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# Heat Transfer in a Room and AIr Conditioning7

## Heat Transfer in a Room and AIr Conditioning

(OP)
This is a new post that started with my original OP over in forum403: HVAC/R engineering, titled, "Removing heat from a room". I am moving the discussion here since I apparently pissed off a couple of responders for not placing this discussion in a non-work-related forum. I think they thought I was some kind-of freeloader wanting free information for a college class assignment or something. Don't know how that could have been, since I had started off the thread with all of my work to date, showing that I was clearly trying to understand the problem for myself.

To make clear my intentions, this is not a college class assignment. This was simply a spare-time project of mine to understand heat xfer across walls and air conditioning a volume, due to my A/C biting the dust. I have already bought two portable A/C's as an interim solution to my cooling needs until I can afford the roughly $8000 I will need to replace my 15 year old, 48000 BTU central air and furnace. So this forum post is purely recreational in nature. It is simply to understand out of curiosity what is going on with heat xfer and air conditioning and accepting a personal challenge to come up with a solution to it. At any rate, I have started from a completely different direction than in my original post. The problem in its entirety is to understand heat xfer across walls from outside, into a simple rectangular room with five exposed surfaces (4 walls and ceiling). Then, to address how an air conditioner can fight that heat load to keep the room temperature to within desired temp + 2 deg. The + 2deg is to allow the air conditioner some down-time so that it is not constantly cycling on and off at an unacceptably high rate. For now, the problem deals with only the heat xfer across the walls problem. The air conditioner part of the problem will be added later once I have the heat transfer problem solved to a good degree of accuracy. My last post started with a too-simplistic solution that had heat xfer rates that were WAY too fast. I also made some errors at the top of the post that were creating havoc. So I went back to square one and rebuilt the problem, basing it on a numerical integration method. As stated above, I decided to not address the cooling part of the problem yet, and, instead, concentrate on the heat entering the room from outside. I will leave the air conditioner's fight with that heat load to later, once I have the external heat entering the room solved first. I already knew that dT/dt = c(deltaT), where (c) is some proportionality constant. So I then researched what (c) could possibly be and ran into this equation: Heat Transfer Rate = k*A*deltaT / L, where (L) = wall thickness. Turning this into a differential equation to get at elapsed time, I came up with: dQ/dt = k*A*deltaT / L which becomes: dt = (L/ k*A) * (1/deltaT) dQ Integrating: t = (L*Q) / (k*A*deltaT) Solving numerically for Q: Q = (k*A*deltaT*TimeStep) / L , where TimeStep = the brief length of time of one iteration in a loop. Since I needed to use thermal conductivity for entire wall surfaces, instead of a unit (k), (k) becomes (U), and the above equation changes to: Q = (U*A*deltaT*TimeStep) During each numerical iteration, I calculated the deltaT that would occur in the room, based on the heat passing through each wall. For convenience sake, here, I only show the contribution for one wall. So the pseudo-code for the loop I created looks something like this: '------------------------------------------------------------------------------------------------------------------- t = -TimeStep Recycle: t = t + TimeStep deltaT = T_e - T_i 'deltaT between outside and room temperatures. 'Heat energy that has crossed wall into room (joules): Q = (U * A * deltaT * TimeStep) 'This section assumes no change in room pressure and any resulting equalization air drafts between the room and the 'outside world): Call separate routine to get specific heat capacity and density of moist air. For dry air: Density_air = P_alt / (R_s * T_i) 'Density of air in room (kg / m^3), P_alt = pressure at altitude, 'R_s = Specific Gas Constant. Mm_air = R / R_s 'Molar mass of dry air in room (kg/mol), R=Gas Constant. V_m = Mm_air / Density_air 'Molar volume of air mass in room (m^3 / mol). n = V_r / V_m 'Moles of air in room (mol). m_r = Mm_air * n 'Total mass of air in room (kg). deltaT_prime = Q / (m_r * c_air) 'Change in room temperature (degK) during current iteration. T_i_old = T_i 'Save the old room temp, for later. T_i = T_i + deltaT_prime If (loop reaches some prescribed end-state) Then Exit Sub End If Print results Goto Recycle '------------------------------------------------------------------------------------------------------------------- As the code above only shows the basic concept, it does not show that I went to the trouble to actually BUILD the walls and ceiling in the program, basing the framing and finishing on UBC codes. The only deviation I made in that regard was to treat the ceiling trusses as solid wood joists to simplify the modeling problem a tad. I also included relative humidity as an input, although I found what many online were saying...Humidity barely affects the results. The image below shows how I handled the simplification of the original downstairs area: [/URL]] When I first ran the program, I was still getting heat-up times per degree temperature, that were about 2 times too fast. The first problem I noted was that I had set my default exterior temperature too high. Once I fixed that I got results that were a lot closer but still too fast. That was when I realized that in my haste, I had defined all four walls as exterior walls when in fact, one of them is an internal wall separating the 18 foot tall living room from the upstairs area. That wall also leads