Question:
>Does someone have the formula to figure out the BTUs needed to heat a house.
Sure. Divide each external surface area Ai by its R-value Ri, and add up each Ai/Ri number to find a Sum of Areas/R-values, SAR = sum(Ai/Ri). Then find the volume V of the house in cubic feet, multiply V by the number of Air Changes per Hour (ACH = 0.2 for a good superinsulated house, 1 for a good new house, 2 for an older house in good condition, and 4 for an old air-leaky house), divide VxACH by 60 to find the number of cubic feet per minute of air leaks, and add that to the SAR to find a thermal conductance U = SAR+VxACH/60 Btu/h-F. Then multiply U by the indoor-outdoor temperature difference, to find the power P = (Tindoors-Toutdoors)U in Btu per hour needed to keep the house warm. Some of that heat power may come from the occupants and their pets and electrical energy consumption, and the sun… For instance, a 10′ R20 cubical "house" with 6" of fiberglass insulation and 1 ACH has 6×10x10ft^/R20 = 30 Btu/h-F of thermal conductance and 10×10x10×1/60 = 16.7 cfm of air leaks, with an effective air infiltration conductance of about 16.7 Btu/h-F, so the total cube conductance U is about 46.7 Btu/h-F, and keeping it 70 F inside when it’s 30 F outside takes about (70-30)46.7 = 1,534 Btu/hour, ie 1,534/3.41 = 450 watts, so this cube can be kept warm with 3 50 watt light bulbs and 3 100 watt people. Replacing one side of the cube with an 10′x10′ double-glazed R2 sliding glass door with a thermal conductance of 10′x10′/R2 = 50 Btu/h-F leaves a conductance of 5×10x10/R20 = 25 for the other 5 walls, and the same 16.7 Btu/h-F for air leaks, making the total 91.7 Btu/h-F, so that cube needs (70-30)91.7 = 3,668 Btu/h to stay warm, just over 1.076 kW, which might come (at night) from 4 100 watt light bulbs, 4 100 watt people, a 50 watt dog, 4 25 watt cats, and 5 10 watt rabbits. If the glass wall faces south, it passes about 750 Btu/ft^2 of sun on an average January day where I live in Pennsylvania, ie it collects 10×10x750 = 75K Btu/day or a 24-hour average of 3,125 Btu/h, or a continuous 916 watts, so with enough thermal mass inside, eg masonry walls with the insulation outside the walls, the glazing can keep the cube warm on an average Jan day with only a 50 watt bulb and a 100 watt person who make an extra 512 Btu/hr. After a few cloudy days, the indoor temperature of that cube would drop to 30F+ 512Btu/h/(91.7Btu/h-F) = 35.6 F. This might be improved by making another masonry wall inside the sliding glass doors, with R20 insulation between the masonry and the glass, and an air gap between the insulation and the glass, with a black window screen in the air gap to help absorb the sun. The glass wall still gathers 75K Btu/day of sun, with warm air circulating between the air gap and the cube during the day, but the air gap behind the glass can quickly get cold, so little heat is lost to the outdoors, IF the air circulation is stopped at night… With an indoor temperature of T degrees F and average 6 hour winter solar collection day, this structure loses about 6h(T-30)100ft^2/R2= 300(T-30) Btu from the south wall during the day, 18h(T-30)100ft^2/R22 = 82(T-30) from the south wall at night, and 24(T-30)(500ft^2/R20+16.7) = 1001(T-30) Btu from the other 5 walls and air leaks over a 24 hour day, ie the cube loses a total of 1383(T-30) Btu. If all this heat comes from the sun, 75K = 1383(T-30), so T = 30+75K/1383 = 84 F. A small exhaust fan with a 75 F thermostat could keep the indoor temperature comfortable. Suppose the 4 walls were made from 30 lb 8×8x16" cement blocks with a specific heat of 0.16 Btu/F, ie 0.16×30x144/(8×16)= 5.4 Btu/F-ft^2, ie the cube has a heat capacitance C = 4×10′x10′x5.4 = 2,160 Btu/F, and the occupants keep it 75 F inside on an average day. Over one cloudy 30 F day, with no air circulating through the south wall, the cube will lose about 24hx1,534Btu/h = 36.8K Btu, so the interior temperature will drop by about 36.8K/2,160 = 17 F. More insulation or thermal mass can reduce the drop: if each 3-hole block has an interior volume of 3×4"x6"x8"/12^3 = 0.33 ft^3, ie 0.33×144/128 = 0.375 ft^3 per ft^2 of wall, we might fill most of the blocks with sand (about 18 Btu/F-ft^3) to raise the cube