Calculating heat loss based on drop in temperature?
TimTuckerCom
| Posted in General Questions on
Just had an issue with the ignitor on our furnace and we were without heat for the last 24 hours or so.
The experience left me wondering if there’s an easy way to do at least a ballpark calculation for heat loss based on before / after indoor temperatures & the hourly outdoor temperatures overnight. (I’ve done some estimates using BEopt & coolcalc, but figured it would be good to compare to real world measurements)
Temperature at the thermostat when the furnace was working last: 72°F
Temperature at the thermostat when I turned it back on: 57°F (upstairs bedrooms stayed significantly warmer)
Overnight lows look like they went down to 12°F
House is ~2000 sq. ft. built in 1995. Overall pretty happy with the performance after spending the time to ensure that the attics were sealed & insulated properly (an ongoing project over the last 3 years or so).
For those curious on project details:
Started out with taking out the whole house fan — it was noisy, positioned poorly to do any good, and spent most of its existence creating a giant chimney effect.
I contracted spray foam for air sealing & walls exposed to the attic, but had to keep going up to point out spots they’d missed or hadn’t sprayed to the quoted depth. Also had them do dense pack cellulose under one bedroom that extended over the garage & foam around the rim joists.
After that I put in Rockwool over the foam in the rim joists (so ~3″ of foam + 3.5″ of mineral wool) and in the attic near outside edges where there was more limited space. Final step was somewhere between 24-36″ of cellulose blown in on top of the old R19 fiberglass.
GBA Detail Library
A collection of one thousand construction details organized by climate and house part
Search and download construction details
Replies
For a basic measure of energy loss via walls/windows/etc you can use the basic formula of: U value x Wall area x Delta T. In your scenario Delta T changes as the outdoor temps drop, but also as your interior temperature drops.
Not accounting for energy loss through air leakage, as your indoor temperature gets closer to outdoor temperature you will lose energy less quickly.
The information you have is how many degrees per hour your house cools at a given temperature. What you are trying to get is how many BTU's per hour it loses at a given temperature. The quantity you are missing is BTU/degree, which is called the heat capacity. (Sometimes erroneously called "thermal mass," don't get me started.)
You could estimate the heat capacity of a building if you knew the specific heat of the materials, and how much the building weighed. Most common building materials -- drywall, lumber, concrete, tile -- have specific heats of an order of magnitude of 0.5. A single family house weighs on the order of 100,000 lbs. So a ballpark guess would be it takes 50,000 BTU to raise a house one degree, or a house loses 50,000 BTU when it cools by one degree.
I would say that estimating the heat capacity of a house in this way is less predictive than estimating the heat loss using Manual J.
Thanks -- most everything I've seen from previous calculations would indicate that it's probably in the neighborhood of 60,000 BTUs needed for heating & 40,000 BTUs for cooling, which is not terribly far off from the existing 88,000 BTU furnace (110,000 BTU @ 80% efficiency) & 3 ton AC.
https://www.greenbuildingadvisor.com/article/out-with-the-old-in-with-the-new
give this method a try!
You should know the basic heat loss equation is Q = - k A (T2-T1), similar to the one agm gave you.. You can use this simply to figure out a quick heat loss buy just knowing the surface Area and the change in indoor temperature over a specific time. k doesn't vary, constant, since it is the properties of the materials/walls/roof etc.
So A would be the surface area of the house, walls, roof etc. Use the inside temperature for T2 and T1 over a period of time, hours.
Q would normally end up being BTUs/hr so when you multiply by the time you end up with a basic overall heat loss.
eg. Q = -k (Btu/ hr*ft^2*F) (1000ft^2) x (70 - 60F) = k x -10,000 Btu/hr x let's say over 5 hrs time = k * -50,000 Btus....which is close to what you came up with. Your area may be larger and the time less. k shouldn't be that big. So pretty close for a quick, overall estimate. If you figure out k, conductivity for portion of a wall you can plug that in but you may have to use the outside temperature for T2.
Tagging on to DC's comment below using this example to solve for k
So we have Q = k(constant) x -10000 BTU/hr and A (Tout -Tin) = (1000ft^2)(40 - 70F) then when you solve for k = Q/ -A (T2 -T1)
k= (-10000 BTU/hr)/( 30000 ft^2*F) =- 0.3 BTU/hr*ft^2*F
How long did the heat have to run to get the house back to temperature? And what is the rated output of your heat source?