into the kitchen (one of the side-volumes), which butts up against the garage. So I added a feature to the program to "turn off" any of the walls, so that they are not included in the exterior wall area. When all is said and done, I found that the program appears to be very accurate, if I accept my interpretation of readings I was getting back from my lousy thermometer with its +- 2deg F accuracy. I turned off the downstairs portable A/C and the upstairs window A/C, and waited downstairs to see how long it would take to raise the temperature 5 deg to get a sufficient temperature spread to work from, logging every minute. I am going to get a precision digital thermometer that will verify if my reasoning has been correct. If so, I will move on to the air conditioning part of the program. Here is an image of the current GUI: [/URL]] ### RE: Heat Transfer in a Room and AIr Conditioning (OP) Also, no provision yet, for incident sunlight or convection due to wind or breezes. I think that the model I used to represent the wall areas and ceiling is fine for what I was attempting to do. To get a real-world idea of how an entire house absorbs heat, I will need to nix this approach and model a house as it truly is built...Where all walls are faithfully modeled, and then do the equivalent of a an electronics network analysis, so that the house is reduced to an equivalent rectangular box, kind-of like I did here, but much more extensive. ### RE: Heat Transfer in a Room and AIr Conditioning There are tables available which give heat transfer values for typical walls. These reduce the calculations to (table value) x (area of the wall) x the (temperature difference across the wall). Don't forget the floor. Don't forget air changes. (Doors opening and air leaks through shoddy construction.) Don't forget occupant load. As a quick check try logging the A\Cs run time. During a period of stable outside temperature, run time per hour x BTU rating of the A/Cs will give a fair indication of the heat load in BTU per hour. The efficiency of an A/C is in part dependent on the temperature difference between the condenser and the evaporator. The bad news is that the measured values of BTU/Hr may not be accurate compared to the real or calculated values. The good news is that the shift in efficiency with the change in temperature difference is a characteristic of the refrigeration cycle and will affect all A/Cs. Using the time logging method may give better results when it comes to selecting a replacement AC based on rated BTU/Hr. Bill -------------------- "Why not the best?" Jimmy Carter ### RE: Heat Transfer in a Room and AIr Conditioning You're missing heat transferred from above (the attic) and from below (as the room tries to exchange heat coming up form below (the basement in this case); both of which will vary through the time-of-day with a significant delay factor through the attic insulation and less delay through the smaller floor wood resistance. ### RE: Heat Transfer in a Room and AIr Conditioning Google " radiant time series method". Follow enough references will eventually lead you to the fundamental development work, which is pretty much what you are trying to independently develop. ### RE: Heat Transfer in a Room and AIr Conditioning (OP) First, thanks to everyone for chiming in on this! #### Quote (waross) There are tables available which give heat transfer values for typical walls. These reduce the calculations to (table value) x (area of the wall) x the (temperature difference across the wall). Do you have a link to the tables? I have spent a lot of time researching this, and have never run into them. They would be nice in order to check the accuracy of my approach. #### Quote (waross) Don't forget the floor. I'm not sure what to do about that yet...I have a cement foundation. My first impression, right or wrong, was that it's pretty well isolated. On the other hand, it's definitely acting as a sink for the surrounding ground heat. #### Quote (waross) Don't forget air changes. (Doors opening and air leaks through shoddy construction.) I was limiting the method to a "closed" volume. Other than that, I have already added the first of what could be multiple inputs. The one for now is a percent loss due to air leaks, which I have arbitrarily set at 10%, for now. #### Quote (waross) Don't forget occupant load. I originally had decided to leave that out, but you're probably right, so I will put it in as an option. #### Quote (waross) As a quick check try logging the A\Cs run time. During a period of stable outside temperature, run time per hour x BTU rating of the A/Cs will give a fair indication of the heat load in BTU per hour. The efficiency of an A/C is in part dependent on the temperature difference between the condenser and the evaporator. The bad news is that the measured values of BTU/Hr may not be accurate compared to the real or calculated values. The good news is that the shift in efficiency with the change in temperature difference is a characteristic of the refrigeration cycle and will affect all A/Cs. Using the time logging method may give better results when it comes to selecting a replacement AC based on rated BTU/Hr. " I was really frustrated that my central air unit died before I took an interest in this. I only have a vague idea of how often the compressor was cycling. And the inability of SEER to reflect accurate conditions will be a big problem. Add to that, any temperature logging I do now is limited to a cheap digital thermometer (+-2degF). I really need to get at least a$20.00 unit with at least +-.1degF accuracy. Then, like you suggest, there is the frustrating (should be illegal) exaggeration of efficiency by manufacturers. Also, for my temporary portable unit (and portables/wall units in general) they usually do not post the fan amperage in their specs. So in order to get an amperage for it, I will probably have to buy an AMP clamp, too.

