computer system in a small room, out of bedroom where it produces too much noise and heat.
I want the system to run 24/7, but room I have for it is really small, and I am worried about the ambient temperature raise during my absence at home...

I would love to get a formula to calculate what time does it take to raise an ambient temperature by 1 degree Centigrade if you put 1000BTU/hr heat source in a 1m/1.5m/2.5m room.

I suspect that a temperature raise it is not linear, so formula would be the best solution.

Perhaps Data Center specialists have some methods of calculating such things?

TIA

You need to provide more data.

Is this a CLOSED room?

Does it have vents?

Does it have windows?

Is the cieling insulated?

Are the walls insulated?

What's the highest temp the house in general gets to?

1,000 BTU's an hour is a fairly small amount of heat, so I'm going to say offhand that this shouldn't be a problem, but without the answers to the above its hard to offer any real advice.

Arthur


--------------------
"We cannot prove that those are in error who tell us that society has reached a turning point; that we have seen our best days. But so said all before us, and with just as much apparent reason. On what principle is it that, when we see nothing but improvement behind us, we are to expect nothing but deterioration before us?"

Thomas B. Macaulay
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Schneibster
Posted: Nov 30 2005, 09:07 PM


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Hmmmm, OK, I'd say that the 'fridge, the laptop, the external HD, and the ps2 are the major sources of heat INSIDE the room, but the major heat source FOR the room is the window that faces the Sun. This makes putting some sort of reflective substance on the windows a priority; adoucette suggested the hardware store, I'm going to suggest the automotive store. You can get mylar film that you put on by cleaning the window and squeegeeing the mylar on with water, and this is known to help with vans and campers.

You should also check the insulation in the walls, and if the attic is above the room, or the roof, then check that too; finally, the suggestion about checking next door for a utility (hot water heater, furnace, whatever) is also a good one.

In terms of addressing the problem, the mylar is a good start, but if you've got an insulation problem, you're going to find it cheaper to solve the problem with your fans than by messing with insulation. You'll want to eject heated air, and inject cooler air. Remembering that hot air rises, you'll want to blow air out high and suck it in low. If you set up to do this air exchange through the room's door with the rest of the house, this will have the added benefit of increasing the temperature and triggering your central A/C sooner; the balancing of the air ducts then becomes another important step that might help (based on personal experience, I don't expect it would be enough by itself). Remember however that if you mess with the air ducts, you must never completely close one off! This can cause long-term problems with your A/C and furnace, due to unbalanced pressures.

None of these solutions is complete; you'll almost certainly have to use more than one of them; but then again, none of them is particularly expensive either, so I'd do one, wait a week and see how it works, and then do another if needed, and so forth (a few days might do it, but be sure to account for cloudy vs. warm days).

Best of luck!
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Issachar
Posted: Dec 2 2005, 09:36 PM


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Well Faber,
I will present a very generalized solution below and then anyone else may supplement, check my logic or improve on it, etc.
Better details about the construction of the room as those mentioned prior would be needed. It would also be helpful to know if it is an interior room. If one wall is facing the exterior, what is the exterior wall composition and do we have a window?

We are looking for the change in energy storage to the room over time:
∆E stored = E in + E generated E out

1.) In the base case, we will assume a uniform room with E in = 0, which is not the case if your room has an exterior wall facing the sun.

2.) E gen = 1000 BTU/hr = 1000* .292875 Watts = 292.875 W
We will assume the initial steady state inside and outside the room to be 'room' temperature, 70F = 21.1C +273 = 294.1K

3.) E out = q conduction = q convection + q radiation

3a.) The rate equation for heat conduction is known as Fourier's law. For a one dimensional plane wall, the rate equation is expressed as:
qx" = -k*(dT)/(dx) ,
where the heat flux q(subscript x) ,unit (W/m^2), is the heat transfer rate in the direction x per unit area perpendicular to the direction of transfer, and it is proportional to the temperature gradient, dT/dx in this direction. k is the transport property proportionality constant called the thermal conductivity (W/m*K) and is a characteristic of the wall material. (The negative is due to the direction of heat flow)
The conduction heat loss q cond = qx* Area = -k(dT)/(dx)*m^2

The thermal conductivity constant k for gypsum sheetrock is 0.17 (W/m*K).
In the base case, it is assumed the ceiling and floor are well insulated and conduction surface area of the walls are of uniform materials (the thermal conductivity difference of a wood door is not greatly different than sheetrock and is ignored).
The inch wall thickness is approx. 0.01905 meters
The total wall surface area is: 2 * (1m +1.5m) * (2.5m) = 12.5m^2
The conduction heat loss would then be:
(-0.17W/m*K)*(T1-T2 K)/(0.01905m)*12.5m^2 =-111.5486 W/K*(T1-T2 K)