capacitance by 400ft^2×0.375×18 = 2700 Btu/F to a total of 4,860, and lower the drop to 36.8K/4,860 = 8 F. (Some of the blocks need to remain empty, to provide sufficient surface for the warm air from the "sunspace" to efficiently heat the masonry as it flows upwards through some vertically lined-up holes in the blocks, with a few holes into the blocks at the top and bottom of the wall, and the aid of 4 small exhaust fans near the top of the wall that blow air from the wall cavities into the room.) The south masonry wall might be thicker, and insulated on the inside as well as the outside, so it can store heat at a higher temperature than the living space, to allow better temperature control of the living space temperature on cloudy days, and make the stored heat last longer… Suppose the south wall is about 16" thick, using the same kind of blocks, with say, 2" of foam insulation on the inside, and 6" of fiberglass on the outside, and a heat capacity C = 100ft^2×13 Btu/F-ft^2 = 1,300 Btu/F. Suppose the heat that isn’t required to keep the cube warm on an average January day ends up in the south wall (using something like a small fan with a 70 F cooling thermostat) : 75K Btu/day arrive from the sun, and the cube needs about 1383(70-30) = 55K to stay 70 F on an average day, so 20K Btu/day go into the south wall, at an average temperature Tc, where 18h(Tc-30)100ft^2/R20 = 90(Tc-30) = 20K Btu, roughly, so Tc = 30+20K/90 = 252 F. I don’t think it will get much warmer than about 130 F without a selective surface, so let’s assume Tc = 130 F. Then the south wall stores about (130-80)C = 65K Btu of useful space heat, enough to keep the indoor temperature exactly 70 F for more than 24 hours. A selective surface would help here, maybe some sort of rusty galvanized iron. Or we might make the walls of the cube out of 55 gallon drums full of water, or use more insulation… If the cube walls are made with 48 55 gallon drums, we have C = 48×55x8 = 21,120 Btu/F, and R = 1/46.7, so RC = 452 hours or 19 days. If that cube starts out a 30 F cloudy week at 75 F, at the end of the week it will be about 30+(75-30)exp(-7/19) = 61 F. At that point, we might burn some wood, or add more insulation. We might get equivalent performance by just replacing the south wall masonry by higher-temperature drums, and heat water for showers too. A houses loses heat through its surface, and living space increases with volume, so bigger houses are easier to solar heat, with a smaller ratio of thermal storage to living space volume. I hope this answers your question. Nick Nicholson L. Pine System design and consulting Pine Associates, Ltd. (610) 489-0545 821 Collegeville Road Fax: (610) 489-7057 Computer simulation and modeling. High performance, low cost, solar heating and cogeneration system design. BSEE, MSEE. Senior Member, IEEE. Registered US Patent Agent. Solar closet paper: http://leia.ursinus.edu/~physics/solar.html Web site: http://www.ece.vill.edu/~nick
Response:
WOW! If I had to go through all that every time I needed to replace a residential furnace, I would have gone out of business. As a small contractor, there is no time, nor the need, for such complex figures. I have found that 4 btu’s per cubic foot works just fine, in the Chicago area anyway. I’ve compared it to engineered drawings throughout my career, and found that this rule of thumb works. For example, some engineer would figure that a house needed 97,000 btu’s. The rule of thumb would come up with anything from 90,000 to 110,000. Either way, you end up installing a 125,000 btu furnace (which will actually only put out about 100,000 btu’s). Because furnaces generally come in 25,000 btu increments, you have a given margin for error. Sure, you might oversize it and cause it to short cycle. However, in the cases where short cycling was a problem, it was always caused by some restriction to the air flow like dirty filters or furniture blocking the air vents. Also, I believe here in Chicago we should use 5 degrees as the average low temp, figuring on a 65 deg temp difference, but on New Years Eve we had a 20 below zero wind chill, without a single complaint. Sam P.S. Pardon the rant, but I am often attacked for not being more scientific. Just practical. 