I'm still shady on understanding how to deal with A/C specs still. If it were as simple as BTU/Hr and SEER, it would be easy.

#### Quote (racookpe1978)

You're missing heat transferred from above (the attic) and from below (as the room tries to exchange heat coming up form below (the basement in this case); both of which will vary through the time-of-day with a significant delay factor through the attic insulation and less delay through the smaller floor wood resistance.

For now (probably out of laziness :)), I have limited the attic to just 10" ceiling joists (standing in for trusses to simplify the calculations), and 10" thick insulation. I have not found any references yet for roof material R-values.

As stated above, there is no basement...Just a cement foundation. But if I want this thing to be more versatile, I will definitely have to allow for basements.

#### Quote (MintJulep)

Thanks for that!!! It's really difficult when you don't know the terminology to use when researching. :)

### RE: Heat Transfer in a Room and AIr Conditioning

Also Google "flywheel effect" this is the phenomena where the building absorbs heat as the day heats up, and gives it back when the day cools down.
This tends to be more prominent with masonry and concrete than studs and insulation.
B.E.

You are judged not by what you know, but by what you can do.

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
Thanks! I'll check it out.

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
That's cool. Thanks. I tried looking at the page source to see if they put the code there, but I couldn't find it. :( So I'll have to use it online. Heheh.

### RE: Heat Transfer in a Room and AIr Conditioning

Look at the Maya web site: It has several on-line calculators for various geometries (vertical wall, horizontal floor (heated from below), horizontal floor (heated from above) that will help with both convection and conduction, thermal mass problems, etc.

You've got a complex problem - but it is a real world problem as well. Attack each "wall" as accurately as possible and you'll get through ... Well,the heat will always "get through" as well also, but at least you'll be able to more accurately predict how much is getting through. We've been trying to solve transient heat transfer problems in housing units ever since the first cave woman complained about the drafts coming in through the her front door.

'Course, she was also always complaining about the kids drawing on the walls, the debris building up outside the door, the animals that keep wandering through the kitchen, and who didn't make the bed furs up that morning ....

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
It seems to me that the problem is not how to get an accurate model. The problem seems to be how to do it efficiently. For instance, all numerical methods are approximations, and a FEM model COULD be built to solve it with a high degree of accuracy (no pun intended), but it might take forever and a day to get a solution that spans a 24 hour period. That having been said, I think the prickly parts of the problem have to do with all of the unknowns that come with humans actually building a structure. It is easy to spec out insulation for example, but how well was it installed? What issues during construction forced a bit of rethinking to decide how to proceed at some point? And how to get manufacturers to hand over unpublished specs about their air conditioners and furnaces. And predicting human traffic patterns will always be, at best, based on assumptions.