You could do a rough calculation of heat out=heat in. IE, if the heat was out for 24 hours and then had to run for 6 to catch up, overall 30 hours of heating required 6 hours of capacity, so your load at that temperature is 20% of capacity. That's a first approximation.
What that leaves out is that heat loss at 57F is going to be less than at 72F. And the outside temperature may not have been constant. So a better approximation is to create a spreadsheet with a column for each hour. For each hour you have outside temperature, which you get from observations or estimate; beginning inside temperature, which is 72F for the first column and the ending temperature of the preceding hour for every other column. Then you calculate heat flow out, which is beginning inside temperature minus outside temperature, times the thermal constant of the house[1], and heat flow in, which is the full capacity of your heating system in the hours that the heat was running. You substract those to get net heat flow, and divide by the heat capacity of the house[2] to get net temperature change, which you then add to the beginning temperature to get the ending temperature.
The heat constant of the house and the heat capacity of the house are the unknowns that you are trying to solve for. You know three points on the curve -- 72F at the beginning, 57F at the middle and 72F at the end -- so there is a unique solution for the two unknowns. The way I would solve[3] it is to plug in guesses and adjust them until the spreadsheet balances. In the second paragraph of this post I showed how to do a rough estimate of heat constant, use that as your first guess and adjust from there.
[1] The heat constant of the house is in BTU/hr/degree, it's kA in the formula Tom gave in post #5, and it's what a Manual J is ultimately trying to figure out. The Manual J process is basically measuring A and estimating k.
[2] The heat capacity of a house is in degrees/BTU, it's how much heat it takes to raise the temperature of the house.
[3] I have an honors degree in applied mathematics and I solve problems like this by plugging numbers into a spreadsheet in successive approximation. It's easier and less error-prone than trying to solve them analytically.
Great stuff here.
You make a good point about heat loss rate being different at lower temperatures. My house is currently empty, holding 60F, with a flat 35F exterior temp. One 4 hour cycle consists of 26 minute runtime of the boiler, which outputs 153k, so a 10.8% duty rate or 16.5k per hour. With a 25 delta, roughly 660 btu/degree hour.
I also have data from the cold snap last year, where at -10F and 71F inside, the boiler was running 50% of each hour, or 980 btu/degree hour. Pretty significant difference.
What base temperature do you use if you are back testing like this and you keep the house at 71F? I’ve been using 68 to account for the slightly higher setpoint. Thanks.
Manual J, which is what by code now must be used to size heating systems, is all about the assumption that heat loss is linear to the difference between inside temperature and outside temperature.
By convention the base temperature is 65F, the assumption is that occupant activity provides about five degrees worth of heating. That's a crude assumption. I believe it's used because heating degree-days -- itself a crude measure -- is reported with a 65F baseline.
The title of this post draw my attention, because I'm also trying to work out the heatloss of my house based on the temperature decrease inside during the night. I have a Ecowitt weather station and continuously measure the inside and outside temperature, but also the windspeed and direction. So I also know the 'feels like' outside temp.
I heat the whole house during the day/evening and my (oil fired) boiler stops every night the same time and the temperature at that point in time is always the same (around 21 degrees C).
I cannot measure the amount of oil I'm using because I have an oil tank. And the boiler is also used to heat a hot water cylinder.
I have just this autumn and winter to figure out if I can switch from an oil boiler to an Air source heat pump (maybe high temperature). I'm running the boiler on a minimum flow temp (55 C) to simulate an ASHP and that works fine most of the time (except on a couple days with a 'feels like' below zero). Location is Montrose Scotland.
My guess would be that the data I collect should be enough to calculate the thermal performance of the house. Any thoughts on that are more than welcome.
You can use the fuel usage method. Just keep oil receipts.
The problem is that we recently moved in and have no way to measure the oil usage. The tank was more than half full and the earliest top-up will be next spring. So we have to think of another method.
There are devices that clamp on to the wiring and measure the time that something is running. Most oil burners are simply on/off, so by measuring the run time you can derive the output. On a cold day measuring the run time and the average temperature over 24 hours should give a pretty good estimate.