– Hide quoted text — Show quoted text ->Does someone have the formula to figure out the BTUs needed to heat a house. > Sure. Divide each external surface area Ai by its R-value Ri, and add up > each Ai/Ri number to find a Sum of Areas/R-values, SAR = sum(Ai/Ri). > Then find the volume V of the house in cubic feet, multiply V by the number > of Air Changes per Hour (ACH = 0.2 for a good superinsulated house, 1 for > a good new house, 2 for an older house in good condition, and 4 for an old > air-leaky house), divide VxACH by 60 to find the number of cubic feet per > minute of air leaks, and add that to the SAR to find a thermal conductance > U = SAR+VxACH/60 Btu/h-F. > Then multiply U by the indoor-outdoor temperature difference, to find the > power P = (Tindoors-Toutdoors)U in Btu per hour needed to keep the house > warm. Some of that heat power may come from the occupants and their pets > and electrical energy consumption, and the sun… > For instance, a 10′ R20 cubical "house" with 6" of fiberglass insulation > and 1 ACH has 6×10x10ft^/R20 = 30 Btu/h-F of thermal conductance and > 10×10x10×1/60 = 16.7 cfm of air leaks, with an effective air infiltration > conductance of about 16.7 Btu/h-F, so the total cube conductance U is about > 46.7 Btu/h-F, and keeping it 70 F inside when it’s 30 F outside takes about > (70-30)46.7 = 1,534 Btu/hour, ie 1,534/3.41 = 450 watts, so this cube > can be kept warm with 3 50 watt light bulbs and 3 100 watt people. > Replacing one side of the cube with an 10′x10′ double-glazed R2 sliding > glass door with a thermal conductance of 10′x10′/R2 = 50 Btu/h-F leaves > a conductance of 5×10x10/R20 = 25 for the other 5 walls, and the same > 16.7 Btu/h-F for air leaks, making the total 91.7 Btu/h-F, so that cube > needs (70-30)91.7 = 3,668 Btu/h to stay warm, just over 1.076 kW, which > might come (at night) from 4 100 watt light bulbs, 4 100 watt people, a > 50 watt dog, 4 25 watt cats, and 5 10 watt rabbits. If the glass wall > faces south, it passes about 750 Btu/ft^2 of sun on an average January > day where I live in Pennsylvania, ie it collects 10×10x750 = 75K Btu/day > or a 24-hour average of 3,125 Btu/h, or a continuous 916 watts, so with > enough thermal mass inside, eg masonry walls with the insulation outside > the walls, the glazing can keep the cube warm on an average Jan day with > only a 50 watt bulb and a 100 watt person who make an extra 512 Btu/hr. > After a few cloudy days, the indoor temperature of that cube would drop > to 30F+ 512Btu/h/(91.7Btu/h-F) = 35.6 F. This might be improved by making > another masonry wall inside the sliding glass doors, with R20 insulation > between the masonry and the glass, and an air gap between the insulation > and the glass, with a black window screen in the air gap to help absorb > the sun. The glass wall still gathers 75K Btu/day of sun, with warm air > circulating between the air gap and the cube during the day, but the > air gap behind the glass can quickly get cold, so little heat is lost > to the outdoors, IF the air circulation is stopped at night… > With an indoor temperature of T degrees F and average 6 hour winter solar > collection day, this structure loses about 6h(T-30)100ft^2/R2= 300(T-30) > Btu from the south wall during the day, 18h(T-30)100ft^2/R22 = 82(T-30) > from the south wall at night, and 24(T-30)(500ft^2/R20+16.7) = 1001(T-30) > Btu from the other 5 walls and air leaks over a 24 hour day, ie the cube > loses a total of 1383(T-30) Btu. If all this heat comes from the sun, > 75K = 1383(T-30), so T = 30+75K/1383 = 84 F. A small exhaust fan with > a 75 F thermostat could keep the indoor temperature comfortable. > Suppose the 4 walls were made from 30 lb 8×8x16" cement blocks with a > specific heat of 0.16 Btu/F, ie 0.16×30x144/(8×16)= 5.4 Btu/F-ft^2, ie > the cube has a heat capacitance C = 4×10′x10′x5.4 = 2,160 Btu/F, and the > occupants keep it 75 F inside on an average day. Over one cloudy 30 F day, > with no air circulating through the south wall, the cube will lose about > 24hx1,534Btu/h = 36.8K Btu, so the interior temperature will drop by about > 36.8K/2,160 = 17 F. More insulation or thermal mass can reduce the drop: > if each 3-hole block has an interior volume of 3×4"x6"x8"/12^3 = 0.33 ft^3, > ie 0.33×144/128 = 0.375 ft^3 per ft^2 of wall, we might fill most of the > blocks with sand (about 18 