#### Quote (racookpe1978)

Look at the Maya web site: It has several on-line calculators for various geometries (vertical wall, horizontal floor (heated from below),

I will be curious how my results jive with theirs, since this whole project was to challenge myself to see if I could tackle the same problem solved by others.

### RE: Heat Transfer in a Room and AIr Conditioning

My calculations are as follows:

150 m3 air =5,296.5 ft3

90F air weighs about 14.2 ft3/lb

total weight of air is 5296.5/14.2 = 373 lbs

90F air has a heat value of 38.3 btu/lb

77F air has a heat value of 35.5 btu/lb

therefore heat to be removed = (38.3-35.5)x373 = 1,044 btu

An air conditioner supplying 48,000 Btu/hr will supply 800 Btu/min so it will take 1.3 minutes or about 78 seconds to cool the room, based on the ideal conditions described.

Defenestration, infiltration, conduction and heat sources within the space will affect this number.

### RE: Heat Transfer in a Room and AIr Conditioning

Interesting calculation.
But most important is the capacity to maintain the desired temperature.
That will be the sum of the followng.
Conducted heat through the walls floor and ceiling,
Cooling infiltrated air,
Heat contribution of appliances and equipment in the cooled space,
And,
Excess capacity will speed the cool-down when the system is initially turned on. Although the air may be cooled quite quickly, it will be continually re-warmed until the temperature of the walls, floor and ceiling are also drawn down.
These results will all be skewed by the physics of the refrigeration cycle.
The BTUs per KiloWatt drop off as the temperature difference increases.
As an example, a heat pump at an evaporator ambient of 32 Deg. F will produce about the same BTUs per kW as electric heat. With a higher ambient on a mild day it may produce three times as many BTUs per kW as electric heating.
The temperature differences are not as extreme with A/C applications but there is enough change to skew the precise results you seem to be striving for.
Take the temperature difference to be the condenser ambient temperature plus 10 degrees for the condenser and the room temperature minus 10 degrees for the evaporator temperature. Probably not exact but ignoring the temperature difference between the ambient and the refrigerant will through your results off more than my WAG.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)

#### Quote (VerneE)

An air conditioner supplying 48,000 Btu/hr will supply 800 Btu/min so it will take 1.3 minutes or about 78 seconds to cool the room, based on the ideal conditions described.

Well, now that I have what I THINK is a fairly accurate heat transfer rate, for my current purposes, I will be interested in seeing what happens compared to your result. But I left out an important detail...Sure, I had a 48000 BTU/Hr unit, but the volume in my sim is only the downstairs area, which is roughly 3/4 of the total. So a 36000 BTU/Hr A/C would be more accurate for the sim, to make a fair comparison to the whole house.

#### Quote (waross)

Take the temperature difference to be the condenser ambient temperature plus 10 degrees for the condenser and the room temperature minus 10 degrees for the evaporator temperature.

So if the conditions are 100 degF outside, and 80 degF inside the house, use 110 degF and 70 degF? Can you steer me to a description of why that is?

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
I see. So it's rather like an electrical circuit in that regard. That keeps popping up...When I was working my parallel R-values, I couldn't figure out why my answers were all whacked out...Until it finally dawned on me that you add parallel thermal resistances just like parallel electrical resistances...product-over-sum.

### RE: Heat Transfer in a Room and AIr Conditioning

The evaporator temperature must be colder than the ambient and the condenser temperature must be hotter than the ambient or there will be no heat flow. If you want to want to calculate the efficiency of the AC at different temperatures based on the latent heats of vaporization of the refrigerant at different operating temperatures you will need to use the actual temperature of the refrigerant, not the ambient temperatures.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: Heat Transfer in a Room and AIr Conditioning

Well, sort of.

But you can't use "the same" "outside temperature" for all of the places that are "outside" the room.