I've had a look a the clamp meters. They seem to measure the current, not the on-time. I could however use a smart plug for the boiler and measure the Kwh usage for a day. Knowing the nominal Kwh value of the boiler I can work out the run time over a day (week/month)
I have to check the boiler specs to find out if it's modulating or always burning at max. I guess the flow temperature does not make a difference. I also need to find out the amount of oil that is burned per time unit. The kerosine we use here has around 10.5 Kwh/liter.
So, I've been able to measure the electricity the boiler is using. The boiler pump uses 50w during 16 hours a day = 0.8 kwh a day. When the boiler is running it uses around 160w on top of the pump. The oil usage is around 1.9 l/h. Which means that every hour it's running I get around 1.9 l x10.3 kwh= 19.6 kwh of heat. On 20 dec for instance I used 0.73 kwh/.160=4.6 hour * 1.9 liter =8.7 liters = 90 Kwh on heating. The indoor temperature was 16.3 at 0.800 hr and 20.5 at 23.00.
What you need to do now is keep the interior temperature constant for say 24 hours, monitor the outside temperature and fuel usage.
I'll pick a day where the outside temp is around zero Celsius, no wind, no sunshine. And keep the indoor temp on 19 celcius for 24 hours. I record the outside temp anyway every five minutes. Would that provide enough data to calculate the heat loss?
Oh and another question: I don't seem to get notifications off new posts, even though I checked the box for it.
I finaly have some accurate data. I had an kerosine delivery on jan 16th of 1000l and made a note of the level before and after. That learned me that 1 cm level change corresponded with 29 liter of kerosine. In the next 17 days (408 hrs) I used 232 liter, equals 7484kBtu (232*35.5kBtu*90% boiler efficiency). This is 7484/408=18.35 kBtu/hr The averages were: Indoor temp 67.64F, outdoor 37.94F (including wind chill) and delta T 29.7F. For a whole year in our Scotland Montrose climate I estimate I need 122500 kBtu to keep the house a the desired temperature of 67.64 F.
My house has a ground floor area of 1005 ft2, a first floor of 435 ft2 and a volume of 12000 cubic ft.
Note: the oil usage included the domestic hot water.
Any thoughts on these results regarding the feasibilty of an air source heat pump ?
The design question for a heat pump is basically: can it keep the house warm on the coldest days?
So the other piece of information you need is what your coldest days look like. Typically you size for the 99th percentile temperature. You already have the information to extrapolate to that. Let's say your 99th percentile temperature is 20F, that's 50F from room temperature. If you're using 18,500 BTU/hr at a delta of 29.7, it's just a straight line approximation, so at a delta of 50F you'd be at 50/29.7*18,500 = 31,100 BTU/hr.
In the US you get the 99% temperature from the National Weather Service, I imagine you have something similar.
Heat pumps tend to produce less as the temperature drops, so the next step is to find one that is capable of 31,100 at 20F, that's probably around the 3 ton range, 36000 BTU/hr. That's well within the range of heat pumps, you'll find a large selection of choices. The next question is what kind of configuration you want, how many heads and how many compressors, that's probably worth a separate thread.
I see that your calculation includes oil used for domestic hot water. Usually the way you correct for that is to measure usage during the summer months and assume that it's about the same. A ballpark estimate is that if each resident of the house uses 80 gallons of hot water per day, and the heater has to raise the temperature by 50F, that's 33,200 BTU/resident per day, or 1383 BTU/resident/hour. Since heat pumps typically come in multiples of 6000 BTU/hour that's probably not going to change your choice of equipment.
Thanks DC. I'm going to work on it.
The outdoor design temperature for Scotland is 23F, the indoor design temperature is 68F. That is a 45F difference. That means 45/29.7*18350 = 27800 BTU/hr. My latest candidate heat pump (Vaillant aroTherm plus 12kw, 41kBTU) would deliver 42675 BTU/hr for the 23F with a designed flow temp of 113F. That would be plenty. I've also planned insulations upgrades for underfloor and loft areas (as part of the design)
REally hard to accurately measure fuel usage in any short period of time. Small errors become large. Oil burners don't burn what their nozzles says they do. Efficiency is not what AFUE says it is. Pre and post purge affect time periods.
In my first post, #7, I was looking for a way to calculate the heat loss based on the pace of cooling down inside. Also measuring outside temp and wind chill factor. Then I got nudged into measuring the oil usage.