Btu/F-ft^3) to raise the cube capacitance by > 400ft^2×0.375×18 = 2700 Btu/F to a total of 4,860, and lower the drop to > 36.8K/4,860 = 8 F. (Some of the blocks need to remain empty, to provide > sufficient surface for the warm air from the "sunspace" to efficiently heat > the masonry as it flows upwards through some vertically lined-up holes in > the blocks, with a few holes into the blocks at the top and bottom of the > wall, and the aid of 4 small exhaust fans near the top of the wall that > blow air from the wall cavities into the room.) > The south masonry wall might be thicker, and insulated on the inside as > well as the outside, so it can store heat at a higher temperature than > the living space, to allow better temperature control of the living space > temperature on cloudy days, and make the stored heat last longer… > Suppose the south wall is about 16" thick, using the same kind of blocks, > with say, 2" of foam insulation on the inside, and 6" of fiberglass on > the outside, and a heat capacity C = 100ft^2×13 Btu/F-ft^2 = 1,300 Btu/F. > Suppose the heat that isn’t required to keep the cube warm on an average > January day ends up in the south wall (using something like a small fan > with a 70 F cooling thermostat) : 75K Btu/day arrive from the sun, and the > cube needs about 1383(70-30) = 55K to stay 70 F on an average day, so 20K > Btu/day go into the south wall, at an average temperature Tc, where > 18h(Tc-30)100ft^2/R20 = 90(Tc-30) = 20K Btu, roughly, so Tc = 30+20K/90 > = 252 F. I don’t think it will get much warmer than about 130 F without a > selective surface, so let’s assume Tc = 130 F. Then the south wall stores > about (130-80)C = 65K Btu of useful space heat, enough to keep the indoor > temperature exactly 70 F for more than 24 hours. > A selective surface would help here, maybe some sort of rusty galvanized > iron. Or we might make the walls of the cube out of 55 gallon drums full > of water, or use more insulation… > If the cube walls are made with 48 55 gallon drums, we have C = 48×55x8 > = 21,120 Btu/F, and R = 1/46.7, so RC = 452 hours or 19 days. If that cube > starts out a 30 F cloudy week at 75 F, at the end of the week it will be > about 30+(75-30)exp(-7/19) = 61 F. At that point, we might burn some wood, > or add more insulation. > We might get equivalent performance by just replacing the south wall > masonry by higher-temperature drums, and heat water for showers too. > A houses loses heat through its surface, and living space increases with > volume, so bigger houses are easier to solar heat, with a smaller ratio > of thermal storage to living space volume. > I hope this answers your question. > Nick > Nicholson L. Pine System design and consulting > Pine Associates, Ltd. (610) 489-0545 > 821 Collegeville Road Fax: (610) 489-7057 > Computer simulation and modeling. High performance, low cost, solar heating and > cogeneration system design. BSEE, MSEE. Senior Member, IEEE. Registered US > Patent Agent. Solar closet paper: http://leia.ursinus.edu/~physics/solar.html > Web site: http://www.ece.vill.edu/~nick
Response:
I don’t know about that. I somehow manage to do Manual J calcs on every job I do and seem to do okay. Makes me look good to the customer since I’m the only contractor that generally has a heat loss to show them. It really only takes about an hour for most houses. – Hide quoted text — Show quoted text ->WOW! If I had to go through all that every time I needed to replace a residential >furnace, I would have gone out of business. As a small contractor, there is no >time, nor the need, for such complex figures. I have found that 4 btu’s per cubic >foot works just fine, in the Chicago area anyway. I’ve compared it to engineered >drawings throughout my career, and found that this rule of thumb works. For >example, some engineer would figure that a house needed 97,000 btu’s. The rule of >thumb would come up with anything from 90,000 to 110,000. Either way, you end up >installing a 125,000 btu furnace (which will actually only put out about 100,000 >btu’s). Because furnaces generally come in 25,000 btu increments, you have a given >margin for error. Sure, you might oversize it and cause it to short cycle. However, >in the cases where short cycling was a problem, it was always caused by some >restriction to the air flow like dirty filters or furniture blocking the