Bear with me a minute: It's easier to show than to describe.

Room is (trying to maintain) 73 F, right?
Outside air = 90 F.
Attic (in the afternoon) = enclosed space, heated by the sun, constant temp = 125 F
Basement = enclosed space, constant temp, cooled by ground = 66 F
North wall = (doesn't get sun's heat), hasn't heated up (yet) from outside air) = 85 F
South wall = does get sun's heat, is 98 F.
East wall = is (now) at outside air temperature = 90 F.
West wall = starting to get sun's heat, now at 93 F
Living room (other side of an interior) wall = other side = 73 F.
Kitchen (other side of an interior) wall is 80 due to heat from the refrigerator and oven and dishwasher heat sources over there.

Now, all of these temperatures are at the "wall surface", so the heat transfer for each is the area of wall/floor/ceiling through the thermal resistance of the wall/floor/ceiling (which will be different for internal and for external walls!).

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)

#### Quote (waross)

The evaporator temperature must be colder than the ambient and the ...

OK, that makes perfect sense. The compressor has to make those differences possible, and the efficiency with how it does it, I guess is the tricky part. For a working system, is it accurate enough to get those temps directly off the lines leading into and out of the outdoor unit and the attic unit? If so, which side of the condenser and evaporator do you use? I'm thinking condenser outlet and evaporator inlet. At any rate, it sounds like from what I am reading, that a clamp thermocouple is really the only way to get accurate readings.

#### Quote (racookpe1978)

Now, all of these temperatures are at the "wall surface", so the heat transfer for each is the area of wall/floor/ceiling through the thermal resistance of the wall/floor/ceiling (which will be different for internal and for external walls!).

That really shouldn't be that hard to do, if I treat all the walls as "external", in that regardless of whether or not the wall is exterior or not, it is still a radiator or a soak. I can also treat each wall with a uniform temperature. Once I know that, I can sum the heat contributions to the room. It could get difficult if I make the problem more complicated where two or more rooms share common internal walls. I don't know yet, but that could turn into a finite element method sort of thing where each room is affecting the other rooms simultaneously. Easy enough in principle to do, just time-consuming to code.

### RE: Heat Transfer in a Room and AIr Conditioning

I think that you may be trying to over engineer this. One hint is your search for a thermometer accurate to 0.1 degrees.
Find some good reference tables. (Sorry, I haven't done much A/C planning in the last few decades and my books are old fashioned paper. Throw in a move from North America to Central America and back and I don't know where the books are.)
Calculate the heat load from the ceiling, floor and each wall and add them. You should be working in BTUs per Hour.
Now add some BTUs for infiltration and air changes. Don't forget to consider any exhaust fans. You should be able to find some guidelines in design tables.
Note: A good set of tables will have allowances for joists and studs built in.
A set of tables may have tips for estimating infiltration and air change allowances.
A good set of design tables makes allowances for field factors and gives tips on estimating values.
Once you consider the cooling effect on the outside walls due to wind action you must also consider the relative humidity. Same for infiltration, and don't forget the effect of wind action on infiltration.
Consider also the effect of strong sunlight on the heat load of a south facing wall, oops a cloud just passed in front of the sun and everything changed again.
I'm of two minds here as to the value of over engineering.
1> It is a valuable learning tool. An in depth, rigorous, consideration of all the factors affecting the air conditioning of a space will give you an overall understanding of air conditioning design. Some factors you will mostly ignore in future, (The effect of clouds is one to ignore. Generally for A/C we are concerned with "Worst case" issues and cloud cover is an improvement on worst case.), some factors will be predominant and may be used as "reality checks" in future designs. Some factors may be important some of the time and should be evaluated and then either included or ignored in calculations depending on the situation.
Some variable factors to consider:
Wind; Imagine a day when the temperature is 98 deg.F, and the wind is blowing.
On the north side of the house the ruling temperature is 95 Deg F regardless of the wind.
On the south side of the house the sun will be adding radiant heat to the wall. The amount will depend on the time of day and the time of year. A wind will remove some or almost all of the heat added by the sun. The relative humidity will make a difference when the wind is weak.
The roof: A simple gable roof may gather a lot of BTUs from the sun. These BTUs may be removed by the wind or may be added to the air trapped in the attic. The air temperature in the attic may rise to quite a high value and so affect the heat load of the ceiling. A good attic ventilation system may remove most of this effect.
The floor; To make this more complete, consider that there is a basement below the the floor. One wall is exposed to the south and is affected by the sunlight and the other walls and the basement floor are in contact with the earth. Now we have one wall that contributes to the heat load and three walls and a floor that act as heat sinks. Start looking for R values and specific heat values for earth and don't forget the moisture content.