3b.) The rate equation for heat convection transfer is:
q conv = h * (T surface -T surrounding), from Newtons law of cooling, where the proportionality constant h (W/m^2*K) is the convection heat transfer coefficient or the film coefficient, Tsurface is T2 at the exterior wall and Tsurrounding is the ambient room temperature. (This is probably not the case for air between 2 layers of sheetrock in a wall, which is what is assumed here, since we are only taking the 1st layer of sheetrock into account).
The convection heat loss is the product of the convection heat flux qcon and the surface area, A.
h for fee gases has a range of 2-25 W/m^2*K, we will use approximate the h surrounding air to be 20.
The convection heat loss would then be: (20 W/m^2*K)*(T2 K - Tsurr)*12.5 m^2
=250 W/K*(T2 - Tsurr K)

3c.) The rate equation for heat radiation is:
q rad = ε * σ * (Tsurface^4 - Tsurrounding^4) ,
where q rad is the radiation heat flux, ε is a radiative property of the surface called emissivity (0.90 for gypsum sheetrock), σ is the Stefan-Boltzmann constant = 5.67x10^-8 (W/m^2*K^4),
Tsurface is T2 at the exterior wall, and Tsurrounding is the ambient room temperature.
Multiplying this by the surface area, A will give the radiation heat loss.
For the base test case, the variance of loss to the studs and their surface area touching the gypsum walls are ignored as negligible.
q rad = 0.90* (12.5 m^2) * (5.67x10^-8 (W/m^2*K^4 ) * (T2^4 - Tsurr^4) K^4
q rad = 6.37875E-07 * (T2^4 - Tsurr^4) W

4.) We know that for steady state system, E in E out = 0,
then on a unit area basis, q cond q conv q rad = 0,
when rearranged with formulas is:
k*(dT)/(dx) = h*(T2 T surr) + ε*σ*(T2^4 - Tsurrounding^4)
or
(0.17W/m*K)*(T1-T2 K)/(0.01905m) = 20W/m^2*K*(T2-Tsurr K) + 5.1030E-08 W/m^2*K^4 *(T24-Tsurr^4).
For a 1 degree temperature rise, T1 = 71F=295.1K, Tsurr is the ambient room temperature of 294.1K.
Solving for the exterior wall surface temperature on my computer, T2 =294.361518

5.) Solving for ΔE stored needed to get to a 1K temperature rise:
ΔEst = Egen - Eout = Egen Econd = 292.875 W -111.5486 W/K*(T1-T2 K)
=292.875W-111.5486W/K*(295.1-294.361518 K) = 210.4984 W
Solving for Btu/hr needed to get a 1K temp rise:
210.4984 W*(1 Btu/hr/.292875W)=718.73 Btu/hr
Since we only need 718 Btu/hr increase the temp but are generating 1000 Btu/hr it would take .718 hr or 43 min to raise the temp 1K or 1C.

What this means: you will experience temperature rise in your room, unless you provide a modest level of air exchange in the room, i.e. vented door or the door to remain open.
Good luck!
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faber
Posted: Dec 7 2005, 02:40 PM


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QUOTE (adoucette @ Nov 30 2005, 08:46 PM)
You need to provide more data.
Is this a CLOSED room?
Does it have vents?
Does it have windows?
Is the cieling insulated?
Are the walls insulated?
What's the highest temp the house in general gets to?
1,000 BTU's an hour is a fairly small amount of heat, so I'm going to say offhand that this shouldn't be a problem, but without the answers to the above its hard to offer any real advice.

Is this a CLOSED room?
-Yes
Does it have vents?
-I was thinking of installing vent grid in the door. But only passive ventilation system.
Does it have windows?
-No.
Is the ceiling insulated?
-Not directly. The room is located on the 1st floor, there are 4 condignations above it wink.gif
Are the walls insulated?
-Just brick/concrete.

Will that information be enough?
BTW, for more general use, let's think of another scenario:
Small server room 3m*3m*2.5m with suspended ceiling without active ventilation system.
Room with one wall facing sun between 10AM and 4PM.
With or without windows.
Hardware consuming 5000W.
QUOTE (Issachar @ Dec 2 2005, 09:36 PM)
Well Faber,
(...)
For a 1 degree temperature rise, T1 = 71F=295.1K, Tsurr is the ambient room temperature of 294.1K.
Solving for the exterior wall surface temperature on my computer, T2 =294.361518

5.) Solving for ?E stored needed to get to a 1K temperature rise:
?Est = Egen - Eout = Egen –Econd = 292.875 W -111.5486 W/K*(T1-T2 K)
=292.875W-111.5486W/K*(295.1-294.361518 K) = 210.4984 W
Solving for Btu/hr needed to get a 1K temp rise:
210.4984 W*(1 Btu/hr/.292875W)=718.73 Btu/hr
Since we only need 718 Btu/hr increase the temp but are generating 1000 Btu/hr it would take .718 hr or 43 min to raise the temp 1K or 1C.


Forgive my ignorance, but this logic leads to a conlusion that the speed of heat transfer is constant and independent of temperature difference value.
Is ambient temperature rise speed in a room really linear?

--
Faber
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