air vents. >Also, I believe here in Chicago we should use 5 degrees as the average low temp, >figuring on a 65 deg temp difference, but on New Years Eve we had a 20 below zero >wind chill, without a single complaint. >Sam >P.S. Pardon the rant, but I am often attacked for not being more scientific. Just >practical. > >Does someone have the formula to figure out the BTUs needed to heat a house. > Sure. Divide each external surface area Ai by its R-value Ri, and add up > each Ai/Ri number to find a Sum of Areas/R-values, SAR = sum(Ai/Ri). > Then find the volume V of the house in cubic feet, multiply V by the number > of Air Changes per Hour (ACH = 0.2 for a good superinsulated house, 1 for > a good new house, 2 for an older house in good condition, and 4 for an old > air-leaky house), divide VxACH by 60 to find the number of cubic feet per > minute of air leaks, and add that to the SAR to find a thermal conductance > U = SAR+VxACH/60 Btu/h-F.
** The problem with the average family today is that it’s impossible to support it and the government on one income.
Response:
> WOW! If I had to go through all that every time I needed to replace a residential > furnace, I would have gone out of business. As a small contractor, there is no > time, nor the need, for such complex figures. I have found that 4 btu’s per cubic > foot works just fine, in the Chicago area anyway.
<Snip> > Sure. Divide each external surface area Ai by its R-value Ri, and add up > each Ai/Ri number to find a Sum of Areas/R-values, SAR = sum(Ai/Ri). > Then find …..
Since the original poster (I am assuming, the original post yet hasn’t appeared on my server yet) didn’t specify the type of house he was wanting to do the calculations for, I would have to say that it cannot be taken for granted that the rule of thumb mentioned above is applicable at all – though by pointing out that the rule of thumb is only claimed to work well in the Chicago area the onus is back on the poster to decide if he’s willing to rely on a Chicago area general rule of thumb for his application. Perhaps he was wanting to convert from electrical heat to gas heat in a typical midwestern suburban house and perhaps he is trying to design a dream vacation cabin deep in the Alaskan wilderness. There’s room for both approaches – and the best person is the one that is comfortable with both and knows when to use each in turn. Knowing some rules of thumb that are applicable in, say, 85% of all situations that you are likely to encounter is great. It saves a lot of time, effort and money. But they can become a severe handicap, or even dangerous, unless you know how to at least identify the other 15% of the cases. A common area where this is true in my field of practice is applying the principles of superposition to the analysis of every electronic circuit that comes along without remembering that superposition is only valid for linear systems. Rules of thumb also provide invaluable sanity checks when using more fundamental approaches. Restricting the discussion to engineers that are being trained to do these kinds of designs and perform these kinds of calculations, I think that their education is lacking if they are not expected to look at the relationships involved and see if there are general rules of thumb and over what realm of conditions they are applicable. In this case, they should have to calculate the BTU/ft^3 as a function of all of these variables and then graph out the results as a function of the dominant variables over a realistic range of values. What they would probably see is that there are large, nearly flat regions that can be identified where they can conclude that if a certain set of conditions is met, then 4 BTU/ft^3 is appropriate and under other conditions some other constant ratio can be used. They can also identify regions where the function changes too rapidly thus requiring a basic approach. Even if they decide to take a more basic approach on a particular design, they can cross-check their work by calculating the BTU/ft^3 and asking if the answer makes sense in view of their awareness of the relationships found above. There have been numerous cases where products, buildings or systems have been poorly designed and people have been injured or even killed where it is obvious that no one ever took a step back and asked, "Does this answer make sense?"