Now let's consider the A/C unit. The 10 degree figures that I quoted were for illustration only to illustrate that the temperature difference seen by the refrigeration side of an A/C unit are greater than the temperature differences seen by the air handling side of the A/C unit. The efficiency of the refrigeration cycle depends on the delta T of the refrigeration side.
Solve as many of these issues as you are able and then check your results with a couple of sets of published tables. You will probably only ever do this once. At the end of the day you will have a pretty good grasp of the factors involved in comfort cooling.
On the other hand;
If you just want to size a replacement unit I would do the following.
Did the existing unit provide acceptable cooling at all times? I will assume that the answer is yes. That becomes the upper limit for a replacement size.
If the answer is yes than that becomes the lower limit for a replacement size.
If the answer is no, you may be able to make a rough estimate of the size needed by considering the effectiveness of the replacements during a worst case event.
A couple of things to consider when considering changing A/C sizes as compared to a direct replacement;
Cost- The cost of an incremental size of the A/C section of your central furnace may be small compared to the overall cost of the repairs. I suspect that the labour cost will be significant.
Electrical supply. If the size of the replacement is smaller than it may be a code requirement to install a smaller breaker in the panel. Add the cost of the breaker and the extra labour.
I realize that an A/C is not a typical motor but never-the-less, electric motors often have a "sweetspot" of greatest efficiency at about 75% of rated load. The saving in operating cost may be marginal but over moderate over-sizing may not be a bad thing.
Thanks to racookpe and others for your suggestions and contributions.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)

#### Quote (waross)

I think that you may be trying to over engineer this. One hint is your search for a thermometer accurate to 0.1 degrees.

Well,...I kind of AM over-engineering it only out of curiosity for all the variables. I think it could all be simplified for generic uses afterwards, but my primary goal was to understand those variables and include them (until I reach a point where I'm tired of looking for more! :)) The thermometer accuracy thing was due to not trusting the accuracy as published for various thermometers. For instance, I have one that displays tenths of a degree for a thermometer that is only accurate to +-2 deg!!! I mean, they might as well just drop the frikkin decimal with that level of inaccuracy. So I would like a thermometer that IS accurate to .1 deg, so that I can trust the rounded values without the decimal.

Another problem I have noted in getting accurate room temps, is that I noticed that both the wall thermostat and the thermometer warmed up WAY too fast for the first degree or so, when all blowers and AC were first turned off, then settled into a more "realistic" delta-time per degree after that. The only thing I can guess from that is that the air is very turbulent and mixed before ventilation is shut off, and that it takes a minute or two to build up a fairly uniform temperature gradient from floor to ceiling.

#### Quote (waross)

Once you consider the cooling effect on the outside walls due to wind action you must also consider the relative humidity.

I put that in last week, but it doesn't really seem to affect anything much, very slight. That is in keeping with what I had read in a couple of forum posts. But that was relative humidity INDOORS, not outdoors where moist air convection is interacting with the outside of the walls.

You mention design tables...Where can I find them?

#### Quote (waross)

Generally for A/C we are concerned with "Worst case" issues

That was my primary motivation in the beginning. I wanted a worst case estimate based on the hottest part of the day applied over 24 hours. That is probably way too liberal, but then again, I would rather be pleasantly surprised by an electric bill that is less than predicted.

I have recorded all of your other notes as a checklist to mark off on.