The energy efficiency of my boiler is around 80%. I forgot to take that into account in my calculation in post#12
I'm currently recording the electricity usage on a daily basis. Not intraday. It might be interesting to know how much kwh is needed to bring the house up to 19.5 celcius in the morning.
The pace of cooling depends upon the heat capacity of the house, the insulation level of the house and the temperature difference between inside and outside. The fuel consumption at a steady temperature just depends upon the insulation level and the temperature difference.
What you are ultimately interested in is the insulation level of the house, or its complement, the heating load at design temperature. The heat capacity of the house is not useful in that calculation, and measuring the heat drop just adds another variable to try to account for without adding any insight.
JM,
You are on to something, here is what I did. On a steady state temperature day I kept track of furnace run times for 8 hours. I waited until the afternoon for the house to come up to temperature after the nightly setback. The day was in January in Minneapolis, no wind, no sun and about 32F all day. My furnace is a 95% natural gas forced air. After timing the furnace run time for the 8 hour period and using the factory Btu output specification I could calculate the fuel burn. Using Heating Degree Days I calculated my heat loss to be 360 Btu per hour per degree F. This comes very close to my long term energy monitoring normalized for HDD. The day I tested was just me in the house, nothing extra turned on, just the computer, was kind of fun.
With regard to internal temperature drop and building efficiency you are on to something there as well. A building scientist friend of mine and I were talking back when I had just made some fairly significant building envelope improvements. I said I noticed the house cooled more slowly after the 10 PM thermostat setback than before the efficiency improvements. The building scientist worked with NYSERDA back in the day when they were doing energy retrofits on existing NY housing. Before doing the weatherization on a home they would note the internal temperature drop for a period of time after turning off the heat source. They would then conduct the same exercise after the weatherization to get a quick snapshot of the efficiency gained on a percentage basis. They would compare the temperature drop before and after factoring in the Delta T.
A residential building has an internal heat storage capacity I call surfaces, I determined mine to be about 20,000 Btu per degree F for my Minneapolis house. This is why a home slowly loses heat with the heat source off and why it takes a lot of heating to bring the house back to the setpoint. You are not just heating air but reheating all of the surfaces that cooled below the thermostat setpoint. All to be taken into account when deciding on HVAC type and sizing.
Happy Holidays,
Doug
Thank you very much Doug for your reply and explanation.
Over the period 3 jan - 7 jan, 5 days, I used 55 liters of oil to heat the house. Which makes 55*10.5 kwh/l * 80% efficiency= 462Kwh. The average outdoor temp was 4.2C and the average indoor temp was 19.7C. I'm not entirely sure about the oil used, because I can vary between 1.9 and 2.9 l/hr when burning. I used 2 l/hr because of the low flow temp (55C). But I don't know if this is good or bad. It looks like a lot of energy.
Putting it into units Americans are more familiar with , 462kWh is 1570 kBTU. Over 96 hours that's 16,400 BTU/hr. A temperature difference between inside and outside of 15.5C is 27.9F. In my climate our design day is 50F below interior, so 16.4kBTU/hr at 27.9F equals 29.4K BTU/hr at 50F.
That's almost exactly what my house produced in a Manual J. I have pretty tight new construction and I consider that a good number.
Hi DC, thanks for the translation to the American units. I'm not sure if we can compare our houses. Mine has a floor area of 1938 ft2 and a volume of 15890 cubic ft, divided over two floors . Actually the figure might even be a bit better because the period was 120 hrs (5 days). On the other hand the oil usage might have been higher. Based on the electricity usage of the boiler I calculated that it ran for 29.13 hours in the 120 hr period. So that's a usage between 55.3l (1.9l/h low) and 84.5l (2.9l/h high). In BTU between 1570 kBTU (13,800 BTU/hr) and 2419 kBTU (20,158 BTU/hr).
Currently looking at the Viessmann Vitocal 150-A/151-A range 13Kw. We have also plans to improve the loft insulation and apply underfloor insulation.
Cheers, Jan
Thanks everyone for their contributions to my case. I'm convinced I have done the right amount of research and measuring. I've ordered the Vaillant aroTherm plus 12kw, 41kBTU.
I might come back for advice on solar PV. small domestic windmills and backup battery storage at a later stage, if that's covered here also
Cheeers, Jan