Response:
I used to be a heating contractor too. I found that I’d make more sales by presenting the customer with a heat loss sheet. I went to the IBR 3 day seminar in 88 and learned thier heat loss methods and boiler / baseboard sizing, then put it on a spreadsheet. All I’d need to do for a preliminary heat loss was to treat each wing of the building as a big room, fudge the percentage of window area with a good guestimate, and punch in the U ( 1 / R factor = U factor) factors. This takes 10 minutes per job. I’d tell the customer that I’d do a room by room detailed heat loss if I got the job. If they insisted on it right away, I’d tell them I’d do it for them for $100.00 per hour, deductable from the price of the install. I had no takers on that. That keeps them from going to somerville lumber and doing it themselves off my heat loss specs. Take the wing by wing (example. only set of calcs for a ranch), size it up for baseboard, add 15% to baseboard for trim, use the entire perimiter for the 3/4" piping order (series loop system only). I could crank out an estimate in a half hour. I’d go back and present it to the customer. If I got the job, I’d put in an hour and a half or so into a room by room but I’d pad the labor into the estimate. My sales from newspaper ads went from 2 out of 10 inquiries to 4 out of 10 by doing this. Mark I don’t know about that. I somehow manage to do Manual J calcs on every job I do and seem to do okay. Makes me look good to the customer since I’m the only contractor that generally has a heat loss to show them. It really only takes about an hour for most houses. – Hide quoted text — Show quoted text ->WOW! If I had to go through all that every time I needed to replace a residential >furnace, I would have gone out of business. As a small contractor, there is no >time, nor the need, for such complex figures. I have found that 4 btu’s per cubic >foot works just fine, in the Chicago area anyway. I’ve compared it to engineered >drawings throughout my career, and found that this rule of thumb works. For >example, some engineer would figure that a house needed 97,000 btu’s. The rule of >thumb would come up with anything from 90,000 to 110,000. Either way, you end up >installing a 125,000 btu furnace (which will actually only put out about 100,000 >btu’s). Because furnaces generally come in 25,000 btu increments, you have a given >margin for error. Sure, you might oversize it and cause it to short cycle. However, >in the cases where short cycling was a problem, it was always caused by some >restriction to the air flow like dirty filters or furniture blocking the air vents. >Also, I believe here in Chicago we should use 5 degrees as the average low temp, >figuring on a 65 deg temp difference, but on New Years Eve we had a 20 below zero >wind chill, without a single complaint. >Sam >P.S. Pardon the rant, but I am often attacked for not being more scientific. Just >practical. > >Does someone have the formula to figure out the BTUs needed to heat a house. > Sure. Divide each external surface area Ai by its R-value Ri, and add up > each Ai/Ri number to find a Sum of Areas/R-values, SAR = sum(Ai/Ri). > Then find the volume V of the house in cubic feet, multiply V by the number > of Air Changes per Hour (ACH = 0.2 for a good superinsulated house, 1 for > a good new house, 2 for an older house in good condition, and 4 for an old > air-leaky house), divide VxACH by 60 to find the number of cubic feet per > minute of air leaks, and add that to the SAR to find a thermal conductance > U = SAR+VxACH/60 Btu/h-F.
** The problem with the average family today is that it’s impossible to support it and the government on one income.
Response:
>might come (at night) from 4 100 watt light bulbs, 4 100 watt people, a >50 watt dog, 4 25 watt cats, and 5 10 watt rabbits. If the glass wall >I hope this answers your question.