### RE: Heat Transfer in a Room and AIr Conditioning

#### Quote:

Another problem I have noted in getting accurate room temps, is that I noticed that both the wall thermostat and the thermometer warmed up WAY too fast for the first degree or so
Try measuring the temperature of the wall surfaces. Heat is flowing in across the thermal resistance of the wall insulation. For heat to flow there must be a temperature difference and the wall surface will be a few degrees hotter than the inside air.
I regards to the thermometers; the exact temperature may not be as important as the temperature difference between inside and outside. The efficiency of the refrigeration cycle is referred to absolute zero.
An accurate thermometer is best but the error from a precise, repeatable thermometer that is a few degrees off true temperature may be something that you can live with. Your choice.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
It's hard to turn off the precision part of my brain! :))

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
It just dawned on me as I'm starting to deal with the AC, that one problem that makes it difficult to calculate cooling time is that every time the unit starts up, it goes through a transient period where the coils and ducts have to build up to their stable operating temps, or a curve representing what the temps would be as the room is cooled. So the air coming out of the registers takes a bit of time to reach its maximum cooling efficiency.

### RE: Heat Transfer in a Room and AIr Conditioning

treddie (Computer)

Think of it as opening the hot faucet in the bath room and waiting for the hot water to get there.
B.E.

You are judged not by what you know, but by what you can do.

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
Exactly. So unless there is some standard approximation(s) to use, that would be a bear to model.

### RE: Heat Transfer in a Room and AIr Conditioning

No.

Use test data when the theory fails/gets too complicated.

When the AC in this house starts, the blower motor turns on (time 0). Air begins flowing, and pressure is felt at the furthest discharge point form the AC blower within 1/2 second. Cool air is felt within 1-1/2 seconds, and final tempearture air) is felt coming out of the register within 2-3 seconds.

Now, the thermal resistance of the thermometer is greater than my "hand" so I actually have a situation where the technically correct "test case" (how long does it take to get cold air of the furthers register) is less than the time it takes to get a measurement of the final temperature.

Net? Ignore it. The thermal delay of the afternoon sun through the walls IS a concern: if local apparent noon is 12:00, then the hottest time of day is 2:00 - 3:00 PM. That heat you "feel" on the interior wall at 2:00 is what hit the outside wall as sunlight and air temperature 30 minutes to an hour before. The heat your AC unit will be removing at 6:00, 7:00 AND 8:00 pm - that heat that my own AC is removing right now at 8:51 PM, is what was "outside" a few hours ago as 93 F air temperatures.

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
That fast? I would have thought maybe something more like 60 seconds or so, to get to the final register temperature.

The time lag across the wall makes sense based on what I've come up with so far.

### RE: Heat Transfer in a Room and AIr Conditioning

I have a two-story house with about a max 40-ft run from the A/C in the attic to the rooms on the first floor. The first time the A/C turns on in the afternoon, there is probably about a 10-15 second lag between the time the air starts moving to the time when cold air is felt. After that, the time lag is less than 5 seconds. If you consider the air to be moving at, say, 5 mph, that's 7 ft/s, so a 40-ft run would be traversed in less than 6 seconds.

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### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
Maybe where I am misunderstanding the time lag then is inside the compressor. I would have thought that there would have been a major lag there where the gas has to get compressed from a cold start, then dropped down to a cold temp then back again, until the lo-side and hi-side reach their operating pressures. But now that I think about it, the second that pressure drops to its operating lo-side pressure, which would almost be immediate, the temperature drop across the evaporator would be just as quick.

### RE: Heat Transfer in a Room and AIr Conditioning

treddie (Computer)
If you ever have a chance to look at an air conditioner that has been fitted with a sight gauge as it turns on, you will see how quickly the thing gets into action. At the instant of turn on the sight guage will have only vapour in it, a fraction of a second later the line will have a mixture of bubbles and liquid, in less than 5 to 10 seconds the line will only have liquid in it, and your expansion valve will be working in your evaporator coil.
B.E.

You are judged not by what you know, but by what you can do.

### RE: Heat Transfer in a Room and AIr Conditioning

(OP)
I'll have to check that out when I get a new system in. I hope it has the sight glass.

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