Nick, I still need help here. I have no rabbits or dogs, and I can’t seem to find the wattage rating on my cats. What now ? Paul >~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~
(remove this part )pobox.com My WWW site is at http://www.pobox.com/~pjm, featuring free HVAC software. The Sci.Engr.Heat-Vent-AC and Alt.HVAC FAQ is at http://www.elitesoft.com/sci.hvac/
Response:
>WOW! If I had to go through all that every time I needed to replace a residential >furnace, I would have gone out of business. As a small contractor, there is no >time, nor the need, for such complex figures. I have found that 4 btu’s per cubic >foot works just fine, in the Chicago area anyway.
So, you find that all houses are constructed exactly the same ? They all have the same insulation, sealing against infiltration, solar exposure, etc ? No difference in construction between a 50 year old house, and a new one ? Why not run a heat load, the right way, like you’re supposed to ? If you find it too much work to do it right, get out of the business. Paul >~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~
(remove this part )pobox.com My WWW site is at http://www.pobox.com/~pjm, featuring free HVAC software. The Sci.Engr.Heat-Vent-AC and Alt.HVAC FAQ is at http://www.elitesoft.com/sci.hvac/
Response:
>I don’t know about that. I somehow manage to do Manual J calcs on every >job I do and seem to do okay. Makes me look good to the customer since >I’m the only contractor that generally has a heat loss to show them. It >really only takes about an hour for most houses.
Many utilities and local codes are requiring them, too. Paul >~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~
(remove this part )pobox.com My WWW site is at http://www.pobox.com/~pjm, featuring free HVAC software. The Sci.Engr.Heat-Vent-AC and Alt.HVAC FAQ is at http://www.elitesoft.com/sci.hvac/
Response:
>>…might come (at night) from 4 100 watt light bulbs, 4 100 watt people, >a 50 watt dog, 4 25 watt cats, and 5 10 watt rabbits… > Nick, I still need help here. I have no rabbits or dogs, and >I can’t seem to find the wattage rating on my cats. What now ?
Get some ASHRAE Standard Cats (as described in Table 7 on page 9.12 of the 1993 Handbook of Fundamentals): Heat Generation, Btu/h per animal -Normally active(b, c)- Animal Weight Basal(a) Sensible Latent Total lb … Mouse 0.046 0.66 1.11 0.54 1.65 Cat 6.61 27.41 39.22 19.31 68.02 … (a) Based on standard metabolic rate M = 6.6W^0.75 watt per animal (Klieber 1961) or appropriate reference (W = animal mass, lb) (b) Referenced according to availability of heat generation data. Otherwise, heat generation is calculated on basis of ATHG = 2.5M (Borton _et al_. 1976.) Latent heat is assumed to be approximately 33% of total; sensible heat, 67% of total heat (Besch 1973, Woods _et al_. 1972.) (c) Data taken from Runkle (1964), Kleiber (1961), Besch (1973), Woods and Besch (1974), Woods _et al_. (1959) and Ota and McNally (1961.) A house requiring 750 gallons of oil burned at 80% efficiency over a 200 day heating season, ie 750/200/24×130,000×0.8 = 16,250 Btu/h might be heated with a herd of C cats and M mice (with a suitable condensing air-air heat exchanger, and appropriate controls to avoid Volterra instabilities), where 68.02C + 1.65M = 16,250, and (assuming the cats eat 5 mice per day, and each mouse produces a litter of 3 once a month) M = 50C, so C(68.02+1.65(50)) = 16,250, and C = 108, and M = 5,400. Nick
Response:
Now Paul…you KNOW that when you buy those low-cost cats without the name plates you’re going to have problems finding the wattage and full-load amp ratings. You’ll just have to crank up that trusty Fluke multimeter and measure each cat under full load conditions. Lew >might come (at night) from 4 100 watt light bulbs, 4 100 watt people, a >50 watt dog, 4 25 watt cats, and 5 10 watt rabbits. If the glass wall >I hope this answers your question. > Nick, I still need help here. I have no rabbits or dogs, and > I can’t seem to find the wattage rating on my cats. What now ? > Paul
– Lew Harriman Mason-Grant Company P.O. Box 6547 57 South St. Portsmouth, NH 03802 USA TEL (603) 431-0635 FAX (603) 427-0015 http://www.MasonGrant.com
Response:
NOTE: Even further off-topic: Reminds me of a cartoon I saw where a father comes home and sees his son standing over a little dog that’s trembling and shaking and says to the boy, "What do mean, ‘I figured out where the batteries go.’? This is A PUPPY!" – Hide quoted text — Show quoted text -> Now Paul…you KNOW that when you buy those low-cost cats without the > name plates you’re going to have problems finding the wattage and > full-load amp ratings. You’ll just have to crank up that trusty Fluke > multimeter and measure each cat under full load conditions. > >might come (at night) from 4 100 watt light bulbs, 4 100 watt people, a > >50 watt dog, 4 25 watt cats, and 5 10 watt rabbits. If the glass wall > Nick, I still need help here. I have no rabbits or dogs, and > I can’t seem to find the wattage rating on my cats. What now ?
Response:
Only if it’ll read micro-amps. :) – Hide quoted text — Show quoted text – >Now Paul…you KNOW that when you buy those low-cost cats without the >name plates you’re going to have problems finding the wattage and >full-load amp ratings. You’ll just have to crank up that trusty Fluke >multimeter and measure each cat under full load conditions. >Lew > >might come (at night) from 4 100 watt light bulbs, 4 100 watt people, a > >50 watt dog, 4 25 watt cats, and 5 10 watt rabbits. If the glass wall > >I hope this answers your question. > Nick, I still need help here. I have no rabbits or dogs, and > I can’t seem to find the wattage rating on my cats. What now ? > Paul
** The problem with the average family today is that it’s impossible to support it and the government on one income.
Response:
I prefer kill-o-amps for cat testing. ;-P Only if it’ll read micro-amps. :) – Hide quoted text — Show quoted text – >Now Paul…you KNOW that when you buy those low-cost cats without the >name plates you’re going to have problems finding the wattage and >full-load amp ratings. You’ll just have to crank up that trusty Fluke >multimeter and measure each cat under full load conditions. >Lew > >might come (at night) from 4 100 watt light bulbs, 4 100 watt people, a > >50 watt dog, 4 25 watt cats, and 5 10 watt rabbits. If the glass wall > >I hope this answers your question. > Nick, I still need help here. I have no rabbits or dogs, and > I can’t seem to find the wattage rating on my cats. What now ?
Response:
Boy oh boy! How easy it is to press some peoples buttons! I was simply talking about a rule of thumb that seems to work, and this guy Paul suddenly assumes that I’m walking around with head up my a*s thinking that this guestimate is the ultimate solution. My mention of this rule was to contrast the extremely complicated process that was described in the message that I originally replied too. Other postings mention Manual J, which of course is the best way to go, and it is against these findings that I have compared my rule too, and found that by either method, I ended up installing the same size furnace. Pretty neat I thought. Of course, it is only a rule of thumb, and it only applied to modest residential applications. I did not bet my business on it. Although, I do know men that would just look at a building and tell you what size you need. Amazingly, they were right most of the time. – Hide quoted text — Show quoted text ->WOW! If I had to go through all that every time I needed to replace a residential >furnace, I would have gone out of business. As a small contractor, there is no >time, nor the need, for such complex figures. I have found that 4 btu’s per cubic >foot works just fine, in the Chicago area anyway. > So, you find that all houses are constructed exactly the same > ? They all have the same insulation, sealing against infiltration, > solar exposure, etc ? No difference in construction between a 50 year > old house, and a new one ? > Why not run a heat load, the right way, like you’re supposed > to ? If you find it too much work to do it right, get out of the > business. > Paul >~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~>~~ > (remove this part )pobox.com > My WWW site is at http://www.pobox.com/~pjm, featuring free HVAC software. > The Sci.Engr.Heat-Vent-AC and Alt.HVAC FAQ is at http://www.elitesoft.com/sci.hvac/
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